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3 years ago

What precautions should be taken on board a vessel during a lightning storm?

Physics

2 answers:

ArbitrLikvidat [17]3 years ago

8 0

Answer:

When huge amount of electrical discharge occurs from the atmosphere, a lightning storm is said to occur.

Following precautions must be taken if one is on board a vessel during a lightning storm:

During a lightning storm, all the plugs should be disconnected and electrical appliances must be switched off.
One should remain inside the cabin and not move outside because the charges remain on the surface of a conductor.
The movement should be avoided. It would be best to lay straight down.

dusya [7]3 years ago

7 0

The precautions which should be taken on board a vessel during a lightning storm are:

(i) All electrical equipments should be unplugged.

(ii) Keep yourself away from all metal parts such as engines, metal tanks, stoves etc.

(iii) Stay inside the boat.

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Guys help...A spring that obeys Hooke's law, with a spring constant k1, is cut into N identical springs, each with a spring cons

Tems11 [23]

Answer:

K2 = N*K1

Explanation:

The force you apply to each section is the same you apply to the whole spring, but the extension of each section is dX/N (if dX is the extension of the entire spring)

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3 years ago

A 0.00275 kg air‑inflated balloon is given an excess negative charge q1 =−3.50×10−8 C by rubbing it with a blanket. It is found

kati45 [8]

Answer:

1) $\rm q_2$ is positive.

2) $\rm q_2=4.56\times 10^{-10}\ C.$

Explanation:

The charged rod is held above the balloon and the weight of the balloon acts in downwards direction. To balance the weight of the balloon, the force on the balloon due to the rod must be directed along the upwards direction, which is only possible when the rod exerts an attractive force on the balloon and the electrostatic force on the balloon due to the rod is attractive when the polarities of the charge on the two are different.

Thus, In order for this to occur, the polarity of charge on the rod must be positive, i.e., $\rm q_2$ is positive.

Given:

Mass of the balloon, m = 0.00275 kg.
Charge on the balloon, $\rm q_1 = -3.50\times 10^{-8}\ C.$
Distance between the rod and the balloon, d = 0.0640 m.
Acceleration due to gravity, $\rm g = 9.81\ m/s^2.$

In order to balloon to be float in air, the weight of the balloom must be balanced with the electrostatic force on the balloon due to rod.

Weight of the balloon, $\rm W = mg = 0.00275\times 9.81=2.70\times 10^{-2}\ N.$

The magnitude of the electrostatic force on the balloon due to the rod is given by

$\rm F_e = \dfrac{1}{4\pi \epsilon_o}\dfrac{|q_1||q_2|}{d^2}.$

$\rm \dfrac{1}{4\pi \epsilon_o}$ is the Coulomb's constant.

For the elecric force and the weight to be balanced,

$\rm F_e = W\\\dfrac{1}{4\pi \epsilon_o}\dfrac{|q_1||q_2|}{d^2}=W\\8.99\times 10^9\times \dfrac{3.50\times10^{-8}\times |q_2| }{0.0640^2}=2.70\times 10^{-2}\\|q_2| = \dfrac{2.70\times 10^{-2}\times 0.00640^2}{8.99\times 10^9\times 2.70\times 10^{-7}}=4.56\times 10^{-10}\ C.$

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3 years ago

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3 years ago

A bumper cart has a mass of 200 kg and has a protective bumper around it that behaves like a spring. The spring constant is 5000

34kurt

Part A:
For this part we’re assuming all the kinetic energy of the moving bumper car is converted into elastic potential energy in the spring since the car is brought to rest. Therefore you can find the total kinetic energy to get your answer:

KE = ½ mv^2
KE = ½ (200)(8)^2
KE = 6400 J

Part B:
Now you can use Hooke’s law to find the force:

F = kx
F = (5000)(0.2)
F = 1000 N

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3 years ago

Which of the following is an example of rolling friction?

il63 [147K]

Answer:

O bike tires on the road as you ride

Explanation:

is the rolling friction

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3 years ago

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