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2 years ago

Between the ball and the player’s head, there are forces. Which of Newton’s laws does this represent? Support your choice.

Physics

2 answers:

Nana76 [90]2 years ago

8 0

Answer:according to newtons third law for every action force there is an equal in size and opposite in direction reaction force forces always comes in pairs

Explanation:

ira [324]2 years ago

7 0

I need help with this one to

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1. Why should you use scientific notation when writing big numbers?

Fynjy0 [20]

Answer:

It helps the answer look clean. It also makes the work easier to work with.

Explanation:

Instead of writing a lot of zeros, all you have to do is add exponents to the number to show how much the decimal moved.

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2 years ago

How does the elbow medical and lateral epicondylitis?

dlinn [17]

Lateral epicondylitis, or “tennis elbow,” is an inflammation of the tendons that join the forearm muscles on the outside of the elbow. The bony bump on the outside (lateral side) of the elbow is called the lateral epicondyle. The ECRB muscle and tendon is usually involved in tennis elbow. 
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3 years ago

Bill and Janet are pulling on opposite sides of a table. Bill pulls with a force of 250 N to the left, and Janet pulls with a fo

Dafna1 [17]

Answer: 75 N to the right

Explanation:

5 0

2 years ago

ns:Select the correct answer. In a video game, a flying coconut moves at a constant velocity of 20 meters/second. The coconut hi

Len [333]

You would need to use the equation a= (v-u)÷t
You need to substitute in the correct numbers.
a= (10-20)÷1
Your answer is -10m/s^2

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3 years ago

Sometimes a person cannot clearly see objects close up or far away. To correct this type of vision, bifocals are often used. The

Rudik [331]

Answer:

1) P₁ = -2 D, 2) P₂ = 6 D

Explanation:

for this exercise in geometric optics let's use the equation of the constructor

$\frac{1}{f} = \frac{1}{p} + \frac{1}{q}$

where f is the focal length, p and q are the distance to the object and the image, respectively

1) to see a distant object it must be at infinity (p = ∞)

$\frac{1}{f_1} = \frac{1}{q}$

q = f₁

2) for an object located at p = 25 cm

$\frac{1}{f_2} = \frac{1}{25} + \frac{1}{q}$

We can that in the two expressions we have the distance to the image, this is the distance where it can be seen clearly in general for a normal person is q = 50 cm

we substitute in the equations

1) f₁ = -50 cm

$\frac{1}{f_2} = \frac{1}{25} + \frac{1}{50}$

$\frac{1}{f_2}$ = 0.06

f₂ = 16.67 cm

the expression for the power of the lenses is

P = $\frac{1}{f}$

where the focal length is in meters

1) P₁ = 1/0.50

P₁ = -2 D

2) P₂ = 1 /0.16667

P₂ = 6 D

4 0

2 years ago