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Alex Ar [27]
3 years ago
15

a pine raft (density = 373 kg/m^3) has a volume of 1.43 m^3. How much of the raft's volume is below the water line? Unit is m^3​

Physics
1 answer:
Slav-nsk [51]3 years ago
7 0

Answer: 0.5334

Explanation:

i got it right on accellus :p

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A 0.85 N force exists between a 7.1 * 10 ^ - 6 * C charge 5.4 m away. What is the magnitude of the second charge ? Please show w
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Answer:

Explanation:

Force between charge is given by the following expression

F = k Q₁ Q₂ / R² , k = 9 x 10⁹ , Q₁ and Q₂ are charges , R is distance between charges .

Putting the given values ,

.85  = 9 x 10⁹ x 7.1 x 10⁻⁶ x Q₂ / 5.4²

Q₂ = .85 x  5.4² / (9 x 10⁹ x 7.1 x 10⁻⁶ )

= .38788  x 10⁻³ C .

= 387.88 x 10⁻⁶ C .

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The magnitude of Earth’s magnetic field is about 0.5 gauss near Earth’s surface. What’s the maximum possible magnetic force on a
vampirchik [111]

Answer:

F = 1.5 \times 10^{-16} N

this force is 1.68 \times 10^{13} times more than the gravitational force

Explanation:

Kinetic Energy of the electron is given as

KE = 1 keV

KE = 1 \times 10^3 (1.6 \times 10^{-19}) J

KE = 1.6 \times 10^{-16} J

now the speed of electron is given as

KE = \frac{1}{2}mv^2

now we have

v = \sqrt{\frac{2 KE}{m}}

v = 1.87 \times 10^7 m/s

now the maximum force due to magnetic field is given as

F = qvB

F = (1.6\times 10^{-19})(1.87 \times 10^7)(0.5 \times 10^{-4})

F = 1.5 \times 10^{-16} N

Now if this force is compared by the gravitational force on the electron then it is

\frac{F}{F_g} = \frac{1.5 \times 10^{-16}}{9.1 \times 10^{-31} (9.8)}

\frac{F}{F_g} = 1.68 \times 10^{13}

so this force is 1.68 \times 10^{13} times more than the gravitational force

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What is the effect on the force of gravity between two objects if the mass of one object remains unchanged while the distance to
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Answer:

The correct answer to the question is

B. It always decreases

Explanation:

To solve the question, we note that the foce of gravity is given by

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G= Gravitational constant

m₁ = mass of first object

m₂ = mass of second object

r = the distance between both objects

If the mass of one object remains unchanged while the distance to the second object and the second object’s mass are both doubled, we have

F_{G2} =\frac{Gm_1(2m_2)}{(2r)^2} = \frac{2}{4} \frac{Gm_1m_2}{r^2}

Therefore the gravitational force is halved. That is it will always decrease

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