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3 years ago

a resistance of 30 ohms is placed in a circuit with 90 volt battery what current flow in the circuit

Physics

1 answer:

mamaluj [8]3 years ago

4 0

Answer:

3amps

Explanation:

voltage v = current I × resistance R

current = voltage÷ resistance = 90÷30= 3amperes

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When you let the air go out of a balloon the balloon moves or shoots like a rocket. Would this happen if there was no surroundin

Pie

Answer:

The balloon would still move like a rocket

Explanation:

The principle of work of this system is the Newton's third law of motion, which states that:

"When an object A exerts a force on an object B (action), object B exerts an equal and opposite force (reaction) on object A"

In this problem, we can identify the balloon as object A and the air inside the balloon as object B. As the air goes out from the balloon, the balloon exerts a force (backward) on the air, and as a result of Newton's 3rd law, the air exerts an equal and opposite force (forward) on the balloon, making it moving forward.

This mechanism is not affected by the presence or absence of surrounding air: in fact, this mechanism also works in free space, where there is no air (and in fact, rockets also moves in space using this system, despite the absence of air).

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3 years ago

Which of the following is the correct description of momentum?

victus00 [196]

Answer:

The product of mass and velocity is the correct answer

Explanation:

Momentum is defined as mass × velocity

p = mv

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3 years ago

if a 3-kg object has a momentum of 33 kg��m/s, what's its velocity? a. 99 m/s b. 36 m/s c. 11 m/s d. 9.1 m/s

horsena [70]

M=3kg
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3 years ago

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ivanzaharov [21]

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3 years ago

A spaceship hovering over the surface of Venus drops an object from a height of 17 m. How much longer does it take to reach the

Paraphin [41]

1.96s and 1.86s. The time it takes to a spaceship hovering the surface of Venus to drop an object from a height of 17m is 1.96s, and the time it takes to the same spaceship hovering the surface of the Earth to drop and object from the same height is 1.86s.

In order to solve this problem, we are going to use the motion equation to calculate the time of flight of an object on Venus surface and the Earth. There is an equation of motion that relates the height as follow:

$h=v_{0} t+\frac{gt^{2}}{2}$

The initial velocity of the object before the dropping is 0, so we can reduce the equation to:

$h=\frac{gt^{2}}{2}$

We know the height h of the spaceship hovering, and the gravity of Venus is $g=8.87\frac{m}{s^{2}}$ . Substituting this values in the equation $h=\frac{gt^{2}}{2}$ :

$17m=\frac{8.87\frac{m}{s^{2} } t^{2}}{2}$

To calculate the time it takes to an object to reach the surface of Venus dropped by a spaceship hovering from a height of 17m, we have to clear t from the equation above, resulting:

$t=\sqrt{\frac{2(17m)}{8.87\frac{m}{s^{2} } }} =\sqrt{\frac{34m}{8.87\frac{m}{s^{2} } } }=1.96s$

Similarly, to calculate the time it takes to an object to reach the surface of the Earth dropped by a spaceship hovering from a height of 17m, and the gravity of the Earth $g=9.81\frac{m}{s^{2}}$ .

$t=\sqrt{\frac{2(17m)}{9.81\frac{m}{s^{2} } }} =\sqrt{\frac{34m}{9.81\frac{m}{s^{2} } } }=1.86s$

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a resistance of 30 ohms is placed in a circuit with 90 volt battery what current flow in the circuit​

a resistance of 30 ohms is placed in a circuit with 90 volt battery what current flow in the circuit