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LenaWriter [7]
3 years ago
9

1) On the way to the moon, the Apollo astro-

Physics
1 answer:
kramer3 years ago
6 0
(1) You must find the point of equilibrium between the two forces,

<span>G * <span><span><span>MT</span><span>ms / </span></span><span>(R−x)^2 </span></span>= G * <span><span><span>ML</span><span>ms / </span></span><span>x^2
MT / (R-x)^2 = ML / x^2

So,

x = R * sqrt(ML * MT) - ML / (MT - ML)
R = is the distance between Earth and Moon.

</span></span></span>The result should be,
x = 3.83 * 10^7m
from the center of the Moon, and 

R - x = 3.46*10^8 m
from the center of the Earth.


(2) As the distance from the center of the Earth is the number we found before,
d = R - x = 3.46*10^8m
The acceleration at this point is
g = G * MT / d^2
g = 3.33*10^-3 m/s^2
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3 years ago
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A vehicle that goes from 5m/s to 45m/s in 8s. what is its acceleration?
GaryK [48]

Answer: 5m/s^2

Explanation:

V= 45m/s

U = 5m/s

t = 8s

a =?

V = u + at

45 = 5 + 8a

8a = 45 — 5

8a = 40

a = 40 / 8

a = 5m/s^2

3 0
3 years ago
A basketball with a mass of 20 kg is accelerated with a force of 10 N. If resisting forces are ignored, what is the acceleration
kupik [55]
I’m pretty sure it would be 10/20= 0.5m/s2
7 0
3 years ago
A nylon guitar string is fixed between two lab posts 2.00 m apart. The string has a linear mass density of μ=7.20 g/m\mu=7.20~\t
vladimir2022 [97]

Answer:

4.6 m

Explanation:

First of all, we can find the frequency of the wave in the string with the formula:

f=\frac{1}{2L}\sqrt{\frac{T}{\mu}}

where we have

L = 2.00 m is the length of the string

T = 160.00 N is the tension

\mu =7.20 g/m = 0.0072 kg/m is the mass linear density

Solving the equation,

f=\frac{1}{2(2.00 m)}\sqrt{\frac{160.00 N}{0.0072 kg/m}}=37.3 Hz

The frequency of the wave in the string is transmitted into the tube, which oscillates resonating at same frequency.

The n=1 mode (fundamental frequency) of an open-open tube is given by

f=\frac{v}{2L}

where

v = 343 m/s is the speed of sound

Using f = 37.3 Hz and re-arranging the equation, we find L, the length of the tube:

L=\frac{v}{2f}=\frac{343 m/s}{2(37.3 Hz)}=4.6 m

4 0
3 years ago
The diagram shows a position-time graph What is the displacement of the object
algol13

The object is moving, so at different times, it has different displacement.  I'm guessing that you probably want to know the displacement at the end of the time on the graph ... 5 seconds.

Displacement is the distance and the direction FROM (the position at the  beginning) TO (the position at the end).

At the beginning ... time=0 ... the position is 1 meter.

At the end ... time=5 ... the position is zero.

The distance FROM the beginning TO the end is (zero - 1m) .  That's  <em>-1m </em>.


5 0
3 years ago
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