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frozen [14]
3 years ago
6

Wegut.

Physics
1 answer:
AVprozaik [17]3 years ago
6 0

We/Wm = ge/gm = 120N/1.2N

or

gm = ge/100 = 0.1 m/s^2

density = mass/volume = 3M/(4pir^3)

Re-arranging this equation, we get

M/r^2 = (4/3)×pi×(density)×r

From Newton's universal law of gravitation, the acceleration due to gravity on the moon gm is

gm = G(M/r^2) = G×(4/3)×pi×(density)×r

Solving for density, we get the expression

density = 3gm/(4×pi×G×r)

= 3(0.1)/(4×3.14×6.67×10^-11×2.74×10^6)

= 130.6 kg/m^3

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if you take one hour drive at an average speed of 65 mph, is it possible for another car with an average speed of 55 mph to pass
SVETLANKA909090 [29]
Nope, this is impossible because in order for a car to pass another, they the 55 mph car would have to be behind the 65 mph car (meaning that starting ahead of the 65 mph doesn't count as pass). Assuming that they both drive for one hour, it is impossible because the first will cover a distance of 65 mi and the second would cover a distance of 55 mi. One is obviously ahead of the other and is therefore impossible for the slow one to pass the first one unless the slow car keeps driving after an hour. In that case, it would take approximately 11 minutes for it to pass the other car. This was found by finding the distance needed to pass the first car : 65 - 55 = 10 mi and converted using 1 hr/ 55 mi = .18181818 hr x 60 min/ 1 hr = 11 seconds
I hope this helps :)
7 0
3 years ago
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A steel sphere sits on top of an aluminum ring. The steel sphere (a= 1.1 x 10^-5/degrees celsius) has a diameter of 4.000 cm at
Mars2501 [29]

Answer:

C

Explanation:

To solve this question, we will need to develop an expression that relates the diameter 'd', at temperature T equals the original diameter d₀ (at 0 degrees) plus the change in diameter from the temperature increase ( ΔT = T):

d = d₀ + d₀αT

for the sphere, we were given

D₀ = 4.000 cm

α = 1.1 x 10⁻⁵/degrees celsius

we have D = 4 + (4x(1.1 x 10⁻⁵)T = 4 + (4.4x10⁻⁵)T             EQN 1

Similarly for the Aluminium ring we have

we were given

d₀ = 3.994 cm

α = 2.4 x 10⁻⁵/degrees celsius

we have d = 3.994 + (3.994x(2.4 x 10⁻⁵)T = 3.994 + (9.58x10⁻⁵)T       EQN 2

Since @ the temperature T at which the sphere fall through the ring, d=D

Eqn 1 = Eqn 2

4 + (4.4x10⁻⁵)T =3.994 + (9.58x10⁻⁵)T, collect like terms

0.006=5.18x10⁻⁵T

T=115.7K

4 0
3 years ago
Q
GalinKa [24]

Answer:

B

Explanation:

It would be diffrent if on a downward slope but assuming your going straight it would be the smallest student.

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Graph the following data on the graph, then use the graph to determine the half-life of this isotope.
Harlamova29_29 [7]

Answer:

4, 56

Explanation:

Hope this helped!

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3 years ago
What is the speed of electromagnetic radiation in free speed
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That's 299,792,458 meters per second.
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