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3 years ago

10

A capacitor is connected to a power source and begins charging. What causes the capacitor to stop charging while it is still con

nected to the power source?

1 answer:

Yuri [45]3 years ago

8 0

Charging stops when the voltage across the capacitor reaches the voltage of the power source.

You might be interested in

If an astronaut had a mass of 30 kg on the moon what would his mass be on earth?

Novay_Z [31]

Explanation:

A one-kilogram mass is still a one-kilogram(as mass is an intrinsic property of the object) but the downward force due to gravity, and therefore it's weight, is only one-sixth of what the object would have on the Earth. So man of mass 180 pounds weights only about 30 pounds-force when visiting the moon

hope it help..... pls add me as brainlist.

Have a nice day

7 0

3 years ago

Closing Summary Questions: Phases of Matter

snow_tiger [21]

Answer:

Explanation:

Particles in all states of matter are in constant motion and this is very rapid at room temperature. A rise in temperature increases the kinetic energy and speed of particles; it does not weaken the forces between them. The particles in solids vibrate about fixed positions; even at very low temperatures.

Even with all of these state changes, it is important to remember that the substance stays the same—it is still water, which consists of two hydrogen atoms and one oxygen atom. Changing states of matter are only physical changes; the chemical properties of the matter stays the same regardless of its physical state!

5 0

3 years ago

A bowling ball with a mass of 9kg is thrown down a lane with a constant speed of 3 m/s. The ball hits the 1.5kg pin, initially a

olasank [31]

Answer:

M1 V1 = M1 V2 + M2 V3 conservation of momentum

V2 = (M1 V1 - M2 V3) / M1 where V2 = speed of M1 after impact

V2 = (3 * 9 - 1.5 * 5) / 9 = (27 - 7.5) / 9 = 2.17 m/s

Note: All speeds are in the same direction and have the same sign

7 0

2 years ago

In April 1974, Steve Prefontaine completed a 10 km race in a time of 27 min, 43.6 s. Suppose "Pre" was at the 8.13 km mark at a

SOVA2 [1]

Answer: $0.084 m/s^2$

Explanation:

Given

Total time=27 min 43.6 s=1663.6 s

total distance=10 km

Initial distance $d_1=8.13 km$

time taken=25 min =1500 s

initial speed $v_1=\frac{8.13\times 1000}{25\times 60}=5.6 m/s$

after 8.13 km mark steve started to accelerate

speed after 60 s

$v_2=v_1+at$

$v_2=5.6+a\times 60$

distance traveled in 60 sec

$d_2=v_1\times 60+\frac{a60^2}{2}$

$d_2=336+1800 a$

time taken in last part of journey

$t_3=1663.6-1560=103.6 s$

distance traveled in this time

$d_3=v_2\times t_3$

$d_3=\left ( 5.6+a\times 60\right )103.6$

and total distance $=d_1+d_2+d_3$

$10000=8.13\times 1000+336+1800 a+\left ( 5.6+a\times 60\right )103.6$

$1870=336+1800 a+\left ( 5.6+a\times 60\right )103.6$

$a=0.084 m/s^2$

5 0

3 years ago

A positive kaon (K+) has a rest mass of 494 MeV/c² , whereas a proton has a rest mass of 938 MeV/c². If a kaon has a total energ

vitfil [10]

Answer:

<em>0.85c </em>

Explanation:

Rest mass of Kaon $M_{0K}$ = 494 MeV/c²

Rest mass of proton $M_{0P}$ = 938 MeV/c²

The rest energy is gotten by multiplying the rest mass by the square of the speed of light c²

for the kaon, rest energy $E_{0K}$ = 494c² MeV

for the proton, rest energy $E_{0P}$ = 938c² MeV

Recall that the rest energy, and the total energy are related by..

$E$ = γ $E_{0}$

which can be written in this case as

$E_{K}$ = γ $E_{0K}$ ...... equ 1

where $E$ = total energy of the kaon, and

$E_{0}$ = rest energy of the kaon

γ = relativistic factor = $\frac{1}{\sqrt{1 - \beta ^{2} } }$

where $\beta = \frac{v}{c}$

But, it is stated that the total energy of the kaon is equal to the rest mass of the proton or its equivalent rest energy, therefore...

$E_{K}$ = $E_{0P}$ ......equ 2

where $E_{K}$ is the total energy of the kaon, and

$E_{0P}$ is the rest energy of the proton.

From $E_{K}$ = $E_{0P}$ = 938c²

equ 1 becomes

938c² = γ494c²

γ = 938c²/494c² = 1.89

γ = $\frac{1}{\sqrt{1 - \beta ^{2} } }$ = 1.89

1.89 $\sqrt{1 - \beta ^{2} }$ = 1

squaring both sides, we get

3.57( 1 - $\beta^{2}$ ) = 1

3.57 - 3.57 $\beta^{2}$ = 1

2.57 = 3.57 $\beta^{2}$

$\beta^{2}$ = 2.57/3.57 = 0.72

$\beta = \sqrt{0.72}$ = 0.85

but, $\beta = \frac{v}{c}$

v/c = 0.85

v = <em>0.85c </em>

7 0

3 years ago