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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.
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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.
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 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.
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).
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.
If a 75kg person and a 200kg crate are each parachuted to earth from a plane which will hit the the ground first?
Provided that the parachute has the same surface area for both of the parachutist's, the 200kg man will hit the ground first due to the extra weight from the heavier man.
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 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 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.
What is the net force of a 200-kg crate sitting at rest on a factory floor?
The net force acting on the crate is zero since it's at rest. According to Newton's First Law of Motion, an object will remain at rest unless acted upon by an external force.
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).
A 75kg person and 200kg crate are each parachuted tto earth from a plane which statement is true?
Both the person and the crate experience the same acceleration due to gravity, regardless of their masses, when they are in free fall under the parachute. However, the person will experience a greater air resistance due to their larger surface area compared to the crate, causing them to slow down more quickly and have a lower terminal velocity.
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.
How much work is done on a 100 kg crate that is 2 m in a time of 4 s?
The work done on the crate is calculated using the equation Work = Force x Distance. Given the time (4 s) and distance (2 m), we would need additional information such as the force applied to the crate to determine the total work done.
If you apply a force of 220 N to the lever how much force is applied to lift the crate?
If the perpendicular distance from the point of application of the force to the fulcrum is x metres and the perpendicular distance from the crate to the fulcrum is y metres, then the force applied on the crate is 220*x/y N.
