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4 years ago

Which is the best example of Newton's Second Law of Motion?

1 answer:

6 0

A small ball and a larger ball with greater mass are dropped from a roof. The larger ball strikes the ground with greater force than the smaller ball.

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Suppose an asteroid orbiting the sun had an orbital period of 7. 5 years. What would its orbital radius be?.

creativ13 [48]

By using the orbital period equation we will find that the orbital radius is r = 4.29*10^11 m

<h3>What is the orbital period?</h3>

This would be the time that a given body does a complete revolution in its orbit.

It can be written as:

$T = \sqrt{\frac{4*\pi ^2*r^3}{G*M} }$

Where:

π = 3.14
G is the gravitational constant = 6.67*10^(-11) m^3/(kg*s^2)
M is the mass of the sun = 1.989*10^30 kg
r is the radius, which we want to find.

Rewriting the equation for the radius we get:

$T = \sqrt{\frac{4*\pi ^2*r^3}{G*M} }\\\\r = \sqrt[3]{ \frac{T^2*G*M}{4*\pi ^2} }$

Where T = 7.5 years = 7.5*(3.154*10^7 s) = 2.3655*10^8 s

Replacing the values in the equation we get:

$r = \sqrt[3]{ \frac{(2.3655*10^8 s)^2*(6.67*10^{-11} m^3/(kg*s^2))*(1.989*10^{30} kg)}{4*3.14 ^2} } = 4.29*10^{11 }m$

So the orbital radius is 4.29*10^11 m

If you want to learn more about orbits, you can read:

brainly.com/question/11996385

7 0

2 years ago

The area of the pond is approximately equal to the area of a circle with radius 297m. Find the mass of the ice. Answer in kilogr

True [87]

Answer:

<em>where T is the value of the thickness omitted in the question.</em>

Explanation:

The ice on Walden Pond is .......... thick. The area of the pond is approximately equal to the area of a circle with radius 297 m. Find the mass of the ice. Answer in kg.

<em>The value of the thickness of the ice T is omitted, but I will show the solution, and the real answer can be gotten by multiplying the final calculated answer here by the thickness of the ice omitted.</em>

Given the radius of the equivalent circle of the ice = 297 m'

the area of the ice can be gotten from area A = $\pi r^{2}$ = $3.142*297^{2}$ = 277152.678 m^2

recall that the density of ice p ≅ 920 kg/m^3

also,

density of ice p = (mass of ice, m) ÷ (volume of ice, v)

i.e p = m/v

and,

m = pv

substituting the value of the density of water p into the equation, we have,

mass of the ice, m = 920v ....... equ 1

The volume of the ice above will be = (area of the ice, A) x (thickness of the ice, T)

i.e v = AT

substituting the value of area A into the equation, we have

v = 277152.678T ......equ 2

substitute value of v into equ 1

mass of the ice, m = 920 x (277152.678T)

mass of the ice, m = 254980463.8T kg

where T is the thickness of the ice

NB: To get the mass, multiply this answer with the thickness T given in the question.

7 0

3 years ago

Fig.4.1

Nutka1998 [239]

Answer:

Explanation:

7 0

4 years ago

Which statement tells a result of the big bang?

Nonamiya [84]

Answer:

Explanation:

Galaxies are moving closer

5 0

3 years ago

Read 2 more answers

The two forces of a 3rd law pair always act on different bodies. The two forces of a 3rd law pair always act on different bodies

svetoff [14.1K]

1. The two forces of a 3rd law pair always act on different bodies.

TRUE

SO above statement is TRUE because Newton's III law is valid for two different objects where one object will exert force and other body will exert reaction force on it.

2. Part F Given that two bodies interact via some force, the accelerations of these two bodies have the same magnitude but opposite directions. (Assume no other forces act on either body.)

FALSE

This is false because the force of action and Reaction is same on two different objects but for finding acceleration we need to divide the force by mass of two objects and since the mass of two bodies may be different so we can say that acceleration may be different.

3. Part G According to Newton's 3rd law, the force on the (smaller) moon due to the (larger) earth is

(i) greater in magnitude and antiparallel to the force on the earth due to the moon.

(ii) greater in magnitude and parallel to the force on the earth due to the moon.

(iii) equal in magnitude but antiparallel to the force on the earth due to the moon.

(iv) equal in magnitude and parallel to the force on the earth due to the moon.

(v) smaller in magnitude and antiparallel to the force on the earth due to the moon.

(vi) smaller in magnitude and parallel to the force on the earth due to the moon.

Since Earth and moon is an isolated system so here Newton's III law is valid due to which we can say that two forces are equal in magnitude but opposite in sign

so correct answer will be

equal in magnitude but antiparallel to the force on the earth due to the moon.

4 0

3 years ago