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vivado [14]
3 years ago
11

Erica throws a tennis ball against a wall, and it bounces back. Which force is responsible for sending the ball back to Erica? t

he force that the ball exerts on the wall the frictional force between the wall and the ball the acceleration of the ball as it approaches the wall the normal force that the wall exerts on the ball
Physics
2 answers:
Leona [35]3 years ago
7 0

actually the answer is D.

:)

Nutka1998 [239]3 years ago
6 0

Answer:

the normal force that the wall exerts on the ball

Explanation:

As Newton's third law states:

"when an object A exerts a force on object B, then object B exerts an equal and opposite force on object A".

If we apply this law to this problem, we can identify the ball as object A, and the wall as object B. As the ball hits the wall, the ball exerts a force on the wall (toward the direction of motion of the ball), so the wall exerts an equal and opposite force on the ball (in the opposite direction). This force is the normal force of the wall, and it is responsible for pushing the ball back towards Erica.

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A slanted vector has a magnitude of 41 N and is at an angle of 23 degrees north of east. What are the magnitude and direction of
jenyasd209 [6]

Answer:

The horizontal component is 37.74 N to east

The vertical component is 16.02 N to North

Explanation:

Any slant vector has two components

→ Horizontal component = R cos Ф

→ Vertical component = R sin Ф

→ R is the magnitude of the vector

→ Ф is the direction of the vector with positive part of the horizontal axis

A slanted vector has a magnitude of 41 N and is at an angle of 23

degrees north of east

23° north of east means the angle between the vector and the

east direction is 23° (east is the positive horizontal direction)

That means R is 41 N and Ф is 23°

→ R = 41 N , Ф = 23°

→ The horizontal component = 41 × cos(23) = 37.74 N east

→ The vertical component = 41 × sin(23) = 16.02 N North

<em>The horizontal component is 37.74 N to east</em>

<em>The vertical component is 16.02 N to North</em>

7 0
3 years ago
Imagine that two balls, a basketball and a much larger exercise ball, are dropped from a parking garage. If both the mass and ra
pashok25 [27]

Here if we assume that there is no air friction on both balls then we can say

F = mg

now the acceleration is given as

F = ma = mg

a = g

so here both the balls will have same acceleration irrespective of size and mass

so we can say that to find out the time of fall of ball we can use

y = \frac{1}{2}gt^2

t = \sqrt{\frac{2y}{g}}

now from above equation we can say that time taken to hit the ground will be same for both balls and it is irrespective of its mass and size

3 0
3 years ago
Can someone spell the word hi for me I’ll give you 30 points
jasenka [17]

Answer:

hi poop

Explanation:

3 0
3 years ago
Read 2 more answers
Which of the following models shows how unpredictable electrons are?
Fantom [35]

D

The exact location of electrons in electron shells of atoms cannot be exactly ascertained. This is why VSPER atomic models represent the position of electrons (s, p, d, & f) using the probability of where they would most be expected to be found.

Explanation:

This is because merely observing electrons changes their behavior. Remember that to observe something one has to shine light on it so it bounces back to the eye. Due to the negligible mass of electrons, mere photons of light will change their direction of movement, spin or other behaviors/properties.

Learn More:

For more on electron clouds check out;

brainly.com/question/12199882

brainly.com/question/11686000

#LearnWithBrainly

7 0
3 years ago
Read 2 more answers
a ball is thrown inclined in to the air with a initial velocity of u. if it reaches the maximum height in 6 seconds , find the r
olga55 [171]

Answer:

11:1

Explanation:

At constant acceleration, an object's position is:

y = y₀ + v₀ t + ½ at²

Given y₀ = 0, v₀ = u, and a = -g:

y = u t − ½g t²

After 6 seconds, the ball reaches the maximum height (v = 0).

v = at + v₀

0 = (-g)(6) + u

u = 6g

Substituting:

y = 6g t − ½g t²

The displacement between t=0 and t=1 is:

Δy = [ 6g (1) − ½g (1)² ] − [ 6g (0) − ½g (0)² ]

Δy = 6g − ½g

Δy = 5½g

The displacement between t=6 and t=7 is:

Δy = [ 6g (7) − ½g (7)² ] − [ 6g (6) − ½g (6)² ]

Δy = (42g − 24½g) − (36g − 18g)

Δy = 17½g − 18g

Δy = -½g

So the ratio of the distances traveled is:

(5½g) / (½g)

11 / 1

The ratio is 11:1.

8 0
3 years ago
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