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notsponge [240]
3 years ago
14

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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Over the past 150 years, what has happened to the amount of forest cover in Minnesota?
Airida [17]
I think D. None of the above

I think that it
4 0
3 years ago
The coil of a generator has a radius of 0.14 m. When this coil is unwound, the wire from which it is made has a length of 10.0 m
adell [148]

Answer:

5.565 V

Explanation:

Radius of coil of generator=r=0.14 m

Length of wire=l=10 m

Magnetic field,B=0.24 T

Angular speed,\omega=34rad/s

We have to find the peak emf of the generator.

N=\frac{l}{2\pi r}=\frac{10}{2\pi\times 0.14}=11

A=\pi r^2=\pi (0.14)^2=0.062m^2

Peak(maximum) induced emf of generator=E_{max}=NBA\omega

Using the formula

E_{max}=11\times 0.24\times 0.062\times 34

E_{max}=5.565 V

3 0
3 years ago
A physics student throws a softball straight up into the air. The ball was in the air for a total of 3.56 s before it was caught
meriva

Answer:

The initial velocity of the softball is 14.711 meters per second.

Explanation:

This is a case of an object which experiments a free fall, that is, an uniform accelerated motion due to gravity and in which effects from air friction and Earth's rotation can be neglected.

From statement we must understand that the student threw the softball upwards and it is caught at original position 3.56 seconds later. Initial and final heights, time and gravitational acceleration are known and initial speed is unknown. The following equation of motion is used:

y = y_{o} + v_{o}\cdot t + \frac{1}{2}\cdot g \cdot t^{2} (Eq. 1)

Where:

y_{o} - Initial height of the softball, measured in meters.

y - Final height of the softball, measured in meters.

v_{o} - Initial velocity of the softball, measured in meters per second.

t - Time, measured in seconds.

g - Gravitational acceleration, measured in meters per square second.

If we know that y = y_{o}, t = 3.56\,s and g = -9.807\,\frac{m}{s^{2}}, the initial velocity of the softball is:

v_{o}\cdot (3\,s)+\frac{1}{2}\cdot (-9.807\,\frac{m}{s^{2}} )\cdot (3\,s)^{2} = 0

3\cdot v_{o} -44.132\,m= 0

v_{o} = 14.711\,\frac{m}{s}

The initial velocity of the softball is 14.711 meters per second.

8 0
3 years ago
You push very hard on a heavy desk, trying to move it. You do work on the desk:
frozen [14]

Answer:

(C) Only if it starts moving

Explanation:

We know that work done is given by

W=F.d=Fdcos\Theta

So there are two case in which work done is zero

First case is that when force and displacement are perpendicular to each other

And other case is that when there is no displacement

So for work to be done there must have displacement, if there is no displacement then there is no work done

So option (c) will be the correct option

3 0
3 years ago
What is the distance of separation between objects of masses 5.6 x 10 5 kg and 8.8 x 10 6 kg if the force of gravity between the
AlladinOne [14]

Answer:

2.87m

Explanation:

Using the law of gravitation to solve this question

F = GMm/r²

G is the gravitational constant

M and m are the masses

r is the distance between the masses

Substitute the given values

G = 6.67×10^-11 m³/kgs²

M =8.8 x 10^6 kg

m = 5.6 x 10^5 kg

F =440N

400 = 6.67×10^-11×8.8 x 10^6 ×5.6 x 10^5/r²

400r² = 328.698×10

400r² = 3286.98

r² = 3286.98/400

r² = 8.21745

r = √8.21745

r = 2.87m

Hence the distance of separation is 2.87m

7 0
3 years ago
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