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3 years ago

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Most ___ do not carry electricity well.

1 answer:

mars1129 [50]3 years ago

6 0

Insulators is the answer

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olya-2409 [2.1K]

Answer:c ITs C ^w^

Explanation:

4 0

3 years ago

A 45.0 g hard-boiled egg moves on the end of a spring with force constant 25.0 N/m. Its initial displacement 0.500 m. A damping

leva [86]

Answer:

0.02896 kg/s

Explanation:

$A_1$ = Initial displacement = 0.5 m

$A21$ = Final displacement = 0.1 m

t = Time taken = 0.5 s

m = Mass of object = 45 g

Displacement is given by

$x=Ae^{-\dfrac{b}{2m}t}cos(\omega t+\phi)$

At maximum displacement

$cos(\omega t+\phi)=1$

$\\\Rightarrow A_2=A_1e^{-\dfrac{b}{2m}t}\\\Rightarrow \dfrac{A_1}{A_2}=e^{\dfrac{b}{2m}t}\\\Rightarrow ln\dfrac{A_1}{A_2}=\dfrac{b}{2m}t\\\Rightarrow b=\dfrac{2m}{t}\times ln\dfrac{A_1}{A_2}\\\Rightarrow b=\dfrac{2\times 0.045}{5}\times ln\dfrac{0.5}{0.1}\\\Rightarrow b=0.02896\ kg/s$

The magnitude of the damping coefficient is 0.02896 kg/s

6 0

3 years ago

Hey I need help do animals take in oxygen and give off carbon dioxide or Plants take in oxygen and give off carbon dioxide or pl

Sophie [7]

Answer:

Animals take in oxygen and give off carbon dioxide.

Explanation:

This is because plants take in our carbon dioxide and give off the oxygen that we as people and animals need to breathe.

3 0

3 years ago

Read 2 more answers

If the distance from a light source triples, how does light intensity change? The intensity will be 3x greater. The intensity wi

Tcecarenko [31]

Answer:

The intensity will be 1/9 as much.

Explanation:

The intensity of the light or any source is inversely related to the square of the distance.

$I\alpha \frac{1}{r^{2} }$

Now according to the question the distance is increased by three times than,

$\frac{I_{2} }{I_{1} }=\frac{r_{1}^{2} }{r_{2}^{2} }$

Therefore,

$\frac{I_{2} }{I_{1} }=\frac{r_{1}^{2} }{(3r_{1})^{2} }\\\frac{I_{2} }{I_{1} }=\frac{1}{9} \\{I_{2}=\frac{1}{9}{I_{1} }$

Therefore the intensity will become 1/9 times to the initial intensity.

3 0

3 years ago

Returning once again to our table top example of a horizontal mass on a low-friction surface with m = 0.254 kg and k = 10.0 N/m

Julli [10]

Explanation:

Given that,

Mass = 0.254 kg

Spring constant [tex[\omega_{0}= 10.0\ N/m[/tex]

Force = 0.5 N

y = 0.628

We need to calculate the A and d

Using formula of A and d

$A=\dfrac{\dfrac{F_{0}}{m}}{\sqrt{(\omega_{0}^2-\omega^{2})^2+y^2\omega^2}}$ .....(I)

$tan d=\dfrac{y\omega}{(\omega^2-\omega^2)}$ ....(II)

Put the value of $\omega=0.628\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-0.628)^2+0.628^2\times0.628^2}}$

$A=0.0198$

From equation (II)

$tan d=\dfrac{0.628\times0.628}{((10.0^2-0.628)^2)}$

$d=0.0023$

Put the value of $\omega=3.14\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-3.14)^2+0.628^2\times3.14^2}}$

$A=0.0203$

From equation (II)

$tan d=\dfrac{0.628\times3.14}{((10.0^2-3.14)^2)}$

$d=0.0120$

Put the value of $\omega=6.28\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-6.28)^2+0.628^2\times6.28^2}}$

$A=0.0209$

From equation (II)

$tan d=\dfrac{0.628\times6.28}{((10.0^2-6.28)^2)}$

$d=0.0257$

Put the value of $\omega=9.42\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-9.42)^2+0.628^2\times9.42^2}}$

$A=0.0217$

From equation (II)

$tan d=\dfrac{0.628\times9.42}{((10.0^2-9.42)^2)}$

$d=0.0413$

Hence, This is the required solution.

5 0

3 years ago