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never [62]
3 years ago
7

In which direction does a Newton's Third Law reaction force act?

Physics
1 answer:
AnnyKZ [126]3 years ago
7 0

Answer:

B. In the opposite direction from the action force

Explanation:

if object A exerts a force on object B, then object B also exerts an equal and opposite force on object A. Notice that the forces are exerted on different objects.

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Resisting force is
Alina [70]

Answer:

D

Explanation:

6 0
3 years ago
Newton’s Second Law of motion defines force in terms of mass and acceleration, F= ma. Which of the following is false? The term
NikAS [45]

Answer:

The forces creating the net force must lie in the same direction.

Explanation:

newton's second law states that the net force acting on the body is equal to the product of mass and the acceleration of the body.

If there are several forces acting on the body in different directions, then we have to find teh net force by using the vector sum and then find the acceleration.

It is not necessary that all the forces acting in the same direction.

if they are in different directions then we have to find the net force by t=using the formula for the vector sum.

5 0
3 years ago
A 3.50 cm tall object is held 24.8 cm from a lens of focal length 16.0 cm. What is the image height?
Mandarinka [93]

Answer:

<h2>6.36 cm</h2>

Explanation:

Using the formula to first get the image distance

1/f = 1/u+1/v

f = focal length of the lens

u = object distance

v = image distance

Given f = 16.0 cm, u = 24.8 cm

1/v = 1/16 - 1/24.8

1/v = 0.0625-0.04032

1/v = 0.02218

v = 1/0.02218

v = 45.09 cm

To get the image height, we will us the magnification formula.

Mag = v/u = Hi/H

Hi = image height = ?

H = object height = 3.50 cm

45.09/24.8 = Hi/3.50

Hi = (45.09*3.50)/24.8

Hi = 6.36 cm

The image height is 6.36 cm

6 0
3 years ago
An interference pattern is produced by light with a wavelength 580 nm from a distant source incident on two identical parallel s
irakobra [83]

Answer:

Explanation:

1 )

Here

wave length used that is λ = 580 nm

=580 x 10⁻⁹

distance between slit d = .46 mm

= .46 x 10⁻³

Angular position of first order interference maxima

= λ / d radian

= 580 x 10⁻⁹ / .46 x 10⁻³

= 0.126 x 10⁻² radian

2 )

Angular position of second order interference maxima

2 x  0.126 x 10⁻² radian

= 0.252 x 10⁻² radian

3 )

For intensity distribution the formula is

I = I₀ cos²δ/2 ( δ is phase difference of two lights.

For angular position of θ1

δ = .126 x 10⁻² radian

I = I₀ cos².126x 10⁻²/2

= I₀ X .998

For angular position of θ2

I = I₀ cos².126x2x 10⁻²/2

=  I₀ cos².126x 10⁻²

8 0
3 years ago
For which question could a testable hypothesis be developed?
adell [148]
Does wood float in water
8 0
3 years ago
Read 2 more answers
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