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Sever21 [200]
2 years ago
13

When current first leaves the battery, how much voltage does it have

Physics
1 answer:
rjkz [21]2 years ago
8 0

Answer:

depends on the voltage of battery

Explanation:

for example if you connect a battery of 6V,6V will be  provided

You might be interested in
A wave with a wavelength of 125 meters is moving at a speed of 20 m/s. What is it’s frequency
IceJOKER [234]

Answer:

0.16Hz

Explanation:

wavelength (λ) = 125 meters

speed (V) = 20 m/s

frequency (F) = ?

Recall that frequency is the number of cycles the wave complete in one

second. And its value depends on the wavelength and speed of the wave.

So, apply the formula V = F λ

Make F the subject formula

F = V / λ

F = 20 m/s / 125 meters

F = 0.16 Hz

8 0
3 years ago
If you went to a planet with an acceleration due to gravity of 20 m/s^2 and twirled a bucket of water, what happens to the angul
wel

Answer:

B

Explanation:

It's Scientifically proven

7 0
3 years ago
A train has a constant velocity of 2 m/s. east what is the magnitude of the horizontal acceleration of the trian?
kupik [55]

Answer:

0 m/s²

Explanation:

The velocity is constant, so there is no acceleration.

4 0
3 years ago
How much work would it take to push two protons very slowly from a separation of 2.00×10−10m (a typical atomic distance) to 3.00
laiz [17]

Answer:

Work= -7.68×10⁻¹⁴J

Explanation:

Given data

q₁=q₂=1.6×10⁻¹⁹C

r₁=2.00×10⁻¹⁰m

r₂=3.00×10⁻¹⁵m

To find

Work

Solution

The work done on the charge is equal to difference in potential energy

W=ΔU

Work=U_{1}-U_{2}\\ Work=-kq_{1}q_{2}[\frac{1}{r_{2}}-\frac{1}{r_{1}} ]\\Work=(-9*10^{9})*(1.6*10^{-19} )^{2}[\frac{1}{3.0*10^{-15} }-\frac{1}{2*10^{-10} } ]\\  Work=-7.68*10^{-14}J

4 0
3 years ago
A trooper is moving due south along the freeway at a speed of 30.1 m/s when a red car passes the trooper. The red car moves with
romanna [79]

Answer:

The correct answer to the following question will be "41.87 m".

Explanation:

The given values are:

The speed of trooper = 30.1 \ m/s

The velocity of red car = 45.4 \ m/s

Now,

A red car goes as far as possible until the speed or velocity of the troops is the same as that of of the red car at

t=\frac{45-30}{2.7}                                                              (∵ time=\frac{distance}{time})

t=5.56 \ sec

then,

The distance covered by trooper,

t1=30\times 5.56+\frac{1}{2}\times 2.7\times (5.56)^2

   =208.33 \ m

The distance covered by red car,

= 45\times 5.56

= 250.2 \ m

Maximum distance = 250.2-208.33

                                = 41.87 \ m                                                        

6 0
3 years ago
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