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Makovka662 [10]
3 years ago
12

COMPAWE APPLIANCENO STABILIZERREQUIRED* For 160V-260V power supply​

Physics
1 answer:
sergey [27]3 years ago
5 0

Answer:

cgyd6rvnfu ygoigvigceu fee duucdubr5vef8cuecyec

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A car increased its velocity to 62m/s in 10s starting from rest. Calculate the distance it covers during this time?
Roman55 [17]

Answer:

distance= velocity ×time

distance= 62×10

distance=620m

hope it helps you mate please mark me as brainliast

6 0
3 years ago
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If two stars are the same size and one is twice the temperature of the other, how much more luminous is the hotter one? quizlit
Dmitriy789 [7]
The hotter star will be 16 times more luminous  - luminosity depends on two things  - the size of the star and the temperature of the star. The hotter a star is, the more energy it will give out. This will give rise to greater luminosity.
3 0
3 years ago
A6 kg mass moving at 10m/s collides with a 4 kg mass moving in the
Nat2105 [25]

Answer:

Explanation:

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4 0
3 years ago
At what distance will the weight of a body be halved , Earth's radius=6.4×10^6
Sholpan [36]

A body of mass m has weight

F = GMm/r²

on the surface of the Earth, where G is the universal gravitational constant, M is the mass of the Earth, and r is it's radius.

If the weight is to be halved, then we have

1/2 F = 1/2 GMm/r² = (1/√2)² GMm/r² = GMm/(√2 r²)

so the distance between the body and the planet's center needs to be

√2 × 6.4 × 10⁶ m ≈ 9.1 × 10⁶ m

5 0
3 years ago
Two identical 9.10-g metal spheres (small enough to be treated as particles) are hung from separate 300-mm strings attached to t
Musya8 [376]

Answer:

n = 1.266\times 10^{12}

Explanation:

Given data:

mass of sphere is 10 g

Angle between string and vertical axis is \theta = 13 degree

thickness of string  300 mm = 0.3 m

sin\theta =\frac{2}{0.3 m}

r =0.3 sin 13 = 0.067 m

Fe = \frac{ kq_1 q-2}{d^2}

Fe = \frac{kq^2}{(2r)^2} = mg tan\theta

q^2 =  mg tan\theta \frac{(2r)^2}{k}

    = 0.0091 \times 9.8 tan13 \times \frac{(2\times 0.067)^2}{9\times 10^9}

q^2 = 4.10\times 10^{-14}

q = 2.026 \times 10^{-7} C

q = ne

n = \frac{1.6\times 10^{-19}}{2.02\times 10^{-7}}

n = 1.266\times 10^{12}

3 0
3 years ago
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