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Andreas93 [3]
3 years ago
10

How do the causes of surface and deepwater currents differ

Physics
1 answer:
zepelin [54]3 years ago
5 0

The temperature is colder, and the water pressure is higher.

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1. Two-point charges, QA = +8 μC and QB = -5 μC, are separated by a distance r = 10 cm. What is the magnitude and direction of t
tiny-mole [99]

Explanation:

Charges,q_1=8\ \mu C=8\times 10^{-6}\ C

q_2=-5\ \mu C=-5\times 10^{-6}\ C

The distance between charges, r = 10 cm = 0.1 m

We need to find the magnitude and direction of the electric force. It is given by :

F=\dfrac{kq_1q_2}{r^2}\\\\F=\dfrac{9\times 10^9\times 8\times 10^{-6}\times 5\times 10^{-6}}{(0.1)^2}\\\\F=36\ N

So, the required force between charges is 36 N and it is towards positive charge i.e. +8 μC.

6 0
3 years ago
Which is the BEST definition of refraction? A) Light or sound waves change direction. B) Light or sound waves bounce off a mediu
melisa1 [442]
The answer to this is A. this is because, refraction with a light or sound wave changing its direction involve propagation,(in which propagation is the change in direction of a light or sound wave)


4 0
3 years ago
Read 2 more answers
a train has an initial velocity of 30 m/s. If the train accelerates uniformly at a rate of 6.3 m/s ^ for 2.8 seconds what is the
Annette [7]

T

Answer:

the velocity is a second final to initial velocity of 39

3 0
2 years ago
A surfer is moving at a constant velocity of 5.2 m/s north relative to a wave which is pushing him west at a constant velocity o
Vika [28.1K]
The speed and distances are directly proportional. Use ratios to solve for vertical y-distance. The ratio of x-distance west to y-distance north equals the x-velocity to y-velocity.

x/y = vx/vy
41/y = 8.6/5.2
41/y = 1.65
41/1.65 = y
24.8 m = y

5 0
3 years ago
A sled with a mass of 20 kg slides along frictionless ice at 4.5 m/s. It then crosses a rough patch of snow that exerts a fricti
uysha [10]

Answer:The sled slides 16.875m before rest.

Explanation:

a=\frac{F}{m} =\frac{12N}{20kg}

a=0.6 m/s²

Vf=0=Vi-a.t

t=\frac{Vi}{a} =t=\frac{4.5m/s}{0.6 m/s2} =t=7.5seg

d= Vi.t - \frac{a.t^{2}}{2}

d= 4.5 * 7.5 - \frac{0.6*7.5^{2} }{2} \\\\d=16.875m

3 0
3 years ago
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