Which is the most abundant element in the Sun?

What is the path that an electric current follows called

Gwar [14]

I’m sure it’s called a circuit:)

5 0

3 years ago

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How many times smaller is the moon than earth?

Yuki888 [10]

The moon has approximately 1/4 of earths diameter, 1/50 of earths volume and 1/80 of earths mass

3 0

3 years ago

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# A cheetah can start from rest and attain the velocity 72km/h in 2 seconds. Calculate the acceleration of cheetah

yan [13]

Answer:

Explanation:

Solution,

When a certain object comes in motion from rest, in the case, initial velocity = 0 m/s

Initial velocity ( u ) = 0 m/s

Final velocity ( v ) = 72 km/h ( Given)

We have to convert 72 km /h in m/s

$72 \: km \: per \: hour$

$= 72 \times \frac{1000}{60 \times 60}$

$= 20$ m/s

Final velocity ( v ) = 20 m/s

Time taken ( t ) = 2 seconds

Acceleration (a) = ?

Now,

we have,

$a = \frac{v - u}{t}$

$a = \frac{20 - 0}{2}$

$a = \frac{20}{2}$

$a = 10$ m/s ^2

Hope this helps...

Good luck on your assignment..

7 0

3 years ago

A roller coaster pushes a 25 kg person upward with a force of 300 N. What is the acceleration?

vlada-n [284]

300N/25 kg= divide them for the answer

8 0

3 years ago

An isolated conducting sphere has a 17 cm radius. One wire carries a current of 1.0000020 A into it. Another wire carries a curr

notsponge [240]

14 ms is required to reach the potential of 1500 V.

<u>Explanation:</u>

The current is measured as the amount of charge traveling per unit time. So the charge of electrons required for each current is determined as the product of current with time.

$Charge = Current \times Time$

As two different current is passing at two different times, the net charge will be the different in current. So,

$\text { Charge }=(1.0000020-1.0000000) \times t=2 \times 10^{-6} \times t$

The electric voltage on the surface of cylinder can be obtained as the ratio of charge to the radius of the cylinder.

$V=\frac{k q}{R}$

Here $k = 9 * 10^9$ , q is the charge and R is the radius. As $q=2 \times 10^{-6} \times t$ and R =17 cm = 0.17 m, then the voltage will be

$V=\frac{9 \times 10^{9} \times 2 \times 10^{-6} \times t}{0.17}$

The time is required to find to reach the voltage of 1500 V, so

$1500 =\frac{9 \times 10^{9} \times 2 \times 10^{-6} \times t}{0.17}$

$\begin{aligned}&t=\frac{1500 \times 0.17}{\left(9 \times 10^{9} \times 2 \times 10^{-6}\right)}\\&t=14.1666 \times 10^{-3} s=14\ \mathrm{ms}\end{aligned}$

So, 14 ms is required to reach the potential of 1500 V.

3 0

3 years ago