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photoshop1234 [79]
3 years ago
8

A baseball has a mass of 0.145 kilograms, and a bowling ball has a mass of 6.8 kilograms. What is the gravitational force betwee

n them if their centers are 0.5 m apart?
1.0 × 10–13 m
1.3 × 10–10 m
2.6 × 10–10 m
1.1 × 10–8 m
Physics
2 answers:
larisa [96]3 years ago
8 0
We can use the equation for Newton's Law of Gravitation
Fg = (Gm₁m₂)/r²
Where gravitational constant = G = 6.674 x 10⁻¹¹ N · m²/kg²
mass m₁ = 0.145 kg
mass m₂ = 6.8 kg
distance between centers of masses = r = 0.5 m

Substitute these values into... 
Fg = (Gm₁m₂)/r²
Fg = ((6.674 x 10⁻¹¹)(0.145)(6.8)) / (0.5)²
Fg = 2.63 x 10⁻¹⁰ N 

Therefore, your answer should be 
<span>2.6 × 10–10</span>
Brums [2.3K]3 years ago
8 0

Answer:

C: 2.6 × 10^–10 N

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adelina 88 [10]

Answer:

<h2>3,343.68kJ </h2>

Explanation:

Heat energy used up can be calculated using the formula:

H = mcΔt

m = mass oof the object (in kg) = 20kg

c = specific heat capacity of water = 4179.6J/kg°C

Δt change in temperature = 80-40 = 40°C

H= 20* 4179.6 * 40

H = 3,343,680Joules

H = 3,343.68kJ

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If this energy were used to vaporize water at 100.0 ∘C, how much water (in liters) could be vaporized? The enthalpy of vaporizat
Zanzabum

Answer:

0.429 L of water

Explanation:

First to all, you are not putting the value of the energy given to vaporize water, so, to explain better this problem, I will assume a value of energy that I took in a similar exercise before, which is 970 kJ.

Now, assuming that the water density is 1 g/mL, this is the same as saying that 1 g of water = 1 mL of water

If this is true, then, we can assume that 1 kg of water = 1 L of water.

Knowing this, we have to use the expression to get energy which is:

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Solving for m:

m = Q / ΔH

Now "m" is the mass, but in this case, the mass of water is the same as the volume, so it's not neccesary to do a unit conversion.

Before we begin with the calculation, we need to put the enthalpy of vaporization in the correct units, which would be in grams. To do that, we need the molar mass of water:

MM = 18 g/mol

The enthalpy in mass:

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3 years ago
A long, East-West-oriented power cable carrying an
Alla [95]

Answer:

200A

Explanation:

Given that

the distance between earth surface and power cable d = 8m

when the current is flowing through cable , the magnitude flux density at the surface is 15μT

when the current flow throught is zero the magnitude flux density at the surface is 20μT

The change in flux density due to the current flowing in the power cable is

B = 20μT - 15μT

B =5μT -----(1)

The expression of magnitude flux density produced by the current carrying cable is

B=\frac{\mu_0I}{2\pi d}-----(2)

Substitute the value of flux density

B from eqn 1 and eqn 2

\frac{\mu_0I}{2\pi d}=5\times 10^-^6\\\\\frac{(4\pi \times 10^-^7)I}{2 \pi (8)} =5\times 10^-^6\\\\I=200A

Therefore, the magnitude of current I is 200A

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storchak [24]

Answer:

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7 0
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