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neonofarm [45]
3 years ago
6

Will is a scientist. He’s designing a spacecraft that would allow people to land on Mars. Will’s mass on Earth is 75 kilograms.

Will knows that the gravitational pull of Mars is less than the gravitational pull of Earth. When he lands on Mars, his mass will be (Less than, Equal to or More than) 75 kilograms. His weight on Mars will be (less than, Equal to, or more than) his weight on Earth.
Physics
2 answers:
s2008m [1.1K]3 years ago
8 0

Mass is the amount of substance present inside the body.

As per Einstein's relativity the mass of the body increases with the increase of the speed of the body.

Mathematically it is given as m=\frac{m_{0} }{\sqrt{1-\frac{v^2}{c^2} } }

where m is the mass of the body during motion and m_{0} is the rest mass.

Here V is the velocity of the body and c is the speed of light.

But during non-relativistic situation i.e the velocity of the body is very very less as compared to the velocity of the light or equal to zero, the mass of the substance remains unchanged.

Hence mass of Will will be the same both in Mars and Earth i.e 75 kg

Again the weight of a body is mathematically given as -

                             weight= mass×acceleration due to gravity

                           i.e W= m×g

The value of g is less in Mars as compared to Earth.

Hence the weight of Will on Mars is less as compared to weight on the Earth.

Vesnalui [34]3 years ago
6 0
Mass will remain unchanged, always.  His weight, which is the gravitational force acting on that mass will be less in this case.
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In the periodic table, elements are arranged in order of increasing atomic number into groups and periods.

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If I pick two representative elements in the same period, their number of valence electrons will differ by 1 but they will all be accommodated in the same energy level.

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7 0
3 years ago
Find the useful power output (in W) of an elevator motor that lifts a 2600 kg load a height of 30.0 m in 12.0 s, if it also incr
Annette [7]

Answer:

P = 251,916.667 W

Cost = 2,267.25 cents

Explanation:

To solve this question we will use the Work Energy Theorem, which is

W = dP + dK\\

Where

dP = Change in Potential Energy

dK = Change in Kinetic Energy

Change in Potential Energy

P_{i} = mgh_{i}\\  P_{f} = mgh_{f}

Where

P_{i} = Initial Potential Energy

P_{f} = Final Potential Energy

m = Mass of System = 10,000 kg

g = Acceleration due to gravity = 9.81 m/s

h_{i} = Initial Height = 0

h_{f} = Final Height = 30 m

Inputting the values we get the answer for dP

dP = P_{f} - P_{i}\\dP= mgh_{f} - mgh_{i}\\ dP= 10000(9.81)(30) - 0\\ dP= 2943000

Change in Kinetic Energy

K_{i} = \frac{1}{2} mv_{i} ^2\\ K_{f} = \frac{1}{2} mv_{f} ^2

Where

K_{i} = Initial Kinetic Energy

K_{f} = Final Kinetic Energy

m = Mass of System = 10,000 kg

g = Acceleration due to gravity = 9.81 m/s

v_{i} = Initial Velocity = 0 m/2

v_{f} = Final Velocity = 4 m/s

Inputting the values we get the answer for dK

dK = K_{f} - K_{i}\\ dK = \frac{1}{2} mv_{f} ^2 - \frac{1}{2} mv_{i} ^2\\ dK = \frac{1}{2} (10000)(4)^2 - 0 \\ dK = 80000

Total Work

W = dP + dK\\

Inputting the values

W = 2943000 + 80000

W = 3,023,000

a) Finding the useful Power Output

P = \frac{W}{t}

Where

P = Power Output

W = Work Done = 3,023,000J

t = Time = 12s

Inputting the values

P = \frac{3,023,000}{12}\\ P = 251,916.667

P = 251,916.667 W

b) Finding the Total Cost

Cost = $0.0900 x P/1000

Cost = $0.0900 x (251,916.667/1000)

Cost = $22.67 or 2,267.25 cents

4 0
3 years ago
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uysha [10]

Answer:

<h3>U_{2}=4U_{1}</h3><h3>Explanation:</h3>

Let U1 and U2 be the gravitational potential energy of the first and second stone  respectively.

U1 is given by:

U_1 = (m)gh\\

and U2 is given by:

U_2 = (4m)gh  

therefore, comparing the energies we have:

\frac{U_2}{U_1} = \frac{(4m)gh}{(m)gh}= 4

therefore:

U_2=4U_1

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