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2 years ago

In a circuit, what is responsible for lighting a bulb?

Physics

2 answers:

Yakvenalex [24]2 years ago

6 0

Answer:

3) flow of electrons, if im correct

icang [17]2 years ago

6 0

Answer:

Explanation:

The flow of electrons. A current means the flow of electrons and without a current there would be no electricity therefore the bulb would not light up.

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I need help for the first 3 plz.

madam [21]

1) hypothesis
2) data
3) method

I think these are correct.

7 0

3 years ago

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PLEASE HELP! thxssss<br>

Cerrena [4.2K]

Answer:

D is correct

Explanation:

5 0

3 years ago

1.33 m^3 of fluid flows out of a pipe in 24.5 s. the fluid leaves the pipe at 3.55 m/s. what is the area of the pipe. (unit=m^2)

Margaret [11]

Answer:

0.015 $meter^{2}$

Explanation:

Total volume of water coming out = 1.33 $meter^{3}$

Also volume = Cross sectional area*Length covered

Length covered = Velocity *time

=24.5*3.55

=86.97 meter

Let the cross sectional area be A.

1.33 = 86.97*A

A =0.015 $meter^{2}$

3 0

3 years ago

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A car's bumper is designed to withstand a 6.84 km/h (1.9-m/s) collision with an immovable object without damage to the body of t

den301095 [7]

Answer:

F = 5.256 x $10^{3} N$

Explanation:

From the work energy theorem we know that:

The net work done on a particle equals the change in the particles kinetic energy:

W = F.d, ΔK = $\frac{1}{2} mv^{2}_{f} - \frac{1}{2} mv^{2}_{i} , F.d = \frac{1}{2}mv^{2}_{f} -\frac{1}{2} mv^{2}_{i}$

where:

W = work done by the force

F = Force

d = Distance travelled

m = Mass of the car

vf, vi = final and initial velocity of the car

kf, ki = final and initial kinetic energy of the car

Given the parameters;

m = 830kg

vi = 1.9 m/s

vf = 0 km/h

d = 0.285 m

Inserting the information we have:

F.d = $\frac{1}{2} mv^{2}_{f} - \frac{1}{2} mv^{2}_{i}$

F = $\frac{\frac{1}{2} mv^{2}_{f} - \frac{1}{2} mv^{2}_{i} }{d}$

F = $\frac{ 0 - \frac{1}{2} X830 X 1.9^{2} }{0.285}$

F = 5.256 x $10^{3} N$

3 0

3 years ago

A bowler throws a bowling ball of radius R = 11 cm along a lane. The ball slides on the lane with initial speed vcom,0 = 6.0 m/s

ankoles [38]

Answer:

Explanation:

Radius of the ball is $R=11cm=0.11m$

Initial speed of the ball is $v_{com0}=6.0m/s$

Initial angular speed of the ball is $\omega = 0$

Coefficient of kinetic friction between the ball and the lane

is $\mu =0.35$

Due to the presence of frictional force, ball moves with

decreasing velocity.

(a)

velocity $v_{com0}$ in terms of $\omega$ is

$V_{com0} = -R\omega\\\\=-(0.11m)\omega\\\\= (-0.11\omega)m/s$

(b)

Ball's linear acceleration is given by

$a=-\mu g\\\\=-(0.35) (9.8 m/s^2)\\\\= -3.43m/s^2$

(c)

During sliding, ball's angular acceleration is calculated as

$\alpha=-\frac{\tau}{I}\\\\-\frac{\mu mgR}{(\frac{2}{5}mR^2)}\\\\-\frac{2}{5}\frac{\mu g}{R}\\\\-\frac{2}{5}\frac{(0.35)(9.8)}{0.11}\\\\-77.95rad/s^2$

(d)

The time for which the ball slides is calculated from the

equation of motion is

$V_{cm}= V_{cm0} + at\\\\V_{cm} = V_{cm0} + (-\mu g )t\\\\-0.11\omega=6.0m/s -(3.43m/s^2 )t\\\\-(0.11) (\alpha t) =6.0 m/s - (3.43 m/s^2)\\\\- (0.11)(-77.95 rad/s^2)t = 6.0m/s - (3.43 m/s^2 )t\\\\8.5745t + 3.43t= 6.0\\\\12.0045t = 6.0\\\\t= 0.4998s$

(e)

Distance traveled by the ball is

$X= V_{com,0}+ \frac{1}{2}at^2\\\\= (6.0m/s)(0.4998 s)+ 0.5(-3.43m/s^2) (0.4998 s)^2\\\\=2.57m$

(for)

The speed of the ball when smooth rolling begins is

$V_{cm} = V_{com, 0}+ at\\\\=6.0 m/s +(-3.43m/s^2 )(0.4998 s)\\\\= 4.29m/s$

5 0

3 years ago

Read 2 more answers