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Leona [35]
3 years ago
8

If a freely falling rock were equipped with a speedometer, by how much would its speed readings increase with each second of fal

l?
Physics
1 answer:
Klio2033 [76]3 years ago
5 0
Its speed reading would increase to 10 m/s every second 
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What does a plant need to take in from environment to live?
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Answer:

Carbon Dioxide.

Explanation:

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2 years ago
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What happens to the density of a given substance if you increase the amount of the substance that you have?
Amanda [17]
Have you ever looked up the density of a substance ?  You ought to try it.  Go ahead. Pick a substance, then go online or open up an actual book and find its density.  You will never see any particular volume mentioned along with the density . . . because it doesn't matter.  The whole idea of density is that it describes the substance, no matter how much or how little you have of it.  The density of a tiny drop of water under a microscope is the same as the density of a supertanker-ful of water.
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3 years ago
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Given a force of 100 N and a acceleration of 5 m/s, what is the mass
SIZIF [17.4K]

Answer:

force = mass \times acceleration \\ 100 = m \times 5 \\ m =  \frac{100}{5}  \\ m = 20 \:  kg

8 0
2 years ago
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7. A block of copper of unknown mass has an initial temperature of 65.4oC. The copper is immersed in a beaker containing 95.7g o
dolphi86 [110]

Answer:

37.34372 kg

Explanation:

m = Mass

\Delta T = Change in temperature

1 denotes water

2 denotes copper

c = Heat capacity

Heat is given by

Q=mc\Delta T

In this case the heat transfer will be equal

m_1c_1\Delta T_1=m_2c_2\Delta T_2\\\Rightarrow m_2=\frac{m_1c_1\Delta T_1}{c_2\Delta T_2}\\\Rightarrow m_2=\frac{95.7\times 4.18(24.2-22.7)}{0.39(65.4-24.2)}\\\Rightarrow m_2=37.34372\ kg

Mass of copper block is 37.34372 kg

5 0
3 years ago
A point charge q is located at the center of a spherical shell of radius a that has a charge −q uniformly distributed on its sur
muminat

Answer:

a) E = 0

b) E =  \dfrac{k_e \cdot q}{ r^2 }

Explanation:

The electric field for all points outside the spherical shell is given as follows;

a) \phi_E = \oint E \cdot  dA =  \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}

From which we have;

E \cdot  A =  \dfrac{{\Sigma Q}}{\varepsilon _{0}} = \dfrac{+q + (-q)}{\varepsilon _{0}}  = \dfrac{0}{\varepsilon _{0}} = 0

E = 0/A = 0

E = 0

b) \phi_E = \oint E \cdot  dA =  \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}

E \cdot  A  = \dfrac{+q }{\varepsilon _{0}}

E  = \dfrac{+q }{\varepsilon _{0} \cdot A} = \dfrac{+q }{\varepsilon _{0} \cdot 4 \cdot \pi \cdot r^2}

By Gauss theorem, we have;

E\oint dS =  \dfrac{q}{\varepsilon _{0}}

Therefore, we get;

E \cdot (4 \cdot \pi \cdot r^2) =  \dfrac{q}{\varepsilon _{0}}

The electrical field outside the spherical shell

E =  \dfrac{q}{\varepsilon _{0} \cdot (4 \cdot \pi \cdot r^2) }= \dfrac{q}{4 \cdot \pi \cdot \varepsilon _{0} \cdot r^2 }=  \dfrac{q}{(4 \cdot \pi \cdot \varepsilon _{0} )\cdot r^2 }

k_e=  \dfrac{1}{(4 \cdot \pi \cdot \varepsilon _{0} ) }

Therefore, we have;

E =  \dfrac{k_e \cdot q}{ r^2 }

5 0
2 years ago
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