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Nady [450]
3 years ago
10

On a caterpillars map all distances are marked in kilometers . The caterpillars map shows the distance between two milkweed plan

ts is 4,012 milimeters what is the distance in kilometers
Physics
1 answer:
frutty [35]3 years ago
5 0

Answer:

The distance in kilometers is 4012 ×10^{-6} km.

Explanation:

We know that the conversion of 1 millimeters is equal to 10^{-3} meter. And then the conversion of 1 meter is equal to 10^{-3} km. Then the conversion of 1 millimeter to km will be

1 mm = 10^{-3} m

1 m = 10^{-3} km

So, 1 mm = 10^{-3}×10^{-3} km = 10^{-6} km.

As here the the distance is 4012 mm, then the distance in km will be

4012 mm = 4012 ×10^{-6} km.

So the distance is 4012 ×10^{-6} km.

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3. Solve for y:<br><br><br> 14=84 over y <br><br><br> what is y
andreyandreev [35.5K]

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2 years ago
A 0.600 kg block is attached to a spring with spring constant 15 N/m. While the block is sitting at rest, a student hits it with
gulaghasi [49]

Answer:

8.8 cm

31.422 cm/s

Explanation:

m = Mass of block = 0.6 kg

k = Spring constant = 15 N/m

x = Compression of spring

v = Velocity of block

A = Amplitude

As the energy of the system is conserved we have

\dfrac{1}{2}mv^2=\dfrac{1}{2}kA^2\\\Rightarrow A=\sqrt{\dfrac{mv^2}{k}}\\\Rightarrow A=\sqrt{\dfrac{0.6\times 0.44^2}{15}}\\\Rightarrow A=0.088\ m\\\Rightarrow A=8.8\ cm

Amplitude of the oscillations is 8.8 cm

At x = 0.7 A

Again, as the energy of the system is conserved we have

\dfrac{1}{2}kA^2=\dfrac{1}{2}mv^2+\dfrac{1}{2}kx^2\\\Rightarrow v=\sqrt{\dfrac{k(A^2-x^2)}{m}}\\\Rightarrow v=\sqrt{\dfrac{15(0.088^2-(0.7\times 0.088)^2)}{0.6}}\\\Rightarrow v=0.31422\ m/s

The block's speed is 31.422 cm/s

4 0
3 years ago
The magnitude of electrical force between a pair of charged particles is ____ as much when the particles are moved half as far a
Gnesinka [82]

The magnitude of the electrical force between a pair of charged particles is 4 Times as much when the particles are moved half as far apart.

This can be easily understood by Columb's law,

F_{new} = \frac{kQ_{1}Q_{2}}{r^{2}}

which state's that the amount of electrical force experienced by two charged particles is inversely proportional to the square of the distance between them.

∴ \frac{F_{new} }{F_{old} } = \frac{Distance_{new}^{2}  }{Distance_{old}^{2}  }

Now, we know the new distance is half the original distance,

F_{new} = \frac{kQ_{1}Q_{2}}{\frac{r}{2}^{2} } \\F_{new} = 4\frac{kQ_{1}Q_{2}}{r^{2}}

F_{new} = 4F_{old}

The electrical force of attraction or electrostatic force of attraction between two charged particles refers to the amount of attractive or repulsive force that exists between the two charges. This can be calculated by Columb's Law.

A charged particle in physics is a particle that has an electric charge. It might be an ion, such as a molecule or atom having an excess or shortage of electrons in comparison to protons. The same charge is thought to be shared by an electron, a proton, or another primary particle.

Learn more about electrical force here

brainly.com/question/2526815

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