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adelina 88 [10]
3 years ago
12

A person of mass m is standing on the surface of the Earth, of mass M E . What is the acceleration that the Earth experiences du

e to the person's gravitational pull?
Physics
1 answer:
Lana71 [14]3 years ago
7 0

Answer:

a_E=\dfrac{Gm}{r^2}

Explanation:

M_E = Mass of the Earth =  5.972 × 10²⁴ kg

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

r = Radius of Earth = 6371000 m

m = Mass of person

The force on the person will balance the gravitational force

M_Ea_E=\dfrac{GmM_E}{r^2}\\\Rightarrow a_E=\dfrac{Gm}{r^2}

The acceleration that the Earth will feel is a_E=\dfrac{Gm}{r^2}

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D, the lithosphere. (CRUST AND UPPER MANTLE)

Explanation:

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You need to make a very dilute solution of sodium phosphate dihydrate (MW 142 g/mol, FW 178 g/mol). How many grams of sodium pho
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Answer:

first question is 178 g of sodium phosphate dihydrate. second question is 0.25 mL

Explanation:

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A car with a mass of 1380 Kg is traveling at 23 m/s to the north. A truck with a mass of 1625 Kg is traveling at 26 m/s to the s
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Answer: -3.49 m/s (to the south)

Explanation:

This problem can be solved by the Conservation of Momentum principle which establishes the initial momentum p_{i} must be equal to the final momentum p_{f}, and taking into account this is aninelastic collision:

Before the collision:

p_{i}=mV_{o}+MU_{o} (1)

After the collision:

p_{f}=(m+M)V_{f} (2)

Where:

m=1380 kg is the mass of the car

V_{o}=23 m/s is the velocity of the car, directed to the north

M=1625 kg is the mass of the truck

U_{o}=-26 m/s is the velocity of the truck, directed to the south

V_{f} is the final velocity of both the car and the truck

p_{i}=p_{f} (3)

mV_{o}+MU_{o}=(m+M)V_{f} (4)

Isolating V_{f}:

V_{f}=\frac{mV_{o}+MU_{o}}{m+M} (5)

V_{f}=\frac{(1380 kg)(23 m/s)+(1625 kg)(-26 m/s)}{1380 kg+1625 kg} (6)

Finally:

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