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umka21 [38]
3 years ago
8

** PLEASE PLEASE HELP A student uses the right-hand rule as shown.

Physics
2 answers:
guapka [62]3 years ago
7 0

As per right hand thumb rule if we put our thumb of right hand along the current in the wire then curl of fingers will show the direction of magnetic field.

So here as we can see the fingers in the above case is curled such that in front position its showing towards right.

So the direction of magnetic field near the student in front position must be towards Right.

elena-14-01-66 [18.8K]3 years ago
5 0

What is the direction of the magnetic field in front of the wire closest to the student?


B) Right

                                                                                                                                           

                                                                                                                                                 

                                                                                                                                               

                                                                                                                                                                                                                                                                                                       

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An experimenter finds that standing waves on a string fixed at both ends occur at 24 Hz and 32 Hz , but at no frequencies in bet
Vera_Pavlovna [14]

Answer:

8 Hz

Explanation:

Given that

Standing wave at one end is 24 Hz

Standing wave at the other end is 32 Hz.

Then the frequency of the standing wave mode of a string having a length, l, is usually given as

f(m) = m(v/2L), where in this case, m could be 1. 2. 3. 4 etc

Also, another formula is given as

f(m) = m.f(1), where f(1) is the fundamental frequency..

Thus, we could say that

f(m+1) - f(m) = (m + 1).f(1) - m.f(1) = f(1)

And as such,

f(1) = 32 - 24

f(1) = 8 Hz

Then, the fundamental frequency needed is 8 Hz

4 0
3 years ago
A box slides down a frictionless incline, gaining speed. The work done by the normal force n is _______.
jeka57 [31]

The work done by the normal force n when the box slides down a frictionless incline and gaining speed is zero.

<h3>What is normal force?</h3>

The force of contact is called the normal force. When the two surfaces are in contact with each other, then the normal force acts.

This force is applied by the solid bodies on each other in order to prevent the passing through each other.

A box slides down a frictionless incline, gaining speed. For this box, the value of work done by normal force has to be found out. Let's analyze the given condition.

  • The body is gaining the speed, which means there is a change in kinetic energy.
  • The change in kinetic energy is equal to the work done.
  • The friction force is the product of coefficient of the friction and normal force.
  • The friction force for the given case is zero. Thus, the normal force must be equal to the zero.

Thus, the work done by the normal force n when the box slides down a frictionless incline and gaining speed is zero.

Learn more about the normal force here;

brainly.com/question/10941832

7 0
2 years ago
Read 2 more answers
One mole of an ideal gas does 3000 J of work on its surroundings as it expands isothermally to a final pressure of 1.00 atm and
taurus [48]

Answer:

(a) Initial volume will be 7.62 L

(b) Final temperature will be 303.85 K

Explanation:

We have given one mole of ideal gas done 3000 J

So work done W = 3000 J

Let initial volume is V_1 and initial pressure P_1=1atm ( As pressure is constant )

Final volume V_2=25L = 0.025 m^3

Number of moles n = 1

(B) From ideal gas of equation we know that PV=nRT

So 1.01\times 10^5\times0.025=1\times 8.31\times T

T = 303.85 Kelvin

(B) For isothermal process work done is equal to

W=nRTln\frac{V_2}{V_1}

3000=1\times 8.314\times 303.85\times ln\frac{0.025}{V_1}

ln\frac{0.025}{V_1}=1.1881

\frac{0.025}{V_1}=3.2808

V_2=0.00846m^3=7.62L

So initial volume will be 7.62 L

5 0
3 years ago
an object 50 cm high is placed 1 m in front of a converging lens whose focal length is 1.5 m. determine the image height (in cm)
Juli2301 [7.4K]

Given :

An object 50 cm high is placed 1 m in front of a converging lens whose focal length is 1.5 m.

To Find :

the image height (in cm).

Solution :

By lens formula :

\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}

Here, u = - 100 cm

f =  150 cm

\dfrac{1}{v} - \dfrac{1}{-100} = \dfrac{1}{150}\\\\v = - 300 \ cm

Now, magnification is given by :

m = \dfrac{v}{u} = \dfrac{h_i}{h_o}\\\\h_i = \dfrac{300}{100}\times 1\\\\h_i = 3 \ m

Therefore, the image height is 3 m or 300 cm.

5 0
3 years ago
A light wave has a wavelength of 0.015 m. What is the frequency of this light wave? (1 point)
Natalija [7]

Answer:

Did you get an answer?

Explanation:

6 0
4 years ago
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