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Nitella [24]
3 years ago
7

If the distance between two objects increased, which of the following things could keep the force of gravity between the two obj

ects the same?
a. increasing the mass of one of the objects.
b. decreasing the mass of one of the objects.
c. increasing the velocity of one of the objects.
d. decreasing the velocity of one of the objects.
Physics
1 answer:
AURORKA [14]3 years ago
7 0
The answer A hope this helped

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A 46.8-g golf ball is driven from the tee with an initial speed of 58.8 m/s and rises to a height of 24.7 m. (a) Neglect air res
Andre45 [30]

Answer:

a) the kinetic energy of the ball at its highest point is 69.58 J

b) its speed when it is 8.11 m below its highest point is 55.97 m/s

Explanation:

Given that;

mass of golf ball m = 46.8 g = 0.0468 kg

initial speed of the ball v₁ = 58.8 m/s

height h = 24.7 m

acceleration due to gravity = 9.8 m/s²

the kinetic energy of the ball at its highest point = ?

from the conservation of energy;

Kinetic energy at the highest point will be;

K.Ei + P.Ei = KEf + PEf

now the Initial potential energy of the ball P.Ei = 0 J

so

1/2mv² + 0 J = KEf + mgh

K.Ef = 1/2mv² - mgh

we substitute

K.Ef = [1/2 × 0.0468 × (58.8 )²] - [0.0468 × 9.8 × 24.7]

K.Ef  = 80.904 - 11.3284

K.Ef = 69.58 J

Therefore, the kinetic energy of the ball at its highest point is 69.58 J

b) when the ball is 8.11 m below the highest point, speed = ?

so our raw height h' will be ( 24.7 m - 8.11 m) = 16.59 m

so our velocity will be v₂

also using the principle of energy conservation;

K.Ei + P.Ei = KEh + PEh

1/2mv² + 0 J = 1/2mv₂² + mgh'

1/2mv₂² = 1/2mv² - mgh'

multiply through by 2/m

v₂² = v² - 2gh'

v₂ = √( v² - 2gh' )

we substitute

v₂ = √( (58.8)² - 2×9.8×16.59 )

v₂ = √( 3457.44 - 325.164 )  

v₂ = √( 3132.276 )

v₂ = 55.97 m/s

Therefore, its speed when it is 8.11 m below its highest point is 55.97 m/s

5 0
3 years ago
1. A cyclist accelerates from 0 m/s to 9 m/s in 3 seconds. What is his<br>acceleration?​
Lapatulllka [165]

The cyclist accelerates from 0 m/s to 9 m/s in 3 seconds with an acceleration of 3 m/s².

Answer:

Explanation:

Acceleration exerted by an object is the measure of change in speed or velocity of that object with respect to time. So the initial and final velocities play a major role in determining the acceleration of the cyclist. As here the initial velocity of the cyclist is the speed at rest and that is given as 0 m/s. Then after 3 seconds, the velocity of the cyclist changes to 9 m/s.

Then acceleration = change in velocity/Time.

Acceleration = \frac{Change in velocity}{Time taken}

Acceleration = (9-0)/3=9/3=3 m/s².

So the cyclist accelerates from 0 m/s to 9 m/s in 3 seconds with an acceleration of 3 m/s².

3 0
3 years ago
What is the distance an object would be from Earth if its parallax were one arcsecond?
ZanzabumX [31]
The correct answer is (a.) a parsec. A parsec is a distance an object would be from Earth if its parallax were one arcsecond. This unit of measurement is usually used in astronomy which makes it easier for astronomers to calculate or measure in space accurately. 
8 0
3 years ago
A girl is sledding down a slope that is inclined at 30º with respect to the horizontal. The wind is aiding the motion by providi
OleMash [197]

Answer:

The sled required 9.96 s to travel down the slope.

Explanation:

Please, see the figure for a description of the problem. In red are the x and y-components of the gravity force (Fg). Since the y-component of Fg (Fgy) is of equal magnitude as Fn but in the opposite direction, both forces get canceled.

Then, the forces that cause the acceleration of the sled are the force of the wind (Fw), the friction force (Ff) and the x-component of the gravity force (Fgx).

The sum of all these forces make the sled move. Finding the resulting force will allow us to find the acceleration of the sled and, with it, we can find the time the sled travel.

The magnitude of the friction force is calculated as follows:

Ff = μ · Fn

where :

μ = coefficient of kinetic friction

Fn =  normal force

The normal force has the same magnitude as the y-component of the gravity force:

Fgy = Fg · cos 30º = m · g · cos 30º

Where

m = mass

g = acceleration due to gravity

Then:

Fgy = m · g · cos 30º = 87.7 kg · 9.8 m/s² · cos 30º

Fgy = 744 N

Then, the magnitude of Fn is also 744 N and the friction force will be:

Ff = μ · Fn = 0.151 · 744 N = 112 N

The x-component of Fg, Fgx, is calculated as follows:

Fgx = Fg · sin 30º = m·g · sin 30º = 87.7 kg · 9.8 m/s² · sin 30º = 430 N

The resulting force, Fr, will be the sum of all these forces:

Fw + Fgx - Ff = Fr

(Notice that forces are vectors and the direction of the friction force is opposite to the other forces, then, it has to be of opposite sign).

Fr = 161 N + 430 N - 112 N = 479 N

With this resulting force, we can calculate the acceleration of the sled:

F = m·a

where:

F = force

m = mass of the object

a = acceleration

Then:

F/m = a

a = 479N/87.7 kg = 5.46 m/s²

The equation for the position of an accelerated object moving in a straight line is as follows:

x = x0 + v0 · t + 1/2 · a · t²

where:

x = position at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

Since the sled starts from rest and the origin of the reference system is located where the sled starts sliding, x0 and v0 = 0.

x = 1/2· a ·t²

Let´s find the time at which the position of the sled is 271 m:

271 m = 1/2 · 5.46 m/s² · t²

2 · 271 m / 5.46 m/s² = t²

<u>t = 9.96 s </u>

The sled required almost 10 s to travel down the slope.

8 0
3 years ago
Suppose a wheel with a tire mounted on it is rotating at the constant rate of 2.59 times a second. A tack is stuck in the tire a
Vesna [10]

Answer:

Tangential speed=5.4 m/s

Radial acceleration=88.6m/s^2

Explanation:

We are given that

Angular speed=2.59 rev/s

We know that

1 revolution=2\pi rad

2.59 rev=2\pi\times 2.59=5.18\pi=5.18\times 3.14=16.27 rad/s

By using \pi=3.14

Angular velocity=\omega=16.27rad/s

Distance from axis=r=0.329 m

Tangential speed=r\omega=16.27\times 0.329=5.4m/s

Radial acceleration=\frac{v^2}{r}

Radial acceleration=\frac{(5.4)^2}{0.329}=88.6m/s^2

6 0
4 years ago
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