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How much power is required to raise a 30 kg crate a vertical distance of 6 m in a time of 4 seconds?
The work done in lifting the crate is equal to its change in potential energy: ( \text{Work} = \text{Force} \times \text{distance} = m \cdot g \cdot h ). The power required is the work done divided by the time taken: ( \text{Power} = \frac{\text{Work}}{\text{time}} ). Plug in the values to calculate the power required.
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How much power is required to raise a 200-kg crate a vertical distance of 2 m in a time of 4 s?
The work done to raise the crate is given by the formula W = mgh, where m is the mass of the crate, g is the acceleration due to gravity, and h is the height raised. In this case, W = (200 kg)(9.81 m/s^2)(2 m) = 3924 J. The power required is P = W/t, so P = 3924 J / 4 s = 981 W.
How much work is done in lifting a 60Kg crate a vertical distance of 10 meters?
The work done in lifting a 60kg crate a vertical distance of 10 meters is given by the formula: work = force x distance x cos(theta), where force = weight x gravitational acceleration = 60kg x 9.8 m/s^2, distance = 10m, theta is the angle between the force and direction of displacement (which is 0 in this case, as it's lifting vertically). Plugging in the values, the work done is approximately 5,880 Joules.
How do you solve a 200-lb crate lies on the floor. The coefficient of static friction between the crate and the floor is 0.60. What is the minimum force required to start the crate sliding?
To calculate the minimum force required to start the crate sliding, you would multiply the weight of the crate by the coefficient of static friction. In this case, 200 lb crate * 0.60 static friction coefficient = 120 lb minimum force needed to start the crate sliding.
If it took 5 seconds to lift the crate using a machine what was the power rating of the machine?
We have no way of knowing what power the machine was rated for, but with the information given in the question, we can calculate the power it delivered during the crate-lift: It was (1.96) x (mass of the crate in kilograms) x (distance the crate was lifted in meters) watts.
How much work is done in lifting a 50kg crate a vertical distance of 10 meters?
Potential Energy is given by the fourmulai PE=MGH where M=mass in kilos, G=the force of gravity in Netwons (9.8N) and H=height in meters. So 50*9.8*10=4900joules. A Watt is a unit of power that =1 joule per second. So 4900 joules divided by 5 seconds = 980 Watts, not allowing for losses due to friction etc.
What is the power rating of a machine to lift a crate in 5 seconds?
That really depends on the weight of the crate. Also, on how high you want to lift it. Calculate the energy required to lift the crate with the formula for gravitational potential energy: PE = mgh (mass x gravity x height) Then divide this by the 5 seconds to get the minimum power required. (The actual power is somewhat larger, for various reasons - the initial acceleration required, and losses due to friction.)
What is the potential energy of a crate lifted with a force of 100 newtons a vertical distance of 2 meters?
Force x distance = 100 x 2 = 200 newton-meters = 200 joules.
How much power is required to raise a 200-kg crate a vertical distance of 2 m in a time of 4 s?
The work done to raise the crate is given by the formula W = mgh, where m is the mass of the crate, g is the acceleration due to gravity, and h is the height raised. In this case, W = (200 kg)(9.81 m/s^2)(2 m) = 3924 J. The power required is P = W/t, so P = 3924 J / 4 s = 981 W.
How much work is done in lifting a 60Kg crate a vertical distance of 10 meters?
The work done in lifting a 60kg crate a vertical distance of 10 meters is given by the formula: work = force x distance x cos(theta), where force = weight x gravitational acceleration = 60kg x 9.8 m/s^2, distance = 10m, theta is the angle between the force and direction of displacement (which is 0 in this case, as it's lifting vertically). Plugging in the values, the work done is approximately 5,880 Joules.
If a 50 crate was lifted to a height of 10 meters and it took 5 seconds to lift the crate using a machine what was the power rating of the machine?
Please use the formula for gravitational potential energy (PE = mgh) to calculate the energy required. Then divide that by the time to get the power.
How do you solve a 200-lb crate lies on the floor. The coefficient of static friction between the crate and the floor is 0.60. What is the minimum force required to start the crate sliding?
To calculate the minimum force required to start the crate sliding, you would multiply the weight of the crate by the coefficient of static friction. In this case, 200 lb crate * 0.60 static friction coefficient = 120 lb minimum force needed to start the crate sliding.
If it took 5 seconds to lift the crate using a machine what was the power rating of the machine?
We have no way of knowing what power the machine was rated for, but with the information given in the question, we can calculate the power it delivered during the crate-lift: It was (1.96) x (mass of the crate in kilograms) x (distance the crate was lifted in meters) watts.
How much work is done in lifting a 50kg crate a vertical distance of 10 meters?
Potential Energy is given by the fourmulai PE=MGH where M=mass in kilos, G=the force of gravity in Netwons (9.8N) and H=height in meters. So 50*9.8*10=4900joules. A Watt is a unit of power that =1 joule per second. So 4900 joules divided by 5 seconds = 980 Watts, not allowing for losses due to friction etc.
If a crate weighing 508 N is resting on a plane inclined 26degrees above the horizontal if acceleration is 4.29 then how fast will the crate be moving after 6 seconds?
To determine the speed of the crate after 6 seconds, we first need to calculate the net force acting on the crate on the inclined plane. This can be done by resolving the weight of the crate into components parallel and perpendicular to the plane. Then, using Newton's second law, F = ma, where F is the net force, m is the mass of the crate, and a is the acceleration, we can find the acceleration down the incline. After finding this acceleration, we can use the kinematic equation v = u + at to calculate the final speed of the crate after 6 seconds, where v is the final velocity, u is the initial velocity (assumed to be 0), a is the acceleration, and t is the time.
If 68 joules of work were necessary to move a 4 newton crate how far was the crate moved?
The work done is given by the formula Work = Force x Distance. Rearranging this formula we find Distance = Work / Force. Plugging in the values given, we get Distance = 68 joules / 4 newtons = 17 meters. So, the crate was moved 17 meters.
A forklift picks up a crate using 400 N It moves the crate 20 meters in 50 seconds How much power is used?
From the question, it's hard to tell whether the 20 meters is the vertical lift, or a horizontal transfer that occurs after the lift.If the 20 meters is the vertical lift (performed by a very large fork-lift in a shop with a very high ceiling):Energy = work = 400 N times 20 m = 8,000 Newton-meters = 8,000 joules8,000 joules in 50 seconds = 8,000 / 50 = 160 joules per second = 160 watts = about 0.214 horsepower.If the 20 meters is a horizontal ride after the lift is complete, then that part of the move consumes nominally no energy or power. No force is required to move an object perpendicular to the force of gravity. Whatever force is applied initially, to get the crate moving, is returned at the end of the 20 meters, when reverse force must be applied to the crate in order to make it stop moving.
How much work is done on a 20-N crate that you lift 2 meters?
The work done on the crate would be 40 joules (work = force x distance).
