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3 years ago

In the parallel circuit, is the current constant throughout the circuit, or does it vary?

1 answer:

Ber [7]3 years ago

8 0

The current in a parallel circuits is not constant throughout the circuit. It varies in each path of a parallel circuit. The components that is place across each other have the same voltage in a parallel circuit. Remember, V=RI.

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A car is moving with 15000J of energy. If the car has a mass of 1500kg, how fast is it moving?

Arte-miy333 [17]

Answer:

V=4.5 m/s

Explanation:

KE=1/2mv^2

1500=1/2(15000)v^2

6 0

3 years ago

the earth's moon has a gravitational field strength of about 1.6 n/kg near its surface. the moon has a mass of 7.35x10^22 kg. wh

Andrew [12]

Answer:

1750km

Explanation:

8 0

3 years ago

Each tire on a car has a radius of 0.330 m and is rotating with an angular speed of 11.7 revolutions/s. Find the linear speed v

Eduardwww [97]

Answer:

The linear speed of the car, v, is 24.26 m/s

Explanation:

Given;

radius of the car's tire, r = 0.330 m

angular speed of the car, ω = 11.7 revolutions/s

The angular speed of the car in radian per second:

$\omega = 11.7 \ \frac{rev}{s} \times \frac{2\pi \ rad}{1 \ rev} \\\\\omega = 73.523 \ rad/s$

The linear speed of the car, v, is calculated as;

v = ωr

v = 73.523 rad/s x 0.33 m

v = 24.26 m/s

Therefore, the linear speed of the car, v, is 24.26 m/s

7 0

3 years ago

A ring with an 18mm diameter falls off a scientist's finger into the solenoid in the lab. The solenoid is 25 cm long, 5.0 cm in

Troyanec [42]

Answer:

The value is $\epsilon = 3.84 *10^{-5} \ V$

Explanation:

From the question we are told that

The diameter of the ring is $d = 18 \ mm = 0.018 \ m$

The length of the solenoid is $l = 25 \ cm = 0.25 \ m$

The diameter of the solenoid is $D = 5.0 \ cm = 0.05 \ m$

The number of turns is N = 1500

The change in current in the solenoid is $\Delta I = 20 \ A$

The time taken is $\Delta t = 1 \ s$

Generally the radius of the ring is

$r = \frac{d}{2}$

=> $r = \frac{0.018 }{2}$

=> $r = 0.009 \ m$

Generally the area of the ring is mathematically represented as

$A = \pi r^2$

=> $A = 3.142 * 0.009^2$

=> $A = 2.545 *10^{-4}\ m^2$

Generally the induced emf is mathematically represented as

$\epsilon = A * \frac{dB}{dt}$

Here

$\frac{dB }{dt} = \mu_o * \frac{N}{l} *\frac{ \Delta I }{\Delta t}$

Here $\mu_o$ is the permeability of free space with value

$\mu_o = 4\pi *10^{-7} \ N/A^2$

$\frac{dB }{dt} = 4\pi * 10^{-7} * \frac{1500}{0.25} *\frac{20 }{1}$

=> $\frac{dB }{dt} = 0.150816\ T/s$

$\epsilon = 0.150816 * 2.545 *10^{-4}$

=> $\epsilon = 3.84 *10^{-5} \ V$

3 0

3 years ago

It is easier to roll a stone up a sloping road than to lift it vertical upwards because.

kvasek [131]

Answer:

Because the force one needs to apply is reduced.

Explanation:

Work can be expressed in terms of force and distance as follows,

Work = Force × distance

To be more precise, the selection of these two parameters should be in a manner that the distance is what the line of action of the force traces.
It is obvious that for a given amount of work when one parameter is changed the other changes to counteract the change of the first one.
Imagine the situation shown in the attached figure. The amount of work that needs to be done during the lifting of the mass to a height of $\small h$ vertically is $E = mg\times h = mgh$ .
On the slope, however, one needs to push the object an $L$ distance up to get the object to that $h$ height ultimately.
The work he does on the slope is $mgh = F\times L$ and since $L > h$ , $F < mg$ .
The force that needs to be applied is less than that applied in the vertical lift. Hence it feels easier. #SPJ4

3 0

1 year ago