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miv72 [106K]
3 years ago
7

The gravitational attraction between two masses of 3kg that are separated by a distance of 1cm is

Physics
1 answer:
Luba_88 [7]3 years ago
6 0

Answer:

6.003×10¯⁶ N

Explanation:

We'll begin by converting 1 cm to m. This can be obtained as follow:

100 cm = 1 m

Therefore,

1 cm = 1 cm × 1 m / 100 cm

1 cm = 0.01 m

Finally, we shall determine the gravitational attraction. This can be obtained as follow:

Mass 1 (M₁) = 3 Kg

Mass 2 (M₂) = 3 Kg

Distance apart (r) = 0.01 m

Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²

Force of attraction (F) =?

F = GM₁M₂ / r²

F = 6.67×10¯¹¹ × 3 × / 0.01²

F = 6.003×10¯¹⁰ / 1×10¯⁴

F = 6.003×10¯⁶ N

Thus the gravitational attraction is 6.003×10¯⁶ N

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A ball A of mass 0.5 kg moving with a Velacity of 10 m/s a head on Collision with a ball B of mass 2kg moving with a Velocity of
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The common velocity v after collision is 2.8m/s²

Explanation:

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2 years ago
A spherical snowball melts at a rate proportional to its surface area. (a) write a differential equation for its volume, v. (use
tresset_1 [31]

Answer:

Explanation:

The rate of change in volume is proportional to the surface area:

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V = kAt + C

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6 0
3 years ago
If the magnitude of the electric field in air exceeds roughly 3 ✕ 106 n/c, the air breaks down and a spark forms. for a two-di
Vlad1618 [11]

Answer: 39.8 μC

Explanation:

The magnitude of the electric field generated by a capacitor is given by:

E = \frac{V}{d}

d is the distance between the plates.

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C =\frac{\epsilon_o A }{d}

where A is the area of the plate and ε₀ is the absolute permittivity.

substituting, we get

E = \frac{Q}{\epsilon_o A}

It is given that the magnitude of the electric field that can exist in the capacitor before air breaks down is, E = 3 × 10⁶ N/C.

radius of the plates of the capacitor, r = 69 cm = 0.69 m

Area of the plates, A = πr² = 1.5 m²

Thus, the maximum charge that can be placed on disks without a spark is:

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8 0
3 years ago
Two taut strings of identical mass and length are stretched with their ends fixed, but the tension in one string is 1.10 times g
ollegr [7]

Answer:

The  beat frequency when each string is vibrating at its fundamental frequency is 12.6 Hz

Explanation:

Given;

velocity of wave on the string with lower tension, v₁ = 35.2 m/s

the fundamental frequency of the string, F₁ = 258 Hz

<u>velocity of wave on the string with greater tension;</u>

v_1 = \sqrt{\frac{T_1}{\mu }

where;

v₁ is the velocity of wave on the string with lower tension

T₁ is tension on the string

μ is mass per unit length

v_1 = \sqrt{\frac{T_1}{\mu} } \\\\v_1^2 = \frac{T_1}{\mu} \\\\\mu = \frac{T_1}{v_1^2} \\\\ \frac{T_1}{v_1^2} =  \frac{T_2}{v_2^2}\\\\v_2^2 = \frac{T_2v_1^2}{T_1}

Where;

T₁ lower tension

T₂ greater tension

v₁ velocity of wave in string with lower tension

v₂ velocity of wave in string with greater tension

From the given question;

T₂ = 1.1 T₁

v_2^2 = \frac{T_2v_1^2}{T_1}  \\\\v_2 = \sqrt{\frac{T_2v_1^2}{T_1}} \\\\v_2 = \sqrt{\frac{1.1T_1*(35.2)^2}{T_1}}\\\\v_2 = \sqrt{1.1(35.2)^2} = 36.92 \ m/s

<u>Fundamental frequency of wave on the string with greater tension;</u>

<u />f = \frac{v}{2l} \\\\2l = \frac{v}{f} \\\\thus, \frac{v_1}{f_1}  =\frac{v_2}{f_2} \\\\f_2 = \frac{f_1v_2}{v_1} \\\\f_2 =\frac{258*36.92}{35.2} \\\\f_2 = 270.6 \ Hz<u />

Beat frequency = F₂ - F₁

                          = 270.6 - 258

                          = 12.6 Hz

Therefore, the  beat frequency when each string is vibrating at its fundamental frequency is 12.6 Hz

6 0
3 years ago
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