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coldgirl [10]
3 years ago
14

At the surface of Mars the acceleration due to gravity is 3.8m/s . A book weighs 34 N at the surface of the earth. What is it’s

mass on the earth’s surface? What are it’s mass and weight on mar’s surface?
Physics
1 answer:
PilotLPTM [1.2K]3 years ago
8 0

Answer:

On the surface of Mars:

Mass, m = 3.47 kg

Weight, W' = 13.186 N

Solution:

As per the question:

Acceleration due to gravity at the surface of Mars, g' = 3.8\ m/s^{2}

Weight of book on the Earth's surface, W = 34 N

Now,

We know that:

Weight, W = mg

where

m = mass of the body

a = acceleration due to gravity on the earth's surface = 9.8\ m/s^{2}

Thus the mass of the body on the surface of the earth, m:

m = \frac{W}{g} = \frac{34}{9.8} = 3.47\ kg

<em>Also, we know that mass is constant and it is the weight of the body that varies with the acceleration due to gravity.</em>

Now,

Mass of the body on the surface of Mars will be same and is equal to m = 3.47 kg

Weight of the book on the surface of Mars, W' = mg'

W' = 3.47\times 3.8 = 13.186\ N

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a_sh-v [17]

Answer:

a) 35.94 ms⁻²

b) 65.85 m

Explanation:

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ρ = 1000kg/m3

a) First, we need to establish the total pressure of the water in the tank. Note the that the tanks is closed. It means that the total pressure, Ptot,  at the bottom of the tank is the sum of the pressure of the water plus the air trapped between the tank rook and water. In other words:

Ptot = Pgas + Pwater

However, the air is the one influencing the water to move, so elimininating Pwater the equation becomes:

Ptot  = Pgas

        = 6.46 × 10⁵ Pa

The change in pressure is given by the continuity equation:

ΔP = 1/2ρv²

where v is the velocity of the water as it exits the tank.

Calculating:

6.46 × 10⁵  =1/2 ×1000×v²

solving for v, we get v = 35.94 ms⁻²

b) The Bernoulli's equation will be applicable here.

The water is coming out with the same pressure, therefore, the equation will be:

ΔP = ρgh

6.46 × 10⁵  = 1000 x 9.81 x h

h = 65.85 meters

7 0
3 years ago
You test a moon buggy on Earth. When the buggy hits a bump, it oscillates up and down on its springs with a period of 4 seconds.
Blizzard [7]

Answer:

Remains same

Explanation:

T = Time period of oscillation

m = mass

k = spring constant

Time period of oscillation is given as

T = 2\pi \sqrt{\frac{m}{k} }

we know that as we move from earth to moon, the value of spring constant "k"  and mass "m" remains unchanged because they do not depend on the acceleration due to gravity.

Time period depends on spring constant inversely and directly on the mass.

hence the time period remains the same.

3 0
4 years ago
A proton and an electron are released from rest in the center of a capacitor.
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Answer:

a)  equal 1, b) less than 1

Explanation:

a) the electric force is given by

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The charge of the electron and proton has the same value, that of the proton is positive and that of the electron is negative

Proton

        Fp = qE

Electron

         Fe = - q E

         Fp / Fe = -1

If we do not take into account the sign the relationship is equal to one (1)

b) to calculate the force we use Newton's second law

           F = ma

           qE = m a

           a = q E / m

The mass of the proton much greater than the mass of the electron

          ap = q E / m_{p}

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          ap / ae =  m_{e} /  m_{p} =  m_{e}/1600  m_{e} =1/1600

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7 0
3 years ago
The driver of a 1000 kg car traveling on a highway at 35 m-s'applies brake to avoid hitting a van in front of him, which had com
puteri [66]

Answer:

Explanation:

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d = mv²/2F = 1000(35²) / (2(8000)) = 76.5625 ≈ 77 m

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6 0
3 years ago
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natita [175]

Answer:

d=20m/sx60s=1200m=1200/1000Km=1.2km

Explanation:

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