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Lubov Fominskaja [6]
2 years ago
6

When the boy crashes his bumper car into the girl's bumper car, the momentum from his car is transferred to hers. What evidence

from the text supports this statement?
Physics
2 answers:
Darya [45]2 years ago
8 0

Answer:

Where is the text?

Explanation:

If you refer to the short sentence you wrote as text, I believe the answer is probably the word "crashes" because it shows how the momentum was transferred.

Guest1 year ago
0 0

idk a,aj fjianamaybe

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yami pours powdered cocoa mix into milk and stirs it. then she microwaves the mixture for three minutes. when she takes the cup
MrRa [10]
Convection can best be observed as she blows the warm steam air that rises.

As the warm steam rises,  she forces displaces it with cool air from her mouth. Because the warm steam is less dense it rises and because the cool air is more dense, it displaces the warm air.

This scenario is an example of convection.
 
6 0
2 years ago
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A research group at Dartmouth College has developed a Head Impact Telemetry (HIT) System that can be used to collect data about
Olin [163]

Answer:

6.05 cm

Explanation:

The given equation is

2 aₓ(x-x₀)=( Vₓ²-V₀ₓ²)

The initial head velocity V₀ₓ =11 m/s

The final head velocity  Vₓ is 0

The accelerationis given by =1000 m/s²

the stopping distance = x-x₀=?

So we can wind the stopping distance by following formula

2 (-1000)(x-x₀)=[0^{2} -11^{2}]

x-x₀=6.05*10^{-2} m

       =6.05 cm

3 0
1 year ago
Nc-1 has the same dimension as​
lisabon 2012 [21]

Answer:

Explanation:

The answer is electric field intensity. Electric field intensity is the force per unit positive charge which the charge exerts at any point.

8 0
2 years ago
1. Do alto de uma plataforma com 15m de altura, é lançado horizontalmente um projéctil. Pretende-se atingir um alvo localizado n
sveta [45]

Answer:

(a). The initial velocity is 28.58m/s

(b). The speed when touching the ground is 33.3m/s.

Explanation:

The equations governing the position of the projectile are

(1).\: x =v_0t

(2).\: y= 15m-\dfrac{1}{2}gt^2

where v_0 is the initial velocity.

(a).

When the projectile hits the 50m mark, y=0; therefore,

0=15-\dfrac{1}{2}gt^2

solving for t we get:

t= 1.75s.

Thus, the projectile must hit the 50m mark in 1.75s, and this condition demands from equation (1) that

50m = v_0(1.75s)

which gives

\boxed{v_0 = 28.58m/s.}

(b).

The horizontal velocity remains unchanged just before the projectile touches the ground because gravity acts only along the vertical direction; therefore,

v_x = 28.58m/s.

the vertical component of the velocity is

v_y = gt \\v_y = (9.8m/s^2)(1.75s)\\\\{v_y = 17.15m/s.

which gives a speed v of

v = \sqrt{v_x^2+v_y^2}

\boxed{v =33.3m/s.}

4 0
1 year ago
The fastest pitched baseball was clocked at 47 m/s. Assume that the pitcher exerted his force (assumed to be horizontal and cons
arsen [322]

Answer:

Explanation:

F ×1 = 0.5×0.145×47×47

F = 160.15 N

4 0
2 years ago
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