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3 years ago

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The earth is rotating on its axis. It will continue to rotate unless acted upon by an outside force. This is an example of Newto

n's _____.
first law of motion
second law of motion
third law of motion

2 answers:

mojhsa [17]3 years ago

7 0

It is an example of Newton’s first law.

agasfer [191]3 years ago

5 0

Answer:

I think it's an example of his first law of motion

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Blocks 1 and 2 of masses m1 and m2, respectively are connected by a light string, as show above. These blocks are further connec

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Lllllllllllllllllits c

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3 years ago

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A ball with a horizontal speed of 1.0m/s rolls off a bench 2.0 m high. (a) how long will the ball take to reach the floor? (b) h

bixtya [17]

The motion of the ball is a composition of two motions:
- on the x (horizontal) axis, it is a uniform motion with initial velocity $v_x = 1.0 m/s$
- on the y (vertical) axis, it is a uniformly accelerated motion with acceleration $g= 9.81 m/s^2$

(a) to solve this part, we just analyze the motion on the vertical axis. The law of motion here is
$y(t) = h - \frac{1}{2} gt^2$
By requiring y(t)=0, we find the time t at which the ball reaches the floor:
$h- \frac{1}{2}gt^2=0$
$t= \sqrt{ \frac{2h}{g} }= \sqrt{ \frac{2\cdot 2.0 m}{9.81 m/s^2} }=0.64 s$

(b) for this part, we can analyze only the motion on the horizontal axis. To find how far the ball will land, we must calculate the distance covered on the x-axis, x(t), when the ball reaches the ground (so, after a time t=0.64 s):
$x(t) = v_x t = (1.0 m/s)(0.64 s)=0.64 m$

4 0

3 years ago

Why does the cannon move backward when the cannon ball is shot?

seropon [69]

Answer:

This came to mind

Explanation:

when a cannon fires (in real life or in the movies) have noticed that the cannon recoils, sliding backwards after the explosion. Again, a non-zero net force on the cannon changes its momentum.

6 0

2 years ago

What is the distance between a 900 kg compact car and a 1600 kg pickup truck if the gravitational force between them is about 0.

elena-14-01-66 [18.8K]

Answer:

The distance is 0.96m

Explanation:

Given

m1= 900kg

m2= 1600kg

Force F= 0.0001nN

G=6.67430*10^-11 Nm^2/kg^2

Required

The distance r

Step two:

the formula for the force is given as

F = Gm1m2/r2

make r subject of the formula

$r= \sqrt{\frac{Gm1m2}{F} }$

$r= \sqrt{\frac{6.67430*10^-11*900*1600}{0.0001} }\\\\r= 0.00009610992/0.0001`}\\\\r= 0.96m$

Answer:

The distance is 0.96m

Explanation:

Given

m1= 900kg

m2= 1600kg

Force F= 0.0001nN

G=6.67430*10^-11 Nm^2/kg^2

Required:

The distance r

Step two:

the formula for the force is given as

F = Gm1m2/r2

make r subject of the formula

$r= \sqrt{\frac{Gm1m2}{F} }$

$r= \sqrt{\frac{6.67430*10^-11*900*1600}{0.0001} }\\\\r= 0.00009610992/0.0001`}\\\\r= 0.96m$

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2 years ago

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A 5.93 kg ball is attached to the top of a vertical pole with a 2.35 m length of massless string. The ball is struck, causing it

Delvig [45]

Answer

given,

mass of ball = 5.93 kg

length of the string = 2.35 m

revolve with velocity of 4.75 m/s

acceleration due to gravity = 9.81 m/s²

T cos θ = mg

T cos θ = $5.93\times 9.81$

T cos θ = 58.17

$T sin \theta =\dfrac{mv^2}{r}$

$T sin \theta =\dfrac{5.93\times 4.75^2}{2.35 sin \theta}$

$T sin^2 \theta =56.93$

$sin^2 \theta = 1 - cos^2 \theta$

$T (1 - cos^2 \theta) =56.93$

$T (1 - (\dfrac{58.17}{T})^2) =56.93$

T² - 56.93T - 3383.75 = 0

T = 93.22 N

$cos \theta = \dfrac{58.17}{93.22}$

θ = 51.39°

6 0

3 years ago