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3 years ago

What has 12 humps and lives at the north pole?

Physics

1 answer:

Rashid [163]3 years ago

8 0

<h2>Camels have 12 humps and lives at the North Pole </h2>

Six lost camels has 12 humps and lives at the north pole. The reason is that animal which possess hump are camels and they live at the north pole. Camel is used for travel purpose and for transfer goods from one place to another place.

Basically, a camel's hump is a large heap of fat. In a normal camel hump can be of 80 pounds that is equal to 35 kilograms on the scale. Human and many animals stocks their fat blended within the muscle tissue or beneath the skin layers. Camels are the sole living organism with a hump on its back.

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The solar system is made up of eight planets, numerous comets, asteroids and moons, and the Sun. The force that holds all of the

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Your answer is Gravity

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3 years ago

Are rocks made of matter?

ololo11 [35]

Answer:

Yes, the rocks are made of matter

Explanation:

Let's remember the definition of matter.

Matter is all that has mass and occupies a place in space. Therefore, if we measure the mass of each rock we will know its mass, the other fact is that rocks like any particular body are occupying a place in an empty space.

The opposite of this is antimatter and can its extent be given by the quantum mechanics.

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3 years ago

Read 2 more answers

With the piston head locked in place, will the volume of the gas increase, decrease, or stay the same whenthe piston is placed a

Nuetrik [128]

Answer:

For real gas the volume of a given mass of gas will increase with increase in temperature.

Explanation:

With the piston head locked in place and place above the fire,the volume of the gas will increase,because the volume of a given mass of gas increases with increase temperature.

6 0

3 years ago

An AC adapter for a telephone answering machine uses a transformer to reduce the line voltage of 120 V to a voltage of 5.00 V. T

Maksim231197 [3]

Answer:

Explanation:

We are given that

Initial voltage, $V_1=120 V$

Final voltage, $V_2=5 V$

Number of tuns in primary coil of the transformer, $N_p=840$

Rms current, $I_{rms}=580mA=580\times 10^{-3} A$

$1 mA=10^{-3} A$

We have to find the number of turns are there on the secondary coil.

We know that

$\frac{N_s}{N_p}=\frac{V_2}{V_1}$

Using the formula

$\frac{N_s}{840}=\frac{5}{120}$

$N_s=\frac{5}{120}\times 840=35$

Hence, there are number of turns on the secondary coil=35

8 0

3 years ago

Two 22.7 kg ice sleds initially at rest, are placed a short distance apart, one directly behind the other, as shown in Fig. 1. A

boyakko [2]

Newton's third law of motion sates that force is directly proportional to the rate of change of momentum produced

(a) The final speeds of the ice sleds is approximately 0.49 m/s each

(b) The impulse on the cat is 11.0715 kg·m/s

The reason for arriving at the above values is as follows:

The given parameters are;

The masses of the two ice sleds, m₁ = m₂ = 22.7 kg

The initial speed of the ice, v₁ = v₂ = 0

The mass of the cat, m₃ = 3.63 kg

The initial speed of the cat, v₃ = 0

The horizontal speed of the cat, v₃ = 3.05 m/s

(a) The required parameter:

The final speed of the two sleds

For the first jump to the right, we have;

By the law of conservation of momentum

Initial momentum = Final momentum

∴ m₁ × v₁ + m₃ × v₃ = m₁ × v₁' + m₃ × v₃'

Where;

v₁' = The final velocity of the ice sled on the left

v₃' = The final velocity of the cat

Plugging in the values gives;

22.7 kg × 0 + 3.63 × 0 = 22.7 × v₁' + 3.63 × 3.05

∴ 22.7 × v₁' = -3.63 × 3.05

v₁' = -3.63 × 3.05/22.7 ≈ -0.49

The final velocity of the ice sled on the left, v₁' ≈ -0.49 m/s (opposite to the direction to the motion of the cat)

The final speed ≈ 0.49 m/s

For the second jump to the left, we have;

By conservation of momentum law, m₂ × v₂ + m₃ × v₃ = m₂ × v₂' + m₃ × v₃'

Where;

v₂' = The final velocity of the ice sled on the right

v₃' = The final velocity of the cat

Plugging in the values gives;

22.7 kg × 0 + 3.63 × 0 = 22.7 × v₂' + 3.63 × 3.05

∴ 22.7 × v₂' = -3.63 × 3.05

v₂' = -3.63 × 3.05/22.7 ≈ -0.49

The final velocity of the ice sled on the right = -0.49 m/s (opposite to the direction to the motion of the cat)

The final speed ≈ 0.49 m/s

(b) The required parameter;

The impulse of the force

The impulse on the cat = Mass of the cat × Change in velocity

The change in velocity, Δv = Initial velocity - Final velocity

Where;

The initial velocity = The velocity of the cat before it lands = 3.05 m/s

The final velocity = The velocity of the cat after coming to rest =

∴ Δv = 3.05 m/s - 0 = 3.05 m/s

The impulse on the cat = 3.63 kg × 3.05 m/s = 11.0715 kg·m/s

(c) The required information

The average velocity

Impulse = $F_{average}$ × Δt

Where;

Δt = The time of collision = The time it takes the cat to finish landing = 12 ms

12 ms = 12/1000 s = 0.012 s

We get;

$F_{average} = \mathbf{\dfrac{Impulse}{\Delta \ t}}$

∴ $F_{average} = \dfrac{11.0715 \ kg \cdot m/s}{0.012 \ s} = 922.625 \ kg\cdot m/s^2 = 922.625 \ N$

The average force on the right sled applied by the cat while landing, $\mathbf{F_{average}}$ = 922.625 N

Learn more about conservation of momentum here:

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brainly.com/question/20568685

brainly.com/question/22257327

8 0

2 years ago