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3 years ago

Which of the following will increase if the voltage in a circuit is increased?

2 answers:

5 0

If the voltage used to operate a circuit is increased, then
the current flowing at every point in the circuit will increase,
and the power radiated by the components in the circuit will
also increase.

If you increase the voltage TOO much, then you may need to
start cooling the circuit to prevent fire, or to prevent components
from failing.

babymother [125]3 years ago

4 0

The answer is A and D.

The relationship between power current and voltage is given by the following equation;

P=IV

P = power

V = voltage

I = current

When the voltage (V is increased and the current (I) is maintained, then the power will definitely increase.

Increasing the voltage also increases the distance that the power can be transmitted with little attenuation or loss. This is why voltage is usually stepped up (which will reduce current) significantly at power stations before it's transmission over long distances.

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Zero, a hypothetical planet, has a mass of 5.3 x 1023 kg, a radius of 3.3 x 106 m, and no atmosphere. A 10 kg space probe is to

Andrej [43]

(a) $3.1\cdot 10^7 J$

The total mechanical energy of the space probe must be constant, so we can write:

$E_i = E_f\\K_i + U_i = K_f + U_f$ (1)

where

$K_i$ is the kinetic energy at the surface, when the probe is launched

$U_i$ is the gravitational potential energy at the surface

$K_f$ is the final kinetic energy of the probe

$U_i$ is the final gravitational potential energy

Here we have

$K_i = 5.0 \cdot 10^7 J$

at the surface, $R=3.3\cdot 10^6 m$ (radius of the planet), $M=5.3\cdot 10^{23}kg$ (mass of the planet) and m=10 kg (mass of the probe), so the initial gravitational potential energy is

$U_i=-G\frac{mM}{R}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{3.3\cdot 10^6 m}=-1.07\cdot 10^8 J$

At the final point, the distance of the probe from the centre of Zero is

$r=4.0\cdot 10^6 m$

so the final potential energy is

$U_f=-G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{4.0\cdot 10^6 m}=-8.8\cdot 10^7 J$

So now we can use eq.(1) to find the final kinetic energy:

$K_f = K_i + U_i - U_f = 5.0\cdot 10^7 J+(-1.07\cdot 10^8 J)-(-8.8\cdot 10^7 J)=3.1\cdot 10^7 J$

(b) $6.3\cdot 10^7 J$

The probe reaches a maximum distance of

$r=8.0\cdot 10^6 m$

which means that at that point, the kinetic energy is zero: (the probe speed has become zero):

$K_f = 0$

At that point, the gravitational potential energy is

$U_f=-G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{8.0\cdot 10^6 m}=-4.4\cdot 10^7 J$

So now we can use eq.(1) to find the initial kinetic energy:

$K_i = K_f + U_f - U_i = 0+(-4.4\cdot 10^7 J)-(-1.07\cdot 10^8 J)=6.3\cdot 10^7 J$

3 0

2 years ago

A boat moves initially at 10 m/s, then accelerates at 2 m/s^2 for 10 s. What is the velocity of the boat after 10 s?

bija089 [108]

Answer:

Explanation:

acceleration is the change in velocity of a body with respect to time

a = v-u/t

v is the final velocity

u is the initial velocity = 10m/s

t is the time taken = 10s

a is the acceleration = 2m/s²

substitute the values into the formula

2 = v-10/10

cross multiply

20 = v-10

v = 20+10

v = 30m/s

Hence the velocity of the body after 10s is 30m/s

6 0

3 years ago

In a physics lab a 90 kg student runs up a 1.5 meter tall flight of stairs in 3.0 seconds. The students power rating is approxim

wariber [46]

The power generated by the students will be .
•Work done is equal to
Mgh = 90*9.8*1.5
1587.6 j
thus power is equal to work done by time
1587.6/3
529.2 j/secs

7 0

3 years ago

Which answer is a scientifically accurate description of velocity?

Lyrx [107]

The boat traveled from the dock north to the 200-meter marker in the bay in less than 5 minutes, giving the passengers several more hours to fish.

Explanation:

Velocity is the rate of change of displacement with time. It is a physical quantity that has both magnitude and direction.

When expressing velocity, we must consider the directional attribute as it distinguishes it from speed.

Statements:

The hiker began hiking the park's five-mile north hiking trail at 8:00 a.m. with plenty of daylight left to complete the hike before dark.

We are given the displacement of the hiker but the duration of traveling was not indicated. The starting time is known and there is nothing we can infer from it

The motorcyclist traveled along a highway at 55 mph and reached the campsite in less than the two-hour estimated time.

55mph is the speed of the motorcyclist and not velocity. There was no directional attribute.

Walking south along the two-mile canyon trail generally takes less time than traveling the same trail north because walking south is all downhill

This is not quantifiable.

The boat traveled from the dock north to the 200-meter marker in the bay in less than 5 minutes, giving the passengers several more hours to fish.

The 200m marker north is the displacement and the time is 5minutes. This is the most scientifically accurate description of velocity.

Learn more:

Motion brainly.com/question/10883914

#learnwithBrainly

4 0

3 years ago

Patients generally report conscious awareness of posthypnotic suggestions<br> true or false

olga55 [171]

False

I hope that helped

5 0

3 years ago

Read 2 more answers