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sertanlavr [38]

1 year ago

13

a student that weighs 436 n is standing on a scale in an elevator and notices that the scale reads 498 n. from this information,

the elevator is:

1 answer:

mash [69]1 year ago

7 0

The elevator has constant acceleration upward

Given that a student that weighs 436 N is standing on a scale in an elevator and notices that the scale reads 498 N.

The formula of force can be derived from the Newton's second law of motion, which states that the force exerted by any object equals mass times the acceleration of that object i.e., F = ma

Net force = mg + ma

498 = 436 + ma

62 = 436 a/ 9.8

a = (62 x 9.8) / 436

a = 1.39 m/s²

Hence, it is clear that the elevator has constant acceleration upward.

Learn more about Elevator force here

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The magnetic field at the center of a wire loop of radius , which carries current , is 1 mT in the direction (arrows along the w

Citrus2011 [14]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The magnetic field is $B_{net} = \frac{1}{4} * mT$

And the direction is $-\r k$

Explanation:

From the question we are told that

The magnetic field at the center is $B = 1mT$

Generally magnetic field is mathematically represented as

$B = \frac{\mu_o I}{2R}$

We are told that it is equal to 1mT

So

$B = \frac{\mu_o I}{2R} = 1mT$

From the first diagram we see that the effect of the current flowing in the circular loop is (i.e the magnetic field generated)

$\frac{\mu_o I}{2R} = 1mT$

This implies that the effect of a current flowing in the smaller semi-circular loop is (i.e the magnetic field generated)

$B_1 = \frac{1}{2} \frac{\mu_o I}{2R}$

and for the larger semi-circular loop is

$B_2 = \frac{1}{2} \frac{\mu_o I}{2 * (2R)}$

Now a closer look at the second diagram will show us that the current in the semi-circular loop are moving in the opposite direction

So the net magnetic field would be

$B_{net} = B_1 - B_2$

$= \frac{1}{2} \frac{\mu_o I}{2R} -\frac{1}{2} \frac{\mu_o I}{2 * (2R)}$

$=\frac{\mu_o I}{4R} -\frac{\mu_o I }{8R}$

$=\frac{\mu_o I}{8R}$

$= \frac{1}{4} \frac{\mu_o I}{2R}$

Recall $\frac{\mu_o I}{2R} = 1mT$

So

$B_{net} = \frac{1}{4} * mT$

Using the Right-hand rule we see that the direction is into the page which is $-k$

3 0

3 years ago

1. Two wires - A and B - with circular cross-sections have identical lengths and are made of the same material. Yet, wire A has

PolarNik [594]

Answer:

Diameter of wire B is 2 times the diameter of wire A

Explanation:

We have given two wire A and B

They are made up of same material and length of both the wire is sane

So $l_A=l_B$

Let the resistivity of both the wire is $\rho$

It is given that wire A has 4 times the resistance as wire B

So $R_A=4R_B$

So $\frac{\rho l_A}{a_A}=4\frac{\rho l_B}{a_B}$ ( As $l_A=l_B$ )

$\frac{a_A}{a_B}=\frac{1}{4}$

$\frac{d_A^2}{d_B^2}=\frac{1}{4}$

$\frac{d_A}{d_B}=\frac{1}{2}$

$d_B=2d_A$

So diameter of wire B is 2 times the diameter of wire A

6 0

2 years ago

How do you calculate acceleration

Eduardwww [97]

Answer: acceleration is equal to the change in velocity per unit time in seconds.

a= ∆v / t = vf - vi / t

Explanation: change in velocity or ∆v can be expressed as (vf - vi)

7 0

3 years ago

The unit of electric current is the _____.<br><br> A) volt<br> B) ohm<br> C) ampere<br> D) faraday

professor190 [17]

C

the unit of electric current is the ampere, often shortened to amp

6 0

3 years ago

Read 2 more answers

A box is being moved with a velocity (v) by a force P (parallel to v) along a level horizontal floor. The normal force is (Fn),

labwork [276]

Answer:

Force (P) : Positive

Normal Force (Fn) : Zero

Weight (mg) : Zero

Kinetic Frictional Force (fk) : Negative

Explanation:

The work done by a force on an object is given by the following formula:

W = F.d

W = F d Cosθ

where,

W = Work Done

f = Force Applied

d = displacement

θ = Angle between force and displacement

<u>FOR FORCE (P)</u>:

Since, force P is parallel to the motion of the box. Therefore, θ = 0°

Hence,

W = P d Cos 0°

W = P d(1)

W = Pd

<u>Therefore, work done by force (P) is Positive.</u>

<u></u>

<u>FOR NORMAL FORCE (Fn) AND WEIGHT (W)</u>:

Since, normal force and weight are perpendicular to the motion of the box. Therefore, θ = 90°

Hence,

W = Fn d Cos 90°= mg d Cos 90°

W = Fn d(0) = mg d (0)

W = 0

<u>Therefore, work done by Normal Force (Fn) and Weight (mg) is Zero.</u>

<u></u>

<u>FOR KINETIC FRICTIONAL FORCE (fk)</u>:

Since, kinetic frictional force acts in the opposite direction of motion of the box. Therefore, θ = 180°

Hence,

W = fk d Cos 180°

W = fk d(-1)

W = -fk d

<u>Therefore, work done by Kinetic Frictional Force (fk) is Negative.</u>

<u></u>

8 0

3 years ago