<h2>
Answer: 1000 J</h2>
The Work
done by a Force
refers to the release of potential energy from a body that is moved by the application of that force to overcome a resistance along a path.
It should be noted that it is a scalar magnitude, and its unit in the International System of Units is the Joule (like energy). Therefore, 1 Joule is the work done by a force of 1 Newton when moving an object, in the direction of the force, along 1 meter:
Now, when the applied force is constant and the direction of the force and the direction of the movement are parallel, the equation to calculate it is:
(1)
When they are not parallel, both directions form an angle, let's call it
. In that case the expression to calculate the Work is:
(2)
For example, in order to push the 200 N box across the floor, you have to apply a force along the distance
to overcome the resistance of the weight of the box (its 200 N).
In this case both <u>(the force and the distance in the path) are parallel</u>, so the work
performed is the product of the force exerted to push the box
by the distance traveled
. as shown in equation (1).
Hence:
>>>>This is the work
Answer:
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Explanation:
<span>The force of static friction F equals the coefficient of friction u times the normal force N the object exerts on the surface: F = uN. N is the centripetal force of the wall on the people; N = ma_N, where m is the mass of the people and a_N is the centripetal acceleration.
The people will not slip down if F is greater than the force of gravitation: F = uma_N > mg, or u > g/a_N.
a_N is the velocity v of the people squared divided by the radius of the room r: a_N = v^2/r.
The circumference of the room is 2 pi r = 28.3 m. So v = 28.3 * 0.8 m/sec = 22.6 m/sec.
So a_N = 114 m/sec^2.
g = 9.81 m/sec^2, so u must be at least 9.81/114 = 0.086.</span>
F=ma
Therefore the net force = 1000kg × 2 metres per second per second
So F=2000 N
Answer:
option (d)
Explanation:
The relation between the rms velocity and the molecular mass is given by
v proportional to \frac{1}{\sqrt{M}} keeping the temperature constant
So for two gases



