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nlexa [21]
3 years ago
14

PLSS HELP ITS TIMED WILL MARK BRAINLIEST

Physics
1 answer:
dlinn [17]3 years ago
4 0

Answer:

tjfd

Explanation:

tuff

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de mategoloalterfsqol

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3 years ago
“A horse pulls on a cart. By Newton’s third law of motion, the cart pulls back on
Studentka2010 [4]
In order to see what's going on, let's put them in empty space to get rid of any other influences, and let's also make it a push instead of a pull. / / / The horse pushes on the cart, so it begins accelerating away from him. At the same time, because of the equal opposite reaction thing, the cart pushes back on the horse, so the horse starts accelerating backwards, away from the cart. They both accelerate in opposite directions from where they started. BUT . . . their common center of mass doesn't move, and the sum of their momentums (which are in opposite directions) remains zero.
8 0
3 years ago
Read 2 more answers
The photoelectric effect tells us that __________.
Ivanshal [37]
<h2>hey </h2>

  1. It describes the process by which surface electron's are emitted from a metal when light is shined on it.
  2. And it tells us that light at a higher intensity must contain more quanta of energy, known as protons.

6 0
3 years ago
Three charged particles are positioned in the xy plane: a 50-nC charge at y = 6 m on the y axis, a –80-nC charge at x = –4 m on
leva [86]

Answer:48 V

Explanation:

Given

Three charged particle with charge

q_1=50\ nC at y=6\ m

q_2=-80\ nC at x=-4\ m

q_3=70\ nC at y=-6\ m

Electric Potential is given by

V=\frac{kQ}{r}

Distance of q_1 from x=8\ m

d_1=\sqrt{6^2+8^2}

d_1=\sqrt{36+64}

d_1=10\ m

similarly d_2=8-(-4)

d_2=12\ m

d_3=\sqrt{(-6)^2+8^2}

d_3=\sqrt{36+64}

d_3=10\ m

Potential at x=8\ m is

V_{net}=\frac{kq_1}{d_1}+\frac{kq_2}{d_2}+\frac{kq_3}{d_3}

V_{net}=k[\frac{q_1}{d_1}+\frac{q_2}{d_2}+\frac{q_3}{d_3}]

V_{net}=9\times 10^9[\frac{50}{10}-\frac{80}{12}+\frac{70}{10}]\times 10^{-9}

V_{net}=9\times 5.33

V_{net}=47.97\approx 48\ V

5 0
3 years ago
A thin 14.7 m long copper rod in a uniform magnetic field has a mass of 49.0 g. When the rod carries a current of 0.317 A, it fl
MatroZZZ [7]
F_B = ILBsin \theta

F force
I current
L length of wire in magnetic field
B magnetic field strength
θ angle between wire and magnetic field

if θ = 90°

B = \frac{F_B}{IL} \\ \\ F_B = F_G = mg \\ \\ B = \frac{mg}{IL} = \frac{0.049* 9.81 }{0.317*14.7}
3 0
3 years ago
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