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3 years ago

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Physics

1 answer:

Drupady [299]3 years ago

7 0

Answer:

An object in space that has no motion has to be shoved to move in the first place because of <u>newton's first law of motion. </u>
A bowling ball will slow down and stop when rolled on a flat surface, like a street because of <u>rolling friction</u> which comes under<u> Newton's first law of motion.</u>
When a 100Kg astronaut throws a 9Kg bowling ball forward in space, she moves backward but slower, this is because of<u> Newton's third law of motion. </u>

Explanation:

<u>This is due to Newton's first law of motion, explaining that an object in space that has no movement has to be moved to move in the first place.</u> There is no external force or gravitational force in space, so the body can stay at rest and be forced to move first.
<u>Explaining why a bowling ball, like a street, can slow down and stop when rolled on a flat surface.</u> This, again, is attributed to <u>Newton's first law of motion.</u> <u>Due to rolling friction, it happens.</u> The electrons on the surface of the ground in the atoms push against the electrons that touch the ground in the atoms on the surface of the ball as you roll a ball on the ground. A rolling ball stops when the surface on which it rolls opposes its motion. A ball rolling, due to friction, ceases.

<u>Explaining that when a 100 kg astronaut throws a 9 kg bowling ball forward in space, she travels backward but slower</u>, due to <u>Newton's third motion rule , </u>which states that when force is applied, each body exhibits equal and opposite reaction. In this case, the astronaut's mass is greater, so the ball's acceleration will be less than the astronaut 's mass (since the body's mass is inversely proportional to the acceleration). Therefore, because of the reverse direction and slowly because of no gravitational force in space, the girl shifts backward while tossing the ball forward in space.

<u> Newton's Law of Motion - </u>

<u>FIRST LAW -</u> The first law of Newton states that if a body is at rest or moving in a straight line at a constant speed, it will stay at rest or continue to travel at constant speed in a straight line unless it is acted upon by a force.

<u>SECOND LAW - </u> The second law of Newton is a quantitative explanation of the changes that a force can cause in a body 's motion. It states that the rate of change of a body's momentum in time is proportional to the force exerted on it in both magnitude and direction.

<u>THIRD LAW - </u> The third law of Newton notes that they apply forces to each other when two bodies interact, which are equal in magnitude and opposite in direction. Often known as the law of action and reaction, the third law is. In analyzing static equilibrium problems, where all forces are balanced, this law is essential, but it also applies to bodies in uniform or accelerated motion.

Hence , all the three explainations consist of newton's first and third laws of motion.

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A 2.2kg block of ice slides across a rough floor. Its initial velocity is 2.5m/s and its final velocity is 0.50m/s. How much of

kirill [66]

Answer:A 2.2kg block of ice slides across a rough floor. Its initial velocity is 2.5m/s and its final velocity is 0.50m/s. How much of the ice block melted as a result of the work done by friction? (Latent Heat of water is 3.3*10^5J/kg)

Explanation:

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3 years ago

Planets are not uniform inside. Normally, they are densest at the center and have decreasing density outward toward the surface.

elena-s [515]

Answer:

$g=13.42\frac{m}{s^2}$

Explanation:

1) Notation and info given

$\rho_{center}=13000 \frac{kg}{m^3}$ represent the density at the center of the planet

$\rho_{surface}=2100 \frac{kg}{m^3}$ represent the densisty at the surface of the planet

r represent the radius

$r_{earth}=6.371x10^{6}m$ represent the radius of the Earth

2) Solution to the problem

So we can use a model to describe the density as function of the radius

$r=0, \rho(0)=\rho_{center}=13000 \frac{kg}{m^3}$

$r=6.371x10^{6}m, \rho(6.371x10^{6}m)=\rho_{surface}=2100 \frac{kg}{m^3}$

So we can create a linear model in the for y=b+mx, where the intercept b= $\rho_{center}=13000 \frac{kg}{m^3}$ and the slope would be given by $m=\frac{y_2-y_1}{x_2-x_1}=\frac{\rho_{surface}-\rho_{center}}{r_{earth}-0}$

So then our linear model would be

$\rho (r)=\rho_{center}+\frac{\rho_{surface}-\rho_{center}}{r_{earth}}r$

Since the goal for the problem is find the gravitational acceleration we need to begin finding the total mass of the planet, and for this we can use a finite element and spherical coordinates. The volume for the differential element would be $dV=r^2 sin\theta d\phi d\theta dr$ .

And the total mass would be given by the following integral

$M=\int \rho (r) dV$

Replacing dV we have the following result:

$M=\int_{0}^{2\pi}d\phi \int_{0}^{\pi}sin\theta d\theta \int_{0}^{r_{earth}}(r^2 \rho_{center}+\frac{\rho_{surface}-\rho_{center}}{r_{earth}}r)$

We can solve the integrals one by one and the final result would be the following

$M=4\pi(\frac{r^3_{earth}\rho_{center}}{3}+\frac{r^4_{earth}}{4} \frac{\rho_{surface}-\rho_{center}}{r_{earth}})$

Simplyfind this last expression we have:

$M=\frac{4\pi\rho_{center}r^3_{earth}}{3}+\pi r^3_{earth}(\rho_{surface}-\rho_{center})$

$M=\pi r^3_{earth}(\frac{4}{3}\rho_{center}+\rho_{surface}-\rho_{center})$

$M=\pi r^3_{earth}[\rho_{surface}+\frac{1}{3}\rho_{center}]$

And replacing the values we got:

$M=\pi (6.371x10^{6}m)^2(\frac{1}{3}13000 \frac{kg}{m^3}+2100 \frac{kg}{m^3})=8.204x10^{24}kg$

And now that for any shape the gravitational acceleration is given by:

$g=\frac{MG}{r^2_{earth}}=\frac{(6.67408x10^{-11}\frac{m^3}{kgs^2})*8.204x10^{24}kg}{(6371000m)^2}=13.48\frac{m}{s^2}$

4 0

2 years ago

A ball is tossed vertically upward. When it reaches its highest point (before falling back downward) Group of answer choices the

nignag [31]

Answer:

the velocity is zero, the acceleration is directed downward, and the force of gravity acting on the ball is directed downward

Explanation:

Is this exercise in kinematics

v = v₀ - g t

where g is the acceleration of the ball, which is created by the attraction of the ball to the Earth.

At the highest point

velocity must be zero.

The acceleration depends on the Earth therefore it is constant at this point and with a downward direction.

The force of the earth on the ball is towards the center of the Earth, that is, down

all other alternatives are wrong

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3 years ago

A bulldozer does 4,500 J of work to push a mound of soil to the top of a ramp that is 15 m high. The ramp is at an

Andreas93 [3]

Answer:

170 N

Explanation:

Given in the question that, work a bulldozer can do = 4500 J

We will use trigonometry identity to find the distance bulldozer will travel up the hill

sin(35) = opp/hypo

sin(35) = 15/hypo

hypo = 15/sin(35)

hypo = 26.15m

Formula to use

work done = force × distance

Plug values in the above formula

4500 = force x 26.15

force = 4500/26.15

force = 172.08

force ≈ 170 N

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3 years ago

Read 2 more answers

A man has been guarding a house for one hour . Why is it not considered work in science

leva [86]

Answer: because there is no displacement or movement in the watchman's work. according to science when displacement or movement take place it is said to be work. hope this helps you.

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3 years ago