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3 years ago

Please Help

2 answers:

4 0

A man pulls on a box with 50 N of applied force while another man pushes on the box from the opposite direction with 60 N of applied force. The box will stand still.

TRUE.

barxatty [35]3 years ago

4 0

Answer:

False statement.

Explanation:

It is given that,

Force with which a man pulls a box, F = 50 N (one side)

Force with which pushes on the box, F' = 60 N (in opposite direction)

The forces acting on the box are unbalanced. It means that the force acting in one side is more or less then force in other side. As a result the box will move.

So, the given statement is false.

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A child sits on the edge of a spinning merry go round that has a radius of 1.5 . The child’s speed is a 2m/s. What is the child’

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Answer: 2.67 m/s^2

Explanation:

Centripetal acceleration is defined as v^2/r; in this case, it's 2^2/1.5, which is 2.67.

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A 6.0 kg mass is placed on a 20º incline which has a coefficient of friction of 0.15. What is the acceleration of the mass down

Leona [35]

Answer:

Explanation:

The form of Newton's 2nd Law that we use for this is:

F - f = ma where F is the Force pulling the mass down the ramp forward, f is the friction trying to keep it from moving forward, m is the mass and a is the acceleration (and our unknown).

We know mass and we can find f, but we don't have F. But we can solve for that by rewriting our main equation to reflect F:

$wsin\theta-\mu F_n=ma$ That's everything we need.

w is weight: 6.0(9.8). Filling in:

6.0(9.8)sin20 - .15(6.0)(9.8) = 6.0a and

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3 years ago

One end of a 34-m unstretchable rope is tied to a tree; the other end is tied to a car stuck in the mud. The motorist pulls side

anyanavicka [17]

Answer:

Fc = 89.67N

Explanation:

Since the rope is unstretchable, the total length will always be 34m.

From the attached diagram, you can see that we can calculate the new separation distance from the tree and the stucked car H as follows:

L1+L2=34m

$L1^2=L2^2=L^2=2^2+(H/2)^2$ Replacing this value in the previous equation:

$\sqrt{2^2+H^2/4}+ \sqrt{2^2+H^2/4}=34$ Solving for H:

$H=\sqrt{52}$

We can now, calculate the angle between L1 and the 2m segment:

$\alpha = atan(\frac{H/2}{2})=60.98°$

If we make a sum of forces in the midpoint of the rope we get:

$-2*T*cos(\alpha ) + F = 0$ where T is the tension on the rope and F is the exerted force of 87N.

Solving for T, we get the tension on the rope which is equal to the force exerted on the car:

$T=Fc=\frac{F}{2*cos(\alpha) } = 89.67N$

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3 years ago

39. A dog runs on a waxed floor at an initial speed of 2 m/s. It slides to a stop with an

Sedbober [7]

Answer:

Explanation:

Use the one-dimensional equation

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3 years ago