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Dmitriy789 [7]
3 years ago
15

The acceleration due to gravity on Mars is less than that on Earth. On Mars, a person will weigh than on Earth.

Physics
2 answers:
pantera1 [17]3 years ago
4 0
On mars people would way less. 
An example of this is that if I weighed 700 pounds (I don't by the way) I would then weigh 500 pounds or less.

Vlad [161]3 years ago
4 0

Answer:  On Mars, a person will weigh <u>less</u> than on Earth.

Explanation:

The mass of the planet Mars is less than Earth. Because of this it pulls the objects on it with less force and thus Mars has less value of acceleration due to gravity.

The weight of the object is given by the product of mass and acceleration due to gravity.

W = m g'

W ∝ g'

The mass remains constant irrespective of the planet but weight varies with the value of acceleration due to gravity. Thus, on Mars, a person will weigh less than on Earth.

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Answer:

2. ( b ) zero

3. ( c ) 10 s

4. Uniform then decreasing

Explanation:

2.

Since the motion is uniform, initial and final velocity will be 0, hence acceleration will be zero.

3.

Initial velocity ( u ) = 5 m/s

Final velocity ( v ) = 35 m/s

Acceleration ( a ) 3 m/s^2

To find : Time ( t )

Formula : -

t = v - u / a

 = 35 - 5 / 3

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t = 10 s

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2 years ago
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What is the strength of electric field EpEp 0.60 mmmm from a proton? Express your answer to two significant figures and include
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Answer:

3.99*10^-3N/C

Explanation:

Using

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Where r = 0.6mm = 0.6*10^-3m

K= 8.9*10^9 and q= 1.6*10^-19

So = 8.9*10^9 * 1.6*10^-19/0.6*10^-3)²

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If in 30 minutes Allex runs to a store that is 4 km away, what is her average speed in km/h? km/h
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Im pretty sure the answer is 8km/h
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Two metal spheres of different sizes are charged such that the electric potential is the same at the surface of each. Sphere A h
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Answer:

Explanation:

Given

Radius of A is twice of B i.e.

R_A=2R_B

Also Potential of both sphere is same

V_A=V_B

V=\frac{kQ}{R}

thus

k\frac{Q_A}{R_A}=K\frac{Q_B}{R_B}

\frac{Q_A}{Q_B}=\frac{R_A}{R_B}

\frac{Q_A}{Q_B}=\frac{2}{1}=2

\frac{Q_B}{Q_A}=\frac{1}{2}

(b)Ratio of \frac{E_B}{E_A}

Electric Field is given by E=\frac{kQ}{R^2}

thus E_A=\frac{kQ_A}{R_A^2}----1

E_B=\frac{kQ_B}{R_B^2}----2

Divide 2 by 1

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\frac{E_B}{E_A}=\frac{1}{2}\times 4=2

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