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2 years ago

PLZ HELP ME!!!!!!!!! WILL GIVE BRAINLY!!!!!!!! Which photo represents a waxing gibbous moon?

Physics

1 answer:

yarga [219]2 years ago

6 0

A phase of moon with more than half sunlit portion visible on the right

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Understanding that if we say something unkind to someone else his or her

OlgaM077 [116]

Answer: Moral

Explanation:

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3 years ago

What is the S.I. unit of drift velocity and electron mobility?

DanielleElmas [232]

Drift velocity is equal to displacement of the moving object per unit time. The SI unit for displacement is meters while that of time is second. Hence the derived SI unit of velocity is meter per second. This also applies to electron mobility which relates to the displacement per unit time of a moving electron

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3 years ago

Work is a transfer of (1 point) energy. force. mass. motion.

jok3333 [9.3K]

Energy is the ability to do work.

So work can not be done without the transfer of energy from one body to another.

Work is the transfer of energy.

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3 years ago

Read 2 more answers

A 1500 kg car traveling 5.0 m/s collides head on with a 3000 kg truck traveling

Setler [38]

m1=1500kg
m_2=3000kg
v_1=5m/s
v_2=7m/s

Using law of conservation of momentum

$\\ \sf\Rrightarrow m_1v_1-m_2v_2=(m1+m2)v_3$

$\\ \sf\Rrightarrow 1500(5)-3000(7)=(1500+3000)v_3$

$\\ \sf\Rrightarrow 7500-21000=4500v_3$

$\\ \sf\Rrightarrow -13500=4500v_3$

$\\ \sf\Rrightarrow v_3=-3m/s$

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2 years ago

A −3.0 nC charge is on the x-axis at x=−9 cm and a +4.0 nC charge is on the x-axis at x=16 cm. At what point or points on the y-

alexdok [17]

Answer:

y = 10.2 m

Explanation:

It is given that,

Charge, $q_1=-3\ nC$

It is placed at a distance of 9 cm at x axis

Charge, $q_2=+4\ nC$

It is placed at a distance of 16 cm at x axis

We need to find the point on the y-axis where the electric potential zero. The net potential on y-axis is equal to 0. So,

$\dfrac{kq_1}{r_1}+\dfrac{kq_2}{r_2}=0$

Here,

$r_1=\sqrt{y^2+9^2} \\\\r_2=\sqrt{y^2+15^2}$

So,

$\dfrac{kq_1}{r_1}=-\dfrac{kq_2}{r_2}\\\\\dfrac{q_1}{r_1}=-\dfrac{q_2}{r_2}\\\\\dfrac{-3\ nC}{\sqrt{y^2+81} }=-\dfrac{4\ nC}{\sqrt{y^2+225} }\\\\3\times \sqrt{y^2+225}=4\times \sqrt{y^2+81}$

Squaring both sides,

$3\times \sqrt{y^2+225}=4\times \sqrt{y^2+81}\\\\9(y^2+225)=16\times (y^2+81)\\\\9y^2+2025=16y^2-+1296\\\\2025-1296=7y^2\\\\7y^2=729\\\\y=10.2\ m$

So, at a distance of 10.2 m on the y axis the electric potential equals 0.

8 0

3 years ago