Ask question

Ask question

All categories

1 year ago

8

A 25-cm rod moves at 5. 0 m/s in a plane perpendicular to a magnetic field of strength 0. 25 t. the rod, velocity vector, and ma

gnetic field vector are mutually perpendicular, as?

1 answer:

Bad White [126]1 year ago

5 0

Length L = 25 cm = 0.25 m, B = 600 G = 0.06 T ( 1G = 0.0001 T)

emf= 10 V

Solution:

emf = vBL

v= emf / BL

5 = emf/ (0.25 T× 0.25 m)

emf = 0.3125 v

Magnetic field

The magnetic influence on moving electric charges, electric currents, and magnetic materials is described by a magnetic field, which is a vector field. A force perpendicular to the charge's own velocity and the magnetic field acts on it when the charge is travelling through a magnetic field.

To learn more about the magnetic field refer here:

brainly.com/question/23096032

#SPJ4

You might be interested in

A box slides down a frictionless plane inclined at an angle θ above the horizontal. The gravitational force on the box is direct

tamaranim1 [39]

Answer:

D) Vertically.

Explanation:

A free body diagram is used to represent all the forces acting in a body. forces like, the force of gravity as a result of the gravitational interaction between the object and the Earth (W), the frictional force opposite to the movement of the object ( $F_{r}$ ), the normal force due to the plane and the object (N) and the force applied to start the movement in a particular direction (F).

As is show in the free body diagram of the system, W, which is the weight of the body as a consequence of the gravitational force, is at an angle $\theta$ below the inclined plane. that angle between the plane and the x axis is the same that the one of the inclined plane with respect to the horizontal, Since its sides are perpendicular.

Notice how W goes always in the direction to the center of mass of Earth in a vertical path (For comparison see figure (a) and (b)).

4 0

3 years ago

Question 1 of 5<br> In which way are electromagnetic waves different from mechanical waves?

Lina20 [59]

Answer:

Electromagnetic waves differ from mechanical waves in that they do not require a medium to propagate. This means that electromagnetic waves can travel not only through air and solid materials, but also through the vacuum of space.

Explanation:

8 0

2 years ago

Read 2 more answers

A third point charge q3 is now positioned halfway between q1 and q2. The net force on q2 now has a magnitude of F2,net = 14.413

natima [27]

Answer:

The value of charge q₃ is 40.46 μC.

Explanation:

Given that.

Magnitude of net force $F=14.413\ N$

Suppose a point charge q₁ = -3 μC is located at the origin of a co-ordinate system. Another point charge q₂ = 7.7 μC is located along the x-axis at a distance x₂ = 8.2 cm from q₁. Charge q₂ is displaced a distance y₂ = 3.1 cm in the positive y-direction.

We need to calculate the distance

Using Pythagorean theorem

$r=\sqrt{x_{2}^2+y_{2}^2}$

Put the value into the formula

$r=\sqrt{(8.2\times10^{-2})^2+(3.1\times10^{-2})^2}$

$r=0.0876\ m$

We need to calculate the magnitude of the charge q₃

Using formula of net force

$F_{12}=kq_{2}(\dfrac{q_{3}}{r_{3}^2}+\dfrac{q_{1}}{r_{1}^2})$

Put the value into the formula

$14.413=9\times10^{9}\times7.7\times10^{-6}(\dfrac{q_{3}}{(0.0438)^2}+\dfrac{-3\times10^{-6}}{(0.0876)^2})$

$(\dfrac{q_{3}}{(4.38\times10^{-2})^2}+\dfrac{-3\times10^{-6}}{(0.0876)^2})=\dfrac{14.413}{9\times10^{9}\times7.7\times10^{-6}}$

$\dfrac{q_{3}}{(0.0438)^2}=207\times10^{-4}+3.909\times10^{-4}$

$q_{3}=0.0210909\times(0.0438)^2$

$q_{3}=40.46\times10^{-6}\ C$

$q_{3}=40.46\ \mu C$

Hence, The value of charge q₃ is 40.46 μC.

5 0

3 years ago

To calibrate your calorimeter cup, you first put 45 mL of cold water in the cup, and measure its temperature to be 24.7 °C. You

drek231 [11]

Answer : The heat change of the cold water in Joules is, $1.6\times 10^3J$

Explanation :

First we have to calculate the mass of cold water.

As we know that the density of water is 1 g/mL. The volume of cold water is 45 mL.

$Density=\frac{Mass}{Volume}$

$Mass=Density\times Volume=1g/mL\times 45mL=45g$

Now we have to calculate the heat change of cold water.

Formula used :

$Q=m\times c\times (T_2-T_1)$

where,

Q = heat change of cold water = ?

m = mass of cold water = 45 g

c = specific heat of water = $4.184J/g^oC$

$T_1$ = initial temperature of cold water = $24.7^oC$

$T_2$ = final temperature = $33.4^oC$

Now put all the given value in the above formula, we get:

$Q=45g\times 4.184J/g^oC\times (33.4-24.7)^oC$

$Q=1638.036J=1.6\times 10^3J$

Therefore, the heat change of cold water is $1.6\times 10^3J$

4 0

3 years ago

A police car is driving down the street with it's siren on. You are standing still on the sidewalk beside the street. If the fre

AleksandrR [38]

Answer:

A) 1568.60 Hz

B) 1437.15 Hz

Explanation:

This change is frequency happens due to doppler effect

The Doppler effect is the change in frequency of a wave in relation to an observer who is moving relative to the wave source

$f_(observed)=\frac{(c+-V_r)}{(C+-V_s)} *f_(emmited)\\$

where

C = the propagation speed of waves in the medium;

Vr= is the speed of the receiver relative to the medium,(added to C, if the receiver is moving towards the source, subtracted if the receiver is moving away from the source;

Vs= the speed of the source relative to the medium, added to C, if the source is moving away from the receiver, subtracted if the source is moving towards the receiver.

A) Here the Source is moving towards the receiver(C-Vs)

and the receiver is standing still (Vr=0) therefore the observed frequency should get higher

$f_(observed)=\frac{C}{C-V_s} *f_(emmited)\\=\frac{343}{343-15}*1500\\ =1568.60 Hz$

B)Here the Source is moving away the receiver(C+Vs)

and the receiver is still not moving (Vr=0) therefore the observed frequency should be lesser

$f_(observed)=\frac{C}{C+V_s} *f_(emmited)\\=\frac{343}{343+15}*1500\\ =1437.15 Hz$

3 0

3 years ago