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likoan [24]
3 years ago
8

Which statement is true of all matter?

Physics
1 answer:
Fittoniya [83]3 years ago
8 0

Answer:

D. It must have mass and volume

Explanation:

In science, matter is referred to as any substance that has weight and occupies space. This means that the substance must have a MASS of its own when weighed and also a VOLUME.

Matter include elements, molecules, humans, etc. In fact, almost every substance on Earth is considered MATTER. Therefore, the fact that a substance must "have mass and volume" is true for all matter.

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3 years ago
Explain why the same side of the moon is always facing Earth.
Ad libitum [116K]

Answer:

The moon keeps the same face pointing towards the Earth because its rate of spin is tidally locked so that it is synchronized with its rate of revolution (the time needed to complete one orbit). In other words, the moon rotates exactly once every time it circles the Earth.

7 0
3 years ago
Many industries are powered via distant power stations. Calculate the current flowing through a 7,300m long 10. copper power lin
Oliga [24]

Answer:

Current, I = 1000 A

Explanation:

It is given that,

Length of the copper wire, l = 7300 m

Resistance of copper line, R = 10 ohms

Magnetic field, B = 0.1 T

\mu_o=4\pi \times 10^{-7}\ T-m/A

Resistivity, \rho=1.72\times 10^{-8}\ \Omega-m

We need to find the current flowing the copper wire. Firstly, we need to find the radius of he power line using physical dimensions as :

R=\rho \dfrac{l}{A}

R=\rho \dfrac{l}{\pi r^2}

r=\sqrt{\dfrac{\rho l}{R\pi}}

r=\sqrt{\dfrac{1.72\times 10^{-8}\times 7300}{10\pi}}

r = 0.00199 m

or

r=1.99\times 10^{-3}\ m=2\times 10^{-3}\ m

The magnetic field on a current carrying wire is given by :

B=\dfrac{\mu_o I}{2\pi r}

I=\dfrac{2\pi rB}{\mu_o}

I=\dfrac{2\pi \times 0.1\times 2\times 10^{-3}}{4\pi \times 10^{-7}}

I = 1000 A

So, the current of 1000 A is flowing through the copper wire. Hence, this is the required solution.

4 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:

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
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