How to calculate ?Help!!!

A wire with a linear mass density of 1.17 g/cm moves at a constant speed on a horizontal surface and the coefficient of kinetic

stira [4]

Answer:

The value is $B = 0.2312 \ T$

The direction is into the surface

Explanation:

From the question we are told that

The mass density is $\mu =\frac{m}{L} = 1.17 \ g/cm =0.117 kg/m$

The coefficient of kinetic friction is $\mu_k = 0.250$

The current the wire carries is $I = 1.24 \ A$

Generally the magnetic force acting on the wire is mathematically represented as

$F_F = F_B$

Here $F_F$ is the frictional force which is mathematically represented as

$F_F = \mu_k * m * g$

While $F_B$ is the magnetic force which is mathematically represented as

$F_B = BILsin(\theta )$

Here $\theta =90^o$ is the angle between the direction of the force and that of the current

So

$F_B = BIL$

So

$BIL = \mu_k * m * g$

=> $B = \mu_k * \frac{m}{L} * [\frac{g}{I} ]$

=> $B = 0.25 * 0.117 * [\frac{9.8}{1.24} ]$

=> $B = 0.2312 \ T$

Apply the right hand curling rule , the thumb pointing towards that direction of the current we see that the direction of the magnetic field is into the surface as shown on the first uploaded image

8 0

3 years ago

A ball thrown by Ginger is moving upward through the air. Diagram A shows a box with a downward arrow. Diagram B shows a box wit

vichka [17]

As the ball is moving in air as well as we have to neglect the friction force on it

So we can say that ball is having only one force on it that is gravitational force

So the force on the ball must have to be represented by gravitational force and that must be vertically downwards

So the correct FBD will contain only one force and that force must be vertically downwards

So here correct answer must be

<em>Diagram A shows a box with a downward arrow. </em>

8 0

3 years ago

In the diagram, q1 = -6.39*10^-9 C and q2 = +3.22*10^-9 C. What is the electric field at point P? pls help

Alexxx [7]

Answer:

Below

Explanation:

First draw the vectors that represent both electric fields.

E1 is the elictric field created by q1, E2 is the one created by q2.

● q1 is negative so E1 will point from P.

● q2 is positive so E2 will point out of P

(Picture below)

■■■■■■■■■■■■■■■■■■■■■■■■■■

The resulting electric field is equal to the sum of the two fields since both vectors are colinear.

Let E be the total field.

● E = E1 + E2

The formula of the electric field intensity is:

● E = K ×(q/d^2)

-K is Coulomb's constant

-d is the distance between the charge and the object ( here P)

-q is the charge

■■■■■■■■■■■■■■■■■■■■■■■■■■

● E1 = K × (q1/d1^2)

The distance between q1 and P is the qum of 0.15 m 0.25 m. (0.4 m)

Coulombs constant is 9×10^9 m^2/C^2

● E1 = 9×10^9 ×[-6.39 × 10^(-9)/ 0.4^2]

● E1 = -359.43 N/C

■■■■■■■■■■■■■■■■■■■■■■■■■■

● E2 = K ×(q2/d^2)

The distance between q2 and P is 0.25 m.

● E2 = 9×10^9×[3.22×10^(-9) /0.25^2]

● E2 = 463.68 N/C

■■■■■■■■■■■■■■■■■■■■■■■■■■

● E = E1 + E2

● E = -359.43+463.68

● E = 105.25 N/C

4 0

3 years ago

Read 2 more answers

If Q = 16 nC, a = 3.0 m, and b = 4.0 m, what is the magnitude of the electric field at point P?

Lynna [10]

In BPC

tan\theta =a/b = 3/4

\theta = tan^-1(0.75)

\theta = 36.87 deg

BP = sqrt(a^2 + b^2) = sqrt((3)^2 + (4)^2) = 5 m

Eb = k Q/BP^2 = (9 x 10^9) (16 x 10^-9)/5^2 = 5.76 N/C

Ea = k Q/AP^2 = (9 x 10^9) (16 x 10^-9)/4^2 = 9 N/C

Ec = k Q/CP^2 = (9 x 10^9) (16 x 10^-9)/3^2 = 16 N/C

Net electric field along X-direction is given as

Ex = Ea + Eb Cos36.87 = (9) + (5.76) Cos36.87 = 13.6 N/C

Net electric field along X-direction is given as

Ey = Ec + Eb Sin36.87 = (16) + (5.76) Sin36.87 = 19.5 N/C

Net electric field is given as

E = sqrt(Ex^2 + Ey^2) = sqrt((13.6)^2 + (19.5)^2) = 23.8 N/C

8 0

3 years ago

Iguanas need to live in a habitat that is very warm, so the pet store warms their enclosures with “basking lights” which act as

Diano4ka-milaya [45]

Sorry I'm late, but here is the answer!

Answer:

Heat can travel from one place to another in three ways: Conduction, Convection, and Radiation. Both conduction and convection require the matter to transfer heat. Conduction is the transfer of heat between substances that are in direct contact with each other. Thermal energy is transferred from hot places to cold places by convection. Radiation is a method of heat transfer that does not rely upon any contact between the heat source and the heated object as is the case with conduction and convection. Heat can be transmitted through empty space by thermal radiation often called infrared radiation.

8 0

3 years ago