Ask question

Ask question

All categories

3 years ago

9

If it takes 320 N of force to push a box up a ramp, yet it would take 1,600 N of force to lift the box straight up, what is the

mechanical advantage of the ramp? O A. 0.2 O B. 4 O C. 5 O D. 40

1 answer:

grin007 [14]3 years ago

5 0

MA = Fwithout help/Fwith machine
MA = 1600/320 = 5

You might be interested in

In this type of polarization a Polaroid filter serves as a device that filters out one-half of the vibrations.

erik [133]

The answer is d.
hope it helped

6 0

3 years ago

Please look at the photo and give me the answer :)))

san4es73 [151]

Answer:

c

Explanation:

efficiency is measured in % hence 0.02% is smaller than 20%

5 0

2 years ago

A rocket being thrust upward as the force of the fuel being burned pushes downward is an example of which of Newton's laws?

RoseWind [281]

The correct answer was

3rd Law

Thanks Mimi55555

4 0

3 years ago

Read 2 more answers

Which examples are a COMPLETE description of motion? Select ALL that apply.

SIZIF [17.4K]

All of them apply because they all are doing some thing to move

8 0

2 years ago

Let the masses of blocks A and B be 4.50 kg and 2.00 kg , respectively, the moment of inertia of the wheel about its axis be 0.4

Free_Kalibri [48]

Answer:

Accelerations of both the sides is 0.6125 $m/s^{2}$ , A moves downwards whereas B moves upwards.

$\alpha$ =6.125 rad/ $s^{2}$

Tension on side A = 4.5 × g= 44.1 m/ $s^{2}$

Tension on side B= 2.0 × g= 19.6 m/ $s^{2}$

Explanation:

As both, the blocks A and B are attached due to the constraint they can only possess a single acceleration a.

Observe the figure attached, let the tension with Block A be $T_{2}$ and the tension attached with Block B be $T_{1}$ .

Tensions will be only be due to the weight of the blocks as no other force is present.

$T_{2}$ = 4.5 × g= 44.1 m/ $s^{2}$

$T_{1}$ = 2.0 × g= 19.6 m/ $s^{2}$

Now, lets make a torque equation about the center of the wheel and find the alpha

$T_{2}$ ×R- $T_{1}$ ×R= MI( Moment of Inertia of Wheel)× Alpha

where, R= Radius of the wheel=0.100m and

Alpha( $\alpha$ )= Angular acceleration of the wheel

MI of the wheel= 0.400 kg/ $m^{2}$

$(44.1-19.6)R=0.400\alpha$

$\alpha = \frac{24.5 * 0.100}{0.400}$

$\alpha$ =6.125 rad/ $s^{2}$

Acceleration = R × $\alpha$

= 0.1 * 6.125

=0.6125 $m/s^{2}$

Accelerations of both the sides is 0.6125 $m/s^{2}$ , A moves downwards whereas B moves upwards.

7 0

3 years ago