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VashaNatasha [74]
3 years ago
8

students make a small elevator machine with 5 kg and 10 kg masses on either side how fast will the masses accelerate once they a

re released?​
Physics
1 answer:
mihalych1998 [28]3 years ago
8 0

Answer:

The acceleration of the elevator machine is, a = 3 m/s²                

Explanation:

Given data,

The mass on the one end, m = 5 kg

The mass on the other end, M = 10 kg

According to the Atwood's machine

                    Ma = Mg - T

                     ma = T - mg

Adding those equations,

                   a (M + m) = g ( M - m)

                       a = (M + m) / ( M - m)

Substituting the values,

                       a = (10 + 5) / (10 - 5)

                          = 3 m/s²

The acceleration of the elevator machine is, a = 3 m/s²                

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Temperature doesn't do anything, the boiling point of stuff decreases. If you put water in a vacuum chainber then it will start to boil
5 0
3 years ago
A 6,000N is applied to a formula one car that weighs 500kg. What is the car's acceleration?
Vesna [10]

Answer:

<h2>12 m/s²</h2>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

a =  \frac{f}{m}  \\

f is the force

m is the mass

From the question

f = 6000 N

m = 500 kg

We have

a =  \frac{6000}{500}  =  \frac{60}{5}  = 12 \\

We have the final answer as

<h3>12 m/s²</h3>

Hope this helps you

8 0
3 years ago
A spacecraft is in a circular Earth orbit at an altitude of 6000 km . By how much will its altitude decrease if it moves to a ne
blagie [28]

Answer:

a) 2148 km = 2150 km

b) 840 km

Explanation:

The force keeping the satellite in circular motion is the force given by Netwon's gravitational law

Centripetal force = (mv²/r)

Force due to Newton's law of gravitation = (GMm/r²)

where m = mass of satellite

M = mass of the earth

G = Gravitational constant

v = velocity of the satellite

r = radius of circular orbit

(mv²/r) = (GMm/r²)

v² = (GM/r)

Meaning that the square of the velocity of orbit is inversely proportional to the radius of circular orbit. (Since G and M are constants)

v² = (k/r)

when v = v₀, r = 6000 + 6378 = 12,378 km (the radius of orbit = 6000 km + radius of the earth)

v₀² = (k/12,378)

K = 12378v₀²

When the velocity increases by 10%, v₁ = 1.1v₀, the square of the new velocity = (1.1v₀)² = 1.21v₀² and the new radius of orbit = r₁

1.21v₀² = (k/r₁)

r₁ = (k/1.21v₀²)

Recall, k = 12378v₀²

r₁ = 12378v₀² ÷ 1.21v₀² = 10,229.75 km

10,229.75 km = (10,229.75 - 6378) km altitude above the Earth's surface

New altitude of orbit = 3851.75 km

Decrease in altitude = 6000 - 3851.75 = 2148 km

b) The period of orbit is related to the radius of orbit through Kepler's Law

T² ∝ R³

T² = kR³

When the period of orbit is T₀, Radius of orbit = R₀ = (6000 + 6378) = 12378 km (Earth's radius = 6378 km)

T₀² = kR₀³

T₀² = k(12378)³

k = (T₀²) ÷ (12378)³

When the period reduces by 10%, T₁ = 0.90T₀ and the new radius of orbit = R₁

T₁² = kR₁³

(0.90T₀)² = kR₁³

0.81T₀² = kR₁³

R₁³ = (0.81T₀²) ÷ k

Recall, k = (T₀²)/(12378)³

R₁³ = (0.81T₀²) ÷ [(T₀²)/(12378)³]

R₁³ = 1,536,160,005,663.1

R₁ = ∛(1,536,160,005,663.1) = 11,538.4 km

New Altitude = R₁ - (Radius of the Earth)

= 11,538.4 - 6378 = 5160.4 km

Decrease in altitude = 6000 - 5160.4 = 839.6 km = 840 km

Hope this Helps!!!

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Answer:

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Explanation:

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two forces act at an angle of 135 degree.the forces are F1 = 50N and F2 = 25N respectively.graphically construct a parallelogram
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