For displacement we need initial velocity (which is zero because it starts from rest) and final velocity (which is calculatee thro acceleration formula

A= (vf - vi)/t

a= vf-0/t

1.25=vf / 7

1.25*7=vf

8.75 = vf

Now for displacement plug all the values in

X = 1/2(vf-vi)/t formula

The displacement (x) is 30.625 m

For part 3, we know new displacement that is 22m , the final and initial velocities are the same so just plug in the values for same formula above

The answer is t = 5.02

Im pretty sure all the answers are correct

8 0

2 years ago

You leave the doctor's office after your annual checkup and recall that you weighed 688 N in her office. You then get into an el

Lapatulllka [165]

Answer:

$a=0.5418\ m.s^{-2}$ upwards

$a=1.283\ m.s^{-2}$ downwards

Explanation:

Given:

weight of the person, $w=688\ N$

So, the mass of the person:

$m=\frac{w}{g}$

$m=\frac{688}{9.81}$

$m=70.132\ kg$

Now if the apparent weight in the elevator, $w_a= 726\ N$

<u>Then the difference between the two weights is :</u>

$\Delta w=w_a-w$

$\Delta w=726-688$

$\Delta w=38\ N$ is the force that acts on the body which generates the acceleration.

Now the corresponding acceleration:

$a=\frac{\Delta w}{m}$

$a=\frac{38}{70.132}$

$a=0.5418\ m.s^{-2}$ upwards, because the normal reaction that due to the weight of the body is increased here.

Now if the apparent weight in the elevator, $w_a= 598\ N$

<u>Then the difference between the two weights is :</u>

$\Delta w=w-w_a$

$\Delta w=688-598$

$\Delta w=90\ N$ is the force that acts on the body which generates the acceleration.

Now the corresponding acceleration:

$a=\frac{\Delta w}{m}$

$a=\frac{90}{70.132}$

$a=1.283\ m.s^{-2}$ downwards, because the normal reaction that due to the weight of the body is decreased here.

7 0

4 years ago

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The focal length of a concave lens 13.3 cm.Calculate its power.​

The focal length of a concave lens 13.3 cm.Calculate its power.