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zloy xaker [14]
2 years ago
13

Given that the period of the Moon's orbit about the Earth is 27.32 days and the nearly constant distance between the center of t

he Earth and the center of the Moon is 3.84 108 m, use T2 = 4 π2 GME a3 to calculate the mass of the Earth.
Physics
1 answer:
Sav [38]2 years ago
4 0

Answer:

M = 6.014x10^24 kg

Explanation:

First of all we need to gather all data:

Period (T) is 27.32 days

Constant distant (a) is 3.84x10^8 m

The expression given is T² = (4π²/GM)*a³

We need to know the value of the constant G which is 6.674x10^-11 Nm² / kg²

Finally M is the mass of the earth.

From that expression, we should solve for M:

M = (4π²/GT²)*a³

In this case, we just trade the T for the M, because it was the only change we needed to do. Now, before we can do the calculations, let's convert the days to second:

T = 27.32 days * 24 h/day * 3600 s/h = 2,360,448 s

Now, let's solve for M replacing all the data in the formula:

M = (4π² / 6.674x10^-11 * 2,360,448²)*(3.84x10^8)³

M = (39.4784 / 371.6891) * 5.662x10^25

M = 6.014x10^24 kg

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Archy [21]

Answer:

a) t = 1.8 x 10² s

b) t = 54 s

c) t = 49 s

Explanation:

a) The equation for the position of an object moving in a straight line at constan speed is:

x = x0 + v * t

where

x = position at time t

x0 = initial position

v = velocity

t = time

In this case, the origin of our reference system is at the begining of the sidewalk.

a) To calculate the time the passenger travels on the sidewalk without wlaking, we can use the equation for the position, using as speed the speed of the sidewalk:

x = x0 + v * t

95 m = 0m + 0. 53 m/s * t

t = 95 m/ 0.53 m/s

t = 1.8 x 10² s

b) Now, the speed of the passenger will be her walking speed plus the speed of th sidewalk (0.53 m/s + 1.24 m/s = 1.77 m/s)

t = 95 m/ 1.77 m/s = 54 s

c) In this case, the passenger is located 95 m from the begining of the sidewalk, then, x0 = 95 m and the final position will be x = 0. She walks in an opposite direction to the movement of the sidewalk, towards the origin of the system of reference ( the begining of the sidewalk). Then, her speed will be negative ( v = 0.53 m/s - 2*(1.24 m/s) = -1.95 m/s. Then:

0 m = 95 m -1.95 m/s * t

t = -95 m / -1.95 m/s = 49 s

3 0
2 years ago
A ball weighing 1 lb is attached to a string 2 feet long and is whirled in a vertical circle at a constant speed of 10 ft/sec.
fredd [130]

Explanation:

It is given that,

Mass of the ball, m = 1 lb

Length of the string, l = r = 2 ft

Speed of motion, v = 10 ft/s

(a) The net tension in the string when the ball is at the top of the circle is given by :

F=\dfrac{mv^2}{r}-mg

F=m(\dfrac{v^2}{r}-g)

F=1\ lb\times (\dfrac{(10\ ft/s)^2}{2}-1\ lb\times 32\ ft/s^2)

F = 18 N

(b) The net tension in the string when the ball is at the bottom of the circle is given by :

F=\dfrac{mv^2}{r}+mg

F=m(\dfrac{v^2}{r}+g)

F=1\ lb\times (\dfrac{(10\ ft/s)^2}{2}+1\ lb\times 32\ ft/s^2)

F = 82 N

(c) Let h is the height where the ball at certain time from the top. So,

T=mg(\dfrac{r-h}{r})+\dfrac{mv^2}{r}

T=\dfrac{m}{r}(g(r-h)+v^2)

Since, v^2=u^2-2gh

T=\dfrac{m}{r}(u^2-3gh+gr)

Hence, this is the required solution.

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A 1,100 kg car comes uniformly to a stop. If the vehicle is accelerating at -1.2 m/s2 , which force is closest to the net force
boyakko [2]

Answer:

Explanation:

D

7 0
1 year ago
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How much power does a machine have that does 15204 J of work in 40 seconds?
pentagon [3]

Explanation:

since P=W/t

P=15204/40

P=380.1 Watt

5 0
1 year ago
A cylindrical specimen of brass that has a diameter listed above, a tensile modulus of 110 GPa, and a Poisson's ratio of 0.35 is
Irina-Kira [14]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The strain experienced by the specimen is 0.00116 which is option A

Explanation:

The explanation is shown on the second uploaded image

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