Ask question

Ask question

All categories

2 years ago

8

While working on her science fair project Venus connected a battery to a circuit that contained a light bulb. Venus decided to c

hange the light bulb to a higher resistance, but she wanted to keep the current the same. What will Venus need to do?

1 answer:

inessss [21]2 years ago

3 0

Answer:

<em>Venus will need to use more voltage to have the same current</em>

Explanation:

<u>Ohm's Law</u>

It states that the current through a conductor is directly proportional to the voltage applied to it and inversely proportional to the resistance of the conductor. This relationship can be written as

$\displaystyle I=\frac{V}{R}$

Solving for V

$V=IR$

Venus originally connected a resistance $R_1$ to a voltage $V_1$ to get some current I

$\displaystyle I=\frac{V_1}{R_1}$

The voltage needed to get that current is

$V_1=IR_1$

If Venus changes the light bulb to another with higher resistance, say, $R_2$ , then the current would change, unless she uses another voltage

$V_2=IR_2$

$Since\ R_2>R_1,\ then\ V_2>V_1$

Venus will need to use more voltage to have the same current

You might be interested in

Shows the position-versus-time graph of a particle in SHM. Positive direction is the direction to the right.

BaLLatris [955]

A) t = 0 s, 4 s, 8 s

B) t = 2 s, 6 s

C) t = 1 s, 3 s, 5 s, 7 s

Explanation:

A)

The figure is missing: find it in attachment.

A particle is said to be in Simple Harmonic Motion (SHM) when it is acted upon a restoring force proportional to its displacement, and therefore, the acceleration of the particle is directly proportional to its displacement (but in the opposite direction):

$a\propto -x$

Also, the displacement of a particle in SHM can be described by a sinusoidal function, as shown in the figure for the particle in this problem.

For the particle in this problem, from the figure we can see that the displacement is described by a function in the form

$x(t)=A sin(\omega t)$

where

A is the amplitude

$\omega=\frac{2\pi}{T}$ is the angular frequency, where T is the period. From the graph, we see that the particle completes 1 oscillation in 4 seconds, so

T = 4 s

So the angular frequency is

$\omega = \frac{2\pi}{4}=\frac{\pi}{2}rad/s$

So

$x(t)=Asin(\frac{\pi}{2}t)$

The velocity of a particle in SHM can be found as the derivative of the displacement, so here we find:

$v(t)=x'(t)=A\omega cos(\omega t) = \frac{A\pi}{2}cos(\frac{\pi}{2}t)$

So, the particle is moving to the right at maximum speed when

$cos(\frac{\pi}{2}t)=+1$

(because this is the maximum positive value for the cosine part). Solving for t,

$\frac{\pi}{2}t=0\\t=0 s$

We also have another time in which this occurs, when

$\frac{\pi}{2}t=2\pi\\\rightarrow t=\frac{4\pi}{\pi}=4 s$

And also when

$\frac{\pi}{2}t=4\pi\\\rightarrow t=8s$

B)

In this case, we want to find the time t at which the particle is moving to the left at maximum speed.

This occurs when the cosine part has the maximum negative value, so when

$cos(\frac{\pi}{2}t)=-1$

Which means that this occurs when:

$\frac{\pi}{2}t=\pi\\\rightarrow t=2 s$

Also when

$\frac{\pi}{2}t=3\pi\\\rightarrow t=6 s$

So, the particle has maximum speed moving to the left at 2 s and 6 s.

C)

The particle is instantaneously at rest when the speed is zero, this means when the cosine part is equal to zero:

$cos(\frac{\pi}{2}t)=0$

This occurs when the argument of the cosine is:

$\frac{\pi}{2}t=\frac{\pi}{2}\\\rightarrow t = 1 s$

Also when

$\frac{\pi}{2}t=\frac{3\pi}{2}\\\rightarrow t = 3 s$

And when

$\frac{\pi}{2}t=\frac{5\pi}{2}\\\rightarrow t = 5 s$

And finally when

$\frac{\pi}{2}t=\frac{7\pi}{2}\\\rightarrow t = 7s$

So, the particle is at rest at t = 1 s, 3 s, 5 s, 7 s.

3 0

2 years ago

Two identical closely spaced circular disks form a parallel-plate capacitor. Transferring 2.1×109 electrons from one disk to the

valkas [14]

Answer:

Two identical closely spaced circular disks form a parallel-plate capacitor. Transferring 2.1×109 electrons from one disk to the other causes the electric field strength between them to be 1.6×105 N/C. What are the diameters of the disks?

Explanation:

Check attachment for solution

7 0

2 years ago

Read 2 more answers

When calculating the mechanical advantage of a lever, what two pieces of information are needed?

DIA [1.3K]

From the items on this list, the only one that allows calculation
of the mechanical advantage is 'B' ... the lengths from the fulcrum
to the effort and the resistance.

The MA can also be calculated when you know the two forces ...
the effort and the resistance ... when the lever is just balanced.

4 0

2 years ago

Read 2 more answers

Bill drives and sees a red light. He slows down to a stop. A graph of his velocity over time is shown below.

antoniya [11.8K]

Answer:

-2 m/s^2

Explanation:

Acceleration is equal to the slope of the graph. You just find the slope of that section. The rise is -20 and the run is 10, so you get -2.

5 0

2 years ago

Read 2 more answers

An green hoop with mass mh = 2.8 kg and radius rh = 0.13 m hangs from a string that goes over a blue solid disk pulley with mass

Otrada [13]

The only force on the system is the mass of the hoop F net = 2.8kg*9.81m/s^2 = 27.468 N The mass equal of the rolling sphere is found by: the sphere rotates around the contact point with the table.
So by applying the theorem of parallel axes, the moment of inertia of the sphere is computed by:I = 2/5*mR^2 for rotation about the center of mass + mR^2 for the distance of the axis of rotation from the center of mass of the sphere.
I = 7/5*mR^2 M = 7/5*m
Therefore, linear acceleration is computed by:F/m = 27.468 / (2.8 + 1/2*2 + 7/5*4) = 27.468/9.4 = 2.922 m/s^2

7 0

2 years ago