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2 years ago

Which purpose does a light bulb serve in a circuit?

Physics

2 answers:

aliina [53]2 years ago

7 0

Answer:

A light bulb is that component of circuit that lights up when electricity flows through it and vice versa. Its main function is to tell whether electricity is being supplied or not. Its function is also to light up dark places.

Explanation:

mars1129 [50]2 years ago

6 0

Answer:

A light bulb is that component of circuit that lights up when electricity flows

Explanation:

You might be interested in

You apply a horizontal force of 25N to push a shopping cart across the parking lot at a constant velocity. a) what is the net fo

AlekseyPX

(a) The net force on the shopping cart is zero.

(b) The the force of friction on the shopping cart is 25 N.

<h3>Net force on the shopping cart</h3>

The net force on the shopping cart is calculated as follows;

F(net) = F - Ff

where;

F is the applied force
Ff is the frictional force

ma = F - Ff

where;

a is acceleration of the cart
m is mass of the cart

at a constant velocity, a = 0

0 = F - Ff

F(net) = 0

F = Ff = 25 N

Net force is zero, and frictional force is equal to applied force.

<h3>On wet surface</h3>

Coefficient of kinetic friction of solid surface is greater than that of wet surface.

Since frictional force limit motion, when the frictional force is smaller, the object tends to move faster.

Thus, the cart will move faster on a wet surface due to decrease in friction.

Learn more about frictional force here: brainly.com/question/24386803

#SPJ1

4 0

2 years ago

Please dont delete this question......

Juli2301 [7.4K]

Answer:

Explanation:

8 0

3 years ago

Read 2 more answers

A car traveling at a speed of 30.0 m/s encounters an emergency and comes to a complete stop. how much time will it take for the

Vedmedyk [2.9K]

Answer:7.5seconds

Explanation:

V=0 u=30m/s a=-4m/s^2

t=(v-u)/a

t=(0-30)/-4=-30/-4

t=7.5second

8 0

3 years ago

The velocity of a particle moving along the x-axis varies with time according to v(t) = A + Bt−1 , where A = 2 m/s, B = 0.25 m,

kondaur [170]

Answer:

$a= -2\ m/s^2$

$a=-12.5\ m/s^2$

x=2.17 m

x=8.4 m

Explanation:

Given that

$v=A+Bt^{-1}$

$v=2+0.25t^{-1}$

To find acceleration :

we know that

$a=\dfrac{dv}{dt}$

$\dfrac{dv}{dt}=0-0.5t^{-2}$

$a=-0.5t^{-2}$

Acceleration at t= 2 s

$a=-0.5\times 2^{-2}$

$a= -2\ m/s^2$

Acceleration at t= 5 s

$a=-0.5\times 5^{-2}$

$a=-12.5\ m/s^2$

We know that

$v=\dfrac{dx}{dt}$

$dx=\left(2+\dfrac{1}{4t}\right)dt$

Position at t= 2 s:

$\int_{0}^{x}dx=\int_{1}^{2} \left(2+0.25\dfrac{1}{t}\right)dt$

$x=\left [2t+0.25\ lnt \right ]_{1}^{2}$

x=2+0.25 ln2

x=2.17 m

Position at t= 5 s:

$\int_{0}^{x}dx=\int_{1}^{5} \left(2+0.25\dfrac{1}{t}\right)dt$

$x=\left [2t+0.25\ lnt \right ]_{1}^{5}$

x=8+0.25 ln5

x=8.4 m

4 0

3 years ago

Two particles with masses 2m and 9m are moving toward each other along the x axis with the same initial speeds vi. Particle 2m i

s2008m [1.1K]

Answer:

The final speed for the mass 2m is $v_{2y}=-1,51\ v_{i}$ and the final speed for the mass 9m is $v_{1f} =0,85\ v_{i}$ .

The angle at which the particle 9m is scattered is $\theta = -66,68^{o}$ with respect to the - y axis.

Explanation:

In an elastic collision the total linear momentum and the total kinetic energy is conserved.

<u>Conservation of linear momentum:</u>

Because the linear momentum is a vector quantity we consider the conservation of the components of momentum in the x and y axis.

The subindex 1 will refer to the particle 9m and the subindex 2 will refer to the particle 2m

$\vec{p}=m\vec{v}$

$p_{xi} =p_{xf}$

In the x axis before the collision we have

$p_{xi}=9m\ v_{i} - 2m\ v_{i}$

and after the collision we have that

$p_{xf} =9m\ v_{1x}$

In the y axis before the collision $p_{yi} =0$

after the collision we have that

$p_{yf} =9m\ v_{1y} - 2m\ v_{2y}$

$p_{xi} =p_{xf} \\7m\ v_{i} =9m\ v_{1x}\Rightarrow v_{1x} =\frac{7}{9}\ v_{i}$

then

$p_{yi} =p_{yf} \\0=9m\ v_{1y} -2m\ v_{2y} \\v_{1y}=\frac{2}{9} \ v_{2y}$

<u>Conservation of kinetic energy:</u>

$\frac{1}{2}\ 9m\ v_{i} ^{2} +\frac{1}{2}\ 2m\ v_{i} ^{2}=\frac{1}{2}\ 9m\ v_{1f} ^{2} +\frac{1}{2}\ 2m\ v_{2f} ^{2}$

$\frac{11}{2}\ m\ v_{i} ^{2} =\frac{1}{2} \ 9m\ [(\frac{7}{9}) ^{2}\ v_{i} ^{2}+ (\frac{2}{9}) ^{2}\ v_{2y} ^{2}]+ m\ v_{2y} ^{2}$

Putting in one side of the equation each speed we get

$\frac{25}{9}\ m\ v_{i} ^{2} =\frac{11}{9}\ m\ v_{2y} ^{2}\\v_{2y} =-1,51\ v_{i}$

We know that the particle 2m travels in the -y axis because it was stated in the question.

Now we can get the y component of the speed of the 9m particle:

$v_{1y} =\frac{2}{9}\ v_{2y} \\v_{1y} =-0,335\ v_{i}$

the magnitude of the final speed of the particle 9m is

$v_{1f} =\sqrt{v_{1x} ^{2}+v_{1y} ^{2} }$

$v_{1f} =\sqrt{(\frac{7}{9}) ^{2}\ v_{i} ^{2}+(-0,335)^{2}\ v_{i} ^{2} }\Rightarrow \ v_{1f} =0,85\ v_{i}$

The tangent that the speed of the particle 9m makes with the -y axis is

$tan(\theta)=\frac{v_{1x} }{v_{1y}} =-2,321 \Rightarrow\theta=-66,68^{o}$

As a vector the speed of the particle 9m is:

$\vec{v_{1f} }=\frac{7}{9} v_{i} \hat{x}-0,335\ v_{i}\ \hat{y}$

As a vector the speed of the particle 2m is:

$\vec{v_{2f} }=-1,51\ v_{i}\ \hat{y}$

8 0

3 years ago