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katovenus [111]
3 years ago
5

If you push on a wall of a building what will happen

Physics
1 answer:
mixas84 [53]3 years ago
8 0

Answer: It will push back with an equal amount of force.

You might be interested in
When a carousel is in motion, the movement of the carousel horse can BEST be described as
lara [203]
Kinetic energy because the horse is in motion
7 0
3 years ago
What is a centripetal acceleration of a point on a bicycle wheel of a radius of 0.70 m when a bike is moving 8.0 m/s
Furkat [3]

Answer:

The acceleration of a point on the wheel is 11.43 m/s² acting radially inward.

Explanation:

The centripetal acceleration acts on a body when it is performing a circular motion.

Here, a point on the bicycle is performing circular motion as the rotation of the wheel produces a circular motion.

The centripetal acceleration of a point moving with a velocity v and at a distance of r from the axis of rotation is given as:

a=\frac{v^2}{r}

Here, v=8\ m/s,r=0.70\ m

∴ a=\frac{8}{0.70}=11.43\ m/s^2

Therefore, the acceleration of a point on the wheel is 11.43 m/s² acting radially inward.

3 0
3 years ago
A 980 kg roller coaster cart is traveling along a track at 17 m/s before it rolls down a 30 m tall hill (Point A). What will be
MrRissso [65]

The kinetic energy halfway the hill is 2.86\cdot 10^5 J

Explanation:

If there are no friction forces acting on the cart, we can apply the law of conservation of energy: the mechanical energy of the cart (which is the sum of potential energy + kinetic energy) must be conserved. So we can write:

U_A +K_A = U_B + K_B

where

U_A=mgh_A is the initial potential energy, at point A, with

m = 980 kg (mass of the cart)

g=9.8 m/s^2 (acceleration of gravity)

h_A = 30 m (height at point A)

K_A=\frac{1}{2}mv_A^2 is the initial kinetic energy, at point A , with

v_A=17 m/s (velocity at point A)

U_B=mgh_B is the final potential energy, at point B, where

h_B = 15 m (height at point B)

K_B=\frac{1}{2}mv_B^2 is the final kinetic energy, at point B, where

v_B is the velocity at point B

Here we are interested in finding K_B, so by re-arranging the equation and substituting we find:

K_B = U_A+K_B-U_B = mg(h_A-h_B)+\frac{1}{2}mv_A^2=(980)(9.8)(30-15)+\frac{1}{2}(980)(17)^2=2.86\cdot 10^5 J

Learn more about kinetic energy:

brainly.com/question/6536722

#LearnwithBrainly

8 0
3 years ago
An object is allowed to fall freely near the surface of a planet. The object has an acceleration due to gravity of 24 m/s2. How
Alborosie

Answer:

12 m

Explanation:

The object is in uniformly accelerated motion, so the distance covered can be found using the following suvat equation:

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

where

s is the distance

u is the initial velocity

t is the time

a is the acceleration

For this problem,

g=24 m/s^2

and

u = 0, since we are considering the first second of motion

So, substituting t = 1 s, we find

s=0+\frac{1}{2}(24)(1)^2=12 m

6 0
3 years ago
A net force, the magnitude of which is 3800 N, accelerates a 1260-kg vehicle for 10.0 s. The vehicle travels 50.0 m during this
Novay_Z [31]

Answer:

SEE EXPLANATION

Explanation:

p =  \frac{fd}{t}  \\ where \: \\p  = power \\  f = force \\ d = distance \\ and \: t = time \\  \\ p =  \frac{3800 \times 50}{10}  \\ p =  \frac{190000}{10}  \\ p = 19000w

7 0
3 years ago
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