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oksian1 [2.3K]
3 years ago
12

The Sojourner Mars rover has a weight of 42.7 N on Mars where the acceleration due to gravity is 3.72 m/s2. What is Sojourner's

mass on mars? What is Sojourner's mass on Earth? What is Sojourner's weight on Earth?
Physics
1 answer:
yulyashka [42]3 years ago
3 0

1) 11.5 kg

2) 11.5 kg

3) 112.7 N

Explanation:

1)

The relationship between the mass of an object and its weight is given by the relationship:

W=mg

where:

W is the weight

m is the mass of the object

g is the acceleration due to gravity (the strength of the gravitational field at the location of the object)

In this problem, the object is on Mars.

Its weight is:

W = 42.7 N

While the acceleration due to gravity is:

g=3.72 m/s^2

Therefore, we can re-arrange the equation to find the mass of the rover:

m=\frac{W}{g}=\frac{42.7}{3.72}=11.5 kg

2)

Here we want to find the mass of the rover on Earth.

We have to identify the main difference between mass and weight:

- The mass of an object is an intrinsec property of the object, and it describes the "amount of matter" contained in the object

- The weight of an object is the force of gravity acting on the object itself

From the definitions above, it is clear that:

- The mass of an object is independent from its location, therefore it does not change if the object moves to another planet

- The weight of an object depends on its location, as it depends on the value of g, the acceleration due to gravity

Therefore, the mass of the rover on the Earth is the same as the mass of the rover on Mars:

m = 11.5 kg

3)

The weight of an object, as stated before, is the force of gravity acting on the object, and it is given by

W=mg

where

m is the mass of the object

g is the acceleration  due to gravity

On the surface of the Earth, the value of the acceleration due to gravity is

g=9.8 m/s^2

While the mass of the rover is

m = 11.5 kg

Therefore, the weight of the rover on Earth is

W=(11.5)(9.8)=112.7 N

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You get circular motion, where the acceleration is pointing towards the center of the circle, as long as they are constant, and not fluctuating.
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If you walk 1.2 km north and then 1.6 km east, what are the magnitude and direction of your resultant displacement?
vodomira [7]

<u>Answer:</u>

 Option A is the correct answer.

<u>Explanation:</u>

  Let the east point towards positive X-axis and north point towards positive Y-axis.

 First walking 1.2 km north,  displacement = 1.2 j km

 Secondly 1.6 km east, displacement = 1.6 i km

 Total displacement = (1.6 i + 1.2 j) km

 Magnitude = \sqrt{1.2^2+1.6^2} = 2 km

 Angle of resultant with positive X - axis = tan^{-1}(1.2/1.6)=36.87^0 = 36.87⁰ east of north.

 

5 0
3 years ago
Two 13 cm -long thin glass rods uniformly charged to +11nC are placed side by side, 4.0 cm apart. What are the electric field st
DedPeter [7]

Answer:

E1  = 10.15 * 10^4 N/C

E2 = 0

E3 = 10.15 *10^4 N/C

Explanation:

Given data:

Two 13 cm-long thin glass rods ( L ) = 0.13 m

charge (Q)  = +11nC

distance between thin glass rods   = 4 cm .

<u>Calculate the electric field strengths </u>

electric charge due to a single glass rod in the question ( E ) = \frac{Q}{2\pi e_{0}rL }

equation 1 can be used to determine E1, E2 and E3 because the points lie within the two rods hence the net electric field produced will be equal to the difference in electric fields produced

applying equation 1 to determine E1

E1 = \frac{Q}{2\pi e_{0}rL } ( \frac{1}{0.01} - \frac{1}{0.03} )    ( distance from 1 rod is 0.01 m and from the other rod is 0.03 )

   = \frac{11*10^{-9} }{2*3.14*8.85*10^{-12}*0.13 } ( 66.67 )

   = 10.15 * 10^4 N/C

applying equation 1 to determine E2

E2 = \frac{Q}{2\pi e_{0}rL }( \frac{1}{0.02} - \frac{1}{0.02} )

therefore E2 = 0

E1 = E3

hence E3 = 10.15*10^4 N/C

4 0
3 years ago
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Zina [86]
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6 0
2 years ago
QUESTION 10
Elena L [17]

The maximum value of θ of such the ropes (with a maximum tension of 5,479 N) will be able to support the beam without snapping is:

\theta =37.01^{\circ}

We can apply the first Newton's law in x and y-direction.

If we do a free body diagram of the system we will have:

x-direction

All the forces acting in this direction are:

T_{1}sin(\theta)-T_{2}sin(\theta)=0    (1)

Where:

  • T(1) is the tension due to the rope 1
  • T(2) is the tension due to the rope 2

Here we just conclude that T(1) = T(2)

y-direction

The forces in this direction are:

T_{1}cos(\theta)+T_{2}cos(\theta)-W=0   (2)

Here W is the weight of the steel beam.

We equal it to zero because we need to find the maximum angle at which the ropes will be able to support the beam without snapping.

Knowing that T(1) = T(2) and W = mg, we have:

T_{1}cos(\theta)+T_{1}cos(\theta)-m_{steel}g=0

2T_{1}cos(\theta)-m_{steel}g=0

2T_{1}cos(\theta)=m_{steel}g

T(1) must be equal to 5479 N, so we have:

cos(\theta)=\frac{m_{steel}g}{2T_{1}}

cos(\theta)=\frac{892*9.81}{2*5479}

cos(\theta)=\frac{892*9.81}{2*5479}

cos(\theta)=0.80

Therefore, the maximum angle allowed is θ = 37.01°.

You can learn more about tension here:

brainly.com/question/12797227

I hope it helps you!

8 0
3 years ago
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