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RoseWind [281]
3 years ago
7

Which of the following statements accurately describes oceanic electric fields?

Physics
2 answers:
tiny-mole [99]3 years ago
6 0

Answer:

ANSWER C

Explanation:

vichka [17]3 years ago
3 0

Answer:

i sure it is D

Explanation:

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Oxana [17]
The object traveled +5 m
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3 years ago
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An ore car of mass 39000 kg starts from rest and rolls downhill on tracks from a mine. At the end of the tracks, 19 m lower vert
cupoosta [38]

Answer:

The compression in the spring is 5.88 meters.                

Explanation:

Given that,

Mass of the car, m = 39000 kg

Height of the car, h = 19 m

Spring constant of the spring, k=4.2\times 10^5\ N/m

We need to find the compression in the spring in stopping the ore car. It can be done by balancing loss in gravitational potential energy and the increase in elastic energy. So,

mgh=\dfrac{1}{2}kx^2

x is the compression in spring

x=\sqrt{\dfrac{2mgh}{k}} \\\\x=\sqrt{\dfrac{2\times 39000\times 19\times 9.8}{4.2\times 10^5}} \\\\x=5.88\ m

So, the compression in the spring is 5.88 meters.                                                                                                                  

6 0
3 years ago
Which of the following objects exerts a gravitational force?
kompoz [17]

the earth exerts a gravitational force

4 0
3 years ago
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Heat is added to a 2kg piece of ice at a rate of 793kW. How long will it take for ice to melt if it was initially 0?
Ede4ka [16]

Answer:

0.84 s

Explanation:

Step 1

Given information:

Mass of the ice (m) = 2.0 kg

Heat transfer rate (Q/T) = 793.0 kW

Latent heat of fusion of ice (Lf) = 334 kJ/kg

\frac{Q}{T}  =  \frac{mLf}{T}

Substituting the corresponding values we have:

793.0 kW=  \frac{2.0kg(334 kJ/kg)}{T} \\  T  =  \frac{2.0kg(334 kJ/kg)}{793.0kW}  =  \frac{668kJ}{793kW}   \\  = 0.84s

8 0
2 years ago
A force of 60 N is applied to a skier to pull him along a horizontal surface so that his speed remains constant. If the coeffici
matrenka [14]

Answer:

1200\ \text{N}

Explanation:

F = Force on the skier = 60 N

\mu = Coefficient of friction = 0.05

w = Weight of skier

Force is given by

F=\mu w

\Rightarrow w=\dfrac{F}{\mu}

\Rightarrow w=\dfrac{60}{0.05}=\dfrac{6000}{5}

\Rightarrow w=1200\ \text{N}

Weight of the skier on which the force is being applied is 1200\ \text{N} .

7 0
3 years ago
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