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3 years ago

Which sentence states Newton’s third law?

Physics

2 answers:

hammer [34]3 years ago

3 0

If two objects collide, each object exerts a force equal to and in the opposite direction of the other.

melamori03 [73]3 years ago

3 0

Explanation:

Sir Issac newton has given three laws of motion i.e.

1. First law of motion

2. Second law of motion

3. Third law of motion

The third law of motion states that, "If two objects collide, each object exerts a force equal to and in the opposite direction of the other". The applied force is called action force and its effect on another object is called reaction force. These two forces are applied on different objects. Mathematically, it is equal to :

$F_{1,2}=-F_{2,1}$

$F_{1,2}$ is the force on object 1 due to object 2

$F_{2,1}$ is the force on object 2 due to object 1

So, the correct option is (b) "If two objects collide, each object exerts a force equal to and in the opposite direction of the other".

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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:

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

A cannon fires a shell straight upward; 2.3 s after it is launched, the shell is moving upward with a speed of 17 m/s. Assuming

PtichkaEL [24]

Answer:

The speed of the shell at launch and 5.4 s after the launch is 13.38 m/s it is moving towards the Earth.

Explanation:

Let u is the initial speed of the launch. Using first equation of motion as :

$u=v-at$

a=-g

$u=v+gt\\\\u=17+9.8\times 2.3\\\\u=39.54\ m/s$

The velocity of the shell at launch and 5.4 s after the launch is given by :

$v=u-gt\\\\v=39.54-9.8\times 5.4\\\\v=-13.38\ m/s$

So, the speed of the shell at launch and 5.4 s after the launch is 13.38 m/s it is moving towards the Earth.

6 0

3 years ago

Two black holes (the remains of exploded stars), separated by a distance of

jolli1 [7]

The largest mass is 4.7 x 10³⁰ kg and the smallest mass is 5 x 10²⁹ kg.

The given parameters;

distance between the two black holes, r = 10 AU = 1.5 x 10¹² m
gravitational force between the two black holes, F = 6.9 x 10²⁵ N.
combined mass of the two black holes = 5.20 x 10³⁰ kg

The product of the two masses is calculated from Newton's law of universal gravitational as follows;

$F = \frac{Gm_1m_2}{r^2} \\\\m_1m_2 = \frac{Fr^2}{G} \\\\m_1m_2 = \frac{(6.9\times 10^{25}) \times (1.5\times 10^{12})^2}{6.67\times 10^{-11}} \\\\m_1m_2 = 2.328 \times 10^{60} \ kg^2$

The sum of the two masses is given as;

m₁ + m₂ = 5.2 x 10³⁰ kg

m₂ = 5.2 x 10³⁰ kg - m₁

The first mass is calculated as follows;

m₁(5.2 x 10³⁰ - m₁) = 2.328 x 10⁶⁰

5.2 x 10³⁰m₁ - m₁² = 2.328 x 10⁶⁰

m₁² - 5.2 x 10³⁰m₁ + 2.328 x 10⁶⁰ = 0

solve the quadratic equation using formula method;

a = 1, b =- 5.2 x 10³⁰, c = 2.328 x 10⁶⁰

$m_1 = \frac{-b \ \ +/- \ \ \sqrt{b^2 - 4ac} }{2a} \\\\m_1 = \frac{-(-5.2\times 10^{20}) \ \ +/- \ \ \sqrt{(-5.2\times 10^{20})^2 - 4(1\times 2.328\times 10^{60})} }{2(1)} \\\\m_1 = 4.7 \times 10^{30} \ kg \ \ or \ \ 4.9 \times 10^{29} \ kg$