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zmey [24]
3 years ago
14

Two piles of heavy boxes must be lifted onto a shelf. Each pile has 20 boxes and each box is the same size and weight. Tom and c

laudia are each responsible for a pile. Tom lifts each box straight up from the pile to the shelf. Claudia places a long gradual ramp that leads from pile to the shelf. She pushes each box up the ramp until it rests on the shelf. Who used more force
Physics
2 answers:
Murrr4er [49]3 years ago
4 0
Tom used more Force but over a shorter distance. Tom and Claudia both did the same amount of work.
Alex787 [66]3 years ago
4 0

In this scenario, we can tell that Tom used more force.


We can tell this is correct because Tom did not use any shortcuts like Claudia did. Claudia placed her heavy box on a ramp that then transferred her box to where the shelves are. After that, all she had to is put the box on the shelf.



- Marlon Nunez

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What r semiconductors?
KengaRu [80]
It is a solid substance that has a conductivity between that of an insulator and that of most metals, either due to the addition of an impurity or because of temperature effects
5 0
3 years ago
F = 50 N<br> m = 72 kg<br> m/s2
lyudmila [28]

Answer:

Explanation:

F=ma

a=F/m

a=50/72=

a=0.694

3 0
3 years ago
A 912-kg car is being driven down a straight, level road at a constant speed of 31.5 m/s. When the driver sees a police cruiser
sergejj [24]

Answer:

786.6 N

Explanation:

mass of car, m = 912 kg

initial velocity of car, u = 31.5 m/s

final velocity of car, v = 24.6 m/ s

time, t = 8 s

Let a be the acceleration of the car

Use first equation of motion

v = u + a t

24.6 = 31.5 + a x 8

a = - 0.8625 m/s^2

Force, F = mass x acceleration

F = 912 x 0.8625

F = 786.6 N

Thus, the force on the car is 786.6 N.

8 0
3 years ago
NEED HELP ASAP!!!!
podryga [215]
The answer is B
I rhink
6 0
3 years ago
A quarterback passes a football from height h = 2.1 m above the field, with initial velocity v0 = 13.5 m/s at an angle θ = 32° a
SOVA2 [1]

Answer:

a)    x = v₀² sin 2θ / g

b)    t_total = 2 v₀ sin θ / g

c)    x = 16.7 m

Explanation:

This is a projectile launching exercise, let's use trigonometry to find the components of the initial velocity

        sin θ = v_{oy} / vo

        cos θ = v₀ₓ / vo

         v_{oy} = v_{o} sin θ

         v₀ₓ = v₀ cos θ

         v_{oy} = 13.5 sin 32 = 7.15 m / s

         v₀ₓ = 13.5 cos 32 = 11.45 m / s

a) In the x axis there is no acceleration so the velocity is constant

         v₀ₓ = x / t

          x = v₀ₓ t

the time the ball is in the air is twice the time to reach the maximum height, where the vertical speed is zero

          v_{y} = v_{oy} - gt

          0 = v₀ sin θ - gt

          t = v_{o} sin θ / g

         

we substitute

       x = v₀ cos θ (2 v_{o} sin θ / g)

       x = v₀² /g      2 cos θ sin θ

       x = v₀² sin 2θ / g

at the point where the receiver receives the ball is at the same height, so this coincides with the range of the projectile launch,

b) The acceleration to which the ball is subjected is equal in the rise and fall, therefore it takes the same time for both parties, let's find the rise time

at the highest point the vertical speed is zero

          v_{y} = v_{oy} - gt

          v_{y} = 0

           t = v_{oy} / g

           t = v₀ sin θ / g

as the time to get on and off is the same the total time or flight time is

           t_total = 2 t

           t_total = 2 v₀ sin θ / g

c) we calculate

          x = 13.5 2 sin (2 32) / 9.8

          x = 16.7 m

5 0
3 years ago
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