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Anuta_ua [19.1K]
3 years ago
14

A car mass of 1.2 x 10 kilograms starts from rest and attains a speed of 20 meters/seconds in 5 seconds. What net force acted on

the car to cause that acceleration
Physics
1 answer:
lorasvet [3.4K]3 years ago
7 0

Net force on the car=F=4.8 x 10³ N

Explanation:

mass of car= 1.2 x 10³ Kg

initial velocity= Vi=0

Final velocity= Vf= 20 m/s

time = t= 5 s

Using kinematic equation,

Vf= Vi + at

20= 0 + a (5)

5 a=20

a= 20/5

a= 4 m/s²

Now force is given by F = ma

F= 1.2 x 10³ (4)

F=4.8 x 10³ N

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mestny [16]

An astronaut inside a spacecraft, which protects her from harmful radiation, is orbiting a black hole at a distance of 120 km from its center. The black hole is 5.00 times the mass of the sun and has a Schwarzschild radius of 15.0 km. The astronaut is positioned inside the spaceship such that one of her 0.030 kg ears is 6.0 cm farther from the black hole than the center of mass of the spacecraft and the other ear is 6.0 cm closer.

What is the tension between her ears?

Would the astronaut find it difficult to keep from being torn apart by the gravitational forces?

Answer:

The tension between the ears = 2.07 KN

The astronaut will find it difficult to keep and will eventually be in trouble because the tension is now greater compared to the tension in the human tissues.

Explanation:

Given that:

Orbital radius of the spacecraft (R) = 120 Km = 120 × 10³ m

Mass of the black hole (m) = 5 \ * (M \ _{sun})

where : M_{sun} = 1.99*10^{33} \ kg

Then; we have:

 m = 5*(1.99*10^{30} \ kg ) \\ = 9.95*10^{30} kg

Schwarzchild radius of the black hole

r - 15.0 km

Mass of each ear m_{ear} = 0.030 \ kg

Farther distance between one ear and the black hole (d) = 6.0 cm

= 0.06 m

Closer distance between the other ear and the black home is (d) 6.0 cm

= 0.6 cm

NOW, If we assume that the tension force should be T; then definitely the two ears will posses the same angular velocity .

The net force on the ear closer to the black hole will be:

\frac{GMm_{ear} }{(R-d)}- T = m_{ear} (R -  d) \omega^2

\frac{GMm_{ear} }{(R-d)^2}- \frac{T}{(R-d)} = m_{ear} \omega^2 \ ----> \ (1)

The net force on the ear farther to the black hole is :

\frac{GMm_{ear} }{(R+d)}- T = m_{ear} (R +  d) \omega^2

\frac{GMm_{ear} }{(R+d)^2}- \frac{T}{(R+d)} = m_{ear} \omega^2 \ ----> \ (2)

Equating equation (1) and (2) & therefore making (T) the subject of the formula; we have:

T = \frac{3GMm_{ear}d}{R^3}

T = \frac{3(6.67*10^{-11}N.m^2/kg^2)(1.95*10^{30}kg)(0.03kg)(0.06m)}{(120*10^3m)^3}

T = 2073.9 N\\T = 2.07 KN

The tension between the ears = 2.07 KN

The astronaut will find it difficult to keep and will eventually be in trouble because the tension is now greater compared to the tension in the human tissues.

3 0
3 years ago
What is the average speed for the entire graph?
Maru [420]

Answer:

Explanation:

Divide the total distance traveled by the total time spent traveling. This will give you your average speed. . So if Ben traveled 150 miles in 3 hours, 120 miles in 2 hours, and 70 miles in 1 hour, his average speed was about 57 mph.

7 0
2 years ago
Thomas and John are carrying a 43kg cylinder head on a 510cm X 510mm board. The cylinder head with dimensions of 43cm X 250mm li
tatyana61 [14]

John carry the heaviest load.

<h3>How to find out who is carrying the heavy load?</h3>

Write down given data from questions:

Board=510cm X 510mm.

Cylinder head with dimensions=43cm X 250mm.

Cylinder lies across the board 210cm from john.

Find out: Who is carry the heaviest load?​

Calculation:

We assume that mass of cylinder head = x kg

Then weight=x x 9*81

                 W=9.81x  Newton.

Weight per unit length= Weight/Total leanth

Weight per unit length= 9.81x/43

(w/l)=0.23x N/cm

From equation contition: (F_{J} +F_{T} =9.81x n)

F_{T} (510)=9.81x  (210+21.5)

F_{T} (510)=9.81 x (231.5)

F_{T} =4.452x N.

F_{J} =9.81x-4.452x

F_{J} =5.358 xN

Therefore F_{J} \geq F_{T}

To learn more about mass per unit length, refer to:

brainly.com/question/24180692

#SPJ9

4 0
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Light incident on a glass sheet is partly reflected and partly refracted. How is the reflected ray different from the refracted
kkurt [141]

Answer:

E. The refracted ray is vertically polarized whereas the reflected ray is horizontally polarized.

Explanation:

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2 years ago
three carts of masses 2 kg, 18 kg, and 9 kg move on a frictionless horizontal track with speeds of 10m/s, 8m/s, and 2m/s. the ca
Elan Coil [88]

Answer:

V = 6.3 m/s

Explanation:

Given:

m₁ = 2 kg

m₂ = 18 kg

m₃ = 9 kg

V₁ = 10 m/s

V₂ = 8 m/s

V₃ = 2 m/s

__________

V - ?

Let us write the momentum conservation law for an inelastic impact:

m₁·V₁ + m₂·V₂ + m₃·V₃ = (m₁ +m₂ + m₃) ·V

Cart speed after interaction:

V = ( m₁·V₁ + m₂·V₂ + m₃·V₃ ) / (m₁ +m₂ + m₃)

V = (2·10 + 18·8 + 9·2) / ( 2 + 18 + 9) = 182 / 29 ≈  6.3 m/s

4 0
1 year ago
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