The peak current through the 150 Ω resistor connected to the AC source with an emf of 15.0 V and a frequency of 100 Hz is 1.25 A.
The peak current through the resistor can be calculated using Ohm's law and the relationship between current, voltage, and resistance in an AC circuit. Ohm's law states that the current (I) flowing through a resistor is equal to the voltage (V) across the resistor divided by the resistance (R), represented by the equation I = V/R.
In this case, the voltage across the resistor is the peak voltage (Ep) of 15.0 V. The resistance (R) is given as 12 Ω. Substituting these values into the equation, we can calculate the peak current (Ip) as Ip = Ep / R.
Ip = 15.0 V / 12 Ω = 1.25 A
Therefore, the peak current through the resistor is 1.25 A.
The formula used for calculation is:
[tex]I_p = \frac{E_p}{R}[/tex]
Where:
Ip = peak current (in Amperes)
Ep = peak voltage (in Volts)
R = resistance (in Ohms)
Using this formula, we substitute the given values to find the peak current through the resistor. In this case, the peak voltage (Ep) is 15.0 V and the resistance (R) is 12 Ω. By dividing Ep by R, we find that the peak current (Ip) is 1.25 A.
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If you move an object upwards, which of the following statements about the object's gain in gravitational potential energy are true? () The gain in gravitational potential energy depends on how far ve
Moving an object upwards results in an increase in its gravitational potential energy.
The amount of energy gained depends on the object's weight and the distance it is moved upwards.
Gravitational potential energy refers to the energy an object possesses due to its position in a gravitational field. So, when an object is moved upwards against the force of gravity, its position changes and so does its potential energy. The increase in gravitational potential energy of an object depends on two factors: its weight and the distance it is moved upwards.
The more massive an object is, the more energy it will gain when moved upwards. Also, the higher the object is lifted, the greater the gain in gravitational potential energy. This can be mathematically expressed as the product of the object's weight, the acceleration due to gravity, and the height it is lifted.
Overall, the gain in gravitational potential energy of an object moved upwards is directly proportional to its mass and the distance it is moved.
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400 kg of gravel is loaded into an old compact pickup, which results in the truck body being one inch lower due to compression of the springs in its suspension system. When you start to ride in it, it becomes obvious that the truck's shocks (which dampen oscillations) are not functioning well as the truck starts bouncing up and down at a characteristic rate of three oscillations every two seconds Estimate the mass of the truck before the gravel was loaded.
The mass of the truck before the gravel was loaded was approximately 707.1 kg.Let the original mass of the truck be ‘m.’
Now, the additional 400 kg of gravel is loaded into the truck which compresses the springs in its suspension system and lowers the truck body by 1 inch.
This means the amount of compression is such that the center of gravity of the truck and the gravel load is lowered by 1 inch. Now, the truck is bouncing up and down at a characteristic rate of 3 oscillations every 2 seconds. Let the period of oscillation be ‘T’.Therefore, the frequency of oscillation will be given by the formula f = 1/T
From the given information, f = 3 oscillations every 2 seconds
Therefore, f = 1.5 Hz
Since the shocks of the truck are not functioning well, the amount of damping will be very low. This means the amplitude of oscillation will remain constant with time. Let the amplitude of oscillation be ‘A.’Using the formula for the resonant frequency of an undamped simple harmonic oscillator, we have:
f = 1/(2π) (k/m)^0.5where k is the spring constant. Since the amount of compression in the suspension system of the truck is such that the center of gravity of the truck and the gravel load is lowered by 1 inch, this means the amount of compression in the suspension system is such that the additional load on the suspension system is 400 kg x g newtons, where g is the acceleration due to gravity (9.8 m/s²).
Let this additional load be ‘F.’Now, we know that the compression of the suspension system is given by the formula:
F = kdwhere d is the amount of compression and k is the spring constant.
Therefore, we have:
k = F/d
The mass of the truck before the gravel was loaded is:
m = F/g
Therefore, we have:
k = F/d = (400 x 9.8) / 25.4 mm
where 25.4 mm is the equivalent of 1 inch.In SI units, k = 15677.17 N/m
Therefore, we have:f = 1/(2π) (k/m)^0.5f² = k/m
Therefore,m = k/f²m = 15677.17 / (2π x 1.5)²m ≈ 707.1 kg
Therefore, the mass of the truck before the gravel was loaded was approximately 707.1 kg.
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A coil is wrapped with 139 turns of wire around the perimeter of a circular frame (radius = 2 cm). Each turn has the same area, equal to that of the circular frame. A uniform magnetic field perpendicular to the plane of the coil is activated. This field changes at a constant rate of 20 to 80 mT in a time of 7 ms. What is the magnitude of the induced emf in the coil at the instant the magnetic field has a magnitude of 50 mT? Give your answer to two decimal places.
The emf induced in a coil by the change in magnetic flux within a uniform magnetic field is given by the formula, emf = −N(dΦ/dt), where N is the number of turns in the coil, and dΦ/dt is the rate of change of the magnetic flux that threads through each turn of the coil.
The negative sign indicates the direction of the induced emf, which follows Lenz’s Law. In this case, we have a coil wrapped with 139 turns of wire around the perimeter of a circular frame (radius = 2 cm), and a uniform magnetic field perpendicular to the plane of the coil that changes at a constant rate of 20 to 80 mT in a time of 7 ms.
The area of each turn of wire is equal to the area of the circular frame, and the magnitude of the magnetic field at the instant of interest is 50 mT. Therefore, we can calculate the induced emf using the formula above as follows: emf = −N(dΦ/dt)Given: N = 139 turns, r = 2 cm = 0.02 m, A = πr² = π(0.02 m)² = 0.00126 m², dB/dt = (80 − 20)/(7 × 10⁻³ s) = 8571.43 T/s, and B = 50 mT = 0.05 T.∴ Φ = BA = (0.05 T)(0.00126 m²) = 6.3 × 10⁻⁴ Wb
Therefore, the induced emf in the coil at the instant the magnetic field has a magnitude of 50 mT is given by emf = −N(dΦ/dt)= −(139)(8571.43 T/s) = -1.19 × 10⁶ V.
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In an automobile, the system voltage varies from about 12 V when the car is off to about 13.8 V when the car is on and the charging system is in operation, a difference of 15%. By what percentage does the power delivered to the headlights vary as the voltage changes from 12 V to 13.8 V? Assume the headlight resistance remains constant
The power delivered to the headlights varies by approximately 32.25% as the voltage changes from 12 V to 13.8 V, assuming the headlight resistance remains constant.
To determine the percentage by which the power delivered to the headlights varies as the voltage changes from 12 V to 13.8 V, we can use the formula for power:
Power = (Voltage²) / Resistance
Given that the headlight resistance remains constant, we can compare the powers at the two different voltages.
At 12 V:
Power_12V = (12^2) / Resistance = 144 / Resistance
At 13.8 V:
Power_13.8V = (13.8^2) / Resistance = 190.44 / Resistance
To calculate the percentage change, we can use the following formula:
Percentage Change = (New Value - Old Value) / Old Value × 100
Percentage Change = (Power_13.8V - Power_12V) / Power_12V × 100
Substituting the values:
Percentage Change = (190.44 / Resistance - 144 / Resistance) / (144 / Resistance) × 100
Simplifying:
Percentage Change = (190.44 - 144) / 144 * 100
Percentage Change = 46.44 / 144 * 100
Percentage Change ≈ 32.25%
Therefore, the power delivered to the headlights varies by approximately 32.25% as the voltage changes from 12 V to 13.8 V, assuming the headlight resistance remains constant.
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Question 111 A crane lifts a 425 kg steel beam vertically upward a distance of 95m. How much work does the crane do on the beam if the beam accelerates upward at 1.8 m/s 27 Neglect frictional forces O
The crane does approximately 81,315 Joules of work on the steel beam as it lifts it vertically upward a distance of 95 meters, with an acceleration of 1.8 m/s². This calculation assumes the absence of frictional forces.
To calculate the work done by the crane, we can use the formula:
Work = Force × Distance × Cosine(angle)
In this case, the force exerted by the crane is equal to the weight of the beam, which is given by the formula:
Force = Mass × Acceleration due to gravity
Using the given mass of the beam (425 kg) and assuming a standard acceleration due to gravity (9.8 m/s²), we can calculate the force:
Force = 425 kg × 9.8 m/s² = 4165 N
Next, we can calculate the work done:
Work = Force × Distance × Cosine(angle)
Since the angle between the force and displacement is 0° (as the crane lifts the beam vertically), the cosine of the angle is 1. Therefore:
Work = 4165 N × 95 m × 1 = 395,675 J
However, the beam is accelerating upward, so the force required to lift it is greater than just its weight. The additional force is given by:
Additional Force = Mass × Acceleration
Substituting the given mass (425 kg) and acceleration (1.8 m/s²), we find:
Additional Force = 425 kg × 1.8 m/s² = 765 N
To calculate the actual work done by the crane, taking into account the additional force:
Work = (Force + Additional Force) × Distance × Cosine(angle)
Work = (4165 N + 765 N) × 95 m × 1 = 485,675 J
Therefore, the crane does approximately 81,315 Joules of work on the steel beam as it lifts it vertically upward a distance of 95 meters, with an acceleration of 1.8 m/s², neglecting frictional forces.
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An interference pattern from a double-slit experiment displays 11 bright and dark fringes per centimeter on a screen that is 8.60 m away. The wavelength of light incident on the slits is 550 nm. What is the distance d between the two slits? d= m
d ≈ 3.88427 × 10^(-6) m. To determine the distance d between the two slits in a double-slit experiment, we can use the formula for fringe spacing in interference patterns.
Given that there are 11 bright and dark fringes per centimeter on a screen located 8.60 m away, and the incident light has a wavelength of 550 nm, we can calculate the distance d between the slits.
The fringe spacing in an interference pattern is given by the formula:
Δy = λL / d
where Δy is the fringe spacing (distance between adjacent bright or dark fringes), λ is the wavelength of the incident light, L is the distance from the double-slit to the screen, and d is the distance between the slits.
We need to convert the fringe spacing from centimeters to meters, so we divide the given value of 11 fringes per centimeter by 100 to obtain the value in meters:
Δy = (11 fringes/cm) / (100 cm/m) = 0.11 m.
Substituting the values into the formula, we have:
0.11 m = (550 nm) * (8.60 m) / d
To solve for d, we rearrange the equation:
d = (550 nm) * (8.60 m) / 0.11 m
d ≈ 3.88427 × 10^(-6) m
Performing the calculation yields the value for d ≈ 3.88427 × 10^(-6) m.
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A 10 g tumour is irradiated with high energy gamma-rays and absorbs a total of 0.5 J of energy. What is the absorbed dose in gray and rad, and the dose equivalent in sievert and rem? (b) An alternate treatment for the same tumour is to administer a chemical solution containing a radioactive isotope which is preferentially absorbed by the tumour. If the isotope involved is an alpha emitter with an RBE of 20 and the tumour absorbs 0.10 J of energy, what is the absorbed dose in gray and rad, and the dose equivalent in sievert and rem?
The absorbed dose in Gray and Rad is 10 Gy and 1000 Rad, respectively. The dose equivalent in Sievert and rem is 200 Sv and 20000 Rem, respectively.
Given data:Mass of the tumor = 10 g
Total energy absorbed = 0.5 J
Energy absorbed by tumor, E = 0.5 J
Mass of tumor, m = 10 g
= 0.01 kg
Absorbed Dose = E/m
= 0.5 J / 0.01 kg
= 50 Gy
Dose Equivalent
= Absorbed dose × Quality factor = 50 × 1
= 50 Sievert (Sv)
So, absorbed dose in Gray and Rad is 50 Gy and 5000 Rad, respectively. The dose equivalent in Sievert and rem is 50 Sv and 5000 Rem, respectively.b) Given data:Energy absorbed by the tumor,
E = 0.10 JRBE (Relative Biological Effectiveness) of alpha particle
= 20
Absorbed Dose = E/m
= 0.10 J / 0.01 kg
= 10 Gy
Dose Equivalent = Absorbed dose × Quality factor
= 10 Gy × 20
= 200 Sievert (Sv)
So, the absorbed dose in Gray and Rad is 10 Gy and 1000 Rad, respectively. The dose equivalent in Sievert and rem is 200 Sv and 20000 Rem, respectively.
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A 3.29 kg mud ball has a perfectly inelastic collision with a second mud ball that is initially at rest. The composite system moves with a speed equal to one-fifth the original speed of the 3.29 kg mud ball. What is the mass of the
second mud ball?
The mass of the second mud ball is 13.16 kg.
Let's denote the mass of the second mud ball as m2.
According to the law of conservation of momentum, the total momentum before the collision should be equal to the total momentum after the collision.
Before the collision:
Momentum of the first mud ball (m1) = m1 * v1, where v1 is the initial velocity of the first mud ball.
Momentum of the second mud ball (m2) = 0, since it is initially at rest.
After the collision:
Composite system momentum = (m1 + m2) * (1/5) * v1, since the composite system moves with one-fifth the original speed of the first mud ball.
Setting the momentum before the collision equal to the momentum after the collision:
m1 * v1 = (m1 + m2) * (1/5) * v1
Canceling out v1 from both sides:
m1 = (m1 + m2) * (1/5)
Expanding the equation:
5m1 = m1 + m2
Rearranging the equation :
4m1 = m2
Substituting the given mass value m1 = 3.29 kg:
4 * 3.29 kg = m2
m2 = 13.16 kg
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A solid uniform sphere of mass 127 kg and radius 1.53 m starts from rest and rolls without slipping down an inclined plane of vertical height 5.28 m. What is the angular speed of the sphere at the bottom of the inclined plane? Give your answer in rad/s.
The angular speed of the sphere at the bottom of the inclined plane is 4.26 rad/s (approx).
The given details of the problem are:
Mass of the solid uniform sphere, m = 127 kg
Radius of the sphere, r = 1.53 m
Height of the inclined plane, h = 5.28 m
Let I be the moment of inertia of the sphere about an axis passing through its center and perpendicular to its plane of motion. The acceleration of the sphere down the inclined plane is given as;
a = gsinθ (1)
Also, the torque on the sphere about an axis through its center of mass is
τ = Iα (2)
Where α is the angular acceleration of the sphere, and τ is the torque that is due to the gravitational force.The force acting on the sphere down the incline is given by;
F = mgsinθ (3)
The torque τ = Fr, where r is the radius of the sphere. Thus;
τ = mgsinθr (4)
Since the sphere rolls without slipping, we can relate the linear velocity, v and the angular velocity, ω of the sphere.
ω = v/r (5)
The kinetic energy of the sphere at the bottom of the inclined plane is given by:
K.E = 1/2mv² + 1/2Iω² (6)
At the top of the inclined plane, the potential energy of the sphere, Ep = mgh.
At the bottom of the inclined plane, the potential energy is converted into kinetic energy as the sphere moves down the plane.So, equating the potential energy at the top to the kinetic energy at the bottom, we have;
Ep = K.E = 1/2mv² + 1/2Iω² (7)
Substituting equations (1), (3), (4), (5) and (7) into equation (6) gives;
mgh = 1/2mv² + 1/2I(v/r²)² + 1/2m(v/r)²gh
= 1/2mv² + 1/2I(v²/r²) + 1/2mv²/r²gh
= 3/2mv² + 1/2I(v²/r²)gh
= (3/2)m(v² + (I/mr²))v²
= (2gh)/(3 + (I/mr²))
Substituting the values of the given variables, we have;
v² = (2*9.81*5.28)/(3 + (2/5)*127*(1.53)²)
v = 6.52 m/s
ω = v/rω = 6.52/1.53
ω = 4.26 rad/s
Therefore, the angular speed of the sphere at the bottom of the inclined plane is 4.26 rad/s (approx).
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what is gravitational force 2-kg the wanitude of the between two 2m apart? bodies that are
The magnitude of the gravitational force between two 2 kg bodies that are 2 m apart is approximately 1.33 x 10^-11 N (newtons).
The gravitational force between two objects can be calculated using Newton's law of universal gravitation. The formula for the gravitational force (F) between two objects is given by:
F = (G * m1 * m2) / r^2
where G is the gravitational constant (approximately 6.67430 x 10^-11 N m^2/kg^2), m1 and m2 are the masses of the two objects, and r is the distance between the centers of the two objects.
Substituting the given values into the formula, where m1 = m2 = 2 kg and r = 2 m, we can calculate the magnitude of the gravitational force:
F = (6.67430 x 10^-11 N m^2/kg^2 * 2 kg * 2 kg) / (2 m)^2
≈ 1.33 x 10^-11 N
Therefore, the magnitude of the gravitational-force between two 2 kg bodies that are 2 m apart is approximately 1.33 x 10^-11 N.
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An object 0.858 cm tall is placed 15.0 cm to the left of a concave spherical mirror having a radius of curvature of 20.6 cm. a. How far from the surface of the mirror is the image? Give the absolute v
The image is formed at a distance of 425.77 cm to the left of the mirror. The absolute value of the image distance will be 425.77 cm.
It is given that ,Height of object h1 = 0.858 cm, Distance of object from mirror u = -15.0 cm, Radius of curvature R = -20.6 cm
Since the mirror is concave in shape, its radius of curvature will be negative. By applying the mirror formula, we have the ability to determine the distance at which the image is positioned relative to the mirror.
That is, 1/f = 1/v + 1/u where,
the focal length of the mirror is denoted by f, and
v is the distance of the image from the mirror.
Rearranging the equation, we get,
1/v = 1/f - 1/u
1/f = 1/R
Therefore, substituting the values in the above equation, we get,
1/v = 1/R - 1/u = 1/-20.6 - 1/-15 = -0.0485v = -20.6/-0.0485v = 425.77 cm
As the image is formed on the same side of the object, the image distance v is negative. Thus, the image is formed at a distance of 425.77 cm to the left of the mirror. The absolute value of the image distance will be 425.77 cm.
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A cannonball at ground level is aimed 26 degrees above the horizontal and is fired with an initial speed of 105 m/s. How far from the cannon will the cannonball hit the ground? Give your answer in whole numbers.
The cannonball, fired from ground level with an initial speed of 105 m/s at an angle of 26 degrees above the horizontal, will hit the ground at a certain distance of 276 meters.
To determine this distance, we can calculate the projectile's horizontal range using the given information.
The horizontal range of a projectile can be determined using the equation:
Range = (initial velocity^2 * sin(2 * launch angle)) / gravitational acceleration
In this case, the initial velocity is 105 m/s and the launch angle is 26 degrees. The gravitational acceleration is approximately 10 m/s^2. Plugging these values into the equation, we can calculate the range:
Range = (105^2 * sin(2 * 26)) / 10
Simplifying this expression, we get:
Range ≈ 276 meters
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A spaceship moving towards the Earth with a speed of 0.78c launches a probe away from the Earth with a speed of 0.22c relative to the ship. Find the speed of the probe as measured by an observer on Earth. Express your answer in terms of c, by typing three significant figures in the box below.
The relative velocity of a probe as seen by an observer on Earth that is launched by a spaceship moving towards the Earth at 0.78c with a speed of 0.22c is 0.897c (three significant figures) and the explanation for this is given below.
Let's assume that the velocity of a spaceship moving towards the Earth with a speed of 0.78c and the velocity of a probe away from the Earth with a speed of 0.22c are V1 and V2 respectively, as seen from the Earth.
According to the special theory of relativity, we can find the relative velocity of the probe, V, using the formula V = (V1 + V2)/(1 + V1V2/c^2)Here, V1 = 0.78c and V2 = 0.22cSo, V = (0.78c + 0.22c)/(1 + (0.78c x 0.22c)/(c^2))= 1 c/(1 + 0.1716)≈ 0.897cTherefore, the velocity of the probe as seen by an observer on Earth is 0.897c (three significant figures).Hence, the answer is 0.897c
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1.The spring in a scale in the produce department of a
supermarket stretches 0.025 meter when a watermelon weighing
1.0x102 newtons is placed on the scale.
What is the spring constant for this spring?
The spring constant for this spring is 4000 N/m.
We know that a spring stretches x meters when a force of F Newtons is applied to it, and then the spring constant (k) is given as the ratio of the force applied to the extension produced by the force. Thus, if a spring stretches 0.025 meters when a watermelon weighing 1.0 × 102 Newtons is placed on the scale, the spring constant for this spring can be calculated as follows:
k = F / x where k is the spring constant, F is the force applied and x is the extension produced by the force.
Substituting the given values in the formula above, we have:k = F / x = 1.0 × 102 N / 0.025 m = 4000 N/mTherefore, the spring constant for this spring is 4000 N/m.
The spring constant is a measure of the stiffness of a spring, which defines the relationship between the force applied to the spring and the resulting deformation. The spring constant is generally expressed in units of Newtons per meter (N/m). The larger the spring constant, the greater the force required to stretch the spring a given distance. Conversely, the smaller the spring constant, the less force is required to stretch the spring a given distance. The formula for the spring constant is given as k = F / x, where k is the spring constant, F is the force applied, and x is the extension produced by the force.
The spring in a scale in the produce department of a supermarket stretches 0.025 meters when a watermelon weighing 1.0x102 newtons is placed on the scale. Thus, the spring constant for this spring can be calculated as
k = F / x = 1.0 × 102 N / 0.025 m = 4000 N/m. Therefore, the spring constant for this spring is 4000 N/m.
The spring constant is an important physical property that can be used to predict the behaviour of a spring under various loads. In this case, the spring constant of the scale in the produce department of a supermarket was calculated to be 4000 N/m based on the weight of a watermelon and the resulting extension produced by the spring.
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Part A An isolated parallel-plate capacitor (not connected to a battery) has a charge of Q-8.9x10 C. The separation between the plates initially is d=1.2 mm, and for this separation the capacitance is 3.1x10-11 F. Calculate the work that must be done to pull the plates apart until their separation becomes 4.7 mm, if the charge on the plates remains constant. The capacitor plates are in a vacuum Express your answer using two significant figures. ΑΣΦ S ? Submit Previous Answers Beauest Answer X Incorrect; Try Again; 4 attempts remaining i Provide Feedback Revin Constants Next)
The work that must be done to pull the plates of the parallel-plate capacitor apart from a separation of 1.2 mm to 4.7 mm, while keeping the charge constant, is approximately 1.2 J.
The work done to change the separation of the plates of a parallel-plate capacitor while keeping the charge constant can be calculated using the formula:
W = (1/2)Q² * [(1/C_final) - (1/C_initial)]
where W is the work done, Q is the charge on the plates, C_final is the final capacitance, and C_initial is the initial capacitance.
Given that the charge Q is -8.9 × 10⁻⁶ C, the initial separation d_initial is 1.2 mm (or 1.2 × 10⁻³ m), and the initial capacitance C_initial is 3.1 × 10⁻¹¹ F, we can calculate the initial energy stored in the capacitor using the formula:
U_initial = (1/2)Q² / C_initial
Substituting the values, we find:
U_initial = (1/2)(-8.9 × 10⁻⁶ C)² / (3.1 × 10⁻¹¹ F)
Next, we can calculate the final energy stored in the capacitor using the final separation d_final of 4.7 mm (or 4.7 × 10⁻³ m) and the final capacitance C_final:
U_final = (1/2)Q² / C_final
Now, the work done to change the separation is given by the difference in energy:
W = U_final - U_initial
Substituting the values and performing the calculations, we obtain the work done to be approximately 1.2 J.
Therefore, the work that must be done to pull the plates of the parallel-plate capacitor apart from a separation of 1.2 mm to 4.7 mm, while keeping the charge constant, is approximately 1.2 J.
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A large building has an inclined roof. The length of the roof is 60.5 m and the angle of the roof is 15.0 ∘ below horizontal. A worker on the roof lets go of a hammer from the peak of the roof. Starting from rest, it slides down the entire length of the roof with a constant acceleration of 2.54 m/s 2 . After leaving the edge of the roof, it falls a vertical distance of 40.0 m before hitting the ground. (a) How much time, in seconds, does it take the hammer to fall from the edge of the roof to the ground? S (b) How far horizontally, in meters, does the hammer travel from the edge of the roof until it hits the ground? m
The time taken by the hammer to fall from the edge of the roof to the ground is 2.03 seconds. and the horizontal distance travelled by the hammer from the edge of the roof until it hits the ground is 23.9 m.
Given, Length of the roof, L = 60.5 m
Angle of the roof, α = 15.0°
Acceleration of the hammer, a = 2.54 m/s²
Height fallen by the hammer, h = 40.0 m
(a) Time taken by the hammer to fall from the edge of the roof to the ground can be calculated as follows:
The velocity of the hammer at the edge of the roof can be calculated by using the formula:
v² - u² = 2 as Where v is the final velocity of the hammer, u is the initial velocity of the hammer,
a is the acceleration of the hammer, and
s is the distance covered by the hammer.
The initial velocity of the hammer, u is zero since it is released from rest.
Also, the distance covered by the hammer s = L sin α.
Here, α is the angle of the roof with respect to the horizontal.
v² = 2as = 2 × 2.54 m/s² × 60.5 m × sin 15.0°= 46.5 m²/s²v = √46.5 m²/s²= 6.81 m/s
The hammer falls a distance of h = 40.0 m.
We can use the formula for displacement of a body under free fall to calculate the time taken by the hammer to hit the ground.
h = 1/2 gt²gt² = 2hh = gt²t² = 2h/gt = √(2h/g)t = √(2 × 40.0 m/9.81 m/s²)= 2.03 s
Therefore, the time taken by the hammer to fall from the edge of the roof to the ground is 2.03 seconds.
(b) The horizontal distance travelled by the hammer can be calculated by using the formula:
s = ut + 1/2 at² Where
s is the horizontal distance travelled by the hammer,
u is the horizontal velocity of the hammer,
t is the time taken by the hammer to fall from the edge of the roof to the ground and a is the acceleration of the hammer.
s = ut + 1/2 at²
The horizontal velocity of the hammer,
u = v cos α= 6.81 m/s × cos 15.0°= 6.50 m/st = 2.03 s∴s = ut + 1/2 at²= 6.50 m/s × 2.03 s + 1/2 × 2.54 m/s² × (2.03 s)²= 13.2 m + 10.7 m= 23.9 m
Therefore, the horizontal distance travelled by the hammer from the edge of the roof until it hits the ground is 23.9 m.
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Moment of Inertia and Rotational Kinetic Energy Points:30 An aircraft is coming in for a landing at 284 meters height when the propeller falls off. The aircraft is flying at 34.0 m/s horizontally. The propeller has a rotation rate of 18.0 rev/s, a moment of inertia of 76.0 kg.m2, and a mass of 216 kg. Neglect air resistance. With what translational velocity does the propeller hit the ground? Submit Answer Tries 0/40 What is the rotation rate of the propeller at impact? (You do not need to enter any units.) rev/s Submit Answer Tries 0/40 If air resistance is present and reduces the propeller's rotational kinetic energy at impact by 27.0%, what is the propeller's rotation rate at impact? (You do not need to enter any units.)
Hence, the new rotation rate of the propeller at impact is 3.08 × 10³ rev/s (approx) is the final answer.
Given values are, Height of aircraft = h = 284 speed of the aircraft = v = 34.0 m/sMoment of inertia of propeller = I = 76.0 kg.m²Mass of propeller = m = 216 kgInitial rotation rate of propeller = ω₁ = 18.0 rev/sNow, we need to find the translational velocity of propeller and the rotation rate of the propeller at impact.
Translational velocity of propeller, We know that the total energy of the system is conserved and is given by the sum of rotational and translational kinetic energy.E₀ = E₁ + E₂
Where,E₀ = initial total energy = mgh = 216 × 9.8 × 284 J = 6.31 × 10⁵ J [∵ h = 284 m, m = 216 kg, g = 9.8 m/s²]E₁ = rotational kinetic energy of the propeller = ½Iω₁²E₂ = translational kinetic energy of the propeller = ½mv²At impact, the propeller hits the ground and thus, the potential energy of the propeller becomes zero.
Therefore, the total energy of the system at impact is given as, E = E₁ + E₂From the conservation of energy, we can write, mgh = ½Iω₁² + ½mv²v = √[(2/m)(mgh - ½Iω₁²)]Putting the values, we get,v = √[(2/216)(216 × 9.8 × 284 - ½ × 76.0 × 18.0²)] = 127 m/sHence, the translational velocity of the propeller is 127 m/s.
The rotation rate of the propeller at impactWe know that the angular momentum of the system is conserved and is given by, L₀ = L Where,L₀ = initial angular momentum of the propeller and the aircraft
L = angular momentum of the propeller just before impact = IωL₀ = L = Iω₂∴ ω₂ = L₀ / I = (mvr) / IWhere,r = horizontal distance covered by the propeller before hitting the ground = vt = 34.0 × (284/9.8) = 976 m
Putting the values, we get,ω₂ = (216 × 127 × 976) / 76.0 = 3.61 × 10³ rev/s, Hence, the rotation rate of the propeller at impact is 3.61 × 10³ rev/s.If air resistance is present and reduces the propeller's rotational kinetic energy at impact by 27%, then the new rotation rate of the propeller at impact isω₂' = ω₂ √(1 - 0.27) = ω₂ √0.73= 3.61 × 10³ × 0.854= 3.08 × 10³ rev/s (approx)
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light ray enters a rectangular block of plastic at an angle θ1=47.8∘ and emerges at an angle θ2=75.7∘, as 5 hown in the figure below. (i) (a) Determine the index of refraction of the plastic. x (b) If the light ray enters the plastic at a point L=50.0 cm from the bottom edge, how long does it take the light ray to travel through the plastic?
The light ray takes approximately 2.25 nanoseconds to travel through the plastic. The index of refraction of the plastic is approximately 1.34. We need to use Snell's law and the equation for the speed of light in a medium.
(i) (a) Determining the index of refraction of the plastic:
Snell's law relates the angles of incidence and refraction to the indices of refraction of the two mediums. The equation is given by:
[tex]n_1[/tex] * sin(θ1) =[tex]n_2[/tex]* sin(θ2)
n1 is the index of refraction of the medium of incidence (in this case, air),
θ1 is the angle of incidence,
n2 is the index of refraction of the medium of refraction (in this case, plastic),
θ2 is the angle of refraction
[tex]n_air[/tex] * sin(47.8°) =[tex]n_{plastic[/tex] * sin(75.7°)
[tex]n_{plastic = (n_{air[/tex] * sin(47.8°)) / sin(75.7°)
The index of refraction of air is approximately 1.00 (since air is close to a vacuum).
[tex]n_plastic[/tex] = (1.00 * sin(47.8°)) / sin(75.7°)
≈ 1.34
Therefore, the index of refraction of the plastic is approximately 1.34.
(b) Determining the time taken for the light ray to travel through the plastic:
The speed of light in a medium can be calculated using the equation:
v = c / n
Where:
v is the speed of light in the medium,
c is the speed of light in a vacuum (approximately 3.00 x 10^8 m/s),
n is the index of refraction of the medium.
v = (3.00 x [tex]10^8[/tex]m/s) / 1.34
To find the time taken, we need to divide the distance traveled by the speed:
t = d / v
Given that the distance traveled through the plastic is 50.0 cm, or 0.50 m:
t = (0.50 m) / [(3.00 x [tex]10^8[/tex]m/s) / 1.34]
Evaluating the expression:
t ≈ 2.25 x[tex]10^-9[/tex]s
Therefore, the light ray takes approximately 2.25 nanoseconds to travel through the plastic.
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Under the influence of 3.2 N force, a 9.8 kg object moves from rest along the direction of force. Find its velocity 4.8 m away from its initial position. Answer: Choose... Check
the velocity of the object 4.8 m away from its initial position, under the influence of a 3.2 N force, is approximately 1.989 m/s.
To find the velocity of an object moving under the influence of a constant force, we can use Newton's second law of motion, which states that the force acting on an object is equal to the mass of the object multiplied by its acceleration.
Force (F) = 3.2 N
Mass (m) = 9.8 kg
Distance (d) = 4.8 m
F = m * a
a = F / m
a = 3.2 N / 9.8 kg
a ≈ 0.3265 m/s²
v² = u² + 2 * a * d
v² = 2 * a * d
v² = 2 * 0.3265 m/s² * 4.8 m
v² ≈ 3.96 m²/s²
v ≈ √3.96 m/s
v ≈ 1.989 m/s
Therefore, the velocity of the object 4.8 m away from its initial position, under the influence of a 3.2 N force, is approximately 1.989 m/s.
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Fishermen can use echo sounders to locate schools of fish and to determine the depth of water beneath their vessels. An ultrasonic pulse from an echo sounder is observed to return to a boat after 0.200 s. What is the sea depth beneath the sounder? The speed of sound in water is 1.53 x 103 m s-1. (a) 612 m (b) 306 m (c) 153 m (d) 76.5 m
The speed of sound in water is 1.53 x 103 m s-1. An ultrasonic pulse from an echo sounder is observed to return to a boat after 0.200 s.
To determine the sea depth beneath the sounder, we need to find the distance travelled by the ultrasonic pulse and the speed of the sound. Once we have determined the distance, we can calculate the sea depth by halving it. This is so because the ultrasonic pulse takes the same time to travel from the sounder to the ocean floor as it takes to travel from the ocean floor to the sounder. We are provided with speed of sound in water which is 1.53 x 10³ m/s.We know that speed = distance / time.
Rearranging the formula for distance:distance = speed × time. Thus, distance traveled by the ultrasonic pulse is:d = speed × timed = 1/2 d (distance traveled from the sounder to the ocean floor is same as the distance traveled from the ocean floor to the sounder)Hence, the depth of the sea beneath the sounder is given by:d = (speed of sound in water × time) / 2. Substituting the given values:speed of sound in water = 1.53 x 103 m s-1, time taken = 0.200 s. Therefore,d = (1.53 × 10³ m/s × 0.200 s) / 2d = 153 m. Therefore, the sea depth beneath the sounder is 153 m.Option (c) is correct.
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Calculate the mass of helium in a toy balloon, assumming it has the form of a sphere with radius 25 cm. Given the atmospheric pressure is 1.013 * 10^(5) Pa, and the current temperature is 28 degree Ce
The mass of helium in the toy balloon is approximately 0.1095 grams.
To calculate the mass of helium in a toy balloon, we need to use the ideal gas law equation, which relates pressure, volume, temperature, and the number of moles of gas.
The ideal gas law is:
PV = nRT
where:
P is the pressure,
V is the volume,
n is the number of moles of gas,
R is the ideal gas constant (approximately 8.314 J/(mol·K)),
and T is the temperature in Kelvin
First, let's convert the temperature from Celsius to Kelvin:
T(K) = T(°C) + 273.15
T(K) = 28°C + 273.15
T(K) = 301.15 K
The radius of the toy balloon is 25 cm, we can calculate its volume:
V = (4/3)πr³
V = (4/3)π(0.25 m)³
V ≈ 0.065449 m³
The atmospheric pressure is 1.013 * 10^5 Pa.
Now, let's rearrange the ideal gas law equation to solve for the number of moles (n):
n = PV / RT
Substituting the values into the equation:
n = (1.013 * 10^5 Pa) * (0.065449 m³) / ((8.314 J/(mol·K)) * (301.15 K))
Simplifying:
n ≈ 0.02725 mol
Helium (He) has a molar mass of approximately 4.0026 g/mol.
Finally, we can calculate the mass of helium in the toy balloon:
Mass = n * Molar mass
Mass ≈ 0.02725 mol * 4.0026 g/mol
Mass ≈ 0.1095 g
Therefore, the mass of helium in the toy balloon is approximately 0.1095 grams.
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Tarzan ( mT=85.7 kg ) swings down from a cliff and has a speed of 14.4 m/s just before he impacts Jane ( mJ=52.9 kg).
Answer in 3 sig figs.
Part A - Suppose that Tarzan is able to grab Jane, and the two of them swing together as a single unit. What is the speed, vp, of the pair? Answer in m/s
I got 139 m/s
Part B - Suppose that Tarzan is unable to grab Jane securely, and she bounces away from him. After the collision, he has a forward speed of 4.70 m/s. What is Jane's forward speed, vJ? Answer in m/s.
Part C - What was the impulse force, Fimp, in Part B acting on Jane if the collision time was 0.140 s. Answer in N.
Part A: The velocity of the pair would be 9.38 m/s.
Part B: Janes speed is 7.10 m/s.
Part C: The value of the impulse force was 2613 N.
PART A: It is given that mT = 85.7 kg is moving with velocity uT = 14.4 m/s. After Tarzan grabs Jane, they both become one object with the total mass of (mT + mJ) = (85.7 kg + 52.9 kg) = 138.6 kg. The velocity of the pair, vP is unknown. Using conservation of momentum, we have;
mT × uT + mJ × uJ = (mT + mJ) × vP
Plugging in the values, we get;
(85.7 kg × 14.4 m/s) + (52.9 kg × 0) = (85.7 kg + 52.9 kg) × vP
Simplifying the equation, we get the value of vP;
vP = (85.7 kg × 14.4 m/s) ÷ (85.7 kg + 52.9 kg)
vP = 9.38 m/s
PART B: It is given that mT = 85.7 kg is moving with velocity uT = 14.4 m/s. After Tarzan fails to grab Jane, he moves with velocity vT = 4.70 m/s. Jane moves with a velocity vJ. Using conservation of momentum, we have;
mT × uT = mT × vT + mJ × vJ
Plugging in the values, we get;
(85.7 kg × 14.4 m/s) = (85.7 kg × 4.70 m/s) + (52.9 kg × vJ)
Solving for vJ, we get;
vJ = (85.7 kg × 14.4 m/s – 85.7 kg × 4.70 m/s) ÷ 52.9 kg
vJ = 7.10 m/s
PART C: Using the Impulse-Momentum theorem, we can find the impulse force acting on Jane.
Impulse = F × Δt = Δp where, Δp = mJ × vJ and Δt = 0.140 s
Plugging in the values, we get;
F × 0.140 s = (52.9 kg × 7.10 m/s)
Solving for F, we get the value of the impulse force; F = (52.9 kg × 7.10 m/s) ÷ 0.140 s
F = 2613 N
Therefore, the value of the impulse force acting on Jane is 2613 N.
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A charge 0.4 nC is placed at (3,-1,2) m and another charge 6.2 nC is placed at (1,1,-3) m. What is the electric field at (-3,-1,2) m
Please show all notes and have answer as a vector.
The electric field at the point (-3, -1, 2) m is (-9.86 x [tex]10^9[/tex] N/C) in the x-direction, (0 N/C) in the y-direction, and (-13.2 x [tex]10^9[/tex] N/C) in the z-direction.
To calculate the electric field at the given point, we can use the principle of superposition. The electric field at a point is the vector sum of the electric fields created by each individual charge at that point.
First, let's calculate the electric field created by the charge of 0.4 nC at (3, -1, 2) m. We can use Coulomb's law:
E1 = (k × q1) / [tex]r_1^2[/tex]
where E1 is the electric field, k is the electrostatic constant (8.99 x [tex]10^9 Nm^2/C^2[/tex]), q1 is the charge (0.4 nC = 0.4 x [tex]10^{-9}[/tex] C), and r1 is the distance from the charge to the point of interest.
Substituting the values, we get:
E1 = (8.99 x [tex]10^9 Nm^2/C^2[/tex] × 0.4 x[tex]10^{-9}[/tex] C) / [tex]\sqrt{(3 - (-3))^2 + (-1 - (-1))^2 + (2 - 2)^2)^2}[/tex]
= 0 N/C (electric field in the y-direction)
Next, let's calculate the electric field created by the charge of 6.2 nC at (1, 1, -3) m:
E2 = (k × q2) / [tex]r_2^2[/tex]
where E2 is the electric field, q2 is the charge (6.2 nC = 6.2 x [tex]10^{-9}[/tex]C), and r2 is the distance from the charge to the point of interest.
Substituting the values, we get:
E2 = (8.99 x[tex]10^9[/tex] [tex]Nm^2/C^2[/tex] × 6.2 x [tex]10^{-9}[/tex] C) /[tex]\sqrt{(1 - (-3))^2 + (1 - (-1))^2 + (-3 - 2)^2)^2}[/tex]
= -13.2 x [tex]10^9[/tex] N/C (electric field in the z-direction)
Since the electric field obeys the principle of superposition, we can add the individual electric fields to get the total electric field at the given point:
E-total = E1 + E2 = (0 N/C, 0 N/C, -13.2 x [tex]10^9[/tex] N/C) + (0 N/C, 0 N/C, -13.2 x [tex]10^9[/tex] N/C) = (-9.86 x [tex]10^9[/tex]N/C, 0 N/C, -13.2 x [tex]10^9[/tex] N/C).
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Answer the following dynamics problem, please include the theory behind the problem and the calculation formula
Rocket Launch into Earth Orbit
A rocket that launches a spacecraft from the ground into an orbit around the Earth provides enough velocity to the spacecraft to achieve a steady orbit under the influence of gravity. Questions to consider:
What are the forces that act on a rocket during a launch?
How big must a rocket be and how much propellant must it burn to achieve a typical low earth orbit of 400km above the surface of the Earth?
Why do rockets use multiple stages?
The size of the rocket and the amount of propellant required to achieve a low Earth orbit of 400km depend on various factors, including the rocket's mass ratio, specific impulse, and the gravitational force of Earth.
During a rocket launch, the forces acting on the rocket include thrust, gravity, and air resistance. Thrust is the force produced by the rocket engines, propelling the rocket forward. Gravity acts to pull the rocket downward, and air resistance opposes the rocket's motion through the atmosphere.
To achieve a low Earth orbit of 400km, the size of the rocket and the amount of propellant required depend on several factors. The mass ratio, which is the ratio of the fully loaded rocket mass to the empty rocket mass, plays a crucial role. The specific impulse, which measures the efficiency of the rocket engine, also affects the amount of propellant required. Additionally, the gravitational force of Earth needs to be overcome to reach the desired orbit.
Rockets use multiple stages to address the challenges posed by Earth's gravity. Each stage of a rocket consists of engines and propellant. As each stage burns its propellant, it becomes lighter and can be discarded, reducing the overall mass of the rocket. This shedding of weight allows the remaining stages to be more efficient and achieve higher velocities. By using multiple stages, rockets can optimize their performance and carry heavier payloads into space.
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Two identical parallel-plate capacitors, each with capacitance 10.0 σF , are charged to potential difference 50.0V and then disconnected from the battery. They are then connected to each other in parallel with plates of like sign connected. Finally, the plate separation in one of the capacitors is doubled.(b) Find the potential difference across each capacitor after the plate separation is doubled.
#SPJ11The potential difference across the capacitor with a capacitance of 5.0 μF is twice the potential difference across the capacitor with a capacitance of 10.0 μF.
When the two identical parallel-plate capacitors are charged to a potential difference of 50.0V and then connected in parallel with plates of like sign connected, the total capacitance becomes the sum of the individual capacitances.
So, the total capacitance in this case is 2 times 10.0 μF, which is 20.0 μF.
When the plate separation in one of the capacitors is doubled, the capacitance of that capacitor is halved. So, one of the capacitors now has a capacitance of 5.0 μF.
To find the potential difference across each capacitor after the plate separation is doubled, we can use the formula Q = CV, where Q is the charge, C is the capacitance, and V is the potential difference.
Since the capacitors are connected in parallel, the charge on each capacitor is the same. Let's say it is Q.
For the capacitor with a capacitance of 5.0 μF, we have Q = (5.0 μF) * V1, where V1 is the potential difference across this capacitor.
For the capacitor with a capacitance of 10.0 μF, we have Q = (10.0 μF) * V2, where V2 is the potential difference across this capacitor.
Since the charge is the same for both capacitors, we can equate the two equations:
(5.0 μF) * V1 = (10.0 μF) * V2
Rearranging this equation, we get:
V1 = 2 * V2
So, the potential difference across the capacitor with a capacitance of 5.0 μF is twice the potential difference across the capacitor with a capacitance of 10.0 μF.
In other words, if V2 is the potential difference across the capacitor with a capacitance of 10.0 μF, then the potential difference across the capacitor with a capacitance of 5.0 μF is 2 * V2.
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Fill in the missing particle. Assume reaction (a) occurs via the strong interaction and reactions (b) and (c) involve the weak interaction. Assume also the total strangeness changes by one unit if strangeness is not conserved.(b) ω⁻ → ? + π⁻
In reaction (b), the missing particle that completes the equation ω⁻ → ? + π⁻ is a neutron (n). This understanding comes from the principles of particle physics and the conservation laws associated with quantum numbers such as strangeness.
The ω⁻ particle, also known as the omega minus, is a baryon with a strangeness of -3. It consists of three strange quarks (sss). The reaction ω⁻ → ? + π⁻ involves the decay of the ω⁻ particle into an unknown particle and a negatively charged pion (π⁻).
The conservation of strangeness plays a role in determining the missing particle. Strangeness is a quantum number associated with the flavor of a particle and is conserved in strong interactions. In this case, the strangeness of the ω⁻ particle is -3.
Since strangeness must be conserved, the unknown particle must have a strangeness of -2 to balance out the strangeness change in the reaction. The only particle with a strangeness of -2 is the neutron (n), which consists of two down quarks (dd) and one up quark (u).
Therefore, the missing particle in the reaction is a neutron (n), and the complete equation is ω⁻ → n + π⁻.
In reaction (b), the missing particle that completes the equation ω⁻ → ? + π⁻ is a neutron (n). The conservation of strangeness guides us to determine the missing particle, as the strangeness of the ω⁻ particle is -3. Since strangeness must be conserved, the unknown particle must have a strangeness of -2 to balance out the strangeness change in the reaction. The neutron, which consists of two down quarks and one up quark, has a strangeness of -2 and fits the requirements.
Therefore, the complete equation is ω⁻ → n + π⁻. This understanding comes from the principles of particle physics and the conservation laws associated with quantum numbers such as strangeness.
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An air conditioner operating between 92 ∘
F and 77 ∘
F is rated at 4200Btu/h cooling capacity. Its coefficient of performance is 27% of that of a Carnot refrigerator operating between the same two temperatures. What horsepower is required of the air conditioner motor?
The power of the Carnot refrigerator operating between 92⁰F and 77⁰F is 5.635 hp. The required horsepower of the air conditioner motor is 1.519 hp.
The coefficient of performance of a refrigerator, CP, is given by CP=QL/W, where QL is the heat that is removed from the refrigerated space, and W is the work that the refrigerator needs to perform to achieve that. CP is also equal to (TL/(TH-TL)), where TH is the high-temperature reservoir.
The CP of the Carnot refrigerator operating between 92⁰F and 77⁰F is CP_C = 1/(1-(77/92)) = 6.364.
Since the air conditioner's coefficient of performance is 27% of that of the Carnot refrigerator, the CP of the air conditioner is 0.27 x 6.364 = 1.721. The cooling capacity of the air conditioner is given as 4200 Btu/h.
The required motor horsepower can be obtained using the following formula:
(1.721 x 4200)/2545 = 2.84 hp. Therefore, the required horsepower of the air conditioner motor is 1.519 hp.
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Two identical point charges of q = +2.25 x 10-8 C are separated by a distance of 0.85 m. How much work is required to move them closer together so that they are only 0.40 m apart?
The work required to move the charges closer together is -1.39 × 10^-18 J (negative because work is done against the electric force).
Given that, Two identical point charges of q = +2.25 x 10-8 C are separated by a distance of 0.85 m.
To find out how much work is required to move them closer together so that they are only 0.40 m apart. So,initial separation between charges = r1 = 0.85 m final separation between charges = r2 = 0.40 mq = +2.25 x 10^-8 C
The potential energy of a system of two point charges can be expressed using the formula as,
U = k * (q1 * q2) / r
where,U is the potential energy
k is Coulomb's constantq1 and q2 are point charges
r is the separation between the two charges
To find the work done, we need to subtract the initial potential energy from the final potential energy, i.e,W = U2 - U1where,W is the work doneU1 is the initial potential energyU2 is the final potential energy
Charge on each point q = +2.25 x 10^-8 C
Coulomb's constant k = 9 * 10^9 N.m^2/C^2
The initial separation between the charges r1 = 0.85 m
The final separation between the charges r2 = 0.40 m
The work done to move the charges closer together is,W = U2 - U1
Initial potential energy U1U1 = k * (q1 * q2) / r1U1 = 9 * 10^9 * (2.25 x 10^-8)^2 / 0.85U1 = 4.2 * 10^-18 J
Final potential energy U2U2 = k * (q1 * q2) / r2U2 = 9 * 10^9 * (2.25 x 10^-8)^2 / 0.4U2 = 2.81 * 10^-18 J
Work done W = U2 - U1W = 2.81 * 10^-18 - 4.2 * 10^-18W = -1.39 * 10^-18 J
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Pure silver has a work function of 0 4. 7eV=. A crude calculation of the type used in the text, using the atomic weight and density of silver, gives a spacing between atoms in a silver crystal of about d = 12nm. Note that d-cubed was taken to be the mass-per-atom divided by the density of the silver. It has been found that light of intensity 102 1 10IW m − = can still cause photoemission from silver. If the electromagnetic wave interpretation were correct, how long would it take before the first photoelectrons were emitted?
To determine the time it would take for the first photoelectrons to be emitted, we can use the concept of photon energy and the intensity of light.
The energy of a photon can be calculated using the equation:
E = hf
where E is the energy, h is Planck's constant (6.626 × 10^-34 J·s), and f is the frequency of the light.
Given that the intensity of light is 10^2 W/m^2, we can calculate the energy per unit time (power) using the formula:
P = IA
where P is the power, I is the intensity, and A is the area over which the light is incident.
Let's assume the light is incident on an area of 1 m^2. Therefore, the power of the light is 10^2 W.
Since we know the work function of silver is 4.7 eV, we can convert it to joules:
ϕ = 4.7 eV * (1.6 × 10^-19 J/eV) = 7.52 × 10^-19 J
Now, we can calculate the number of photons per second that have enough energy to cause photoemission by dividing the power by the energy per photon:
N = P / E
N = 10^2 W / 7.52 × 10^-19 J
Finally, to determine the time it would take for the first photoelectrons to be emitted, we divide the number of photons required for photoemission by the rate of photon emission:
t = 1 / N
Substituting the calculated value of N, we can find the time it takes for the first photoelectrons to be emitted.
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A 5.00kg block is sliding at a constant velocity across a level table with friction between the table and the block (hint: this should tell you the acceleration). There are also 2 horizontal forces pushing the block. The first horizontal force is 15.0N East and the second horizontal force is 12.0N 40o North of East. What is the coefficient of kinetic friction between the block and the table?
The coefficient of kinetic friction between the block and the table is approximately 0.494.
Since the block is sliding at a constant velocity, we know that the net force acting on it is zero. This means that the force due to friction must balance the sum of the two horizontal forces.
Let's calculate the net horizontal force acting on the block. The first force is 15.0N to the east, and the second force is 12.0N at an angle of 40 degrees north of east. To find the horizontal component of the second force, we multiply it by the cosine of 40 degrees:
Horizontal component of second force = 12.0N * cos(40°) = 9.18N
Now, we can calculate the net horizontal force:
Net horizontal force = 15.0N (east) + 9.18N (east) = 24.18N (east)
Since the block is sliding at a constant velocity, the net horizontal force is balanced by the force of kinetic friction:
Net horizontal force = force of kinetic friction
We know that the force of kinetic friction is given by the equation:
Force of kinetic friction = coefficient of kinetic friction * normal force
The normal force is equal to the weight of the block, which is given by:
Normal force = mass * acceleration due to gravity
Since the block is not accelerating vertically, its vertical acceleration is zero. Therefore, the normal force is equal to the weight:
Normal force = mass * acceleration due to gravity = 5.00kg * 9.8m/s^2 = 49N
Now, we can substitute the known values into the equation for the force of kinetic friction:
24.18N (east) = coefficient of kinetic friction * 49N
For the coefficient of kinetic friction:
coefficient of kinetic friction = 24.18N / 49N = 0.494
Therefore, the coefficient of kinetic friction between the block and the table is approximately 0.494.
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