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2 years ago

8

If you want to conduct an electrical current, which situation would produce a solution capable of this? A) Dissolving water in o

il. B) Dissolving sugar in water. C) Dissolving solid NaBr in oil. D) Dissolving solid NaF in water.

2 answers:

Phoenix [80]2 years ago

8 0

D) Dissolving solid NaF in water.

svp [43]2 years ago

6 0

D dissolving solid NaF in water

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A car moves round a circular track of radius 0.3m of two revolution per/sec find its angular velocity.

Pie

Answer:

the angular velocity of the car is 12.568 rad/s.

Explanation:

Given;

radius of the circular track, r = 0.3 m

number of revolutions per second made by the car, ω = 2 rev/s

The angular velocity of the car in radian per second is calculated as;

From the given data, we convert the angular velocity in revolution per second to radian per second.

$\omega = 2 \ \frac{rev}{s} \times \frac{2\pi \ rad}{1 \ rev} = 4\pi \ rad/s = 12.568 \ rad/s$

Therefore, the angular velocity of the car is 12.568 rad/s.

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2 years ago

What role does carbon dioxide play for all living things

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Answer:

carbon dioxide plays an important part in vital plant and animal process, such as photosynthesis and respiration.

Explanation:

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What does a mechanical wave always travel through?

jarptica [38.1K]

A mechanical wave<span> requires an initial energy input. Once this initial energy is added, the </span>wave travels through<span> the medium until all its energy is transferred.</span>

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2 years ago

How much work is required to accelerate a proton from rest up to a speed of 0.993 c? Express your answer with the appropriate un

Keith_Richards [23]

Answer:

(a). The work done is 7001 MeV.

(b). The momentum of this proton is $4.20\times10^{8}\ kg-m/s$ .

Explanation:

Given that,

Speed = 0.993 c

We need to calculate the work done

Using work energy theorem

The work done is equal to the kinetic energy relative to the proton

$W=K.E$

$W=\dfrac{m_{p}c^2}{\sqrt{1-\dfrac{v^2}{c^2}}}-m_{p}c^2$

Put the value into the formula

$W=\dfrac{1.67\times10^{-27}\times(3\times10^{8})^2}{\sqrt{1-(\dfrac{0.993c}{c})^2}}-1.67\times10^{-27}\times(3\times10^{8})^2$

$W=\dfrac{1.67\times10^{-27}\times(3\times10^{8})^2}{\sqrt{1-(0.993)^2}}-1.67\times10^{-27}\times(3\times10^{8})^2$

$W=1.122\times10^{-9}\ J$

$W=7001\ MeV$

(b). We need to calculate the momentum of this proton

Using formula of momentum

$p=\dfrac{m_{0}v}{\sqrt{1-\dfrac{v^2}{c^2}}}$

Put the value into the formula

$p=\dfrac{1.67\times10^{-27}\times0.993c}{\sqrt{1-(\dfrac{0.993c}{c})^2}}$

$p=\dfrac{1.67\times10^{-27}\times0.993c}{\sqrt{1-(0.993)^2}}$

$p=1.404\times10^{-26}c$

$p=4.20\times10^{8}\ kg-m/s$

Hence, (a). The work done is 7001 MeV.

(b). The momentum of this proton is $4.20\times10^{8}\ kg-m/s$ .

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Answer:

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Explanation:

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