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3 years ago

15

What is the current if 3600 C of charge passes a point in 15 min.

1 answer:

Butoxors [25]3 years ago

5 0

I believe the correct answer is B

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Matter that has a fixed composition is a ________. Which is formed by_______.

Alenkasestr [34]

A) Pure substances, Chemical changes

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4 years ago

Which of the following is an example of energy changing from one form to another?

Anastaziya [24]

The correct option is

a fire

In fact, a fire is a conversion of chemical energy (contained in the molecules of the initial substances) into thermal energy (the heat released by the fire itself), therefore this is an example of energy changing from one form to another. All the other objects, instead, do not represent any form of energy conversion.

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4 years ago

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Which of these is an organism??

elena55 [62]

You need to add a picture or answers!

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3 years ago

A baseball is thrown 50.0m in 3.0s, what is the average speed of the baseball? Calculated answer is 16.666667! What is the corre

castortr0y [4]

Answer:

20m/s

Explanation:

Given parameters:

Distance of throw = 50m

Time = 3s

Unknown:

Average speed = ?

Solution:

Average speed is distance divided by time;

Average speed = $\frac{distance }{time}$

Insert the parameters and solve;

Average speed = $\frac{50}{3}$ = 16.66667m/s = 20m/s

5 0

3 years ago

A proton is accelerated from rest through a potential difference V0 and gains a speed v0. If it were accelerated instead through

Svet_ta [14]

Answer:

The speed is $\sqrt{2}v_{0}$ .

(a) is correct option.

Explanation:

Given that,

Potential difference $V= V_{0}$

Speed $v = v_{o}$

If it were accelerated instead

Potential difference $V'=2V_{0}$

We need to calculate the speed

Using formula of initial work done on proton

$W = q V$

We know that,

$\Delta W=\Delta K.E$

$q V=\dfrac{1}{2}mv^2$

Put the value into the formula

$q V_{0}=\dfrac{1}{2}mv_{0}^2$

$v_{0}^2=\dfrac{2qV_{0}}{m}$ ....(I)

If it were accelerated instead through a potential difference of $2 V_{0}$ , then it would gain a speed will be given as :

Using an above formula,

$v_{0}'^2=\dfrac{2qV_{0}}{m}$

Put the value of $V_{0}$

$v_{0}'^2=\dfrac{2q\times2V_{0}}{m}$

$v_{0}'=\sqrt{\dfrac{4qV_{0}}{m}}$

$v_{0}'=\sqrt{2}v_{0}$

Hence, The speed is $\sqrt{2}v_{0}$ .

6 0

3 years ago