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3 years ago

Which circuits are parallel circuits?

Physics

1 answer:

ANEK [815]3 years ago

6 0

Answer:

The bottom two lines.

Explanation:

They need their own line of voltage quantity. A parallel circuit has the definition of 'two or more paths for current to flow through.' The voltage does stay the same in each line.

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I need help finding the answer

miss Akunina [59]

Speed = (distance) / (time)

Speed = (2.3 m) / (3 sec)

Speed = (2.3/3) (m/s)

<em>Speed = 0.766... m/s</em>

5 0

3 years ago

A plastic cup weighing 100 g floats on water so that 1/4 of the volume of the cup is immersed in water. How much volume of oil c

k0ka [10]

Answer:

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2 years ago

4A. How high is a 12 kg monkey in a tree if it has 509 J of gravitational potential Energy?

True [87]

4A. PE = MxGxH. (You can consider g as 9.8 / 10m/s as well)

509 J = 12x10xH

509 J = 120xH

H = 509/120

H = 4.24 m

Hope u got the answer....pls rate the answer if it is helpful for u....and I'm sorry I could not understand B part so I didn't do it.

Thank you

8 0

3 years ago

A machine lifts a 35 kg object doing 6860 J worth of work. How much power is produced by the machine if it lifts the object in 4

timofeeve [1]

Answer:

Explanation:

We need the power equation for this which is

P = Work/time

We have everything we need to solve this (the mass of the object is extra information):

P = 6860/4

P = 1715W

5 0

3 years ago

The current supplied by a battery as a function of time is I(t) = (0.88 A) e^(-t*6 hr). What is the total number of electrons tr

Degger [83]

Answer:

e. 1.2 x 10²³

Explanation:

According to the problem, The current equation is given by:

$I(t)=0.88e^{-t/6\times3600s}$

Here time is in seconds.

Consider at t=0 s the current starts to flow due to battery and the current stops when the time t tends to infinite.

The relation between current and number of charge carriers is:

$q=\int\limits {I} \, dt$

Here the limits of integration is from 0 to infinite. So,

$q=\int\limits {0.88e^{-t/6\times3600s}}\, dt$

$q=0.88\times(-6\times3600)(0-1)$

q = 1.90 x 10⁴ C

Consider N be the total number of charge carriers. So,

q = N e

Here e is electronic charge and its value is 1.69 x 10⁻¹⁹ C.

N = q/e

Substitute the suitable values in the above equation.

$N= \frac{1.9\times10^{4} }{1.69\times10^{-19}}$

N = 1.2 x 10²³

8 0

3 years ago