Help with the one above and the three below. I WILL GIVE BRAINLIEST!

Physics

1 answer:

kodGreya [7K]4 years ago

8 0

1. The rocket would experience a downwards force due to gravity from the moon. Also, the rocket would experience an upwards force from the acceleration of the rocket thrusters. Assuming you are neglecting the force of air friction, there would be 2 forces. If you are considering air friction as well, there would be also a downwards force of kinetic air friction acting down on the rocket as it flies, making it 3 forces. If the rocket has already taken off, then there will be no normal force. However, if the rocket is still on the surface of the moon, the rocket would experience a normal force from the moon. Since the question isn't specific I can't provide a concrete answer, but it would range from 2-4 forces depending on the above factors.

2. Since we are not sure if the person in the question is actively lifting the crate, we have to determine the downwards force of the crate due to gravity and compare it to the normal force.

F = ma

F = (15.3)(-9.8)

F = -150N

Since the downwards force of the crate is equivalent to the normal force, it means the person is applying no force in picking up the object. So to pick up a 150N object from scratch, you would have to exert more force than the weight of the object, so the answer is 294N.

3. Same idea as question 2. First determine the weight of the object:

F = ma

F = (30)(-9.8)

F = -294N

The crate in question is not moving, so the magnitudes of the forces in the upwards and downwards direction has to equal to 0.

-294 + 150N + x = 0

x = 144N

So the person is exerting 144 N. (Which I hope is a choice that you forgot to type out on the question)

5. Image 3 has the least amount of tension since both tension forces exerted on the cables are shared amongst the two cables and they are parallel to the downwards force of gravity that acts on the object and against them.

You might be interested in

A lowly high diver pushes off horizontally with a speed of 2.00 m/s from the platform edge 10.0 m above the surface of the water

VladimirAG [237]

Answer:

(a) 1.6m

(b) 6.86m

Explanation:

This is a horizontal shooting problem, so we will use the following equations.

For horizontal distance x we use:

$x(t)=x_{0}+v_{0}t$

Where $x(t)$ is the horizontal distance at a time $t$ , $x_{0}$ is the initial horizontal position which in this case will be zero: $x_{0}=0$ . And $v_{0}$ is the initial speed: $v_{0}=2m/s$

So to solve (a):

(a) At what horizontal distance from the edge is the diver 0.800 s after pushing off?

we have that the time is: $t=0.8s$ so the horizontal distance is:

$x(0.8)=(2m/s)(0.8s)1.6m$

The answer for (a) is 1.6m

Now, to solve (b) we need the equation for vertical distance:

$y(t)=y_{0}-\frac{1}{2}gt^2$

Where $y(t)$ is the vertical distance at a time $t$ , $y_{0}$ is the initial vertical distance: $y_{0}=10m$ And $g$ is the gravitational acceleration: $g=9.81m/s^2$ }