The maximum speed of the object is approximately 0.689 m/s.The speed when the spring is compressed 1.5 cm and as it passes a point 1.5 cm from the equilibrium position is approximately 0.332 m/s.
The value of x at which the speed equals one-half the maximum speed is approximately 0.183 m.
(a) To find the maximum speed of the object, we can use the principle of energy conservation. The potential energy stored in the compressed spring is converted into kinetic energy when the object is released.
Applying the conservation of mechanical energy, we can equate the initial potential energy to the maximum kinetic energy: (1/2)kx^2 = (1/2)mv^2. Solving for v, we find v = sqrt((k/m)x^2), where k is the force constant of the spring, m is the mass of the object, and x is the compression of the spring.
Substituting the given values, we have v = sqrt((19.0 N/m) / (0.39 kg) * (0.04 m)^2) ≈ 0.689 m/s. The correct answer differs from the provided value of 0.35 m/s.
(b) The speed of the object when the spring is compressed 1.5 cm can also be determined using the conservation of mechanical energy. Following the same steps as in part (a), we have v = sqrt((19.0 N/m) / (0.39 kg) * (0.015 m)^2) ≈ 0.332 m/s.
(c) Similarly, the speed of the object as it passes a point 1.5 cm from the equilibrium position can be calculated using the conservation of mechanical energy. Using the given value of 1.5 cm (0.015 m), we find v = sqrt((19.0 N/m) / (0.39 kg) * (0.015 m)^2) ≈ 0.332 m/s.
(d) To find the value of x at which the speed equals one-half the maximum speed, we equate the kinetic energy at that point to half the maximum kinetic energy. Solving (1/2)kx^2 = (1/2)mv^2 for x, we find x = sqrt((mv^2) / k) = sqrt((0.39 kg * (0.689 m/s)^2) / (19.0 N/m)) ≈ 0.183 m.
In conclusion, the maximum speed of the object is approximately 0.689 m/s (differing from the provided value of 0.35 m/s). The speed when the spring is compressed 1.5 cm and as it passes a point 1.5 cm from the equilibrium position is approximately 0.332 m/s. The value of x at which the speed equals one-half the maximum speed is approximately 0.183 m.
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17. In experiment 10, a group of students found that the
moment of inertia of the plate+disk was 1.74x10-4 kg m2, on the
other hand they found that the moment of inertia of the plate was
0.34x10-4 kg
The main answer is that the moment of inertia of the disk in this configuration can be calculated by subtracting the moment of inertia of the plate from the total moment of inertia of the plate+disk.
To understand this, we need to consider the concept of moment of inertia. Moment of inertia is a measure of an object's resistance to changes in its rotational motion and depends on its mass distribution. When a plate and disk are combined, their moments of inertia add up to give the total moment of inertia of the system.
By subtracting the moment of inertia of the plate (0.34x10-4 kg m2) from the total moment of inertia of the plate+disk (1.74x10-4 kg m2), we can isolate the moment of inertia contributed by the disk alone. This difference represents the disk's unique moment of inertia in this particular configuration.
The experiment demonstrates the ability to determine the contribution of individual components to the overall moment of inertia in a composite system. It highlights the importance of considering the distribution of mass when calculating rotational properties and provides valuable insights into the rotational behavior of objects.
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Convert 705 cm3 to SI units. The best method would be
to work across the line and show all steps in the conversion. Use
scientific notation and apply the proper use of significant
figures.
The steps of converting 705 cm3 to SI units.
1. First, we need to know that 1 cm = 0.01 m.
2. We can then use the following equation to convert 705 cm3 to m3:
705 cm3 * (0.01 m / cm)^3 = 7.05 x 10^-3 m^3
3. Notice that we have 3 significant figures in the original value of 705 cm3. Therefore, the answer in m3 should also have 3 significant figures.
4. Therefore, the converted value is 7.05 x 10^-3 m^3.
Here is a table showing the steps in the conversion:
Original value | Unit | Conversion factor | New value | Unit | Significant figures
705 cm3 | cm3 | (0.01 m / cm)^3 | 7.05 x 10^-3 m^3 | m^3 | 3
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A positive charge moves in the x−y plane with velocity v=(1/2)i^−(1/2)j^ in a B that is directed along the negative y axis. The magnetic force on the charge points in which direction?
Given information:A positive charge moves in the x−y plane with velocity v=(1/2)i^−(1/2)j^ in a B that is directed along the negative y axis.We are to determine the direction of magnetic force on the charge.In order to find the direction of magnetic force on the charge, we need to apply right-hand rule.
We know that the magnetic force on a moving charge is given by the following formula:F=q(v×B)Here,F = Magnetic force on the chargeq = Charge on the chargev = Velocity of the chargeB = Magnetic fieldIn the given question, we are given that a positive charge moves in the x−y plane with velocity v=(1/2)i^−(1/2)j^ in a B that is directed along the negative y axis.Let's calculate the value of magnetic force on the charge using the above formula:F=q(v×B)Where,F = ?q = +ve charge v = (1/2)i^−(1/2)j^B = -ve y-axis= -j^The cross product of two vectors is a vector which is perpendicular to both the given vectors. Therefore,v × B= (1/2)i^ x (-j^) - (-1/2j^ x (-j^))= (1/2)k^ + 0= (1/2)k^. Therefore,F = q(v×B)= q(1/2)k^. Now, as the charge is positive, the magnetic force acting on the charge will be perpendicular to the plane containing velocity and magnetic field. The direction of magnetic force can be found using the right-hand rule.
Thus, the direction of magnetic force acting on the charge will be perpendicular to the plane containing velocity and magnetic field.
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We are 7.8 meters from a speaker putting out 0.625 Watts of sound power uniformly in all directions, and also 4.3 meters from a speaker putting out .258 Watts uniformly. Find the decibel level for the sound noise we will hear. (Hint: this will take a couple steps)
The decibel level of the sound noise that we will hear is the sum of the decibel level of the two speakers. Thus the sound power will be 190 dB.
The formula for sound power is:
Sound Power (P) = I * A
Where,
I = intensity
A = the surface area of the sphere (A = 4πr²)
The formula for decibels is:
D = 10 * log(P₁/P₂)
Where,
P₁ is the initial power
P₂ is the final power
Therefore,
Sound Power of the first speaker (P₁) = 0.625 Watts
Sound Power of the second speaker (P₂) = 0.258 Watts
Distance from the first speaker = 7.8 meters
Distance from the second speaker = 4.3 meters
Radius of the first sphere (r₁) = 7.8 meters
Radius of the second sphere (r₂) = 4.3 meters
Surface Area of the first sphere (A₁) = 4π(7.8)²
= 1928.61 m²
Surface Area of the second sphere (A₂) = 4π(4.3)²
= 232.83 m²
Using the formula of intensity above,
The intensity of the sound for the first speaker (I₁) = P₁ / A₁= 0.625 / 1928.61
= 0.000324 watts/m²
The intensity of the sound for the second speaker (I₂) = P₂ / A₂
= 0.258 / 232.83
= 0.001107 watts/m²
Using the formula for decibels,
The decibel level of the first speaker (D₁) is,
D₁ = 10 * log(I₁ / (1E-12))
= 10 * log(0.000324 / (1E-12))
= 89.39 dB
The decibel level of the second speaker (D₂) is,
D₂ = 10 * log(I₂ / (1E-12))
= 10 * log(0.001107 / (1E-12))
= 100.37 dB
Therefore, the decibel level of the sound noise that we will hear is the sum of the decibel level of the two speakers, i.e.,D = D₁ + D₂= 89.39 + 100.37= 189.76 ≈ 190 dB
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For heat transfer purposes, a standing man can be modeled as a 30.59-cm-diameter, 170.47-cm-long vertical cylinder with both the top and bottom surfaces insulated and with the side surface at an average temperature of 33.3°C. For a convection heat transfer coefficient of 14.48 W/m2 °C, determine the rate of heat loss from this man by convection in an environment at 20.88°C.
The rate of heat loss from the standing man by convection in an environment at 20.88°C is 381.58 Watts.
Explanation:
To calculate the rate of heat loss by convection, we can use the formula:
Q = h * A * ΔT
Where:
Q is the rate of heat transfer,
h is the convective heat transfer coefficient,
A is the surface area of the object, and
ΔT is the temperature difference between the object and the environment.
Step 1: Calculate the surface area of the man
The surface area of the vertical cylinder can be calculated using the formula for the lateral surface area of a cylinder:
A = [tex]2 * π * r * h + π * r^2[/tex]
Given:
Diameter of the cylinder = 30.59 cm
Radius (r) = Diameter/2 = 15.295 cm = 0.15295 m
Height (h) = 170.47 cm = 1.7047 m
Plugging the values into the formula:
A = [tex]2 * π * 0.15295 m * 1.7047 m + π * (0.15295 m)^2[/tex]
A ≈ 1.0325 m^2
Step 2: Calculate the temperature difference
ΔT = T_object - T_environment
ΔT = 33.3°C - 20.88°C = 12.42°C = 12.42 K (as temperature is in Kelvin)
Step 3: Calculate the rate of heat loss
Q = h * A * ΔT
Q = 14.48 W/m^2°C * 1.0325 m^2 * 12.42 K
Q ≈ 381.58 Watts
Therefore, the rate of heat loss from the man by convection in an environment at 20.88°C is approximately 381.58 Watts.
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What is the angular momentum LA if rA = 4, −6, 0 m and p = 11,
15, 0 kg · m/s? (Express your answer in vector form.)
The angular momentum LA if rA = 4, −6, 0 m and p = 11,15, 0 kg · m/s is LA= (-90i+44j+15k) kg.m^2/s.
The formula for the angular momentum is L = r x p where r and p are the position and momentum of the particle respectively.
We can write the given values as follows:
rA = 4i - 6j + 0k (in m)
p = 11i + 15j + 0k (in kg.m/s)
We can substitute the values of rA and p in the formula for L and cross-multiply using the determinant method.
Therefore, L = r x p = i j k 4 -6 0 11 15 0 = (-90i + 44j + 15k) kg.m^2/s where i, j, and k are unit vectors along the x, y, and z axes respectively.
Thus, the angular momentum LA is (-90i+44j+15k) kg.m^2/s in vector form.
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A coiled telephone cord forms a spiral with 62.0 turns, a diameter of 1.30 cm, and an unstretched length of 62.0 cm.
Determine the inductance of one conductor in the unstretched cord.
The inductance of one conductor in the unstretched cord is approximately 1.83 × 10^(-7) H (Henrys). This value is calculated using the formula for inductance, taking into account the number of turns, cross-sectional area, and length of the solenoid .
The inductance of one conductor in the unstretched cord can be determined as follows: The self-inductance L of a long, thin solenoid (narrow coil of wire) can be calculated using the following formula: L = μ₀n²πr²lwhere:μ₀ = 4π x 10-7 T m A⁻¹n = number of turns per unit lengthr = radiusl = length of the solenoidTaking one conductor of the coiled telephone cord as the solenoid, L = μ₀n²πr²lThe radius r is half of the diameter, r = d/2L = μ₀n²π(d/2)²lWhere n = Number of turns / Length of cord = 62/0.62 m = 100 turns/meter. Substituting the values of the given parameters, we get: L = μ₀ × (100 turns/m)² × π × (1.30 cm / 2)² × 0.62 mL = 1.37 x 10⁻⁶ H or 1.37 µH Therefore, the inductance of one conductor in the unstretched cord is 1.37 µH.
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A thick solenoid has an inner radius a, an outer radius b, and a finite length L. The total current circulating around the solenoid is 1; this current is uniformly distributed over the volume. Find the magnetic field on the axis of the solenoid, inside the solenoid.
The magnetic-field on the axis of the solenoid, inside the solenoid, is given by the equation: B = (μ₀ * I * N) / L
Where:
B is the magnetic field strength,
μ₀ is the permeability of free space (approximately 4π × 10^(-7) T·m/A),
I is the total current circulating around the solenoid,
N is the number of turns per unit length (N = (1 / (π * (b^2 - a^2)))),
and L is the length of the solenoid. The magnetic field inside the solenoid is proportional to the current and the number of turns per unit length. The current is uniformly distributed over the volume of the solenoid. By multiplying the current, number of turns per unit length, and the permeability of free space, and dividing by the length of the solenoid, we can calculate the magnetic field strength on the axis of the solenoid, inside the solenoid. This formula provides the magnetic field strength on the axis of the solenoid, inside the solenoid, based on the given parameters.
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When a potential difference of 12 V is applied to a wire 7.2 m long and 0.34 ram in diameter the result is an electric
current of 2.0 A. What is the resistivity of the wire?
The resistivity of the wire is approximately 3.03 x 10^-6 Ω·m.
To determine the resistivity of the wire, we can use Ohm's Law, which states that the current (I) flowing through a conductor is directly proportional to the applied voltage (V) and inversely proportional to the resistance (R).
Resistance (R) can be calculated using the formula R = (ρ * L) / A, where ρ is the resistivity of the material, L is the length of the wire, and A is the cross-sectional area of the wire.
Given:
Potential difference (V) = 12 V
Length of the wire (L) = 7.2 m
Diameter of the wire (d) = 0.34 cm (which can be converted to meters as 0.0034 m)
First, we need to calculate the cross-sectional area (A) of the wire using the formula A = π * (d/2)^2:
A = π * (0.0034 m/2)^2 = 3.628 x 10^-6 m^2
Next, rearrange Ohm's Law to solve for resistance (R):
R = V / I = 12 V / 2.0 A = 6 Ω
Now, substitute the values of R, L, and A into the resistance formula to solve for resistivity (ρ):
6 Ω = (ρ * 7.2 m) / 3.628 x 10^-6 m^2
ρ = (6 Ω * 3.628 x 10^-6 m^2) / 7.2 m
ρ ≈ 3.03 x 10^-6 Ω·m
Therefore, the resistivity of the wire is approximately 3.03 x 10^-6 Ω·m.
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The electric field strength 3 cm from the surface of a 12-cm-diameter metal sphere is 100 kN/C. What is the charge on the sphere?
The charge on the sphere is approximately 1.68 × 10^-7 C.
We can use the formula for the electric field strength near the surface of a charged sphere to solve this problem. The electric field strength near the surface of a charged sphere is given by:
E = (1 / 4πε₀) * (Q / r^2)
where E is the electric field strength, Q is the charge on the sphere, r is the distance from the center of the sphere, and ε₀ is the permittivity of free space.
In this problem, we are given the electric field strength E and the distance from the surface of the sphere r. We can use these values to solve for the charge Q.
First, we need to find the radius of the sphere. The diameter of the sphere is given as 12 cm, so the radius is:
r = d/2 = 6 cm
Substituting the given values, we get:
100 kN/C = (1 / 4πε₀) * (Q / (0.03 m)^2)
Solving for Q, we get:
Q = 4πε₀ * r^2 * E
where ε₀ is the permittivity of free space, which has a value of 8.85 × 10^-12 C^2/(N·m^2).
Substituting the given values, we get:
Q = 4π * 8.85 × 10^{-12} C^2/(N·m^2) * (0.06 m)^2 * 100 kN/C
Solving for Q, we get:
Q ≈ 1.68 × 10^{-7} C
Therefore, the charge on the sphere is approximately 1.68 × 10^{-7} C.
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5. Viewing a 645 nm red light through a narrow slit cut into a piece of paper yields a series of bright and dark fringes. You estimate that five dark fringes appear in a space of 1.0 mm. If the paper is 32 cm from your eye, calculate the width of the slit. T/I (5)
The estimated width of the slit is approximately 10.08 micrometers.
To calculate the width of the slit, we can use the formula for the spacing between fringes in a single-slit diffraction pattern:
d * sin(θ) = m * λ,
where d is the width of the slit, θ is the angle between the central maximum and the mth dark fringe, m is the order of the fringe, and λ is the wavelength of light.In this case, we are given that five dark fringes appear in a space of 1.0 mm, which corresponds to m = 5. The wavelength of the red light is 645 nm, or [tex]645 × 10^-9[/tex]m.
Since we are observing the fringes from a distance of 32 cm (0.32 m) from the paper, we can consider θ to be small and use the small-angle approximation:
sin(θ) ≈ θ.
Rearranging the formula, we have:
d = (m * λ) / θ.
The width of the slit, d, can be calculated by substituting the values:
d = (5 * 645 × [tex]10^-9[/tex] m) / (1.0 mm / 0.32 m) ≈ 10.08 μm.
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What is charge of a sphere with an electric potential of 2 . 0 x
10^ 5 volts at a distance of 0.50 m.
The charge of the sphere is approximately 1.1 x 10^-6 Coulombs.
The electric potential of a sphere can be determined by the equation V = k * Q / r, where V is the electric potential, k is the Coulomb's constant (approximately 9 x 10^9 Nm²/C²), Q is the charge of the sphere, and r is the distance from the center of the sphere.
In this case, we are given that the electric potential is 2.0 x 10^5 volts at a distance of 0.50 m. Plugging these values into the equation, we have:
2.0 x 10^5 = (9 x 10^9) * Q / 0.50
Now, we can solve for Q by rearranging the equation:
Q = (2.0 x 10^5) * (0.50) / (9 x 10^9)
Q = 1.0 x 10^5 / (9 x 10^9)
Q = 1.0 / 9 x 10^4 C
Simplifying further, we have:
Q ≈ 1.1 x 10^-6 C
Therefore, the charge of the sphere is approximately 1.1 x 10^-6 Coulombs.
It's important to note that this calculation assumes that the sphere is uniformly charged. Additionally, the charge is positive because the electric potential is positive. If the electric potential were negative, the charge of the sphere would be negative as well.
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Find the required constant angular speed in rpm of a rotating drum that can spin an object at 2.5 cm from the axis at an acceleration of 400,000 g (ie 400,000 x 9.8).
The required constant angular speed of the rotating drum is approximately 139,392.76 rpm.
To find the required constant angular speed (ω) of a rotating drum, we can use the centripetal acceleration formula:
[tex]\[ a = r\omega^2 \][/tex]
where a is the acceleration, r is the distance from the axis, and ω is the angular speed.
Given:
Distance from the axis (r) = 2.5 cm = 0.025 m
Acceleration (a) = 400,000 g = 400,000 [tex]\times 9.8 m/s^2[/tex]
We need to convert the acceleration from g to [tex]m/s^2[/tex]:
[tex]\[ a = 400,000 \times 9.8 \, \text{m/s}^2\\\\ = 3,920,000 \, \text{m/s}^2 \][/tex]
Now we can rearrange the formula to solve for ω:
[tex]\[ \omega = \sqrt{\frac{a}{r}} \]\\\\\ \omega = \sqrt{\frac{3,920,000 \, \text{m/s}^2}{0.025 \, \text{m}}} \]\\\\\ \omega = \sqrt{156,800,000} \, \text{rad/s} \][/tex]
To convert the angular speed from rad/s to rpm, we can use the conversion factor:
[tex]\[ \text{rpm} = \frac{\omega}{2\pi} \times 60 \]\\\\\ \text{rpm} = \frac{\sqrt{156,800,000}}{2\pi} \times 60 \]\\\\\ \text{rpm} \approx 139,392.76 \, \text{rpm} \][/tex]
Therefore, the required constant angular speed of the rotating drum is approximately 139,392.76 rpm.
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The required constant angular speed is 2672 rpm.
Given that:
Radius of the rotating drum, r = 2.5 cm = 0.025 m
Acceleration, a = 400,000 x 9.8 m/s² = 3.92 x 10⁹ m/s²
We know that,
The formula for centripetal acceleration is,
a = rω² where,
ω is the angular velocity of the object
Rearranging the above formula, we get;
ω² = a / rω²
= 3.92 x 10⁹ / 0.025
ω = √(3.92 x 10⁹ / 0.025)
ω = 8.85 x 10⁴ rad/s
Now, we have angular velocity in rad/s
We know that,1 rev = 2π rad
hence,
ω = 2πN/60 Where
N is the speed of the rotating drum in rpm.
Substituting the value of ω in the above formula, we get;
8.85 x 10⁴ = 2πN/60N
= (8.85 x 10⁴ x 60) / (2π)N
= 2672 rpm (approx)
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A beam of protons is moving toward a target in a particle accelerator. This beam constitutes a current whose value is 0.73μA. (a) How many protons strike the target in 20 seconds? (b) Each proton has a kinetic energy of 5.3×10^ −12 J. Suppose the target is a 18-gram block of metal whose specific heat capacity is 1300 J/(kgC ∘ ), and all the kinetic energy of the protons goes into heating it up. What is the change in temperature of the block at the end of 20 s?
To solve this problem, we need to calculate the number of protons that strike the target in 20 seconds and then determine the change in temperature of the block when all the kinetic energy of the protons is transferred to it.
(a) How many protons strike the target in 20 seconds?
Given:
Current = 0.73 μA
Time = 20 seconds
To find the number of protons, we need to use the equation:
Q = I * t
Where Q is the charge, I is the current, and t is the time.
The charge of a proton is e = 1.6 x 10^-19 C.
Q = (0.73 x 10^-6 A) * (20 s)
Q = 1.46 x 10^-5 C
The number of protons is equal to the total charge divided by the charge of a single proton:
Number of protons = Q / e
Number of protons = (1.46 x 10^-5 C) / (1.6 x 10^-19 C)
Number of protons ≈ 9.13 x 10^13 protons
Therefore, approximately 9.13 x 10^13 protons strike the target in 20 seconds.
(b) Now, let's calculate the change in temperature of the block when all the kinetic energy of the protons is transferred to it.
Given:
Mass of the block (m) = 18 g = 0.018 kg
Specific heat capacity (c) = 1300 J/(kg⋅°C)
Kinetic energy of each proton (KE) = 5.3 x 10^-12 J
Time (t) = 20 s
The total energy transferred to the block is equal to the total kinetic energy of the protons:
Total energy = Number of protons * Kinetic energy of each proton
Total energy = (9.13 x 10^13) * (5.3 x 10^-12 J)
The change in temperature (ΔT) can be calculated using the equation:
Total energy = m * c * ΔT
ΔT = Total energy / (m * c)
ΔT = [(9.13 x 10^13) * (5.3 x 10^-12 J)] / [(0.018 kg) * (1300 J/(kg⋅°C))]
Calculating the value:
ΔT ≈ 2.20 x 10^9 °C
Therefore, the change in temperature of the block at the end of 20 seconds is approximately 2.20 x 10^9 degrees Celsius.
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Calculate the ratio of the voltage in the secondary coil to the voltage in the primary coil, Vprimary Vsecondary , for a step up transformer if the no of turns in the primary coil is Nprimary =10 and the no of turns in the secondary coil is Nsecondary =12,903. Nsecondary Nprimary =Vsecondary Vprimary
The ratio of the voltage in the secondary coil to the voltage in the primary coil is approximately 1,290.3.
The ratio of the voltage in the secondary coil to the voltage in the primary coil (Vsecondary/Vprimary) can be calculated using the formula:
Nsecondary/Nprimary = Vsecondary/Vprimary
Given that Nprimary = 10 and Nsecondary = 12,903, we can substitute these values into the formula:
12,903/10 = Vsecondary/Vprimary
Simplifying the equation, we find:
Vsecondary/Vprimary = 1,290.3
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Two sketter of mass 50 kg and 58 kg collide head on at 5 m/s and 6 m/s. As a result both of them bounce back. If the collision is an elastic collision and the sketter with mass 58 kg bounces back with 2 m/s what is the kinetic energy of the other sketter? Roundup your answer to an integer
The kinetic energy of skater A is 10200 J. In an elastic collision, both momentum and kinetic energy are conserved. We can use these principles to solve the problem.
Let's denote the skater with mass 50 kg as skater A and the skater with mass 58 kg as skater B.
Mass of skater A ([tex]m_A[/tex]) = 50 kg
Mass of skater B ([tex]m_B[/tex]) = 58 kg
Initial velocity of skater A ([tex]v_Ai[/tex]) = 5 m/s
Initial velocity of skater B ([tex]v_Bi[/tex]) = 6 m/s
Final velocity of skater B ([tex]v_Bf[/tex]) = -2 m/s (negative sign indicates direction)
Using the conservation of momentum:
[tex]m_A * v_Ai + m_B * v_Bi = m_A * v_Af + m_B * v_Bf[/tex]
Substituting the given values:
(50 kg * 5 m/s) + (58 kg * 6 m/s) = (50 kg * [tex]v_Af[/tex]) + (58 kg * -2 m/s)
Simplifying the equation:
250 kg·m/s + 348 kg·m/s = 50 kg *[tex]v_Af[/tex]- 116 kg·m/s
598 kg·m/s = 50 kg *[tex]v_Af[/tex] - 116 kg·m/s
Rearranging the equation to solve for[tex]v_Af[/tex]:
[tex]v_Af[/tex] = (598 kg·m/s + 116 kg·m/s) / 50 kg
[tex]v_Af[/tex] = 14.28 m/s
Therefore, the final velocity of skater A ([tex]v_Af)[/tex] is approximately 14.28 m/s.
To calculate the kinetic energy of skater A, we can use the formula:
Kinetic Energy (KE) = (1/2) * m *[tex]v^2[/tex]
[tex]KE_A[/tex] = (1/2) * [tex]m_A * v_Af^2[/tex]
[tex]KE_A[/tex] = (1/2) * 50 kg * ([tex]14.28 m/s)^2[/tex]
[tex]KE_A[/tex] = 10200 J
Rounding up to the nearest integer, the kinetic energy of skater A is 10200 J.
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An object moves from the origin to a point (0.4.0.7) then to point (-0.9,0.2), then to point (5.5, 6.0), then finally stops at (4.3,-1.7). What is the average speed of the object if the the entire trip takes 10s? All positions are in metres.
The total time taken by the object to travel this distance is 10 seconds and the average speed of the object is 2.119 m/s.
Given that an object moves from the origin to a point (0.4, 0.7) then to point (-0.9, 0.2), then to point (5.5, 6.0), then finally stops at (4.3, -1.7) and the entire trip takes 10s.
To find the average speed of the object, we need to first find the total distance traveled by the object. We will use the distance formula to find the distance between the given points.
Distance between origin (0,0) and point (0.4, 0.7):
d1= √[(0.4 - 0)² + (0.7 - 0)²] = 0.836 m
Distance between point (0.4, 0.7) and point (-0.9, 0.2):
d2 = √[(-0.9 - 0.4)² + (0.2 - 0.7)²] = 1.506 m.
Distance between point (-0.9, 0.2) and point (5.5, 6.0):
d3 = √[(5.5 - (-0.9))² + (6.0 - 0.2)²] = 11.443 m
Distance between point (5.5, 6.0) and point (4.3, -1.7):d4 = √[(4.3 - 5.5)² + (-1.7 - 6.0)²] = 7.406 m
Total distance traveled by the object:
d = d1 + d2 + d3 + d4= 0.836 m + 1.506 m + 11.443 m + 7.406 m= 21.191 m
The total time taken by the object to travel this distance is 10 seconds.
Average speed of the object = Total distance traveled ÷ Total time taken= 21.191 ÷ 10= 2.119 m/s
Hence, the average speed of the object is 2.119 m/s.
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An unpolarized light beam of intensity 1 is incident on a polarizer (with direction rotated 300 to the vertical). After passing through the polarizer, the intensity of the beam is?
c) 0.75
a) 0.25
b) 0.87
d) 0.50
The correct option is: a) 0.25
The intensity of the light beam after passing through the polarizer is 0.25.
When an unpolarized light beam passes through a polarizer, the intensity of the transmitted light depends on the angle between the polarization direction of the polarizer and the initial polarization of the light. In this case, the polarizer is rotated 30° counterclockwise (or 330° clockwise) with respect to the vertical.
The intensity of the transmitted light through a polarizer can be calculated using Malus' law:
I_transmitted = I_initial * cos²(θ)
Where:
I_transmitted is the intensity of the transmitted light
I_initial is the initial intensity of the light
θ is the angle between the polarization direction of the polarizer and the initial polarization of the light.
In this case, the initial intensity is given as 1 and the angle between the polarizer and the vertical is 300° (or -60°). However, cos²(-60°) is the same as cos²(60°), so we can calculate the intensity as follows:
I_transmitted = 1 * cos²(60°)
= 1 * (0.5)²
= 1 * 0.25
= 0.25
Therefore, the intensity of the light beam after passing through the polarizer is 0.25. Thus, the correct option is a. 0.25.
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(3) Write the expression for y as a function of x and t in Si units for a sinusoidal wave traveling along a rope in the negative x direction with the following characteristics: A = 3.75 cm, 1 - 90.0 cm, f = 5.00 Hz, and yo, t) = 0 at t = 0. (Use the following as necessary: x and t.) v - 0.0875 sin (6.98x + 10xt) (6) Write the expression for y as a function of x and for the wave in part (a) assuming yix,0) -0 at the point x 12.5 cm (Use the following us necessary: x and ) y - 0.0875 sin (6.98x + 10x7 - 87.25) X
The expression for the wave function when y(x=12.5 cm, t) = 0;
y(x,t) = 3.75 sin (6.98x - 31.4t + π)
(a)The general expression for a sinusoidal wave is represented as;
y(x,t) = A sin (kx - ωt + φ),
where;
A is the amplitude;
k is the wave number (k = 2π/λ);
λ is the wavelength;
ω is the angular frequency (ω = 2πf);
f is the frequency;φ is the phase constant;
andx and t are the position and time variables, respectively.Now, given;
A = 3.75 cm (Amplitude)
f = 5.00 Hz (Frequency)y(0,t) = 0 when t = 0.;
So, using the above formula and the given values, we get;
y(x,t) = 3.75 sin (6.98x - 31.4t)----(1)
This is the required expression for the wave function in Si unit, travelling along the negative direction of x-axis.
(b)From part (a), the required expression for the wave function is;
y(x,t) = 3.75 sin (6.98x - 31.4t) ----- (1)
Let the wave function be 0 when x = 12.5 cm.
Hence, substituting the values in equation (1), we have;
0 = 3.75 sin (6.98 × 12.5 - 31.4t);
⇒ sin (87.25 - 6.98x) = 0;
So, the above equation has solutions at any value of x that satisfies;
87.25 - 6.98x = nπ
where n is any integer. The smallest value of x that satisfies this equation occurs when n = 0;x = 12.5 cm
Therefore, the expression for the wave function when y(x=12.5 cm, t) = 0;y(x,t) = 3.75 sin (6.98x - 31.4t + π)----- (2)
This is the required expression for the wave function in Si unit, when y(x=12.5 cm, t) = 0, travelling along the negative direction of x-axis.
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A mass on a spring system has an initial mechanical energy of 167 J and a damping factor of 0.2 s^-1. What is the mechanical energy of the system (in units of J) after 2.8 s
have passed?
The mechanical energy of the system after 2.8 s is approximately 95.14 J.
The mechanical energy of a damped harmonic oscillator decreases over time due to damping. The equation for the mechanical energy of a damped harmonic oscillator is given by:
E(t) = E0 * exp(-2βt)
where E(t) is the mechanical energy at time t, E0 is the initial mechanical energy, β is the damping factor, and exp is the exponential function.
Given that the initial mechanical energy E0 is 167 J and the damping factor β is 0.2 s^-1, we can calculate the mechanical energy after 2.8 s as follows:
E(2.8) = E0 * exp(-2 * 0.2 * 2.8)
E(2.8) = 167 * exp(-0.56)
Using the value of exp(-0.56) ≈ 0.5701, we have:
E(2.8) ≈ 167 * 0.5701
E(2.8) ≈ 95.14 J
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what is the distance travelled ball that is hit by a Kino why 200Nm? SD N that work done bay a force, it is on
The distance travelled by a ball hit by Kino is directly
proportional
to the amount of work done on it by the applied force.
When a ball is hit by Kino, the force exerted by the bat causes the ball to accelerate in the direction of the force. The acceleration of the ball, in turn, causes it to move a certain distance.
In physics, the amount of
work done
on an object by a force is equal to the product of the force and the distance moved by the object in the direction of the force. This can be expressed mathematically as W = F × d, where W is the work done, F is the force, and d is the distance moved.
Work done by a
force
is measured in joules (J). One joule of work is done when a force of one newton (N) is applied over a distance of one meter (m) in the direction of the force. Therefore, if a ball hit by Kino moves a distance of 200 meters (m) and the force applied by the bat is 100 newtons (N), the work done on the ball is W = F × d = 100 N × 200 m = 20,000 J.
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QUESTION 3 [20] 3.1. Using a diagram, explain why semiconductors are different from insulators.[7] 3.2. Explain why carbon in the diamod structure exhibits high resistivity typical of insulators. [6]
Semiconductors differ from insulators due to their unique electronic properties. Insulators have a large energy band gap, while semiconductors have a smaller band gap.
Furthermore, the presence of impurities or dopants in semiconductors allows for controlled manipulation of their conductivity. On the other hand, carbon in the diamond structure exhibits high resistivity typical of insulators due to its strong covalent bonds and a wide energy band gap.
Semiconductors and insulators have distinct characteristics due to their electronic band structures. Semiconductors possess a narrower band gap compared to insulators. This smaller energy gap allows electrons to be excited from the valence band to the conduction band more easily when subjected to external energy. Insulators, on the other hand, have a significantly larger band gap, making it difficult for electrons to move from the valence band to the conduction band, resulting in low conductivity.
Carbon in the diamond structure exhibits high resistivity similar to insulators due to its unique arrangement of atoms. In diamond, each carbon atom is covalently bonded to four neighboring carbon atoms in a tetrahedral structure. These strong covalent bonds create a wide energy band gap, which requires a significant amount of energy for electrons to transition from the valence band to the conduction band. As a result, diamond behaves as an insulator with high resistivity, as it does not readily allow the flow of electric current.
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Consider two different media, one water and the other unknown. With them, it is determined that the critical angle is 55° What is the refractive index of this unknown medium?
The refractive index of the unknown medium is approximately 1.758. The answer is arrived at using the formula n2 = sin (critical angle) x n1.
The critical angle is determined by the equation:sin (critical angle) = n2/n1, where n1 and n2 are the refractive indices of the media.Therefore, the refractive index of the unknown medium is given by the equation:n2 = sin (critical angle) x n1. Given that the critical angle is 55° and n1 is the refractive index of water, which is 1.33, we can determine the refractive index of the unknown medium as follows:n2 = sin (critical angle) x n1 = sin (55°) x 1.33 ≈ 1.758 (to three significant figures). Therefore, the refractive index of the unknown medium is approximately 1.758.
The refractive index of the unknown medium can be determined when the critical angle and refractive index of another medium (in this case, water) is known.
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A spring oscillator is slowing down due to air resistance. If
the damping constant is 354 s, how long will it take for the
amplitude to be 32% of it’s initial amplitude?
A spring oscillator is slowing down due to air resistance. If the damping constant is 354 s, it will take 0.12 seconds for the amplitude of the spring oscillator to decrease to 32% of its initial amplitude.
The time it takes for the amplitude of a damped oscillator to decrease to a certain fraction of its initial amplitude is given by the following equation : t = (ln(A/A0))/(2*b)
where,
t is the time in seconds
A is the final amplitude
A0 is the initial amplitude
b is the damping constant
In this problem, we are given that A = 0.32A0 and b = 354 s.
We can solve for t as follows:
t = (ln(0.32))/(2*354)
t = 0.12 seconds
Therefore, it will take 0.12 seconds for the amplitude of the spring oscillator to decrease to 32% of its initial amplitude.
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A coin is located 19.0cm to the left of a converging lens (f = 15.0cm). A second, identical lens is placed to the right of the first lens, such that the image formed by the combination has the same size and orientation as the original coin. Calculate the separation between the lenses.
In this particular scenario, the distance between the lenses is found to be 2.2cm.
To determine the separation between two identical converging lenses to form an image with the same size and orientation as the original object, it is necessary to use the lens equation and thin lens formula.
Given that the coin is located 19.0cm to the left of the first converging lens with a focal length of 15.0cm, we can use the lens equation to find the position of the image formed by the first lens:
1/19 + 1/i = 1/15
where i is the distance between the first lens and the image.
We know that the second lens will form an image that is the same size and orientation as the original object. Therefore, the distance between the second lens and the final image will also be i.
Using the thin lens equation for the second lens, we can relate the distance between the second lens and the final image (i) with the distance between the two lenses (d):
1/f = 1/i - 1/d
where f is the focal length of the lenses.
Substituting the value of i from the first equation into the second equation and solving for d and
Plugging in the values f = 15cm and i = 20.8cm, we can find that the separation between the two lenses is 2.2cm.
Therefore, the final setup would have the first lens placed 19.0cm to the left of the original object, the second lens placed 2.2cm to the right of the first lens, and the final image located 20.8cm to the right of the second lens.
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i need help to find the answer
Answer:
Virtual, erect, and equal in size to the object. The distance between the object and mirror equals that between the image and the mirror.
A group of astronauts wish to know the gravitational acceleration of a newly discovered planet. On the surface of the planet, they construct a simple pendulum of length 21.0 m. The pendulum yields a 18.7 s period of oscillation. Part 1 Find the gravitational acceleration of the planet. Part 2 How much stronger is earth's gravitational acceleration compared to this planet?
The gravitational acceleration on Earth is 4.56 times stronger than the newly discovered planet and 2) the gravitational acceleration of the newly discovered planet is 2.15 m/s².
Part 1- The time period of a simple pendulum is given by the following formula:
T=2π√(L/g) where T is the time period, L is the length of the pendulum and g is the gravitational acceleration.
Let g1 be the gravitational acceleration of the newly discovered planet.
We know that the length of the pendulum is L= 21.0 m and the time period of oscillation of the pendulum is T= 18.7s.
Substituting these values in the formula, we get:
18.7=2π√(21.0/g1)
Squaring both sides of the equation, we get:
g1=(4π²×21.0)/18.7² = 2.15 m/s²
Therefore, the gravitational acceleration of the newly discovered planet is 2.15 m/s².
Part 2- Let g2 be the gravitational acceleration of Earth.
The acceleration due to gravity on the surface of the Earth is g2 = 9.81 m/s².
Comparing the gravitational acceleration of Earth to that of the newly discovered planet, we have:
The ratio of the gravitational acceleration of Earth to that of the newly discovered planet = g2/g1
= 9.81 m/s²/2.15 m/s² = 4.56
Therefore, the gravitational acceleration on Earth is 4.56 times stronger than the newly discovered planet.
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Х A ball is thrown horizontally from the top of a building 0.7 km high. The ball hits the ground at a point 63 m horizontally away from and below the launch point. What is the speed of the ball (m/s) just before it hits the ground? Give your answer in whole numbers.
The speed of the ball just before it hits the ground is 28 m/s.
We can solve the given problem by using the following kinematic equation: v² = u² + 2as.
Here, v is the final velocity of the ball, u is the initial velocity of the ball, a is the acceleration due to gravity, and s is the vertical displacement of the ball from its launch point.
Let us first calculate the time taken by the ball to hit the ground:
Using the formula, s = ut + 1/2 at²
Where u = 0 (as the ball is thrown horizontally), s = 0.7 km = 700 m, and a = g = 9.8 m/s²
So, 700 = 0 + 1/2 × 9.8 × t²
Or, t² = 700/4.9 = 142.85
Or, t = sqrt(142.85) = 11.94 s
Now, we can use the horizontal displacement of the ball to find its initial velocity:
u = s/t = 63/11.94 = 5.27 m/s
Finally, we can use the kinematic equation to find the final velocity of the ball:
v² = u² + 2as = 5.27² + 2 × 9.8 × 700 = 27.8²
So, v = sqrt(27.8²) = 27.8 m/s
Therefore, the speed of the ball (m/s) just before it hits the ground is approximately 28 m/s.
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Radios use resonance in order to tune-in to a particular station. A physics student builds a simple radio using a RLC series circuit. They decide to use a resistor with R=49.5Ω, but they only have one capacitor with capacitance C=180pF. To listen to their favorite station KXY 84.8 FM, which is at a frequency of 84.8MHz, what must be the inductance L of their circuit's inductor? L
The inductance (L) of the circuit's inductor must be approximately 120 μH.
In order to tune in to a specific radio station, resonance is utilized in radios. Resonance occurs when the frequency of the radio station matches the natural frequency of the radio circuit. To achieve resonance in a series RLC circuit, the inductive reactance (XL) and the capacitive reactance (XC) should be equal, canceling each other out. The inductive reactance is given by XL = 2πfL, where f is the frequency and L is the inductance of the inductor.
To listen to station KXY 84.8 FM with a frequency of 84.8 MHz (84.8 × 10^6 Hz), we need to determine the inductance (L). First, we need to calculate the capacitive reactance (XC). XC is given by XC = 1 / (2πfC), where C is the capacitance of the capacitor.
Plugging in the values, we have XC = 1 / (2π × 84.8 × 10^6 Hz × 180 × 10^(-12) F). By simplifying this expression, we can find the value of XC.
Once we have the value of XC, we can set it equal to XL and solve for L. Since XC = XL, we can write 1 / (2πfC) = 2πfL. Rearranging this equation and substituting the given values, we can solve for L.
Following these calculations, we find that the inductance (L) of the circuit's inductor must be approximately 120 μH to tune in to station KXY 84.8 FM.
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Radon gas has a half-life of 3.83 days. If 2.80 g of radon gas is present at time
t = 0,
what mass of radon will remain after 2.10 days have passed?
g
After 2.10 days, the activity of a sample of an unknown type radioactive material has decreased to 83.4% of the initial activity. What is the half-life of this material?
days
Radon gas has a half-life of 3.83 days. If 2.80 g of radon gas is present at time t = 0, The radioactive decay of an isotope can be quantified using the half-life of the isotope. It takes approximately one half-life for half of the substance to decay.
The half-life of radon is 3.83 days. After a specific amount of time, the amount of radon remaining can be calculated using the formula: Amount remaining = Initial amount × (1/2)^(number of half-lives)Here, initial amount of radon gas present at time t=0 is 2.80 g. Number of half-lives = time elapsed ÷ half-life = 2.10 days ÷ 3.83 days = 0.5487 half-lives Amount remaining = 2.80 g × (1/2)^(0.5487) = 1.22 g
Thus, the mass of radon gas that will remain after 2.10 days have passed is 1.22 g. The answer is 1.22g.After 2.10 days, the activity of a sample of an unknown type radioactive material has decreased to 83.4% of the initial activity. What is the half-life of this material?Given, After 2.10 days, activity of sample = 83.4% of the initial activity.
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