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3 years ago

6

The force of Earth's gravity pulls down on a stone after someone tosses it off a cliff. What is the "equal and opposite force" d

uring this interaction, according to Newton's third law?

1 answer:

Lapatulllka [165]3 years ago

3 0

The equal and opposite force is the force of the stone's gravity pulling up on thy Earth.

<u>Explanation</u>:

$F_{AB} = - F_{BA}$

When someone tosses a stone off a cliff, the force of gravity of Earth pulls down and the force of the stone's gravity pulls up on the Earth due to the equal and opposite reaction exerted between the Earth and the stone.

In Newton's third law, two important features are present. One is the force exerted on two bodies will have the same magnitude but opposite direction. The other feature is force is acting on two different objects.

This exerted force is equal in magnitude but opposite in direction whereas the gravity exerts a downward force on a stone and upward force on an Earth which is opposite in direction.

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Which particles in an atom are acted upon by the strong force? A. Electrons and neutrons B. Neutrons and protons O C. Protons an

BARSIC [14]

$\large \sf \pmb{B) \: Neutrons \: and \: Protons}$

<em>Neutrons And Protons</em> is a particles in an atom and are acted upon by the strong force.

6 0

2 years ago

0. A 3.00-kg block is dropped from rest on a vertical spring whose spring constant is 750 N/m. The block hits the spring, compre

Dmitry_Shevchenko [17]

Answer:

the spring compressed is 0.1878 m

Explanation:

Given data

mass = 3 kg

spring constant k = 750 N/m

vertical distance h = 0.45

to find out

How far is the spring compressed

solution

we will apply here law of mass of conservation

i.e

gravitational potential energy loss = gain of eastic potential energy of spring

so we say m×g×h = 1/2× k × e²

so e² = 2×m×g×h / k

so

we put all value here

e² = 2×m×g×h / k

e² = 2×3×9.81×0.45 / 750

e² = 0.0353

e = 0.1878 m

so the spring compressed is 0.1878 m

5 0

3 years ago

A water main pipe of diameter 10 cm enters a house 2 m below ground. A smaller diameter pipe carries water to a faucet 5 m above

Lemur [1.5K]

Explanation:

Given that,

Diameter = 10 cm

Distance = 2 m

Speed $v_{1}= 2\ m/s$

Speed $v_{2}=7\ m/s$

Pressure in main pipe $P_{1}=2\times10^{5}\ Pa$

(I). We need to calculate the diameter

Using equation of continuity

$Av_{1}=Av_{2}$

$\pi(\dfrac{d_{1}}{2})^2\times v_{1}=\pi(\dfrac{d_{2}}{2})^2\times v_{2}$

$(\dfrac{10}{2})^2\times2=(\dfrac{d_{2}}{2})^2\times7$

$d_{2}=\sqrt{\dfrac{25\times2\times4}{7}}$

$d_{2}=5.345\ cm$

(II). We need to calculate the pressure the gauge pressure

Using Bernoulli equation

$P_{1}+\dfrac{1}{2}\rho v_{1}^2+\rho gh_{1}=P_{2}+\dfrac{1}{2}\rho v_{2}^2+\rho g h_{2}$

$P_{2}=P_{1}+\dfrac{1}{2}\rho(v_{1}^2-v_{2}^2)-\rho g(h_{1}-h_{2})$

$P_{2}=2\times10^{5}+\dfrac{1}{2}\times1000(4-49)-1000\times 9.8\times(5)$

$P_{2}=1.28500\times10^{5}\ Pa$

(III). If it is possible to carry water to a faucet 17 m above ground,

Using Bernoulli equation

$P_{1}+\dfrac{1}{2}\rho v_{1}^2+\rho gh=P_{3}+\dfrac{1}{2}\rho v_{3}^2+\rho g h_{3}$

$P_{3}=P_{1}+\dfrac{1}{2}\rho v_{1}^2-\rho g(h_{1}-h_{3})$

Here, $h_{3}=0$

Put the value in the equation

$P_{3}=2\times10^{5}+\dfrac{1}{2}\times1000\times4-1000\times 9.8\times17$

$P_{3}=3.5400\times10^{5}\ Pa$

Hence, This is required solution.

7 0

3 years ago

A truck speeds up from velocity of 6m/s to 14m/s in 4 seconds. what is the trucks acceleration?

Tanzania [10]

To calculate the acceleration of the truck, we use the equation of motion

$v = u + at$

Here, v is final velocity of the truck, u is initial velocity of truck, a is acceleration of the truck and t is time taken.

Given, u = 6 m/s, v = 14 m/s and t = 4 s.

Substituting these values in above equation, we get

$14 \ m/s = 6 \ m/s + a \times 4 \ s \\\\ a = \frac{14 \ m/s - 6 \ m/s}{ 4 \ s } = 2 \ m /s^2$ .

Thus, the truck acceleration is $2 \ m /s^2$

3 0

3 years ago

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FinnZ [79.3K]

Answer:

ax = 4.01

Explanation:

6 0

3 years ago