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alisha [4.7K]
3 years ago
12

If all else stays the same, which would cause an increase in the gravitational force on a space shuttle?

Physics
2 answers:
kirill [66]3 years ago
6 0

Answer:

decreasing the distance of the space shuttle from Earth

increasing the universal gravitational constant acting on the space shuttle*

Explanation:

The magnitude of the gravitational force acting between the Earth and the shuttle is given by:

F=G\frac{Mm}{r^2}

where

G is the gravitational constant

M is the mass of the Earth

m is the mass of the shuttle

r is the distance between the shuttle and the Earth's center

By looking at the formula, we see check the effect of each statement given:

- increasing the distance of the space shuttle from Earth  --> FALSE: this means increasing r in the formula, so the gravitational force would decrease

- decreasing the distance of the space shuttle from Earth  --> TRUE: this means decreasing r in the formula, so the gravitational force would increase

- increasing the universal gravitational constant acting on the space shuttle  --> TRUE: this means increasing G in the formula, so the gravitational force would increase. *However, technically is not possible to change the value of G, since it is a universal constant and it has the same value everywhere.

- decreasing the universal gravitational constant acting on the space shuttle: FALSE --> Decreasing G in the formula would decrease the magnitude of the force.

maks197457 [2]3 years ago
5 0
"decreasing the distance of the space shuttle from Earth" 

F = Gm(1)m(2)/R²
where R is the distance between the 2 objects, as it decreases, the force increases. 

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If transpiration didn't take place, would water be able to enter the roots of the plant? Explain your reasoning
s2008m [1.1K]

Answer:

If transpiration didn't take place water would still be able to enter the roots of a plant

Explanation:

transpiration is the process of water leaving from living organisms to the atmosphere, therefore, if transpiration didn't occur the water would not transpire to the atmosphere and would remain in the root but water absorption would not change because it is a biological need for the living organism as such

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3 years ago
An oceanographer is studying how the ion concentration in seawater depends on depth. She makes a measurement by lowering into th
Black_prince [1.1K]

Answer:

a)  R = ρ₀ L /π(r_b² - R_a²) , b)  ρ₀ = V / I    π (r_b² - R_a²) / L

Explanation:

a) The resistance of a material is given by

          R = ρ l / A

where ρ is the resistivity, l is the length and A is the area

the length is l = L and the resistivity is ρ = ρ₀

the area is the area of ​​the cylindrical shell

           A = π r_b² - π r_a²

           A = π (r_b² - r_a²)

we substitute

         R = ρ₀ L /π(r_b² - R_a²)

b) The potential difference is related to current and resistance by ohm's law

         V = i R

         

we subsist the expression of resistance

          V = I ρ₀ L /π (r_b² - R_a²)

           ρ₀ = V / I    π (r_b² - R_a²) / L

6 0
3 years ago
The boundary between the crust and mantle is marked by a seismic-velocity discontinuity called
Minchanka [31]

The boundary between the crust and mantle is marked by a seismic-velocity discontinuity is called Mohorovicic discontinuity.

Mohorovicic discontinuity was discovered by Andrija Mohorovicic in 1909 who was a Croatian seismologist. He realized that the velocity of a seismic wave is related to the material's density where it is moving through. He decoded that the acceleration of the seismic waves that are observed within outer shell of the earth is a compositional change. Thus, the acceleration should be caused by a material of higher density.

7 0
3 years ago
Read 2 more answers
Show that Ns is the same as kgmls.<br>​
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1 N = 1kgm/s²

then,

1Ns = 1kgm/s² * s

1Ns = 1kgm/s

Hence shown.

3 0
2 years ago
The center of a moon of mass m is a distance D from the center of a planet of mass M. At some distance x from the center of the
Eddi Din [679]

Answer:

x = D (M/M-m) 2.41

Explanation:

a) Let's apply Newton's second law to find the summation of force, where each force is given by the law of universal gravitation

        F = g m₁m₂ / r²

        Σ F = 0

       F1- F2 = 0

       F1 = F2

We set the reference system in the body of greatest mass (M) the planet

       F1 = g m₁ M / x²

       F2 = G m1 m / (D-x)²

      G m₁ M / x² = G m₁ m / (D-x)²

      M (D-x)² = m x²

      MD² -2MD x + M x² = m x²

     x² (M-m) -2MD x + MD² = 0

We solve the second degree equation

     x = [2MD  ±√ (4M²D² - 4 (M-m) MD²)] / 2 (M-m)

     x = {2MD ± 2D √ (M² + (M-m) M)} / 2 (M-m)

     x = D {M  ±  Ra (2M²-mM)} / (M-m)

    x = D (M ± M √ (2-m/M)) / (M-m)

    x = D (M / (M-m)) (1 ±√ (2-m/M)

Let's analyze this result, the value of M-m >> 1, so if we take the negative root, the value of x would be negative, it is out of the point between the two bodies, so the correct result must be taken with the positive root

 

    x = D (M / (M-m)) (1 + √2)

     x = D (M/M-m) 2.41

b) X = 2/3 D

     x = D (M/M-m) 2.41

     2/3 D = D (M/(M-m)) 2.41

     2/3 (M-m) = M 2.41

     2/3 M - 2/3 m = 2.41 M

     1.743 M = 0.667 m

     M/m = 0.667/1.743

     M/m =  0.38

3 0
3 years ago
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