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charle [14.2K]
3 years ago
7

Help please i will mark brainlist!!

Physics
2 answers:
Dmitrij [34]3 years ago
7 0
The answer is A.


p=m/v
p= 240/60
p= 4 g/cm^3
Shalnov [3]3 years ago
3 0

Answer:

A is correct a.p.e.x approved

Explanation:

have a nice day

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The inductance of a closely packed coil of 560 turns is 8.9 mH. Calculate the magnetic flux through the coil when the current is
Y_Kistochka [10]

Answer:

0.11\times 10^{-6}weber

Explanation:

We have given number of turns N = 560

Inductance L = 8.9 mH

Current through the coil = 7 mA

Inductance of the coil is given as L=\frac{N\Phi }{I}

Where N  is number of turns I is current and \Phi is flux

So \Phi =\frac{LI}{N}=\frac{8.9\times 10^{-3}\times 7\times 10^{-3}}{560}=0.11\times 10^{-6}weber

6 0
3 years ago
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Acetone is a flammable solvent used in the making of plastics. The density is 790kg/m3. to convert 790kg/m3 to g/cm3​
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Answer: 0.790 g/cm3

Explanation:

The density of acetone is 790 Kg/m3.

To convert from Kg to g we multiply by 1000 (1 Kg = 1000 g)

To convert from m3 to cm3 we multiply by 10∧6

So, The density of acetone in (g/cm3) = (790 x 1000) / (10∧6) = 0.79 g/cm3

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3 years ago
The sound wave passes from the sea-water into the air. State what happens to; the speed of the waves
dusya [7]
Answer. Explanation: Frequency of the sound decreases and the speed of sound becomes 346m/s from near about 1500 m/s.
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2 years ago
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What is -0.000000698 in scientific notation
dangina [55]

-6.98 × 10-^7 is the answer <3

6 0
3 years ago
A 51.0 kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 225 N. For the first 10.0 m th
shepuryov [24]

Answer:

The final speed of the crate is 12.07 m/s.

Explanation:

For the first 10.0 meters, the only force acting on the crate is 225 N, so we can calculate the acceleration as follows:

F = ma

a = \frac{F}{m} = \frac{225 N}{51.0 kg} = 4.41 m/s^{2}

Now, we can calculate the final speed of the crate at the end of 10.0 m:

v_{f}^{2} = v_{0}^{2} + 2ad_{1}                  

v_{f} = \sqrt{0 + 2*4.41 m/s^{2}*10.0 m} = 9.39 m/s    

For the next 10.5 meters we have frictional force:

F - F_{\mu} = ma

F - \mu mg = ma

So, the acceleration is:

a = \frac{F - \mu mg}{m} = \frac{225 N - 0.17*51.0 kg*9.81 m/s^{2}}{51.0 kg} = 2.74 m/s^{2}

The final speed of the crate at the end of 10.0 m will be the initial speed of the following 10.5 meters, so:

v_{f}^{2} = v_{0}^{2} + 2ad_{2}  

v_{f} = \sqrt{(9.39 m/s)^{2} + 2*2.74 m/s^{2}*10.5 m} = 12.07 m/s  

Therefore, the final speed of the crate after being pulled these 20.5 meters is 12.07 m/s.  

I hope it helps you!                              

7 0
3 years ago
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