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Zinaida [17]
3 years ago
13

Identify the smallest unit of an element

Physics
2 answers:
baherus [9]3 years ago
4 0

An atom. In most cases, an atom consists of protons, neutrons, and electrons.

marysya [2.9K]3 years ago
3 0

atom ................................................... sorry for the periods

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A train car has a mass of 10,000 kg and is moving at +3.0 m/s. It strikes an identical train car that is at rest. The train cars
shtirl [24]
In order to compute the final velocity of the trains, we may apply the principle of conservation of momentum which is:
initial momentum = final momentum
m₁v₁ = m₂v₂

The final mass of the trains will be:
10,000 + 10,000 = 20,000 kg
Substituting the values into the equation:

10,000 * 3 = 20,000 * v
v = 1.5 m/s

The final velocity of the trains will be 1.5 m/s
3 0
3 years ago
The speed of light (c) is:
Radda [10]

Answer:

A

Explanation:

Speed of light is 299 792 458 m / s. So option A is answer.

7 0
2 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:

35

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 coils of wire are placed close together. Initially, a current of 3.04 A exists in one of the coils, but there is no current
Alchen [17]

Answer:

M=0.0247H

Explanation:

Given data

V_{voltage}=4.29V\\I_{current}=3.04A\\t_{time}=1.75*10^{-2}s

To find

Mutual inductance of the two-coil system

Solution

The mutual inductance given as:

M= (-VΔt)/ΔI

Substitute the given values

So

M=-\frac{4.29V*1.75*10^{-2}s}{(0-3.04A)}\\ M=0.0247H

4 0
3 years ago
8. Il An 8.00 kg package in a mail-sorting room slides 2.00 m down a
Vitek1552 [10]

Answer:

See below

Explanation:

Normal force = m g cos 53 = 8 kg * 9.8 m/s^2 * cos 53 = 47.1823 N

  no work is done by this force

Force friction = coeff friction * force normal = .4 * 47.1823 = 7.55 N

   work of friction = 7.55 * 2 m = 15.1 j

Force Downplane = mg sin 53 = 62.61  N

    work = 62.61 * 2 = 125.22 j

Net Force downplane =   force downplane - force friction = 55.06 N

net Work = force * distance = 55.06 N * 2 M = 110.12 j

3 0
2 years ago
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