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3 years ago

Compare the gravitational force on a 4 kg

Physics

1 answer:

Eddi Din [679]3 years ago

6 0

Answer:

The gravitational force by the Earth on the object, and by the object on the Earth is

F = GMm/R²

= 6.674×10−11 m3⋅kg−1⋅s−2 × 6 × 10^24 kg × 44.5 kg/(6.4 × 10^6 m)²

= 435 N

Please note that the ration between the gravitation force 435 and the mass 44.5

should be gravitational acceleration

435/44.5 = 9.78

I attribute the discrepancy between 9.78 and the usual 9.81 to rounding off in the

Earth's weight and radius.

The mass of the Moon is M / 81.3.

The radius of the Moon is R × 0.27.

The gravitational force on the moon would be

G(M/81.3)m/(R×0.27)² = 0.17×GMm/R²

The gravitational force on the moon is smaller by the factor of about 0.17.

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In the circuit seen here, the resistor has a resistance of 3 ohms. If no change in the battery size occurs, what will happen to

Bumek [7]

The current will decrease as the resistance has now increased, meaning less current will be 'let through' the resistor. (assuming it's in series, there's no image)

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4 years ago

Read 2 more answers

An isolated conducting sphere has a 17 cm radius. One wire carries a current of 1.0000020 A into it. Another wire carries a curr

notsponge [240]

14 ms is required to reach the potential of 1500 V.

Explanation:

The current is measured as the amount of charge traveling per unit time. So the charge of electrons required for each current is determined as the product of current with time.

$Charge = Current \times Time$

As two different current is passing at two different times, the net charge will be the different in current. So,

$\text { Charge }=(1.0000020-1.0000000) \times t=2 \times 10^{-6} \times t$

The electric voltage on the surface of cylinder can be obtained as the ratio of charge to the radius of the cylinder.

$V=\frac{k q}{R}$

Here $k = 9 * 10^9$ , q is the charge and R is the radius. As $q=2 \times 10^{-6} \times t$ and R =17 cm = 0.17 m, then the voltage will be

$V=\frac{9 \times 10^{9} \times 2 \times 10^{-6} \times t}{0.17}$

The time is required to find to reach the voltage of 1500 V, so

$1500 =\frac{9 \times 10^{9} \times 2 \times 10^{-6} \times t}{0.17}$

$\begin{aligned}&t=\frac{1500 \times 0.17}{\left(9 \times 10^{9} \times 2 \times 10^{-6}\right)}\\&t=14.1666 \times 10^{-3} s=14\ \mathrm{ms}\end{aligned}$

So, 14 ms is required to reach the potential of 1500 V.

3 0

3 years ago

A 12.0 g sample of gas occupies 19.2 L at STP. what is the of moles and molecular weight of this gas?

lubasha [3.4K]

At STP, 1 mole of an ideal gas occupies a volume of about 22.4 L. So if n is the number of moles of this gas, then

n / (19.2 L) = (1 mole) / (22.4 L) ==> n = (19.2 L•mole) / (22.4 L) ≈ 0.857 mol

If the sample has a mass of 12.0 g, then its molecular weight is

(12.0 g) / n ≈ 14.0 g/mol

4 0

3 years ago

Tenses of write= read=

PilotLPTM [1.2K]

Answer:

present

Explanation:

read doesn't change but write is in present tense

7 0

3 years ago

A spring with spring constant of 26 N/m is stretched 0.22 m from its equilibrium position. How much work must be done to stretch

Evgen [1.6K]

Answer:

1.503 J

Explanation:

Work done in stretching a spring = 1/2ke²

W = 1/2ke²........................... Equation 1

Where W = work done, k = spring constant, e = extension.

Given: k = 26 N/m, e = (0.22+0.12), = 0.34 m.

Substitute into equation 1

W = 1/2(26)(0.34²)

W = 13(0.1156)

W = 1.503 J.

Hence the work done to stretch it an additional 0.12 m = 1.503 J

8 0

3 years ago