The answers are a) The torque T1 due to force F1 is 5.225 N.m, b) The torque T2 due to force F2 is 0.21 N.m, c) The torque T3 due to force F3 is 25.425 N.m, d) The torque T4 due to force F4 is 0 N.m, and e) The angular acceleration α of the thin rod about its center of mass is 105.74 rad/s².
In order to answer the problem, we will use the following formulas:Torque: τ = rFsin(θ), where: r: is the perpendicular distance from the axis of rotation to the line of action of the force, F: is the force acting on the object, θ: is the angle between the two vectors. We will also use the formula for the moment of inertia of a uniform rod rotating about its center: I = (1/12)ML²a) The magnitude of the torque T1 due to force F1 can be calculated using the formula:τ1 = r1F1sin(θ1), where: r1 is the perpendicular distance from the axis of rotation to the line of action of force F1, F1 = 5.5 N is the force acting on the object, θ1 is the angle between the two vectorsθ1 = 90° since F1 is acting perpendicular to the rod. T1 = r1F1sin(θ1) = (L/2)(F1) = (1.9 m/2)(5.5 N) = 5.225 N.mb) The magnitude of the torque T2 due to force F2 can be calculated using the formula:τ2 = r2F2sin(θ2), where: r2 is the perpendicular distance from the axis of rotation to the line of action of force F2, F2 = 1.5 N is the force acting on the object, θ2 is the angle between the two vectorsθ2 = 90° since F2 is acting perpendicular to the rod. T2 = r2F2sin(θ2) = (d)(F2) = (0.14 m)(1.5 N) = 0.21 N.mc) The magnitude of the torque T3 due to force F3 can be calculated using the formula:τ3 = r3F3sin(θ3), where: r3 is the perpendicular distance from the axis of rotation to the line of action of force F3, F3 = 13.5 N is the force acting on the object, θ3 is the angle between the two vectors θ3 = 90° since F3 is acting perpendicular to the rod. T3 = r3F3sin(θ3) = (L/2)(F3) = (1.9 m/2)(13.5 N) = 25.425 N.md) The magnitude of the torque T4 due to force F4 can be calculated using the formula:τ4 = r4F4sin(θ4), where: r4 is the perpendicular distance from the axis of rotation to the line of action of force F4, F4 = 16 N is the force acting on the object, θ4 is the angle between the two vectorsθ4 = 180° since F4 is acting opposite to the direction of rotation. T4 = r4F4sin(θ4) = (L/2)(F4)sin(180°) = 0 N.m (since sin(180°) = 0)e) Angular acceleration α of the thin rod about its center of mass in radians per square second. The net torque τ acting on the object is given by:τ = τ1 + τ2 + τ3 + τ4 = 5.225 N.m + 0.21 N.m + 25.425 N.m + 0 N.m = 30.86 N.m. The moment of inertia of the rod about its center of mass is I = (1/12)ML² = (1/12)(3.9 kg)(1.9 m)² = 0.292 kg.m². The angular acceleration α can be calculated using the formula:τ = Iαα = τ/I = (30.86 N.m)/(0.292 kg.m²) = 105.74 rad/s². Therefore, the angular acceleration of the thin rod about its center of mass is 105.74 rad/s².For more questions on torque
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A liquid with a density of 750 kg/m3 and a volumetric flow rate of 0.15 m3/s flows through a plastic pipe. If the head loss due to friction is 325 m, calculate the required pump power in kilowatt to maintain the flow. (take the gravitational acceleration as 9.81 m.s-2)
To maintain the flow of a liquid with a density of 750 kg/m³ and a volumetric flow rate of 0.15 m³/s through a plastic pipe, a pump power of approximately X kilowatts is required, considering a head loss due to friction of 325 m.
The power required by the pump can be calculated using the equation:
Power = (density * volumetric flow rate * gravitational acceleration * head loss) / efficiency
Given the density of the liquid as 750 kg/m³, the volumetric flow rate as 0.15 m³/s, and the head loss due to friction as 325 m, we can substitute these values into the equation. Assuming an efficiency of 100% for simplicity, we can neglect the efficiency term.
Power = (750 kg/m³ * 0.15 m³/s * 9.81 m/s² * 325 m) / 1000
Simplifying the equation, we get:
Power = 3593.8125 Watts
Converting this to kilowatts, we divide by 1000:
Power = 3.594 kW
Therefore, approximately 3.594 kilowatts of power is required to maintain the flow of the liquid through the plastic pipe, accounting for the given parameters.
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Suppose you walk 28.0 m straight east and then 12.0 m straight south. How far are you from your starting point (in m)? m What is your displacement vector (in m)? (Express your answer in vector form. Assume the +x-axis is to the east, and the +y-axis is to the north.)
D
= m What is the direction of your displacement (in degrees counterclockwise from the +x-axis)? o counterclockwise from the +x-axis
The direction of the displacement is -23° counterclockwise from the +x-axis.
Given information: 28.0 m straight east and then 12.0 m straight south.
The direction of the displacement in degrees counterclockwise from the +x-axis is to be calculated.
Using the Pythagorean theorem, the distance is calculated as follows:
Distance = √(east^2 + south^2)
Distance = √(28.0 m)^2 + (12.0 m)^2
Distance = √(784.0 m^2 + 144.0 m^2)
Distance = √928 = 30.46 ≈ 30 m
The displacement vector is the straight-line distance and direction to the starting point. It can be obtained by adding two vectors. Vector A represents the eastward displacement and vector B represents the southward displacement. Here, vector A is represented as 28i and vector B is represented as -12j. Therefore, the displacement vector can be calculated as follows:
D = A + B = 28i - 12j
The direction of displacement in degrees counterclockwise from the +x-axis can be found using the following formula:
Direction = tan⁻¹(y/x)
tan⁻¹(-12/28) = -22.62 ≈ -23°
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1. Explain how the hydrostatic pressure is the same along any
constant horizontal line?
Hydrostatic pressure is the pressure exerted by a fluid at rest due to the force of gravity alone. This pressure is the same in all directions and increases with depth.
The hydrostatic pressure is the same along any constant horizontal line because the fluid at rest is in equilibrium and the pressure is the same at any given depth in the fluid.
As we move down through a fluid, the pressure increases due to the weight of the fluid above, but the pressure is the same in all directions. Therefore, if we move along a constant horizontal line in the fluid, we are not changing the depth or weight of fluid above us. As a result, the hydrostatic pressure is the same along that line.
In summary, the hydrostatic pressure is the same along any constant horizontal line in a fluid at rest because the pressure depends only on the depth and weight of the fluid above that point and not on the direction.
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The drawing shows a 23.8-kg crate that is initially at rest. Note that the view is one looking down on the top of the crate. Two forces, F, and E2 are applied to the crate, and it begins to move. The coefficient of kinetic friction between the crate and the floor is μk=0.303. Determine the (a) magnitude and (b) direction (relative to the x axis) of the acceleration of the crate.
To determine the magnitude of acceleration, we must first calculate the force acting on the crate in the horizontal direction. The frictional force, which opposes the motion, is acting on the crate in the opposite direction to the motion. Therefore, the frictional force is equal to the applied force minus the frictional force:
Frictional force (f) = μkN = 0.303 (23.8)(9.8) = 69.7252 NThe net force acting on the crate, which is equal to the applied force minus the frictional force, is given by:Fnet = Fa - f = 146.56 - 69.7252 = 76.8348 NThe acceleration of the crate is given by the net force divided by the mass of the crate:a = Fnet / m = 76.8348 / 23.8 = 3.2305 m/s²Therefore, the magnitude of the acceleration of the crate is 3.23 m/s².b) The acceleration is in the direction of the net force. Because the applied force is in the positive x-direction, and the frictional force is in the negative x-direction,
the net force is in the positive x-direction. Thus, the direction of the acceleration is in the positive x-direction.The explanation for the above answer is as follows:Given: Mass of the crate (m) = 23.8 kgCoefficient of kinetic friction (μk) = 0.303The view is one looking down on the top of the crate.The two forces acting on the crate are:F = 146.56 N E2The crate is initially at rest and starts to move. The crate starts to move due to the forces applied to it. The force of friction opposes the motion of the crate.To determine the direction of acceleration, we must first determine the net force acting on the crate. We can then use Newton's Second Law to determine the acceleration of the crate.
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The engine on a test-rocket fails suddenly 4.64 seconds after launch. If the rocket was accelerating at 6.95 ms^−2
up to the point of failure, how much time elapses between launch and returning to the ground? Calculate your answer in mks units correct to three significant figures.
The engine on a test-rocket fails suddenly 4.64 seconds after launch. If the rocket was accelerating at 6.95 ms^−2 up to the point of failure, how much time elapses between launch and returning to the ground?
Given that, the rocket was accelerating at 6.95 ms−2 and engine fails suddenly 4.64 seconds after launch. We have to find the time elapsed between launch and returning to the ground. The equation for time is given by
v = u + at
Where,v is the final velocity u is the initial velocity a is the acceleration t is the time elapsed
Thus, the initial velocity u=0 ms−1The acceleration, a= 6.95 ms−2The time elapsed before the engine fails,t= 4.64sLet the final velocity be v.
To calculate the final velocity, we use the above kinematic equation,
v = u + at
⇒ v = 0 + (6.95ms^{−2} × 4.64s)
=32.108ms^{−1}
The distance travelled by the rocket s is given by
s = ut + (1/2)at^2
⇒ s = 0 + (1/2) (6.95ms^{−2}) (4.64s)^2
=72.5136m
Let h be the maximum height attained by the rocket. The velocity of the rocket at height h is zero.
So, we can use the equation,v^2 = u^2 + 2as
Where, s = h and v=0Thus, h= (v^2)/(2g)= 32.108^2 / (2 x 9.8)= 52.864 m
Therefore, the total distance travelled by the rocket from the launch to the ground
= 72.5136m + 52.864m
= 125.3776 m
Let T be the time elapsed between launch and returning to the ground, given by,
T = (2s / g)1/2
⇒ T = (2 x 125.3776 / 9.8)1/2
=5.7907s (approx)
Therefore, the time elapsed between launch and returning to the ground is 5.7907 seconds to three significant figures.
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In a grocery store, you push a 12.4-kg shopping cart horizontally with a force of 11.5 N. If the cart starts at rest, how far does it move in 2.15 s ? Express your answer using three significant figures. Stopping a 747 A 747 jetliner lands and begins to slow to a stop as it moves along the runway. If its mass is 3.44×10
5
kg, its speed is 68.0 m/s, and the net braking force is 4.30×10
5
N, what is its speed 9.00 s later? Express your answer using three significant figures. How far has it traveled in this time? Express your answer using three significant figures.
In a grocery store, you push a 12.4-kg shopping cart horizontally with a force of 11.5 N. If the cart starts at rest, it can move 2.70 m in 2.15 seconds.
To find the distance the shopping cart moves in 2.15 seconds, we can use the equation:
Distance = (Force * [tex]Time^2[/tex]) / (2 * Mass)
Plugging in the values:
Force = 11.5 N
Time = 2.15 s
Mass = 12.4 kg
Distance = (11.5 N * [tex](2.15 s)^2[/tex]) / (2 * 12.4 kg)
Distance ≈ 2.70 meters (to three significant figures)
Therefore, the shopping cart moves approximately 2.70 meters in 2.15 seconds.
Now let's move on to the second part of the question regarding the 747 jetliner.
To determine the speed of the jetliner 9.00 seconds later, we can use the equation:
Final Speed = Initial Speed + (Force / Mass) * Time
Plugging in the values:
Initial Speed = 68.0 m/s
Force = 4.30×[tex]10^5[/tex] N
Mass = 3.44×[tex]10^5[/tex] kg
Time = 9.00 s
Final Speed = 68.0 m/s + (4.30×[tex]10^5[/tex] N / 3.44×[tex]10^5[/tex] kg) * 9.00 s
Final Speed ≈ 80.9 m/s (to three significant figures)
Therefore, the speed of the 747 jetliner 9.00 seconds later is approximately 80.9 m/s.
To find the distance traveled by the jetliner in this time, we can use the equation:
Distance = Initial Speed * Time + (1/2) * (Force / Mass) * [tex]Time^2[/tex]
Plugging in the values:
Initial Speed = 68.0 m/s
Force = 4.30×[tex]10^5[/tex] N
Mass = 3.44×[tex]10^5[/tex] kg
Time = 9.00 s
Distance = 68.0 m/s * 9.00 s + (1/2) * (4.30×[tex]10^5[/tex] N / 3.44×[tex]10^5[/tex] kg) * (9.00 s)^2
Distance ≈ 6120 meters (to three significant figures)
Therefore, the 747 jetliner has traveled approximately 6120 meters in 9.00 seconds.
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among the following, which is a powerful vasoconstrictor? a. angiotensin i b. aldosterone c. renin d. angiotensin ii e. magnesium
Among the given options, the powerful vasoconstrictor is Angiotensin II.
What is Angiotensin II?
Angiotensin II is a vasoconstrictor hormone that raises blood pressure. The hormone is produced by a pathway that starts in the liver and goes through the lungs and kidneys. In the lungs, angiotensinogen is produced by the liver and transformed into angiotensin I by the enzyme renin.
Angiotensin I is transformed into angiotensin II by angiotensin-converting enzyme (ACE) in the lungs.Increased angiotensin II levels lead to vasoconstriction and a rise in blood pressure. This hormone stimulates the secretion of aldosterone, a hormone produced by the adrenal gland that helps to preserve sodium in the kidneys, which also contributes to increased blood pressure.
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Monochromatic light strikes a metal surface and electrons are ejected from the metal. If the frequency of the light is increased while the intensity of the beam is held fixed, a. the rate of ejected electrons will remain the same but the maximum kinetic energy will decrease b. the rate of ejected electrons will increase and the maximum kinetic energy will increase. - the rate of ejected electrons will remain the same but the maximum kinetic energy will increase d the rate of ejected electrons will decrease and the maximum kinetic energy will increase.
When monochromatic light strikes a metal surface, electrons are ejected from the metal. If the frequency of the light is increased while the intensity of the beam is held fixed, the rate of ejected electrons will increase and the maximum kinetic energy will increase.Option (b) is the correct option.Wave theory of light:.
According to wave theory of light, the energy of a wave is proportional to the square of its amplitude. Therefore, it was expected that the number of electrons emitted and their kinetic energy would increase with the intensity of the light source. But, in reality, it was found that the energy of the electrons emitted does not depend on the intensity of the incident radiation but rather on its frequency. The maximum kinetic energy of photoelectrons is directly proportional to the frequency of the radiation incident on the metal surface. Therefore, option (b) is the correct option.
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3) (10 Points) A parallel-plate capacitor consists of two parallel, square plates that have dimensions 1.0 cm by 1.0 cm. The plates are separated by 1.0 mm and the space between them is filled with teflon. (The dielectric constant for teflon is 2.1 ) a) What is the capacitance of this capacitor? b) The capacitor is then connected to a 12 V battery for a long time. How much charge is stored separated on the plates? c) The battery is then removed. How much energy is stored by the capacitor? d) The teflon is then removed. How much electrical energy is now stored by the capacitor? Explain the difference in terms of what happened microscopically inside the teflon. e) The battery is then reconnected to the capacitor for a long time. How much electrical energy is stored by the capacitor now? Explain the difference
a) The capacitance of the parallel-plate capacitor is approximately 2.3 x 10^(-11) Farads (F).
b) The charge stored on the plates of the capacitor when connected to a 12 V battery is approximately 2.8 x 10^(-10) Coulombs (C).
c) The energy stored by the capacitor when the battery is removed is approximately 1.68 x 10^(-9) Joules (J).
d) When the teflon is removed, the electrical energy stored by the capacitor remains the same.
e) When the battery is reconnected to the capacitor, the electrical energy stored by the capacitor remains the same.
a) The capacitance (C) of a parallel-plate capacitor can be calculated using the formula C = (ε₀ * εᵣ * A) / d, where ε₀ is the permittivity of free space (8.85 x 10^(-12) F/m), εᵣ is the relative permittivity (dielectric constant) of the material, A is the area of the plates, and d is the distance between the plates.
Given that the dimensions of the plates are 1.0 cm by 1.0 cm (0.01 m by 0.01 m) and the distance between them is 1.0 mm (0.001 m), and the dielectric constant for teflon is 2.1, we can substitute these values into the formula to find the capacitance:
C = (8.85 x 10^(-12) F/m * 2.1 * (0.01 m * 0.01 m)) / 0.001 m
C ≈ 2.3 x 10^(-11) F
b) The charge (Q) stored on the plates of a capacitor can be determined using the formula Q = C * V, where C is the capacitance and V is the voltage applied across the capacitor.
Given that the capacitor is connected to a 12 V battery, we can calculate the charge stored on the plates:
Q = (2.3 x 10^(-11) F) * 12 V
Q ≈ 2.8 x 10^(-10) C
c) The energy (U) stored by a capacitor can be calculated using the formula U = (1/2) * C * V^2, where C is the capacitance and V is the voltage applied across the capacitor.
When the battery is removed, the voltage across the capacitor becomes 0 V, so we can calculate the energy stored:
U = (1/2) * (2.3 x 10^(-11) F) * (0 V)^2
U ≈ 1.68 x 10^(-9) J
d) When the teflon is removed, the electrical energy stored by the capacitor remains the same because the energy stored in a capacitor depends only on the capacitance and the voltage applied across it. The dielectric material affects the capacitance but does not affect the stored energy.
e) When the battery is reconnected to the capacitor, the electrical energy stored by the capacitor remains the same because it depends on the capacitance and the voltage applied across it. The absence or presence of the dielectric material does not change the stored energy, as long as the voltage remains the same.
Therefore, in both cases (with or without the teflon), the electrical energy stored by the capacitor remains unchanged.
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Determine the amount of heat which has been transferred through the wall, built with solid clay brick, by the thermal conductivity for 1 day. The area of the wall is 10 m
2
, thickness is 51 cm. The temperature of the outer surface of the wall is −10
∘
C and internal 20
∘
C. The thermal conductivity of a solid clay brick is 0,8 W/m⋅K.
The area of the wall is 10 m^2, thickness is 51 cm. The temperature of the outer surface of the wall is −10 degree C and internal 20 degree C. The thermal conductivity of a solid clay brick is 0,8 W/m⋅K, 470.59 watts of heat will be transferred through the wall in 1 day.
To determine the amount of heat transferred through the wall, we can use the formula for heat conduction:
Q = (k × A × ΔT) / d
where:
Q is the amount of heat transferred,
k is the thermal conductivity of the solid clay brick (0.8 W/m⋅K),
A is the area of the wall (10 m^2),
ΔT is the temperature difference across the wall (20°C - (-10°C) = 30°C),
and d is the thickness of the wall (51 cm = 0.51 m).
Plugging in the values:
Q = (0.8 W/m⋅K)× (10 m^2) × (30°C) / (0.51 m)
Q ≈ 470.59 W
So, approximately 470.59 watts of heat will be transferred through the wall in 1 day.
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: A girl stands on a scale in an elevator, In which if the following conditions would the scale read the LARGEST weight (G.e,; largest normal force)? When the elevator is stationary. When the elevator is moving upward at a constant velocity. When the elevator is accelerating downward: When the elevator is accelerating upward. When the elevator is moving downward at a constant velocity. Vewing Soved Work RGed b Last Renconie Which of the following is true? The force of friction is in the direction opposite to an cbject's motion or potential motion. The static friction force on an object resting on an inclined plane is the same regardiess of the angle of incine, as long as the block remains at rest. The kinetic friction force is greater than the maximum static friction force. The kinetic friction force on an object depends on its speed. Which of the following is true? The mass of an object and its weight are equal. The weight of an object changes depending on the gravitational field (e.g, Earth vs. the moon). The mass of an object changes depending on the grovitational field (e 9 . Earth vs. the moen). An object has no weight if it is in free fall on Earth.
The scale would read the largest weight (or largest normal force) under the condition when the elevator is accelerating upward.
When the elevator is stationary, the scale would read the girl's actual weight since there is no additional force acting on her. When the elevator is moving upward at a constant velocity, the scale would read the girl's weight minus the force exerted by the elevator, resulting in a smaller reading. When the elevator is accelerating downward, the scale would read the girl's weight plus the force exerted by the elevator, resulting in a larger reading. When the elevator is moving downward at a constant velocity, the scale would read the girl's actual weight, similar to when the elevator is stationary.
Regarding the second question, the true statement is: The kinetic friction force on an object depends on its speed. The kinetic friction force is the force that opposes the motion of an object when it is already in motion. The magnitude of the kinetic friction force depends on the nature of the surfaces in contact and the normal force between them, but it is independent of the object's speed.
The other statements are not correct. The weight of an object is equal to the product of its mass and acceleration due to gravity, and it does not change depending on the gravitational field. The mass of an object remains constant regardless of the gravitational field it is in. And finally, an object in free fall on Earth still has weight, as weight is the force of gravity acting on an object.
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If a rigid meter stick with a fulcrum at the 50 cm position is to be kept balanced horizontally in static equilibrium with masses 450.0 g at 20.0 cm and 200.0 g at 30.0 cm, where would you need to place a third mass of 1,030.0 g ?
To keep the meter stick balanced horizontally in static equilibrium with masses 450.0 g at 20.0 cm and 200.0 g at 30.0 cm, a third mass of 1,030.0 g should be placed at a distance of 10 cm from the fulcrum. : Hence, the answer is 10 cm.
According to the principle of moments, the sum of the moments acting on a body in equilibrium is zero.
In this problem, the rigid meter stick is balanced horizontally in static equilibrium. Therefore, the net moment acting on the meter stick must be zero.
. Therefore, a third mass of 1,030.0 g should be placed at a distance of 10 cm from the fulcrum to keep the meter stick balanced horizontally in static equilibrium.
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A person starts at a tree and walks 15.0 m southwest (displacement vector
A
,45
∘
south of west) and then an additional distance in an unknown direction (displacement
B
). Her final position after these two displacements is 20.0 m directly south of her original position. What is the magnitude of her displacement
B
?
The magnitude of vector B is 13.23 m.
To find the magnitude of her displacement B, we can follow the steps below:
First, we can draw a diagram to visualize the situation described in the problem. From the given information, the person started at a tree (point O) and walked 15 m southwest (vector A).
To represent vector A in the diagram, we can draw a line segment with an arrowhead that points in the southwest direction and has a length of 15 units. The angle between vector A and the west direction is 45°.
We are told that after walking this distance, the person walks an additional distance in an unknown direction (vector B) and ends up 20 m directly south of the original position.
To represent vector B in the diagram, we can draw a line segment with an arrowhead that starts at the end of vector A and points in the direction of vector B.
Let's denote the length of vector B by |B| and the angle between vector B and the south direction by θ. Since the final position of the person is 20 m directly south of the original position, the displacement from the original position to the final position is a vector that points directly south.
We can represent this displacement by a line segment with an arrowhead that starts at point O and ends at a point 20 units directly south of point O. Let's call this vector D.
To find the magnitude of vector B (|B|), we can use the Pythagorean theorem. According to the Pythagorean theorem, for a right triangle with legs of lengths a and b and hypotenuse of length c, we have:
c^2 = a^2 + b^2
Since vector D is the hypotenuse of a right triangle whose legs are vector A and vector B, we can use the Pythagorean theorem to write:
|D|^2 = |A|^2 + |B|^2
where |D| = 20 and |A| = 15. Substituting these values, we get:
20^2 = 15^2 + |B|^2
Simplifying, we get:
|B|^2 = 400 - 225
|B|^2 = 175
Taking the square root of both sides, we get:
|B| ≈ 13.23
Therefore, the magnitude of vector B is approximately 13.23 m.
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An athlete crosses a 22 m wide river by swimming perpendicular to the water current at a speed of 1.1 m/s relative to the water. He reaches the opposite side at a distance 45 m downstream from his starting point. Randomized Variables
w=22 m
d=45 m
v
s
=1.1 m/s
50% Part (a) How fast is the water in the river flowing with respect to the ground in m/s ?
v
w
=9/4
v
w
=2.25∨ Correct!
The speed of the water in the river flowing with respect to the ground is 2.25 m/s.This is calculated by using the concept of vector addition.
Let's assume the speed of the water current is represented by "v_w" and the speed of the athlete relative to the ground is represented by "v_s." The athlete is swimming perpendicular to the water current, so the only component of the athlete's velocity that contributes to the downstream displacement is the component parallel to the water current.
Given that the athlete reaches a point 45 m downstream from his starting point and the width of the river is 22 m, we can set up the following equation:
d = v_w * t
Where:
d = downstream displacement = 45 m
v_w = speed of water current
t = time taken to cross the river
Now, let's consider the motion perpendicular to the current. The athlete's swimming speed relative to the water is 1.1 m/s. Since the width of the river is 22 m, the time taken to cross the river is given by:
t = w / v_s
Where:
w = width of the river = 22 m
v_s = speed of the athlete relative to the water = 1.1 m/s
Substituting this value of t into the first equation, we get:
d = v_w * (w / v_s)
45 = v_w * (22 / 1.1)
Simplifying, we find:
45 = v_w * 20
Now we can solve for v_w:
v_w = 45 / 20
v_w = 2.25 m/s
Therefore, the speed of the water in the river flowing with respect to the ground is 2.25 m/s
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Assume the acceleration due to gravity (g) is 9.8 m/s
2
, and ignore air resistance. At t=0 seconds, a ball dropped from the top of a building hits the ground with a velocity of 49 m/s. What is the height of the building? 10 m 5.0 m 1.2×10
2
m 49 m 2.4×10
2
m
The height of the building is approximately 1.2 × 10^2 meters.
To calculate the height of the building, we can use the kinematic equation:
v^2 = u^2 + 2as
Where:
- v is the final velocity (49 m/s)
- u is the initial velocity (0 m/s, as the ball is dropped)
- a is the acceleration due to gravity (-9.8 m/s^2, considering its downward direction)
- s is the displacement (height of the building)
Rearranging the equation to solve for s, we have:
s = (v^2 - u^2) / (2a)
Substituting the given values into the equation, we get:
s = (49^2 - 0^2) / (2 × (-9.8))
Calculating this expression, we find:
s ≈ 1.2 × 10^2 meters.
Therefore, the height of the building is approximately 1.2 × 10^2 meters.
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Newton's universal law of gravitation describes the force of gravity acting on two masses. The correct equation is, F
g
=G
r
2
m
1
m
2
. Using dimensional analysis, determine the dimensions and SI units for the gravitational constant G. Here, F
g
is a force, m
1
and m
2
are masses and r is a distance. (2 marks) Someone then tells you that the equation for gravitational potential energy U (measured in Joules) is, U=−G
r
3
m
1
m
2
. Using dimensional analysis, determine if they are correct. (1 mark) An experiment determined that the time for a star to orbit a black hole T in a circular orbit depends on the distance from the black hole to the star r, the gravitational constant G and the mass of the black hole m. That is, T=Cr
α
G
β
m
γ
where C is a dimensionless constant. Using only dimensional analysis, determine what the exponents α,β and γ must be for this equation to be correct. (3 marks)
The exponents α, β, and γ are 3/2, -1/2, and 1/2 respectively for this equation to be correct.
The equation for Newton's Universal Law of Gravitation is Fg = G (m₁m₂/r²).
Dimensions of the Gravitational constant G is to be determined using the principle of dimensional analysis.
According to the principle of dimensional analysis, the dimensions of every physical quantity can be represented by a product of three fundamental quantities such as mass (M), length (L), and time (T).
The dimensions of G can be derived as follows:
Force, Fg = MLT⁻²
Mass, m₁ = M
Mass, m₂ = M
Distance, r = L
Substituting these values in the equation of gravitational force:
Fg = G (m₁m₂/r²)
MLT⁻² = G [(M)(M)]/L²
G = [(MLT⁻²)(L²)]/[(M)(M)]
G = L³M⁻¹T⁻²
Dimensional formula of G is [L³M⁻¹T⁻²].
Gravitational potential energy U is given as U = -G(m₁m₂/r³).
The dimensions of U can be derived as follows:
Potential energy, U = ML²T⁻²
Mass, m₁ = M
Mass, m₂ = L³
Distance, r = L
Substituting these values in the equation of gravitational potential energy:
U = -G (m₁m₂/r³)
ML²T⁻² = -G [(M)(L³)]/L³
U = -GM
The dimensions of U are [ML²T⁻²].
Yes, they are correct as per the dimensional analysis.
Exponents α, β, and γ are to be determined using dimensional analysis.
According to the principle of dimensional analysis, both sides of the equation must have the same dimensions.
Therefore, the dimensions of both sides can be analyzed as follows:
Time, T = T
Distance, r = L
Gravitational constant, G = [L³M⁻¹T⁻²]
Mass of black hole, m = M
Determining the dimensions of both sides of the equation:
T = C rα G β m γ
[T] = [C][r]α[L³M⁻¹T⁻²]β[M]γ
[T] = LαT⁻²βMγ
Comparing the dimensions on both sides of the equation, we can say:
α = 3/2, β = -1/2, and γ = 1/2.
Therefore, the exponents α, β, and γ are 3/2, -1/2, and 1/2 respectively for this equation to be correct.
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A mass hanging on a spring is given a positive initial velocity V from equilibrium. The ensuing displacement x=x(t) from equilibrium (x=0) is governed by
mx
′′
x(0)
=−ax∣x
′
∣−kx
=0,x
′
(0)=V
where −ax∣x
′
∣ is a nonlinear damping force and −kx is a linear restoring force of the spring. First make a table of the quantities and their dimensions. What are possible time scales, and on what physical processes are they based. What are possible length scales? If the restoring force is small compared to the damping force, choose appropriate time and length scales and nondimensionalize the model so that the small term appears in the damping force.
The table of quantities and their dimensions for the given scenario is given below: Quantity Dimensions Displacement[x]L Velocity[V]LT−1 Mass[m]M L-3 Force[ax∣x′∣+kx]MLT−2
The possible time scales and their based physical processes for the given scenario are given below: Time Scales Physical ProcessesT1 = (m/a)^(1/2) Viscous damping timescale.T2 = (m/k)^(1/2)Natural frequency of the spring-mass system.T3 = (a/k)^(1/2)Critical damping timescale. The possible length scales are given below: Length Scales Description L1 = v0T1 Initial kinetic energy. L2 = x0 Initial displacement from equilibrium.L3 = cInitial position when the damping force becomes equal to the spring force. Now we need to nondimensionalize the given model, by choosing appropriate time and length scales. Let's select the natural frequency of the spring-mass system (T2) as the characteristic time and the equilibrium position of the mass (L2) as the characteristic length. From this, we can define the following dimensionless variables:t = t/T2x = x/L2. The dimensionless model is as follows:mL2x′′+aTkL2x′−k/mL2x = 0, x′(0) = V/L2T2x(0) = L2Let us assume that the damping force is small compared to the restoring force. Then the modified equation will be: mL2x′′+aTkL2x′ = 0, x′(0) = V/L2T2x(0) = L2For this model, the time scale is T2 and the length scale is L2.
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The first harmonic frequency for a particular organ pipe is 330 Hz. The pipe is closed at one end but open at theyother. What is the frequency of its second harmonic?
When a pipe is closed at one end and open at the other, it supports a specific set of harmonic frequencies. Therefore, the frequency of the second harmonic of the particular organ pipe is 660 Hz.
The first harmonic, also known as the fundamental frequency, is the lowest frequency that can be produced. In this case, the first harmonic frequency is given as 330 Hz.
For a closed-open pipe, the frequency of the second harmonic is twice the frequency of the first harmonic. The second harmonic is the next resonant frequency that can be produced in the pipe.
To calculate the frequency of the second harmonic, we multiply the frequency of the first harmonic by 2:
Frequency of the second harmonic = 2 * Frequency of the first harmonic
Frequency of the second harmonic = 2 * 330 Hz
Frequency of the second harmonic = 660 Hz
Therefore, the answer is 660 Hz.
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When a pipe is closed at one end and open at the other, it supports a specific set of harmonic frequencies. Therefore, the frequency of the second harmonic of the particular organ pipe is 660 Hz.
The first harmonic, also known as the fundamental frequency, is the lowest frequency that can be produced. In this case, the first harmonic frequency is given as 330 Hz.
For a closed-open pipe, the frequency of the second harmonic is twice the frequency of the first harmonic. The second harmonic is the next resonant frequency that can be produced in the pipe.
To calculate the frequency of the second harmonic, we multiply the frequency of the first harmonic by 2:
Frequency of the second harmonic = 2 * Frequency of the first harmonic
Frequency of the second harmonic = 2 * 330 Hz
Frequency of the second harmonic = 660 Hz
Therefore, the answer is 660 Hz.
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the volume of a cylinder is measured in cubic units
The volume of a cylinder is measured in cubic units.
What is the formula for the volume of a cylinder?
To begin with, let us understand what a cylinder is. A cylinder is a three-dimensional object that has two parallel circular bases of equal radii and is connected by a curved lateral surface. The distance between the two bases is referred to as the height of the cylinder.
A cylinder is formed by combining a rectangle and a circle, which are the shapes of the top and bottom faces, respectively.
The formula for calculating the volume of a cylinder is as follows:
V = πr²hWhere, V is the volume of the cylinder, r is the radius of the circular base, and h is the height of the cylinder. In this formula, π represents a constant value that is roughly equivalent to 3.14. The unit of measurement for the volume of a cylinder is cubic units.
Let us solve an example to better understand the concept.
Example:
What is the volume of a cylinder with a radius of 5 units and a height of 6 units?
V = πr²hV = 3.14 × 5² × 6V = 471 cubic units
Therefore, the volume of the given cylinder is 471 cubic units.
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You are standing in air and are looking at a flat piece of glass (n=1.52) on which there is a layer of transparent plastic (n=1.61). Light whose wavelength is 589 nm in vacuum is incident nearly perpendicularly on the coated glass and reflects into your eyes. The layer of plastic looks dark. Find the smallest possible nonzero value for the thickness of the layer.
the thickness of the layer of transparent plastic (n = 1.61) that makes it dark is 147.25 nm.
The optical path length is given by:Optical path length = ndμ = 1.61d
The wavelength of light in the air is λ1 = 589 nmPhase change at the air-plastic interface is φ1 = 0
Phase change at the plastic-glass interface is φ2 = π
There is no phase change upon reflection at the air-glass interface since the refractive index of air and glass is the same.
Hence, the total phase change is:Δφ = φ1 + φ2 + 2π (2n + 1)/λ= π + 2π (2n + 1)/λwhere n = 0, 1, 2, 3, ..... is an integer
The condition for the plastic layer to appear dark is that the reflected light waves from the two interfaces are exactly out of phase i.e. the total phase change is (n + 1/2)λ. Hence,Δφ = π + 2π (2n + 1)/λ = (n + 1/2)λ
For the smallest thickness of the plastic layer, we need to take n = 0. Then, we have:π + 2π(2n + 1)/λ = (n + 1/2)λπ = λ/4 = 0.14725 μm = 147.25 nm
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Your car's wheels are 65.0 cm in diameter, and the wheels are spinning at an angular velocity of 141 rad/s. How fast is your car moving (assume no slippage)? km/h
The car is traveling at an exact speed of approximately 164.97 kmph.
To find the speed of the car, we can relate the angular velocity of the wheels to the linear velocity of the car using the formula:
v = ω × r
where
v is the linear velocity of the car,
ω is the angular velocity of the wheels, and
r is the radius of the wheels.
Given:
Diameter of the wheels = 65.0 cm
Radius of the wheels = (65.0 cm) / 2 = 32.5 cm = 0.325 m
Angular velocity of the wheels (ω) = 141 rad/s
Substituting these values into the formula, we have:
v = ω × r
v = (141 rad/s) × (0.325 m)
Calculating:
v ≈ 45.825 m/s
Converting the speed to kilometers per hour:
v = 45.825 m/s × (3.6 km/h) / (1 m/s)
Calculating:
v ≈ 164.97 km/h
Therefore, the car is moving at approximately 164.97 km/h.
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Two soccer players start from rest, 35 m apart. They run directly toward each other, both players accelerating. The first player's acceleration has a magnitude of 0.55 m/s
2
. The second player's acceleration has a magnitude of 0.40 m/5
2
. (a) How much time passes before the players collide? (b) At the instant they collide, how far has the first player run? (a) Number Units (b) Number Units
The time passes before the players collide is 6.0 seconds and the first player run 37.3 meters.
(a) Time to collide
Let's first find the total distance that the two players will travel before they collide. This distance is equal to the initial distance between the players, 35 m.
Let's also let v be the final speed of each player when they collide. Since the players are accelerating, we know that their final speed is equal to their initial speed plus their acceleration multiplied by time.
v = v0 + at
where:
v0 is the initial speed of each player (which is zero since they start from rest)
where,
a is the acceleration of each player
t is the time it takes for the players to collide
We can also write the total distance that the players travel as the product of their final speed and the time it takes them to collide.
d = v[tex]t[/tex]
Equating these two expressions for d, we get:
35 m = v[tex]t[/tex]
Substituting the expression for v into this equation, we get:
35 m = (v0 + a[tex]t[/tex])t
Since v0 is zero, we can simplify this equation to:
35 m = at^2
We can now solve for t:
t =√(35 m / a)
Substituting the acceleration of each player, we get:
t = √(35 m / 0.55 m/s) = 6.0 seconds
(b) Distance traveled by first player
The distance traveled by the first player is equal to the product of the first player's acceleration, the time it takes for the players to collide, and half of the total distance.
d = (1/2)a[tex]t[/tex]^2
Substituting the values for a, t, and d, we get:
d = (1/2)(0.55 m/s) (6.0 seconds)^2 = 37.3 m
Therefore, the distance traveled by the first player when they collide is 37.3 meters.
Answers
(a) 6.0 seconds
(b) 37.3 meters
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The position of a particle moving long the x axis is given in centimeters by x =9.75 + 1.50t3, where t is in seconds. Calculate the average velocity during the time interval t=2.00s to t=3.00s;
The average velocity during the time interval t = 2.00 s to t = 3.00 s for the particle moving along the x-axis is 28.50 cm/s. This is calculated by finding the displacement of the particle during that time interval, which is 28.50 cm, and dividing it by the duration of 1.00 s. The average velocity represents the overall rate of change in position over the given time interval.
To calculate the average velocity during the time interval t = 2.00 s to t = 3.00 s, we need to find the displacement of the particle during that time interval and divide it by the duration.
Given:
Position equation: x = 9.75 + 1.50t^3
Initial time: t1 = 2.00 s
Final time: t2 = 3.00 s
First, let's find the position of the particle at the initial and final times:
x1 = 9.75 + 1.50(2.00^3)
x1 = 9.75 + 1.50(8.00)
x1 = 9.75 + 12.00
x1 = 21.75 cm
x2 = 9.75 + 1.50(3.00^3)
x2 = 9.75 + 1.50(27.00)
x2 = 9.75 + 40.50
x2 = 50.25 cm
The displacement of the particle during the time interval is:
Δx = x2 - x1
Δx = 50.25 cm - 21.75 cm
Δx = 28.50 cm
The duration of the time interval is:
Δt = t2 - t1
Δt = 3.00 s - 2.00 s
Δt = 1.00 s
Now, we can calculate the average velocity:
Average velocity = Δx / Δt
Average velocity = 28.50 cm / 1.00 s
Average velocity = 28.50 cm/s
Therefore, the average velocity during the time interval t = 2.00 s to t = 3.00 s is 28.50 cm/s.
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When you switch on an electric horn it begins to continuously emit a tone of 600 Hz. The horn is quite loud, held 1.0 m from your ear, the horn has sound level 100 dB. You cannot figure out how to turn it off, so you take it to the edge of a cliff and drop it. Recall from previous course work that, as the horn falls its speed, v, increases as its distance from you, y, increases according to v=
2gy
. (a) How far from you is the horn when the sound it emits reaches you with sound level 60 dB ? (b) What is the frequency of the sound that reaches you with sound level 60 dB ?
Therefore, the frequency of the sound that reaches you with sound level 60 dB is 100 × 600 = 60,000 Hz.
(a)To calculate the distance from you at which the horn is when the sound it emits reaches you with sound level 60 dB is calculated as follows
Given Initial sound level (I0) = 100 dB Final sound level (I) = 60 dB The difference in sound level (ΔI) = I0 − I = 100 − 60 = 40 dB
From the sound level formula,
I = I0 - 10 log (I/I0)I = I0 - 10 log (I/I0)40 = 100 - 10 log (I/I0)log (I/I0) = 60/10log (I/I0) = 6I/I0 = antilog (6)I/I0 = 1.0 x 106
When the horn is 1 m away from you, the sound intensity level of the horn, I0 = 100 dB.
So, the sound intensity level of the horn at a distance x from you is given by,
I = I0 - 20 log (x/1)60 = 100 - 20 log (x/1)20 log (x/1) = 100 − 60 = 4020 log (x/1) = 40log (x/1) = 40/20log (x/1) = 2x/1 = antilog (2)x = 100 cm = 1.0 m
Therefore, the distance from you at which the horn is when the sound it emits reaches you with sound level 60 dB is 1.0 m.
(b)The frequency of the sound that reaches you with a sound level of 60 dB can be calculated using the below formula
Given Initial sound level (I0) = 100 dB Final sound level (I) = 60 dB The difference in sound level (ΔI) = I0 − I = 100 − 60 = 40 dB
Frequency (f) is proportional to the square root of the sound intensity level (I).
I/I0 = (f/f0)2I/I0 = antilog (ΔI/10)I/I0 = antilog (4)I/I0 = 10 × 10 × 10 × 10 = 10,000f/f0 = √(I/I0)f/f0 = √10,000f/f0 = 100
Therefore, the frequency of the sound that reaches you with sound level 60 dB is 100 × 600 = 60,000 Hz.
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If the iron core of a dying star initially spins with a rotation frequency of ω0 = 3.20 rad/s, and if the core’s radius decreases during the collapse by a factor of 22.7, what is the angular veloci
The angular velocity of the iron core at the end of the collapse is approximately 0.00602 rad/s.
To find the angular velocity of the iron core at the end of the collapse, we can use the principle of conservation of angular momentum. The initial rotation frequency (ω₀) is given as 3.20 rad/s, and the radius of the core decreases by a factor of 22.7 during the collapse.
The formula for the conservation of angular momentum is:
I₁ ω₁ = I₂ ω₂
Where: I₁ and I₂ are the moments of inertia of the core before and after collapse respectively, and ω₁ and ω₂ are the initial and final angular velocities respectively.
The moment of inertia (I) for a rotating sphere is given by:
I = (2/5) m r²
Where: m is the mass of the sphere, and r is the radius of the sphere. Since the iron core is a uniform sphere before and after the collapse, the mass remains constant. Let's denote the initial radius of the core as r0 and the final radius as rf.
According to the given information,
rf = r₀/22.7.
Substituting the moment of inertia formulas into the conservation of angular momentum equation, we have:
(2/5) m r₀² ω₀ = (2/5) m (r₀/22.7)² ω₂
We can simplify this equation by cancelling out the mass and the (2/5) term:
r₀² ω₀ = (r₀/22.7)² ω₂
To find ω2, we need to solve for it. Rearranging the equation, we have: ω₂ = (r₀² ω₀) / (r₀/22.7)²
Simplifying further, we get:
ω₂ = (r₀² ω₀) / (r₀² / (22.7)²)
The r₀² terms cancel out, and we are left with:
ω₂ = ω₀ / (22.7)²
Now, we can calculate the angular velocity of the iron core at the end of the collapse. Substituting the given value of ω₀ = 3.20 rad/s into the equation, we have:
ω₂ = 3.20 / (22.7)²
Calculating this expression gives:
ω₂ ≈ 0.00602 rad/s
Therefore, the conclusion of the collapse, the iron core's angular velocity was 0.00602 rad/s.
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Complete question is,
If the iron core of a dying star initially spins with a rotation frequency of ω0 = 3.20 rad/s, and if the core’s radius decreases during the collapse by a factor of 22.7, what is the angular velocity of the iron core at the end of the collapse, (assuming that the iron core is a uniform sphere before collapse and is a uniform sphere after collapse)? Use the fact that the rotational inertia of a rotating sphere is given by:
Calculate the neutron flux at a distance of 0.3 m from the neutron source which emits 3×10
∧/7n/s
.
The neutron at a distance of 0.3 m from the neutron source is approximately 2.655×10^8 n/m²/s.
The neutron flux at a certain distance from a neutron source can be calculated using the formula:
Neutron flux = Neutron emission rate / (4πr^2),
where Neutron emission rate is the number of neutrons emitted per second by the source, and r is the distance from the source.
Given that the neutron source emits 3×10^7 neutrons per second, and the distance from the source is 0.3 m, we can substitute these values into the formula:
Neutron flux = (3×10^7 n/s) / (4π * (0.3 m)^2).
Calculating the denominator:
(4π * (0.3 m)^2) ≈ 0.113 m^2.
Now, we can calculate the neutron flux:
Neutron flux ≈ (3×10^7 n/s) / 0.113 m^2.
Evaluating the expression, we find that the neutron flux at a distance of 0.3 m from the neutron source is approximately 2.655×10^8 neutrons per square meter per second.
Therefore, the neutron flux at a distance of 0.3 m from the neutron source is approximately 2.655×10^8 n/m²/s.
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Exercise 4 - Velocity Addition (10\%) Two particles are created in a high-energy accelerator and move off in opposite directions. The speed of one particle, as measured in the laboratory, is +0.650c, and the speed of each particle relative to the other is −0.950 c. 1. [5\%] Draw a sketch of the experiment. 2. [5\%] What is the speed (in unit of c) of the second particle as measured in the laboratory?
Relativistic velocity is of the order of 1/10th of the velocity of light
Sketch of the Experiment:
```
^ y-axis
|
|
-------------------
\ | /
\|\theta/
\ /
\ /
-------O----------
/
/
/|\
/ | \
```
In the sketch above, the horizontal line represents the laboratory frame of reference. The point "O" represents the location of the high-energy accelerator where the particles are created. The particle moving in the positive x-direction is labeled as "Particle 1," and the particle moving in the negative x-direction is labeled as "Particle 2." The angle θ represents the angle between the velocity vectors of the two particles.
2. Calculation of the Speed of Particle 2 in the Laboratory Frame:
Let's assume the speed of light, c, as the unit for velocity.
Given:
- Speed of Particle 1 in the laboratory frame, v₁(lab) = +0.650c
- Relative speed of Particle 2 with respect to Particle 1, v₂(1) = -0.950c
To calculate the speed of Particle 2 in the laboratory frame, we can use the relativistic velocity addition formula:
v₂(lab) = (v₂(1) + v₁(lab)) / (1 + (v₂(1) * v₁(lab)) / c²)
Substituting the given values into the formula:
v₂(lab) = (-0.950c + 0.650c) / (1 + (-0.950c * 0.650c) / c²)
Simplifying the expression:
v₂(lab) = (-0.300c) / (1 - 0.6175)
v₂(lab) = (-0.300c) / (0.3825)
v₂(lab) ≈ -0.784c
Therefore, the speed of Particle 2, as measured in the laboratory frame, is approximately -0.784 times the speed of light (c)
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1.) A Mass of 1 kg on the ground is thrown vertically in the air with a velocity 14 m/s. The mask slows down as it goes up: what will be the height of the mass when the velocity is only 9.3 m/s. Use G=9.8m/s squared and keep 3 significant digits.
2.) A brick has a mass of 0.5 kg and is thrown vertically from the ground with an initial speed of 20 m/s. what will be the speed of the brick when the brick reaches a height of 6.8 m. (use g=10m/s squared and keep three significant digits)
The height of the mass when the velocity is 9.3 m/s is approximately 5.59 meters. The speed of the brick when it reaches a height of 6.8 m is approximately 16.2 meters per second.
To determine the height of the mass when the velocity is 9.3 m/s, we can use the equation of motion for vertical motion with constant acceleration.
Given:
Initial velocity (u) = 14 m/s
Final velocity (v) = 9.3 m/s
Acceleration (g) = 9.8 m/s^2
Using the equation v^2 = u^2 + 2as, where s is the displacement (height in this case), we can solve for s:
(9.3)^2 = (14)^2 + 2 * (-9.8) * s
86.49 = 196 - 19.6s
19.6s = 196 - 86.49
19.6s = 109.51
s ≈ 5.59 m
Therefore, the height of the mass when the velocity is 9.3 m/s is approximately 5.59 meters.
To find the speed of the brick when it reaches a height of 6.8 m, we can again use the equation of motion for vertical motion with constant acceleration.
Given:
Initial velocity (u) = 20 m/s
Height (s) = 6.8 m
Acceleration (g) = 10 m/s^2
Using the equation v^2 = u^2 + 2as, we can solve for v:
v^2 = (20)^2 + 2 * (-10) * 6.8
v^2 = 400 - 136
v^2 = 264
v ≈ 16.2 m/s
Therefore, the speed of the brick when it reaches a height of 6.8 m is approximately 16.2 meters per second.
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proton is located at <0,4×10
−8
,0>m. What is the force on the proton, due to the dipole?
F
=N An electron is located at <−4×10
−8
,0,0>m. What is the force on the electron, due to the dipole?
F
=⟨,
N
N (Hint: Make a diagram! Note that one approach is to calculate magnitudes, then figure out directions from your diagram.)
The magnitude of the force on the proton, due to the dipole formed by the electron and its location, is equal to approximately 8.99 × [tex]10^9[/tex] N. The magnitude of the force on the electron, due to the dipole, is also approximately 8.99 × [tex]10^9[/tex] N.
To calculate the magnitude of the force on the proton and the electron, we can use Coulomb's law. Coulomb's law states that the force between two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.
In this scenario, the proton and the electron form a dipole with a separation of [tex]8*10^{-8[/tex] m. The magnitude of the force on a charge due to a dipole is given by the formula F = (1 / 4πε₀) * [(2p * r) / [tex]r^3[/tex]], where F is the force, p is the dipole moment, r is the distance, and ε₀ is the permittivity of free space.
For the proton, the distance (r) is [tex]8*10^{-8[/tex] m, and the dipole moment (p) is the product of the electron's charge and the separation between the charges. Plugging in the values, we get F = (1 / 4πε₀) * [(2 * e * [tex]8*10^{-8[/tex]) / [tex](8*10^{-8)^3[/tex]], where e is the elementary charge. Simplifying this expression gives us the magnitude of the force on the proton.
Similarly, for the electron, the distance (r) remains the same, but the dipole moment (p) is the negative of the product of the proton's charge and the separation between the charges. By substituting the values into the formula, we can calculate the magnitude of the force on the electron.
Both calculations result in approximately [tex]8.99 * 10^9[/tex] N, which represents the magnitudes of the forces on the proton and the electron due to the dipole formed by their respective charges and locations.
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If the electric field E is 102 V/m at a distance of 2.3 m (in air) from a point charge Q. Find the value of Q in nanocoulomb
The value of the point charge Q is approximately 5.92 nanocoulombs (nC).
To find the value of the point charge Q, we can use the relationship between electric field and point charge given by Coulomb's law.
Coulomb's law states that the electric field (E) created by a point charge (Q) at a distance (r) from the charge is given by:
E = k * Q / r^2
where:
E is the electric field,
k is the electrostatic constant (k ≈ 8.99 x 10^9 Nm^2/C^2),
Q is the point charge, and
r is the distance from the charge.
Electric field (E) = 102 V/m
Distance (r) = 2.3 m
We can rearrange the equation to solve for Q:
Q = E * r^2 / k
Substituting the given values:
Q = (102 V/m) * (2.3 m)^2 / (8.99 x 10^9 Nm^2/C^2)
Q ≈ 5.92 x 10^-9 C
To express the charge in nanocoulombs (nC), we divide the value by 10^-9:
Q ≈ 5.92 nC
Therefore, the value of the point charge Q is approximately 5.92 nanocoulombs.
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