All categories

3 years ago

All describe the relationship of the parts of an electrical current except: _________.

2 answers:

6 0

<span>As we know that
I = V / R
because electric current is measured in amperage
so it will be
</span> Electric current decreases as amperage increases
so correct option is A
HOPE IT HELPS

ehidna [41]3 years ago

5 0

Choice-A is not true. B and C are.

You might be interested in

8. How many calories of heat energy (Q) would be released as a 2,500 g iron (Fe) frying pan at 190˚C cools to 25˚C?

Sedaia [141]

Answer:

$Q=197505J$

Explanation:

From the question we are told that:

Mass $m=2500g=2.5kg$

Initial heat of pan $i_h=190 \tetxtdegree$

Final heat of pan $f_h=25˚C \tetxtdegree$

Generally the equation for Heat energy released is mathematically given by

$Q=mC \triangle T\\\\Q=2.5*462 *(190-26)\\\\Q=1155 *(190-26)\\\\Q=1155 *(171)\\$

$Q=197505J$

3 0

3 years ago

Where did the universe come from?

Leokris [45]

No one is entirely sure of where the universe came from. Scientists have come up with several different theories relating to your question, including the Big Bang Theory, but there is no definite answer.

5 0

4 years ago

An athlete swings a ball, connected to the end of a chain, in a horizontal circle. The athlete is able to rotate the ball at the

garik1379 [7]

(a) 6.04 rev/s

The speed of the ball is given by:

$v=\omega r$

where

$\omega$ is the angular speed

r is the distance of the ball from the centre of the circle

In situation 1), we have

$\omega=8.13 rev/s \cdot 2\pi = 51.0 rad/s$

r = 0.600 m

So the speed of the ball is

$v=(51.0 rad/s)(0.600 m)=30.6 m/s$

In situation 2), we have

$\omega=6.04 rev/s \cdot 2\pi = 37.9 rad/s$

r = 0.900 m

So the speed of the ball is

$v=(37.9 rad/s)(0.900 m)=34.1 m/s$

So, the ball has greater speed when rotating at 6.04 rev/s.

(b) $1561 m/s^2$

The centripetal acceleration of the ball is given by

$a=\frac{v^2}{r}$

where

v is the speed

r is the distance of the ball from the centre of the trajectory

For situation 1),

v = 30.6 m/s

r = 0.600 m

So the centripetal acceleration is

$a=\frac{(30.6 m/s)^2}{0.600 m}=1561 m/s^2$

(c) $1292 m/s^2$

For situation 2 we have

v = 34.1 m/s

r = 0.900 m

So the centripetal acceleration is

$a=\frac{v^2}{r}=\frac{(34.1 m/s)^2}{0.900 m}=1292 m/s^2$

5 0

3 years ago

After the ball reaches the top and begins its return back down, by how much does it speed increase each second?

FrozenT [24]

After the ball reaches the top and begins its return back down, it's just a falling object that's been dropped.

The amount that its downward speed increases every second is just the acceleration of gravity on Earth.

That's <em>9.81 meters per second</em> every second.

7 0

4 years ago

What is the purpose of a circuit breaker?

sashaice [31]

It’s designed to protect an electrical circuit from damage caused by overcurrent, usually resulting from an overload or short circuit. Its basic function is to interrupt current flow after a fault is detected.

That’s really just the basic purpose.
Happy to help!
~Brooke❤️

8 0

4 years ago