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Anna35 [415]
2 years ago
14

A bullet with a mass of 0.040 kg collides inelastically with a wooden block of mass 1.5 kg, initially at rest. After the collisi

on, the bullet along with the block has a speed of 1.0 m/s. Calculate the initial speed of the bullet.
Physics
1 answer:
miss Akunina [59]2 years ago
3 0

Answer:12.43 m/s

Explanation:

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Answer: An electric circuit is a representation of how current moves from the source of the current( example a battery or a cell) through resistors and other devices before entering the source.

Explanation:

The two important types of electrical circuit includes:

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A circuit is said to be open when the electrical source, such as the battery or the cell, is not connected to any external conductor (or resistance). In this situation, any voltmeter connected across the terminal of the cell measures the total driving force of the cell.

In this type of circuit, current cannot flow from one end of power source to the other due to interruptions.

A circuit is said to be closed if the source of electricity is connected to an external conductor through which current is passed.

In a closed circuit, there is complete electrical connection which allows current to flow or circulate. Here, part of the total driving force of the source is used to drive current through the external resistance and the difference is used to overcome the internal resistance of the battery.

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3 years ago
The diagram shows two balls before they collide.
Ronch [10]

Answer:

The Answer is B)0.2 kg • m/s

Explanation:

I made a 100 on my test. Sorry if I'm late but hope I helped.

4 0
2 years ago
Read 2 more answers
Match the following Help please ​
maxonik [38]
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3 years ago
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IgorLugansk [536]

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7 0
3 years ago
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Alex_Xolod [135]

The expression for the magnitude of the electric field between two uniform conducting plates is

E=\frac{V}{d}

Here, V is potential difference between plates and d is separation between plates.

As the potential 6.00 cm from the zero volt plate (and 4.00 cm from the other) is 420 V.

Therefore,

E = \frac{420 \ V}{6 \times 10^{-2} m } = 70 \times 10^2 \ V/m

Thus, the electric field strength between the plates is 7000 V/ m

8 0
3 years ago
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