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neonofarm [45]
3 years ago
9

If the speed of a car is 20m/s .How long does it take to cover a distance of 1km?​

Physics
2 answers:
3241004551 [841]3 years ago
7 0

Answer: 50 seconds

Explanation:

1km=1000m

Distance/speed=time

1000/20=50

50 seconds

Neporo4naja [7]3 years ago
6 0

Explanation:

Hey, there!!

Here,

speed (s)= 20m/s

distance (d)= 1km

= 1000m

now,

we have formula,

t =  \frac{d}{s}

putting value,

t =  \frac{1000m}{20m/ s }

cancelling the like terms,

t = 50s

Therefore, the time is 50s.

<em><u>Hope</u></em><em><u> </u></em><em><u>it helps</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em>

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Natasha_Volkova [10]

Answer:

Part a)

E = \frac{\lambda}{2\pi \epsilon_0 r}

Part b)

E = \frac{\lambda}{2\pi \epsilon_0 r}

Part d)

As we know that due to induction of charge there will be same charge appear on the inner and outer surface of the cylinder but the sign of the charge must be different

On the inner side of the cylinder there will be negative charge induce on the inner surface and on the outer surface of the cylinder there will be same magnitude charge with positive sign.

Explanation:

Part a)

By Guass law we know that

\int E. dA = \frac{q}{\epsilon_0}

E. 2\pi rL = \frac{\lambda L}{\epsilon_0}

E = \frac{\lambda}{2\pi \epsilon_0 r}

Part b)

Outside the outer cylinder we will again use Guass law

\int E. dA = \frac{q}{\epsilon_0}

E. 2\pi rL = \frac{\lambda L}{\epsilon_0}

E = \frac{\lambda}{2\pi \epsilon_0 r}

Part d)

As we know that due to induction of charge there will be same charge appear on the inner and outer surface of the cylinder but the sign of the charge must be different

On the inner side of the cylinder there will be negative charge induce on the inner surface and on the outer surface of the cylinder there will be same magnitude charge with positive sign.

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Without being able to look into the earth , a scientist would not be able to determine if this region has karst topography
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An astronaut floating in space throws a wrench forward with the force of 10 N.
Jobisdone [24]

Answer:

10N

Explanation:

1. Every Action has an equal and opposite reAction.

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(a) Calculate the magnitude of the gravitational force exerted by Mars on a 80 kg human standing on the surface of Mars. (The ma
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Answer:

a) F=1.044\times 10^9\ N

b)F'=1.044\times 10^9\ N

c) F_p=1.0672\times10^{-7}\ N

d) Treat the humans as though they were points or uniform-density spheres.

Explanation:

Given:

  • mass of Mars, M=6.4\times 10^{23}\ kg
  • radius of the Mars, r=3.4\times 10^{6}\ m
  • mass of human, m=80\ kg

a)

Gravitation force exerted by the Mars on the human body:

F=G.\frac{M.m}{r^2}

where:

G=6.67 \times 10^{-11}\ m^3.kg^{-1}.s^{-2} = gravitational constant

F=6.67\times10^{-11}\times \frac{6.4\times 10^{23}\times 80}{(3.4\times 10^{6})^2}

F=1.044\times 10^9\ N

b)

The magnitude of the gravitational force exerted by the human on Mars is equal to the force by the Mars on human.

F'=F

F'=1.044\times 10^9\ N

c)

When a similar person of the same mass is standing at a distance of 4 meters:

F_p=6.67\times10^{-11}\times \frac{80\times 80}{4}

F_p=1.0672\times10^{-7}\ N

d)

The gravitational constant is a universal value and it remains constant in the Universe and does not depends on the size of the mass.

  • Yes, we have to treat Mars as spherically symmetric so that its center of mass is at its geometric center.
  • Yes, we also have to ignore the effect of sun, but as asked in the question we have to calculate the gravitational force only due to one body on another specific body which does not brings sun into picture of the consideration.
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Answer:

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