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Sav [38]
4 years ago
11

If the man on the left pulls on the object with a force of 500 N and the man on the right pulls on the object with a force of 75

0 N, what is the net force exerted on the object?
A.
1250 N to the right
B.
1250 N to the left
C.
250 N to the right
D.
250 N to the left
Physics
1 answer:
Igoryamba4 years ago
3 0
Force is a vector quantity
so pulling from opposite side will be negative
so
750+(-500)= 250N
C is the right answer
becauseause the man on the right applies greater force.
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A woman lifts a 300 newton child a distance of 1.5 meters in 0.75 seconds. What is her power output in lifting the child?
Nitella [24]
Power = Work done / time

Work done = Force * Distance
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7 0
3 years ago
Stan is driving north on his scooter at 8m/s, accelerates 11m/s (North) in 4s, drives a constant velocity for the next 15s, and
kow [346]

A) Acceleration: a_1 = 0.75 m/s^2, a_2 = 0, a_3 = -1.57 m/s^2

B) The total displacement is 209.5 m north

C) The average velocity is 8.06 m/s north

Explanation:

A)

Acceleration is defined as:

a=\frac{v-u}{t}

where

v is the final velocity

u is the initial velocity

t is the time taken for the velocity to change from u to v

Here we have:

- In the first  segment,

u = 8 m/s north

v = 11 m/s north

t = 4 s

So the acceleration is

a_1 = \frac{11-8}{4}=0.75 m/s^2 (north)

- In the second segment, Stan drives at a constant velocity: so the final velocity is equal to the initial velocity,

u = v

Therefore, the acceleration is zero: a_2 = 0

- In the third segment,

u = 11 m/s (north)

v = 0 (he comes to a stop)

t = 7 s

So the acceleration is

a=\frac{0-11}{7}=-1.57 m/s^2

And the negative sign means the acceleration is south, opposite to the direction of motion.

B)

In a uniformly accelerated motion, the displacement can be calculated as:

s=ut+\frac{1}{2}at^2

where

u is the initial velocity

a is the acceleration

t is the time

- For the first segment, we have

u = 0\\a = 0.75 m/s^2\\t=4 s

So the displacement is

s_1 = 0+\frac{1}{2}(0.75)(4)^2=6 m

- For the second segment, we have

u = 11 m/s\\a = 0\\t=15 s

So the displacement is

s_2 = (11)(15)+0=165 m

- For the third segment, we have

u = 11\\a = -1.57 m/s^2\\t=7 s

So the displacement is

s_3 = (11)(7)+\frac{1}{2}(-1.57)(7)^2=38.5 m

So the total displacement is:

s = 6 m + 165 m + 38.5 m = 209.5 m

In the north direction (positive direction)

C)

The average velocity is given by:

v=\frac{d}{t}

where

d is the total displacement

t is the total time

Here we have:

d = 209.5 m

t = 26 s

Therefore, the average velocity is

v=\frac{209.5}{26}=8.06 m/s (north)

Learn more about accelerated motion:

brainly.com/question/9527152

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