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ipn [44]
2 years ago
5

An unstoppable object is heading right toward an unmovable object. What's going to happen?

Physics
1 answer:
marshall27 [118]2 years ago
8 0

Answer:

In my opinion the unstoppable object will hit the unmovable object and stop but the wheels will still be rolling and trying to move but can't.

<h3>Hope this helps.</h3><h3>Good luck ✅.</h3>
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An automobile engine develops a torque of 280 NxM at 3800 rpm. What is the power in watts and horsepower?
kirza4 [7]
280 nitrous times the miles at 3800 rides per minutie equals the power in watts is 5 seconds to go 100 miles per hour.the power in horsepower is 4 seconds to get to 110 miles per hour.
7 0
3 years ago
Two blocks of ice, one four times as heavy as the other, are at rest on a frozen lake. A person pushes each block the same dista
ANEK [815]

Answer:

The kinetic energy of the heavier block after the push is equal to the kinetic energy of the lighter block

Explanation:

Kinetic energy of smaller object

K=(1/2)mv^2

where m= mass of smaller object and v= velocity of smaller object

Also, it is given that heavier object is four times the mass of lighter object and consider its velocity as V

kinetic energy of heavier block K '= (1/2) (4m) V^2

Now, For smaller block , v^2 - u^2=2aS

[by Newtons laws of motion]

Also, v^2 = 2(F/m)S

Where S= displacement, F= force, u= initial velocity

So, K=(1/2)m[2(F/m)]S

\Rightarrow  K = FS

For heavier block ,

V^2 - u^2 = 2a'S

or,  V^2=2(F /4m)S

So,K '= (1/2)(4m)[2(F/4m)S

\Rightarrow  K'= FS

Therefore the kinetic energy of the heavier block after the push is equal to the kinetic energy of the lighter block

5 0
4 years ago
Problem 3: A basketball player is running at 4.75 m/s toward the basket when he jumps into the air to dunk the ball. He maintain
LekaFEV [45]

Answer

given,

speed of the basketball player, v_x = 4.75 m/s

height of the jump = ?

a) initial vertical velocity = 0 m/s

   final vertical velocity = ?

       height, h = 0.75 m

 using equation of motion

   v² = u² +  2 g h

   v² = 0² +  2 x 9.8 x  0.75

   v² = 14.7

   v = 3.83 m/s

b) let horizontal distance= x m

 maximum height at time , t s

 time taken to reach maximum height

    t = \dfrac{3.83}{9.8}

         t = 0.39 s

    horizontal distance

             = v_x  × t

             = 4.75 × 0.39

             = 1.85 m

horizontal distance is equal to 1.85 m

8 0
3 years ago
Part of being a good team worker is
andrey2020 [161]
Team work and a can do attitude is very important when being a team worker, as well as being open minded!
I hope this helps!
6 0
3 years ago
Read 2 more answers
3) If a ball launched at an angle of 10.0 degrees above horizontal from an initial height of 1.50 meters has a final horizontal
Irina-Kira [14]

Answer:

35.6 m

Explanation:

3 0
4 years ago
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