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3 years ago

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A student made the claim that a 4 gram paintball fired from a paintball gun at 90 m/s could have about the same kinetic energy a

s a 1 gram BB pellet fired from a BB gun at 180 m/s. Do you agree or disagree with the student's claim? Use evidence and mathematical reasoning to support your response.

1 answer:

Fiesta28 [93]3 years ago

5 0

Answer:

I do agree with the student. Because imagene you have a tissue and a stick. And you throw them at the same speed , the stick will go farther because it' s heavier but the tissue would just blow around and not go forward because it is not heavy enought.

Explanation:

Hope this is correct have a terrific day.

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Ultraviolet light from a distant star is traveling at 3.0 × 108 m/s. How long will it take for the light to reach Earth if it mu

Deffense [45]

Answer : Time taken for the light to reach earth is $3.7\times 10^4\ Hours$

Explanation :

Given that,

Speed of Ultraviolet light, $v = 3\times 10^8\ m/s$

Distance covered by the light, $d = 4\times 10^{13}\ km=4\times 10^{16}\ km$

We have to find the time taken (t) by the light to reach on the surface of earth.

$t=\dfrac{d}{v}$

$t=\dfrac{4\times 10^{16}\ km}{3\times 10^8\ m/s}$

t = 133333333.333 seconds

$t=37037.03$

or

$t=3.7\times 10^4\ Hours$

Hence, this is the required solution.

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4 years ago

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An object of mass m = 2.9 g and charge Q = +42 µC is attached to a string and placed in a uniform electric field that is incline

Alisiya [41]

(a)
<span>F= qE </span>

<span>F sin 30.0° = mg </span>
<span>= 0.0026(10) </span>
<span>= 0.026 N </span>
<span>F = 0.052 N </span>

<span>E = F/q = 0.052 / 60µ = 867 N/C </span>

<span>(b) </span>
<span>T = F cos 30.0° </span>
<span>= 0.0450 N</span>

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3 years ago

A meter stick is place in a very high-speed spaceship. What length would the astronauts say the meter stick was? What would the

Marysya12 [62]

Answer:

a) The astronauts would see the real length of the meter stick, i.e. L₀

b) The length of the meter stick as measured by the stationary observer will be $L = L_{0} }{\sqrt{(1-(\frac{v}{c} )^{2} } }$

Explanation:

a) Let the proper length of the meter stick be L₀

The meter stick and the astronauts on the on the space ship are on the same moving frame, therefore, they will see the exact length of the meter stick, that is, L₀

b) A stationary observer watching the space ship and meter stick travel past them will see a contracted length of the meter stick

The original length = L₀

Let the speed of the space ship = v

The contracted length, L, is related to the original length in the frame of rest by

L = L₀/γ......................(1)

Where γ = $\frac{1}{\sqrt{(1-(\frac{v}{c} )^{2} } }$ ....................(2)

Substituting equation (2) into (1)

$L = L_{0} }{\sqrt{(1-(\frac{v}{c} )^{2} } }$

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3 years ago

A neuron that is activated when a mosquito lands on your arm

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3 years ago

A parallel plate capacitor is constructed with a dielectric slab with κ = 1.5 inserted between the plates. The area of each plat

maksim [4K]

Answer:

The value is $V = 4.533 \ V$

Explanation:

From the question we are told that

The dielectric constant is k = 1.5

The area of each plate is $A = 5 \ cm^2 = 0.0005 m^2$

The distance between the plates is $d= 1 \ mm = 0.001 \ m$

The charge on the capacitor is $Q = 6*10^{-11} \ C$

Generally the electric field in a vacuum is mathematically represented as

$E_o = \frac{V_o}{d}$

Generally $V_o$ is the voltage of the capacitor which is mathematically represented as

$V_o = \frac{Q}{C_o}$

Here $C_o$ is the capacitance of the capacitor in a vacuum which is mathematically represented as

$C_o = \frac{\epsilon_o * A}{d}$

Here $epsilon_o$ is a constant with value $epsilon_o= 8.85*10^{-12} C^2 \cdot N^{-1} \cdot m^{-2}$

=> $C_o = \frac{8.85*10^{-12} * 0.0005 }{0.001}$

=> $C_o = 4.425 *10^{-12} \ F$

So

$V_o = \frac{6*10^{-11} }{ 4.425 *10^{-12}}$

$V_o = 13.6 \ V$

So

$E_o = \frac{ 13.6}{0.001}$

=> $E_o = 13600 \ V/m$

The electric field when the dielectric slab is inserted is mathematically represented as

$E = \frac{E_o}{k}$

=> $E = \frac{13600}{1.5}$

=> $E =9067 \ V/m$

Generally the electric field between the plates is mathematically evaluated as

$E_{N} = E_o - E$

=> $E_{N} = 13600- 9067$

=> $E_{N} = 4533 \ V/m$

Generally the potential difference between the plates is

$V = 4533 * 0.001$

=> $V = 4.533 \ V$

6 0

4 years ago