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3 years ago

13

What If? The coil and applied magnetic field remain the same, but the circuit providing the current in the coil is now changed.

The new circuit has an emf of e m f = 20.0 V, and this allows a mass of 51.0 g added to the right side to balance the system. What is the value of the resistance R (in Ω)?

1 answer:

RUDIKE [14]3 years ago

5 0

Answer:

Using equations

R will be 1.25 Kilo ohm

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7-10/2<br><br> A.2<br> B.-13<br> C.12

vagabundo [1.1K]

=7-10/2

=7- 5

= 2

A. 2

Hope this helps

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3 years ago

Read 2 more answers

What do the arrows at point three indicate

Maru [420]

Maybe show a picture ? I don’t get the question .

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3 years ago

Which situations might cause two observers (A and B) to measure different frequencies for the same vibrating object? Select the

Alex787 [66]

We want to explain why two different observes may measure different frequencies for the same vibrating object.

We will see that the two correct options are:

<em>Observer A is stationary and Observer B is moving.</em>
<em>Observer A and Observer B are moving at different speeds relative to each other.</em>

<em />

Let's assume that the vibrating object is a guitar string. Thus, the string makes a noise, and from that noise, we can estimate the frequency at which the string vibrates.

Now there appears a really cool effect, called the Doppler Effect. It says that the apparent change of frequency is <u>due to the motion of the observer or the source of the frequency (or both).</u>

For example, if you move towards the vibrating string, the perceived frequency will be larger, and you will hear a "higher" sound.

While if you move away from the string, the opposite happens, and you will hear a "lower" sound.

Then the only thing that impacts in how we perceive the frequency is our velocity relative to the source.

So, why do observers A and B measure different frequencies?

The two correct answers are:

<em>Observer A is stationary and Observer B is moving.</em>
<em>Observer A and Observer B are moving at different speeds relative to each other.</em>

If you want to learn more, you can read:

brainly.com/question/17107808

6 0

3 years ago

A thin nonconducting rod with a uniform distribution of positive charge Q is bent into a complete circle of radius R. The centra

Dmitriy789 [7]

Answer:

(a). If z = 0, The electric field due to the rod is zero.

(b). If z = ∞, The electric field due to the rod is $E\propto\dfrac{1}{z^2}$ .

(c). The positive distance is $\dfrac{R}{\sqrt{2}}$

(d). The maximum magnitude of electric field is $1.54\times10^{4}\ N/C$

Explanation:

Given that,

Radius = 2.00 cm

Charge = 4.00 mC

(a). If the radius and charge are R and Q.

We need to calculate the electric field due to the rod

Using formula of electric field

$E=\dfrac{1}{4\pi\epsilon_{0}}\dfrac{Qz}{(z^2+R^2)^{\frac{2}{3}}}$

Where, Q = charge

z = distance

If z = 0,

Then, The electric field is

$E=0$

(b). If z = ∞, z>>R

So, R = 0

Then, the electric field is

$E=\dfrac{1}{4\pi\epsilon_{0}}\dfrac{Q}{z^2}$

$E\propto\dfrac{1}{z^2}$

(c). In terms of R,

We need to calculate the positive distance

If $E\rightarrow E_{max}$

Then, $\dfrac{dE}{dz}=0$

$\dfrac{Q}{4\pi\epsilon_{0}}(\dfrac{(z^2+R^2)^\frac{3}{2}-\dfrac{3z}{2}(z^2+R^2)^\dfrac{1}{2}}{(z^2+R^2)^2})=0$

Taking only positive distance

$z=\dfrac{R}{\sqrt{2}}$

(d). If R = 2.00 and Q = 4.00 mC

We need to calculate the maximum magnitude of electric field

Using formula of electric field

$E_{max}=\dfrac{1}{4\pi\epsilon_{0}}\dfrac{Qz}{(z^2+R^2)^{\frac{2}{3}}}$

$E_{max}=9\times10^{9}\times\dfrac{4.0\times10^{-6}\times\dfrac{2.00}{\sqrt{2}}}{((\dfrac{2.00}{\sqrt{2}})^2+(2.00)^2)^{\frac{2}{3}}}$

$E_{max}=15418.7\ N/C$

$E_{max}=1.54\times10^{4}\ N/C$

Hence, (a). If z = 0, The electric field due to the rod is zero.

(b). If z = ∞, The electric field due to the rod is $E\propto\dfrac{1}{z^2}$ .

(c). The positive distance is $\dfrac{R}{\sqrt{2}}$

(d). The maximum magnitude of electric field is $1.54\times10^{4}\ N/C$

6 0

3 years ago

A mass m0 is attached to a spring and hung vertically. The mass is raised a short distance in the vertical direction and release

iragen [17]

Answer:

The frequency of the oscillations in terms of fo will be f2=fo/3

E xplanation:

T= $2pie\frac sqrt {m}{k}$

$\frac {{f2}/times {fo}}$ =1:3

⇒f2=fo\3

Here frequency f is inversely poportional to square root of mass m.

so the value of remainder of frequency f2 and fo is equal to 1:3.

⇒ $\frac{f2} {f1}$ = $\frac sqrt{m1}[m2}$

⇒ $\frac{f2}{fo}$ = 1:3

⇒f2= $\frac{fo} {3}$

6 0

3 years ago