Answer:
To calculate the average vertical velocity for both Tracker experiments, we need to consider only the magnitude of the displacement and the time period from t = 0.00 seconds to t = 1.00 second.
The formula to calculate average velocity is:
Average Velocity = Displacement / Time
Given the magnitudes of displacement for the small ball and large ball:
For the small ball: Displacement = 5.039
For the large ball: Displacement = 4.810
The time period for both is 1.00 second.
Calculating the average vertical velocity for each ball:
For the small ball: Average Velocity = 5.039 / 1.00 = 5.039 m/s (rounded to three significant figures)
For the large ball: Average Velocity = 4.810 / 1.00 = 4.810 m/s (rounded to three significant figures)
Comment: During the first second of the fall, the small ball drops faster than the large ball, as it has a greater average vertical velocity.
Explanation:
To calculate the average vertical velocity for the time period between t = 0.00 s and t = 1.00 s, considering only the magnitude of the displacement, we can use the formula:
Average vertical velocity = Magnitude of displacement / Time interval
For the small ball, we have:
Magnitude of displacement = |(-5.039 m) - 0 m| = 5.039 m
Average vertical velocity = 5.039 m / 1.00 s = 5.039 m/s
For the large ball, we have:
Magnitude of displacement = |(-4.810 m) - 0 m| = 4.810 m
Average vertical velocity = 4.810 m / 1.00 s = 4.810 m/s
Therefore, the small ball drops faster during the first second of the fall, as it has a higher average vertical velocity than the large ball. This result is consistent with the analysis of the magnitude of the displacement alone, where we found that the small ball had a larger displacement than the large ball.
1. A boy lifts a 5.1-kg block vertically 6.0 m at constant speed. The work done (in Joules) by the boy is Round of your answer to 1 decimal place. Do not include the unit.
2. A 54.2 kg diver jumps from a height of 2.2 m with an initial speed of 2.5 m/s. What is his speed (in m/s) entering the water?
(1)A boy lifts a 5.1-kg block vertically 6.0 m at constant speed the work done by the boy is approximately 299.9 J.(2) the speed of the diver entering the water is approximately 2.5 m/s.
1): The work done by the boy can be calculated using the formula:
Work = Force × Distance
Since the block is lifted vertically at a constant speed, the force applied by the boy must be equal to the weight of the block.
Weight = mass × acceleration due to gravity
Weight = 5.1 kg × 9.8 m/s^2 (acceleration due to gravity)
Weight ≈ 49.98 N
Therefore, the work done by the boy is:
Work = Force × Distance
Work = 49.98 N × 6.0 m
Work ≈ 299.9 Joules
Rounded to 1 decimal place, the work done by the boy is approximately 299.9 J.
(2) To find the speed of the diver entering the water, we can use the principle of conservation of energy. The initial potential energy of the diver at the top of the dive can be converted into kinetic energy just before entering the water.
The potential energy at the top of the dive is given by:
Potential energy = mass × gravity × height
Potential energy = 54.2 kg × 9.8 m/s^2 × 2.2 m
Potential energy ≈ 1198.36 J
The initial kinetic energy just before entering the water can be calculated as:
Kinetic energy = 0.5 × mass × velocity^2
Kinetic energy = 0.5 × 54.2 kg × (2.5 m/s)^2
Kinetic energy ≈ 169.75 J
According to the conservation of energy, the potential energy at the top should be equal to the kinetic energy just before entering the water. Therefore:
Potential energy = Kinetic energy
1198.36 J = 169.75 J
To find the speed (velocity) of the diver, we can rearrange the equation:
Velocity^2 = (2 × Kinetic energy) / mass
Velocity^2 = (2 × 169.75 J) / 54.2 kg
Velocity^2 ≈ 6.254
Taking the square root of both sides, we find:
Velocity ≈ 2.5 m/s
Therefore, the speed of the diver entering the water is approximately 2.5 m/s.
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A heat pump defrost control must do three things: turn on auxiliary heat, turn off the condenser fan motor, and:
Answer:
Shift the reversing valve from heat to cool.
Question 4 (1 point) Diffraction gratings provide much brighter interference patterns since more light passes through them compared with double slits. O True O False Question 5 (1 point) A When the th
False statement regarding Diffraction and True statement regarding reflected light
4) False. **Diffraction gratings** do not provide much brighter interference patterns compared to double slits. In fact, more light passes through double slits than through diffraction gratings. Diffraction gratings consist of multiple closely spaced slits that diffract and spread out the light, resulting in individual interference maxima and minima that are less intense. On the other hand, double slits allow more light to pass through, resulting in brighter interference patterns.
5) True. When the thickness of a film in air is such that reflected light undergoes destructive interference, the statement is true. Destructive interference occurs when the path length difference between the two reflected rays is an odd multiple of half the wavelength of light. This leads to the cancellation of certain wavelengths of light, resulting in reduced or no reflected light. Therefore, if the film thickness satisfies the condition for destructive interference, the reflected light will be significantly diminished.
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2. a car traveling at 27 m/s runs out of gas while traveling up a slope. if the car coasts 85 m up the slope before starting to roll back down, what is the angle of incline?
The angle of inclination is 25.92°. The gravitational force is equal and opposite to the force component parallel to the inclined surface on a block placed on an inclined surface.
We can find the angle of inclination by using this information. This can be illustrated with the following formula;
mg sin θ = f
Here, m = Mass of car = 1,000 kg
g = acceleration due to gravity = 9.8 ms⁻²
sin θ = Opposite / Hypotenuse
= Height / Length
f = Force component parallel to slope
= Weight × sin θ
= mg sin θ
We'll need to utilize the following data:
Initial velocity, u = 27 m/s
Displacement, S = 85 m
Acceleration, a = -9.8 m/s²
By using the kinematic equation of motion, the time it takes to travel up the slope can be calculated;
v² = u² + 2as,
Where, v = Final velocity = 0 m/s
u = Initial velocity = 27 m/s
a = Acceleration = -9.8 m/s²
s = Displacement = 85 m.
After calculating for t, we can use this time value to calculate the angle of inclination by utilizing the aforementioned formula.
mg sin θ = f
Here, m = 1,000 kg,
g = 9.8 ms⁻²,
f = mgsinθsinθ = f / mg.
Now, solve for f, f = ma
Therefore, f = 1,000 kg × 9.8 m/s² × sin θ
The angle of inclination can now be calculated:
sinθ = f / mg
= 1,000 kg × 9.8 m/s² × sin θ / 1,000 kg × 9.8 m/s²
= sin θ= (u² - v²) / 2as
= (27 m/s)² / 2(-9.8 m/s²)(85 m)
= 25.92°
Therefore, the angle of inclination is 25.92°.
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Consider a cell that is running under standard conditions: nissduni21saqduucu1saqducussd.
a) Is this cell a voltaic or an electrolytic cell, and how can you differentiate between the two?
b) Does current flow spontaneously in this cell under standard conditions?
c) What is the maximum potential of this cell?
d) If the cell is connected to a voltmeter, what would you observe in terms of initial voltage and changes over time?
e) What is the initial free energy of this cell at the point of construction?
f) Does the free energy of the cell change over time as the cell runs, and if so, how does it change?
The given information is incomplete and does not specify whether the cell is voltaic or electrolytic. Differentiating between the two requires understanding their fundamental characteristics.
a) To differentiate between a voltaic and an electrolytic cell, additional information such as the direction of electron flow, the presence of an external power source, and the nature of the electrode reactions is required.
b) The spontaneity of current flow in the cell depends on the overall cell potential, which is not given in the provided information.
c) The maximum potential of the cell cannot be determined without knowledge of the specific redox reactions occurring and the concentrations of species involved.
d) The behavior of the voltmeter connected to the cell would depend on the cell potential and how it changes over time, which is not fundamental provided.
e) The initial free energy of the cell at the point of construction cannot be calculated without more information about the chemical reactions and standard free energy changes.
f) Without further details, it is impossible to determine how the free energy of the cell changes over time as the cell operates. The specific reactions and concentrations involved would be necessary to make such an assessment.
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A wildlife researcher is tracking a flock of geese. The geese fly 5.0 km due west, then turn toward the north by 50 ∘ and fly another 4.5 km .PART A .How far west are they of their initial position? dw=? PART B What is the magnitude of their displacement? d=?
Answer:
PART A:
The geese initially fly 5.0 km due west. This movement is entirely in the west direction, so the west component (dw) is equal to 5.0 km.
Therefore, the geese are 5.0 km west of their initial position.
PART B:
After flying 5.0 km due west, the geese turn toward the north by 50° and fly another 4.5 km.
To determine the displacement (d), we need to find the resultant of their west and north components.
West component: The initial movement of 5.0 km due west does not change after turning north. So the west component of displacement remains the same at 5.0 km.
North component: The geese fly 4.5 km in the north direction. Since they turn by 50° from west, we can use trigonometry to find the north component (dn).
dn = 4.5 km * sin(50°)
dn ≈ 3.454 km
The displacement (d) is the magnitude of the resultant of the west and north components. We can use the Pythagorean theorem to find the magnitude:
d = sqrt(dw^2 + dn^2)
d = sqrt((5.0 km)^2 + (3.454 km)^2)
d ≈ 6.076 km
Therefore, the geese are approximately 6.076 km away from their initial position, and their displacement is approximately 6.076 km.
The flock of geese is initially 5.0 km west of their starting position. Their displacement, considering both the westward and northward movements, is approximately 5.2 km.
In the given scenario, the geese first fly 5.0 km due west. This indicates a purely westward displacement. Therefore, the distance west of their initial position is 5.0 km.
Afterward, the geese turn toward the north by an angle of 50 degrees and continue flying for another 4.5 km. This northward displacement can be broken down into its vertical and horizontal components. The vertical component can be found by multiplying the distance flown (4.5 km) by the sine of the angle (50 degrees). The horizontal component can be found by multiplying the distance flown (4.5 km) by the cosine of the angle (50 degrees).
Calculating the vertical component: 4.5 km × sin(50°) ≈ 3.42 km
Calculating the horizontal component: 4.5 km × cos(50°) ≈ 2.90 km
To find the magnitude of the total displacement, we can use the Pythagorean theorem. The total displacement is the square root of the sum of the squares of the horizontal and vertical components.
Calculating the magnitude of displacement: √(5.0 km² + 2.90 km² + 3.42 km²) ≈ √39.97 km² ≈ 6.32 km
Therefore, the geese are approximately 5.0 km west of their initial position, and their displacement is approximately 6.32 km.
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VP 28.2.2 ▼ Part E A proton (charge +1.60 × 10-¹⁹ C) and an electron (charge -1.60 × 10-¹⁹ C) are both moving in the xy-plane with the same speed, 4.20 × 105 m/s. The proton is moving in the +y-direction along the line x = 0, and the electron is moving in the -y-direction along the line x = +4.00 mm. At the instant when the proton and electron are at their closest approach, what is the magnitude of the magnetic force that the proton exerts on the electron? Express your answer with the appropriate units. μA ? Fon e- = Value Units
a) The distance of closest approach is2.00 mm
b) The magnitude of the magnetic force that the proton exerts on the electron: 1.60 × 10⁻¹⁹ N
a) To find the distance of closest approach between the proton and electron, we need to determine the y-coordinate of the electron when it is at x = 0.
Given that the electron is moving along the line x = +4.00 mm, at the instant of closest approach, the y-coordinate of the electron will be equal to the negative of its x-coordinate.
Therefore, the distance of closest approach is 4.00 mm or 2.00 mm (taking the absolute value).
b) The magnetic force between two moving charges can be calculated using the formula
F = (|q₁| * |q₂| * v * B) / r,
where F is the force,
|q₁| and |q₂| are the magnitudes of the charges,
v is the velocity of the charge,
B is the magnetic field, and
r is the distance between the charges.
In this case, the proton and electron have equal magnitudes of charge |q₁| = |q₂| = 1.60 × 10⁻¹⁹ C, and they are moving with the same speed v = 4.20 × 10⁵ m/s. The magnetic force depends on the magnetic field B, which is not given in the question.
Therefore, we cannot calculate the exact magnitude of the magnetic force without knowing the value of the magnetic field.
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c) what is the angle of incidence from glass when the reflected light in glass is linearly polarized?
The angle of incidence from the glass when the reflected light in the glass is linearly polarized is called Brewster's angle.
Brewster's angle is the angle of incidence at which light is polarized when it is reflected from a transparent surface, such as glass. The reflected light at this angle is entirely polarized and has no parallel component.What is polarization of light?When light waves propagate through space, the electric and magnetic fields at each point in the wave can oscillate in different directions.
The polarization of light is the orientation of the electric field vector that produces the electromagnetic wave as it propagates. The reflected light from a transparent surface, such as glass, is entirely polarized when the angle of incidence is Brewster's angle. When the angle of incidence is greater than Brewster's angle, both polarizations are reflected, and the reflected light is no longer linearly polarized.
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What are advantages of charge-coupled devices (CCDs) over photographic film? Choose all that apply. a. The large CCD cameras that astronomers use can collect almost all the photons of incoming light that strike the chip, which greatly reduces exposure times. b. CCD chips can be designed to detect wavelengths of infrared and ultraviolet light, allowing astronomers to gather data on some astronomical objects that are more visible at these wavelengths than in visible light. c. CCD chips can be designed to block out infrared and ultraviolet light, allowing astronomers to gather data on some astronomical objects that are more visible at these wavelengths than in visible light. d.The large CCD cameras that astronomers use can collect almost all the photons of incoming light that strike the chip, which greatly increases exposure times. e. CCD chips can be designed to detect wavelengths of X-ray and microwave light, allowing astronomers to gather data on some astronomical objects that are more visible at these wavelengths than in visible light.
Charge-coupled devices (CCDs) are electronic devices used for detecting electromagnetic radiation (light) and converting it into an electrical signal. The advantages of CCDs over photographic film include the following:
(a) The large CCD cameras that astronomers use can collect almost all the photons of incoming light that strike the chip, which greatly reduces exposure times. This option a is correct.
(b) CCD chips can be designed to detect wavelengths of infrared and ultraviolet light, allowing astronomers to gather data on some astronomical objects that are more visible at these wavelengths than in visible light is also correct. Thus option b is correct. Additionally, option d is incorrect because large CCD cameras can't increase exposure times but instead, reduce them. Option e is incorrect because CCD chips are not designed to detect wavelengths of X-ray and microwave light but instead for infrared and ultraviolet light as stated in option b. Thus, the options that are correct include options a and b.
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When a single charge q is placed on one corner of a square, the electric field at the center of the square is F/q. If three other equal charges are placed on the other corners, the electric field at the center of the square due to these four equal charges is
a) F/(2q)
b) F/(4q)
c) 4F/q
d) F/q
e) zero
Please explain how you came to the correct answer.
The answer to the question is option (c) 4F/q.
The electric field due to a single charge q at the center of a square is given by
E = F/qWhere F = 1/4πε₀ * q / r²,
where r is the distance between the charge and the center of the square. The electric field of the single charge q has a magnitude of F/q when it is at the center of the square.
If three other charges, each of the same magnitude, are placed on the other corners of the square, the resultant electric field at the center of the square is the vector sum of the electric fields due to the four charges.Let the distance between the charges and the center of the square be a.
The force due to each charge is given by
F' = 1/4πε₀ * q / a²
The electric field due to each charge at the center of the square is given by
E' = F'/q = 1/4πε₀ * q / a²q.
The electric field at the center of the square due to these four charges is the vector sum of the electric fields due to the four charges. Since the charges are placed at the corners of a square and are equidistant from the center, the angle between any two fields is 90°.
Hence, the resultant electric field is given by
E = 2E' sin 45° + 2E' sin 135°= 2E' /√2 + 2E' /-√2= 4E' /√2= 4 (1/4πε₀ * q / a²q) /√2= 4F /q.
Hence, the option (c) is the correct answer.
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car mass 1000kg is travelling along a straight horizontal road at a speed of 20m/s when it brakes sharply then skids. Friction brings the car to
,rest. If the friction force between the tires and road is 9000N. Calculate the distance travelled by car befor it comes to rest
A car mass 1000kg is traveling along a straight horizontal road at a speed of 20m/s when it brakes sharply then skids. Friction brings the car to,rest. If the friction force between the tires and road is 9000N. The distance traveled by the car before it comes to rest (while skidding due to braking) is approximately 22.22 meters.
To calculate the distance traveled by the car before it comes to rest, we can use the equations of motion.
First, we need to find the acceleration of the car when it brakes sharply. The friction force acting on the car is equal to the product of the mass of the car and its acceleration:
Friction force = mass × acceleration
9000 N = 1000 kg × acceleration
acceleration = 9000 N / 1000 kg
acceleration = 9 m/s^2
Next, we can use the equation of motion that relates initial velocity, final velocity, acceleration, and distance:
v^2 = u^2 + 2as
Where:
v = final velocity (0 m/s, as the car comes to rest)u = initial velocity (20 m/s)a = acceleration (-9 m/s^2, negative because it acts in the opposite direction to the car's motion)s = distance traveledPlugging in the values, we get:
0^2 = 20^2 + 2(-9)s
0 = 400 - 18s
18s = 400
s = 400 / 18
s ≈ 22.22 m
Therefore, the distance traveled by the car before it comes to rest (while skidding due to braking) is approximately 22.22 meters.
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use kepler's law, which states that the square of the time, t, required for a planet to orbit the sun varies directly with the cube of the mean distance, a, that the planet is from the sun. using the earth's distance of 1 astronomical unit (a.u.), determine the time, in earth years, for a planet to orbit the sun if its mean distance is 9.58 a.u. (round your answer to two decimal places.) t
The time for the planet to orbit the Sun is 1 Earth year. use kepler's law, which states that the square of the time, t, required for a planet to orbit the sun varies directly with the cube of the mean distance.
Using Kepler's law, which states that the square of the time required for a planet to orbit the Sun varies directly with the cube of the mean distance, we can determine the time for a planet with a mean distance of 9.58 astronomical units (a.u.) to orbit the Sun.
According to Kepler's law, the relationship period between the time (t) and the mean distance (a) is expressed as:
[tex]t^2 = k * a^3[/tex]
where k is a constant.
Given that the mean distance of the planet is 9.58 a.u. and the Earth's distance is 1 a.u., we can set up the following equation:
(1)² = k × (9.58)³
Simplifying the equation, we have:
1 = k × (9.58)³
To solve for the constant k, we divide both sides by (9.58)³:
k = 1 / (9.58)³
Now, we can substitute the mean distance of the planet (9.58 a.u.) into the equation to calculate the time (t):
t² = (1 / (9.58)³) × (9.58)³
Simplifying further, we get:
t² = 1
Taking the square root of both sides, we find:
t = 1
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A tank contains 100 kg of salt and 2000 L of water. Pure water enters a tank at the rate 12 L/min. The solution is mixed and drains from the tank at the rate 13 L/min.
Let y be the number of kg of salt in the tank after t minutes.
The differential equation for this situation would be:
dy
y(0) =
A tank contains 60 kg of salt and 1000 L of water. A solution of a concentration 0.03 kg of salt per liter enters a tank at the rate 9 L/min. The solution is mixed and drains from the tank at the same rate.
Let y be the number of kg of salt in the tank after t minutes.
Write the differential equation for this situation
dy =
y(0) = 60
y' + ty^1/3 = tan(t), y(3) = – 5
a) Rewrite the differential equation, if necessary, to obtain the form y' = f(t, y)
F(t, x) = _______
b) Compute the partial derivative of f with respect to y. Determine where in the ty-plane both f(t, y) and its derivative are continuous.
c) Find the largest open rectangle in the ty-plane on which the solution of the initial value problem above is certain to exist for the initial condition. (Enter oo for infinity)
t interval is
y interval is
The interval of y is (-5,∞), as the solution exists until the concentration of salt reaches zero. The solution to the differential equation is given by,dy/dt = rate in - rate out. There is initially 100 kg of salt and 2000 L of water in the tank.
This means that the initial concentration of salt in the tank is:
100 kg / (1000 L + 2000 L)
= 0.03333 kg/L
As pure water is added to the tank at the rate of 12 L/min, and the mixed solution is drained out of the tank at the rate of 13 L/min. The volume of the solution in the tank after t minutes is given by,
(2000 + 12t) - 13t = 2000 - t.
The concentration of salt in the tank after t minutes is given by,
y = (100 - t)(0.03333)
The differential equation for this situation would be,
dy/dt = (0.03)(9) - (y/2000)(13)dy/dt + (13/2000)y
= 0.27
Rewrite the differential equation, if necessary, to obtain the form
y' = f(t, y):
dy/dt + (13/2000)y
= 0.27 - y(0)
= 60
The given differential equation is,
y' + ty^(1/3) = tan(t),
y(3) = -5a)
Rewriting the differential equation, we gety' = -ty^(1/3) + tan(t) - y...[1]b) The partial derivative of f with respect to y is,
df/dy = (1/3)ty^(-2/3) - 1
f and its derivative are continuous everywhere except for y = 0.
c) The largest open rectangle in the ty-plane on which the solution of the initial value problem above is certain to exist for the initial condition is obtained as follows:
The interval of t is (3,∞), as the initial condition is given at t = 3.
The interval of y is (-5,∞), as the solution exists until the concentration of salt reaches zero.
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A toy boat made of solid wood has a mass of 4.78 kgkg and density of 0.5 g/cm3g/cm3 . It is placed in water. Some steel weights are placed on top of it so it floats with 10 %% of its volume above the surface. Find the mass of steel placed on top of the boat
Therefore, the mass of steel placed on top of the boat is 0.000956 kg or 0.956g
To solve this problem, we need to determine the volume of the toy boat and the volume of water it displaces when floating.
Since the boat floats with 10% of its volume above the water surface, we can calculate the total volume of the boat using the given density.
Density is defined as mass divided by volume.
Rearranging the equation, we can find the volume of the boat:
Volume of the boat = Mass of the boat / Density of the boat
Volume of the boat = 4.78 kg / (0.5 g/cm³ * 1000 cm³/g) [Converting the density from g/cm³ to kg/cm³]
Volume of the boat = 4.78 kg / 0.5 kg/cm³
Volume of the boat = 9.56 cm³
Now, since 10% of the boat's volume is above the water surface, we can calculate the volume of water it displaces:
Volume of water displaced = 10% * Volume of the boat
Volume of water displaced = 10% * 9.56 cm³
Volume of water displaced = 0.1 * 9.56 cm³
Volume of water displaced = 0.956 cm³
To find the mass of the steel placed on top of the boat, we need to consider that the density of water is 1 g/cm³.
Mass of steel = Density of water * Volume of water displaced
Mass of steel = 1 g/cm³ * 0.956 cm³
Mass of steel = 0.956 g
Since we want the mass in kilograms, we convert grams to kilograms:
Mass of steel = 0.956 g / 1000 g/kg
Mass of steel = 0.000956 kg
Therefore, the mass of steel placed on top of the boat is 0.000956 kg or .0956g (rounded to three decimal places).
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a 500 g ball swings in a vertical circle at the end of a 1.5-m-long string. When the ball is at the bottom of the circle the tension in the string is 15 N. What is the speed of the ball at that point?
A 500 g ball swings in a vertical circle at the end of a 1.5-m-long string. When the ball is at the bottom of the circle the tension in the string is 15 N. the speed of the ball at the bottom of the vertical circle is approximately 9.49 m/s.
To determine the speed of the ball at the bottom of the vertical circle, we can make use of the tension in the string and the gravitational force acting on the ball. At the bottom of the circle, the tension in the string provides the centripetal force required to keep the ball moving in a circular path.
The centripetal force Is given by the formula:
F_c = m * (v^2 / r)
Where F_c is the centripetal force, m is the mass of the ball, v is the velocity of the ball, and r is the radius of the circle (1.5 m).
In this case, the tension in the string (F_c) is given as 15 N. Therefore, we can set up the equation:
15 N = (0.5 kg) * (v^2 / 1.5 m)
Simplifying the equati the speed of the ball at the bottom of the vertical circle is approximately 9.49 m/s.on, we find:
V^2 = (15 N * 1.5 m) / 0.5 kg
V^2 = 45 N*m / 0.5 kg
V^2 = 90 m^2/s^2
Taking the square root of both sides of the equation, we obtain:
V = √(90 m^2/s^2) ≈ 9.49 m/s
Hence, At this point, the tension in the string provides the necessary centripetal force to keep the ball moving in its circular path.
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Force (MxLxT) (MXL T-²) XLXT 7. Force is equal to mass x acceleration and is typically expressed in units of Newtons (kg m s). Acceleration is the rate of change of velocity. If gravitational acceleration is equal to 9.8 ms, then what is the gravitation force experienced by the mass of air in the box from Figure 1 (pg. 5) (see question 5)? (Note: again, this question does not involve a conversion but rather use of an equation) kg m s²- Newtons (N) 9.8m5² (kg) mg-2) Pressure (L¹ x M x T²) 8. Pressure is equal to force divided by the area over which the force is applied and is typically expressed in units of N m2 (Pa). If the box in Figure 1 (pg. 5) rests on the Earth's surface, what is the pressure exerted by the gravitational force over the bottom area of the box equal to 4 m²? Nm² = Pa 9. Pressure on weather maps is usually expressed in units of bars, where one bar (100,000 Pa) approximates the average sea-level pressure (101,325 Pa or nearly 100,000 Pa). Using this and other aids (see appendix), convert the following: 1 mb= Pa 101, 325
The mass of the air is 1000 kg and the gravitational acceleration is 9.8 m/s² so the gravitational force is 9800 N.
How to explain the informationThe gravitational force experienced by the mass of air in the box is equal to the mass of the air times the gravitational acceleration.
The pressure exerted by the gravitational force over the bottom area of the box is equal to the force divided by the area. The force is 9800 N and the area is 4 m², so the pressure is 2450 Pa.
1 mb = 101,325 Pa. This can be found by converting 1 mb to Pa using the conversion factor 1 mb = 101,325 Pa.
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Penetration capabilities in...
- Radio Waves
- Microwaves
- Infrared
- Visible light
- Ultraviolet
- X-rays
- Gamma rays
Ultra = X-ray
Explanation:
how are things going to paint the roof for beginners painting and decorating the whole house so I'm going away with a few sports mates for a couple nights for beginners but will have a look painting at a time and see what is going to be Strong enough
a sim,ple elctrical ciurucuiot contains a battery a light bulb and a properly copnnected ammeter the ammeter has a very low internal resistance because it is connected in
In a simple electrical circuit, if it contains a battery, a light bulb, and a properly connected ammeter, the ammeter has a very low internal resistance because it is connected in series with the circuit.
An electrical circuit is made up of a combination of resistors, voltage sources, and current sources that are interconnected in a closed loop. It is used to generate an electric current in a complete circuit and can be as straightforward as a battery connected to a bulb or as complicated as a full-scale electronic circuit.
Ammeters are measuring devices that are used to measure current in a circuit. The ammeter should be connected in series with the circuit to allow current to flow through it. An ammeter with a very low internal resistance should be used since any extra resistance in the ammeter can change the current being measured.
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I n it in 20 minutes
JL.61 A producer of refrigerator compressors wants to implement
a just-in-time production line to support demand from a neighboring
appliance manufacturer. Demand from the applian
The producer of refrigerator compressors needs to determine the number of kanbans required for implementing a just-in-time production line to meet the demand for 150 compressors per day from a neighboring appliance manufacturer.
To calculate the number of kanbans required, we need to consider the production lead time, safety stock factor, and optimal production quantity. The production lead time is 5 days, which means that it takes 5 days to produce a batch of compressors once the production process starts.
The safety stock factor is 17%, indicating that the producer wants to maintain an additional 17% of the daily demand as safety stock to mitigate any unforeseen fluctuations. The optimal production quantity is 95 units, which is the batch size that minimizes setup costs.
To determine the number of kanbans, we first need to calculate the total demand during the production lead time. Since the demand is 150 compressors per day and the lead time is 5 days, the total demand during the lead time is 150 compressors/day * 5 days = 750 compressors. Adding the safety stock, the total demand becomes 750 compressors + (17% * 150 compressors) = 750 compressors + 25.5 compressors = 775.5 compressors.
Next, we divide the total demand by the optimal production quantity to get the number of kanbans required. The number of kanbans is calculated as 775.5 compressors / 95 compressors per kanban = 8.16 kanbans. Since we cannot have a fraction of a kanban, we round up to the nearest whole number. Therefore, the producer of compressors requires 9 kanbans to meet the demand of 150 compressors per day from the neighboring appliance manufacturer.
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The complete question is:
A producer of refrigerator compressors wants to implement a just-in-time production line to support demand from a neighboring appliance manufacturer. Demand from the appliance manufacturer is for 150 compressors a day. The production lead time is 5 days and the producer wants to have a 17% safety stock factor. This producer has also cut setup costs such that the optimal production quantity is 95 units. How many kanbans does this producer of compressors require?
A vertical straight wire 35.0 cm in length carries a current. You do not know either the magnitude of the current or whether the current is moving upward or downward. If there is a uniform horizontal magnetic field of 0.0350 T that points due north, the wire experiences a horizontal magnetic force to the west of 0.0180 N. Find the magnitude of the current. Express your answer with the appropriate units. HÅ B ? I= Value Units Submit Request Answer Part D Find the direction of the current. The current is traveling horizontally leftward. O The current is traveling vertically upward. The current is traveling horizontally rightward. The current is traveling vertically downward. Submit Request Answer
A vertical straight wire 35.0 cm in length carries a current. You do not know either the magnitude of the current or whether the current is moving upward or downward. If there is a uniform horizontal magnetic field of 0.0350 T that points due north, the wire experiences a horizontal magnetic force to the west of 0.0180 N. The magnitude of the current in the wire is approximately 0.073 A.The direction of the current is vertically downward.
To find the magnitude of the current in the wire, we can use the formula for the magnetic force on a current-carrying wire in a magnetic field:
F = B * I * L * sin(theta)
where F is the magnetic force, B is the magnetic field strength, I is the current, L is the length of the wire, and theta is the angle between the wire and the magnetic field.
Given:
B = 0.0350 T (magnetic field strength)
L = 35.0 cm = 0.35 m (length of the wire)
F = 0.0180 N (magnetic force)
Rearranging the formula, we can solve for the current I:
I = F / (B * L * sin(theta))
Since the wire experiences a horizontal magnetic force to the west, the angle theta between the wire and the magnetic field is 90 degrees (perpendicular).
I = 0.0180 N / (0.0350 T * 0.35 m * sin(90°))
Using the given values and evaluating the expression:
I ≈ 0.073 A (amperes)
Therefore, the magnitude of the current in the wire is approximately 0.073 A.
To determine the direction of the current, we can use the right-hand rule. When the magnetic force is directed to the west and the magnetic field is pointing due north, the current in the wire must be traveling vertically downward.
Thus, the direction of the current is vertically downward.
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a rock hits the ground at a speed of 15 m/s and leaves a hold 50 cm deep. after it hits the ground, what is the magnitude of the rock's (assumed) uniform acceleration?
The magnitude of the rock's (assumed) uniform acceleration is v² - 225.
Initial speed, u = 15 m/s
Displacement, s = 50 cm = 0.5 m
Magnitude of acceleration, a = ?
We know, v² - u² = 2as
Let's substitute the given values into the above formula. v² - u² = 2as (v is the final velocity)
Final velocity, v = ?u = 15 m/s (Initial velocity)
s = 0.5 m (Displacement)
a = ?
v² - u² = 2as (v² - u²)/2s = a(v+u)/2(a = (v² - u²)/2s)
(a = (v² - u²)/2s)(a = (v² - (15 m/s)²)/2(0.5 m))(a = (v² - 225)/1)(a = v² - 225)
Therefore, the magnitude of the rock's uniform acceleration is v² - 225, given that a rock hits the ground at a speed of 15 m/s and leaves a hold 50 cm deep after it hits the ground.
The magnitude of the rock's uniform acceleration is v² - 225.
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you are driving a vehicle at 55 mph on dry pavement, about how much total stopping distance do you need to bring it to a stop?
When driving at 55 mph on dry pavement, you'll need approximately 420 feet of total stopping distance to bring your vehicle to a stop.
The total stopping distance that you need to bring a vehicle to a stop when driving at 55 mph on dry pavement depends on many variables, including the following:
Reaction time, vehicle weight, braking system efficiency, tires, road conditions, and so on.
According to a generally accepted formula, the total stopping distance can be calculated as follows:
TSD = (DR + BD) + (RD × DD)Where:TSD = Total Stopping DistanceDR = Distance Traveled during Driver Reaction TimeBD = Braking Distance (Distance required to stop a vehicle, from the point where brakes are applied)RD = Rolling Distance (Distance traveled by the vehicle while its brakes are in operation)
DD = Distance Lost Due to Perception Time or Delay Time
For a vehicle traveling at 55 mph on dry pavement, the driver's reaction time is around 1.5 seconds (according to some sources).
DR is estimated to be approximately 203 feet (60 m), which is equivalent to 1.5 seconds of time at 55 mph. If the brakes on the vehicle are well-maintained, the BD for a typical car is around 216 feet (66 m).
Rolling distance is determined by the friction between the tires and the road surface and is determined by the tire tread and type, as well as the road surface texture.
It ranges from 155 to 175 feet (47 to 53 m) on dry pavement in most cases. If we use the upper limit of 175 feet for rolling distance, the total stopping distance can be calculated as follows:TSD = (DR + BD) + (RD × DD)TSD = (203 + 216) + (175 × 1.5)TSD = 419.5 feet
Therefore, when driving at 55 mph on dry pavement, you'll need approximately 420 feet of total stopping distance to bring your vehicle to a stop.
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a circuit has three resistors connected in series. resistor r2 has a resistance of 200 ohms and a voltage drop of 30 volts. what is the current in resistor r3?
The current through resistor R3 is 0.33 A. A circuit with three resistors connected in series is shown below: Circuit diagram of three resistors connected in series As per the given information, R2 has a resistance of 200 ohms and a voltage drop of 30 volts.
Therefore, the voltage drop across R1 is V1 = V - V2 - V3V = voltage supplied to the circuit = voltage drop across R1 + voltage drop across R2 + voltage drop across R3R1 = Resistance of resistor R1.R2 = Resistance of resistor R2 = 200 Ω.V3 = Voltage drop across resistor R3.I3 = Current through resistor R3.To calculate the current in resistor R3, let's follow the steps given below.Step 1: Find the voltage drop across R1.Using Ohm's Law, the voltage drop across R2 is V2 = IR2Substitute the values of V2 and R2 to get the value of current I.I = V2/R2I = 30/200I = 0.15 A
Using Kirchhoff's voltage law, the voltage drop across R1 isV1 = V - V2 - V3V = V1 + V2 + V3Substitute the values of V, V2, and V3 to get the value of V1.V1 = V - V2 - V3V1 = 100 - 30 - V1V1 = 70 VStep 2: Find the current through R3.Using Ohm's Law, the voltage drop across R3 is V3 = I3R3.Substitute the values of V3 and R3 to get the value of current I3.I3 = V3/R3I3 = (V - V1 - V2)/R3I3 = (100 - 70 - 30)/R3I3 = 0.33 A
Therefore, the current through resistor R3 is 0.33 A.
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2. (6 points] The energy levels for the hydrogen atom are given by the formula E-136V, At what temperature would a gas of hydrogen have, on average, one atom in the first excited state for every milli
The energy levels for the hydrogen atom are given by the formula E-136V. The temperature that a gas of hydrogen have, on average, one atom in the first excited state for every million atoms in the ground state is 121,600 K.
The formula for energy levels for the hydrogen atom is given by: E=-13.6/n²
where,
n is the principal quantum number of the energy level
Here, the first excited state is n = 2 and the ground state is n = 1.
Therefore, the energy difference between the first excited state and the ground state can be calculated by:
E(2) - E(1) = (-13.6 / 2²) - (-13.6 / 1²) = -3.4 eV
The probability that an atom will be in the first excited state is proportional to the Boltzmann factor:
exp(-ΔE/kT) = 10⁶ / 1
We know, ΔE = 3.4 eV.
k is the Boltzmann constant which is equal to 8.617×10⁻⁵ eV/K.
Let's solve for the temperature, T:
T = ΔE / (k ln(10⁶))= (3.4 eV) / (8.617×10⁻⁵ eV/K ln(10⁶))= 121,600 K
Therefore, at a temperature of 121,600 K, a gas of hydrogen would have, on average, one atom in the first excited state for every million atoms in the ground state.
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Find the center of mass of the following particles (drawn large so they can be seen)
It is possible to suspend an object at rest from its center of gravity, and gravity won't induce it to begin rotating no matter how it is positioned.
Thus, The center of gravity of an object will be located somewhere along a vertical line that passes through the point of suspension if you suspend it from any point and allow it to come to rest.
The gravitational acceleration is (almost) constant near the surface of the earth, where the center of mass also lies.
A place that represents the average or typical location of an object's mass, as if all of the mass were contained there, is known as the center of mass (CM). The geometric center of a uniform sphere serves as its center of mass. The barycenter is another name for the CM.
Thus, It is possible to suspend an object at rest from its center of gravity, and gravity won't induce it to begin rotating no matter how it is positioned.
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A jumbo jet of mass 4 x 10² kg travelling at a speed 5000 m/s lands on the airport. It takes 2 minutes t come to rest. Calculate the average force applied by th ground on the aeroplane. 12 Ans: -1.67x107
Answer:
I got 16680N
Explanation:
We can use this formula; F = ma
F = ma
F = (4 × 10²) × ((0-5000)/120)
F = (4 × 10²) × (-41.7)
F = -16680N
Therefore, Fₐᵥ = 16680N.
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the figure and show complete solution on each problem. please
answer all
IV. Find the momentum of a 60-g bullet whose kinetic energy is 270 J. V. A 60-ton car moving at 1.2 mile per hour is instantaneously coupled to a stationary 40-ton car. What is the speed of the couple
The momentum of the car is 5.7 Kgm/s
The final velocity is 0.3 m/s
What is the momentum?
Momentum is a fundamental concept in physics that describes the motion of an object. It is defined as the product of an object's mass and its velocity.
p = m * v
Where;
KE = 1/2mv^2
v = √2KE/M
v = √2 * 270/0.06
v = 95 m/s
Then;
p = mv
p = 0.06 * 95 m/s
= 5.7 Kgm/s
V = 1.2 m/hr or 0.5 m/s
mass = 60 ton and 40 ton or 54431.1 Kg and 36287.4 Kg
Momentum before collision = Momentum after collision
(54431.1 * 0.5 m/s) + 0 = (54431.1 + 36287.4 )v
v = 27215.55/90718.5
v = 0.3 m/s
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in an electrical charge, some charges would repel one another. which is an example of electrical charges that would repel one another?
In an electrical charge, like charges would repel one another.
An example of electrical charges that would repel one another is two positive charges or two negative charges. Like charges have the same polarity, that is, they have the same charge, and they would repel one another. It's essential to understand the concept of electrical charges in order to understand this concept.
Electric charge is a fundamental property of matter, and it can exist in two types: positive and negative. Like charges repel one another, whereas opposite charges attract one another. The reason why charges attract or repel is that they produce electric fields that interact with other charges.
It is important to note that the magnitude of the electric field is inversely proportional to the square of the distance between two charges.
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Cosmos: The Electric Boy Assignment
1. What was Humphry Davy's experiment and how did it go wrong? What could he have done differently?
2. What did Humphry Davy notice about a wire with electricity running through it as he brought it near a compass?
3. What was Humphry Davy's next project for Michael Faraday and why did he give him that particular project?
4. What did Michael Faraday create as a result of his efforts? How did it work?
5. What did Michael Faraday notice when he moved a magnet in and out of a wire?
6. What were some of the materials Michael Faraday used to see if light would be affected by magnets? What ended up working in the end? What did it mean?
7. What did Michael Faraday notice when he sprinkled iron filings around current carrying wires? What did he think ultimately meant?
8. Why did Michael Faraday's contemporaries in science not believe his hypothesis about field forces? What did he need in order to convince them?
9. How do the effects of Michael Faraday's invention shape society even today?
Humphry Davy's most famous experiment involved the use of a voltaic pile (an early battery) to conduct electrolysis on various substances.
1. Humphry Davy's most famous experiment involved the use of a voltaic pile (an early battery) to conduct electrolysis on various substances. One of his experiments involved passing an electric current through a mixture of potassium and water. The experiment went wrong when he increased the power of the battery, causing a violent explosion due to the release of hydrogen gas. The explosion was caused by the high reactivity of potassium with water. To prevent this mishap, Davy could have used a smaller amount of potassium or diluted the solution to reduce the reactivity.
2. When Humphry Davy brought a wire with electricity running through it near a compass, he noticed that the needle of the compass was deflected from its usual north-south orientation. This observation indicated that an electric current produces a magnetic field around the wire, leading to the deflection of the compass needle.
3. Humphry Davy assigned Michael Faraday the task of finding a way to liquefy chlorine gas. He gave Faraday this project because he recognized Faraday's experimental skills and believed that his ingenuity and dedication would lead to a successful outcome.
4. As a result of his efforts, Michael Faraday succeeded in liquefyingchlorine gas and several other gases. He developed a method using high pressure and low temperatures to condense the gases into liquid form. This breakthrough allowed for further investigation and study of these substances.
5. Michael Faraday noticed that when he moved a magnet in and out of a wire, it induced an electric current in the wire. This phenomenon is known as electromagnetic induction, and it demonstrated the relationship between magnetism and electricity.
6. Michael Faraday experimented with various materials to test their response to magnetic fields. He tried substances such as glass, copper, and sulfur, but they did not show any significant effects. However, when he used a coil of wire, he observed that a current was induced in the wire when exposed to a changing magnetic field. This discovery led to the development of the concept of electromagnetic induction and its practical applications.
7. When Michael Faraday sprinkled iron filings around current-carrying wires, he observed that the filings arranged themselves in a pattern, forming circles around the wires. He realized that these patterns represented the lines of magnetic force around the wires. This finding suggested the existence of magnetic fields and provided evidence for Faraday's theory of field forces.
8. Michael Faraday's contemporaries in science initially did not believe his hypothesis about field forces because it went against the prevalent understanding of action at a distance. They adhered to the idea that forces acted only through direct contact between objects. To convince them, Faraday needed to provide experimental evidence and develop a coherent theoretical framework to explain the observed phenomena. He achieved this through his extensive experiments and the formulation of field theory, which established the concept of field forces acting at a distance.
9. Michael Faraday's inventions and discoveries in electromagnetism and electrochemistry have shaped society to this day. They laid the foundation for the development of modern electrical technology and power generation. Faraday's work led to the invention of electric motors, generators, and transformers, which are essential components of our electrical infrastructure. His principles of electromagnetic induction and field theory also underpin technologies such as wireless communication, electric lighting, and the functioning of modern electronics. Additionally, Faraday's emphasis on experimental investigation and his dedication to sharing scientific knowledge contributed to the advancement of scientific methodology and the popularization of science education.
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(a) A 19.0 kg child is riding a playground merry-go-round that is rotating at 40.0 rev/min. What centripetal force (in N) must she exert to stay on if she is 2.50 m from its center? (Enter a number.)
A 19.0 kg child is rotating at 40.0 rev/min in a merry-go-round. The centripetal force that the child must exert to stay on the merry-go-round is 832.8 N.
The formula to determine the centripetal force,
Fc = mv^2/r
where,
m is the mass,
v is the velocity,
r is the radius of the rotation
Here,
mass of the child = 19 kg
distance of the child from the center of the merry-go-round = 2.5 m
The velocity can be determined using the given frequency of 40.0 rev/min, which is the same as 40.0/60 = 0.67 revolutions per second.
The circumference of the circular path is given by 2πr.
Therefore, the velocity v is given by,
v = 2πr * f
where,
f is the frequency
v = 2π * 2.5 * 0.67 = 10.5 m/s
Substituting the given values in the formula to determine the centripetal force, we get,
Fc = mv^2/r
Fc = 19 * 10.5^2/2.5
Fc = 832.8 N
Therefore, the centripetal force that the child must exert to stay on the playground merry-go-round is 832.8 N.
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