PHYSICS COLLEGE

If a single circular loop of wire carries a current of 61 A and produces a magnetic field at its center with a magnitude of 1.70 10-4 T, determine the radius of the loop.

Answers

Answer 1

Answer:

Answer:

The radius is $R = 0.22 5 \ m$

Explanation:

From the question we are told that

The current is $I = 61 \ A$

The magnetic field is $B = 1.70 *10^(-4) \ T$

Generally the magnetic field produced by a current carrying conductor is mathematically represented as

$B = (\mu_o * I)/(2 * R )$

=> $R = (\mu_o * I )/( 2 * B )$

Here $\mu_o$ is the permeability of free space with value $\mu_o = 4\pi * 10^(-7) N/A^2$

=> $R = ( 4\pi * 10^(-7) * 61 )/( 2 * 1.70 *10^(-4) )$

=> $R = 0.22 5 \ m$

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Before 1960, people believed that the maximum attainable coefficient of static friction for an automobile tire on a roadway was ?s = 1. Around 1962, three companies independently developed racing tires with coefficients of 1.6. This problem shows that tires have improved further since then. The shortest time interval in which a piston-engine car initially at rest has covered a distance of one-quarter mile is about 4.43 s. (A) Assume the car's rear wheels lift the front wheels off the pavement as shown in the figure above. What minimum value of ?s is necessary to achieve the record time?

Answers

Answer:

4.18

Explanation:

Givens

The car's initial velocity $v_(i)$ = 0 and covering a distance Δx = 1/4 mi = 402.336 m in a time interval t = 4.43 s.

Knowns

We know that the maximum static friction force is given by:

$f_(s_max) =$ μ_s*n (1)

Where μ_s is the coefficient of static friction and n is the normal force.

Calculations

(a) First, we calculate the acceleration needed to achieve this goal by substituting the given values into a proper kinematic equation as follows:

Δx= $v_(i) +(1)/(2) at^2$

a=41 m/s

This is the acceleration provided by the engine. Applying Newton's second law on the car, so in equilibrium, when the car is about to move, we find that:

$f_(y)=n-mg=0\n n=mg\nf_(x)=F-f_(s,max) =0\n f_(s,max)=F=ma\n$

Substituting (3) into (1), we get:

$f_(s,max)=$ μ_s*m*g

Equating this equation with (4), we get:

ma= μ_s*m*g

μ_s=a/g

=4.18

A wheel 2.45 m in diameter lies in a vertical plane and rotates about its central axis with a constant angular acceleration of 4.30 rad/s2. The wheel starts at rest at t = 0, and the radius vector of a certain point P on the rim makes an angle of 57.3° with the horizontal at this time. At t = 2.00 s, find the following.a. the angular speed of the wheel and, for point P
b. the tangential speed.
c. the total acceleration.
d. the angular position.

Answers

(a) The angular speed of the wheel at point P is 8.6 rad/s.

(b) The tangential speed of the wheel is 10.54 m/s.

(d) The angular position of the wheel is 87 ⁰.

The given parameters;

diameter of the wheel, d = 2.45 m
radius of the wheel, r = 1.225 m
angular acceleration of the wheel, α = 4.3 rad/s²
angular displacement of the wheel, θ = 57.3⁰
time of motion, t = 2.0 s

The angular speed of the wheel at point P is calculated as follows;

$\omega_f = \omega _i + \alpha t\n\n\omega _f = 0 + 4.3 * 2\n\n\omega _f = 8.6 \ rad/s$

The tangential speed of the wheel is calculated as follows;

$v = \omega _f r\n\nv = 8.6 * 1.225 \n\nv = 10.54 \ m/s$

The centripetal acceleration of the wheel is calculated as follows;

$a_c = (v^2)/(r) \n\na_c = ((10.54)^2)/(1.225) \n\na_c = 90.69 \ m/s^2$

The total acceleration of the wheel is calculated as follows;

$a_t = √(a_c^2 + a_r^2) \n\na_t = √(90.69^2 + 4.3^2) \n\na_t = 90.8 \ m/s^2$

The angular position is calculated as follows;

$\theta = tan^(-1) ((a_c)/(a_r) )\n\n\theta = tan^(-1) ((90.69)/(4.3) )\n\n\theta = 87 \ ^0$

Learn more here: brainly.com/question/14508449

Answer:

Explanation:

Radius of wheel R = 1.225 m

For angular motion of wheel

ω = ω ₀ + α t

= 0 + 4.3 x 2

= 8.6 rad / s

This is angular speed of wheel and point P .

b )

Tangential speed = ωR

8.6 x 1.225

= 10.535 m / s

c )

radial acceleration

a_r = v² / r

= 10.535² / 1.225

= 90.6 m / s²

tangential acceleration = radius x angular acceleration

a_t = 4.3 x 1.225

= 5.2675

Total acceleration = √ 90.6² + 5.2675²

= √ 8208.36 + 27.7465

= 90.75 m/s²

d ) angle of rotation

= 1/2 α t²

= .5 x 4.3 x 4

= 8.6 radian

= (8.6/3.14) x 180

= 499 degree

= 499 + 57.3

= 556.3

556.3 - 360

= 196.3 degree

Point p will rotate by 196.3 degree

The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and the process is reversible and adiabatic. Use constant specific heat at 300 K to find the

Answers

Complete question:

Answer:

The exit velocity is 629.41 m/s

Explanation:

Given;

initial temperature, T₁ = 1200K

initial pressure, P₁ = 150 kPa

final pressure, P₂ = 80 kPa

specific heat at 300 K, Cp = 1004 J/kgK

k = 1.4

Calculate final temperature;

$T_2 = T_1((P_2)/(P_1))^{(k-1 )/(k)$

k = 1.4

$T_2 = T_1((P_2)/(P_1))^{(k-1 )/(k)}\n\nT_2 = 1200((80)/(150))^{(1.4-1 )/(1.4)}\n\nT_2 = 1002.714K$

Work done is given as;

$W = (1)/(2) *m*(v_i^2 - v_e^2)$

inlet velocity is negligible;

$v_e = \sqrt{(2W)/(m) } = √(2*C_p(T_1-T_2)) \n\nv_e = √(2*1004(1200-1002.714))\n\nv_e = √(396150.288) \n\nv_e = 629.41 \ m/s$

Therefore, the exit velocity is 629.41 m/s

Which if, any, of these statements are true? (More than one may be true.) Assume the batteries are ideal. Check all that apply. A battery supplies the energy to a circuit. A battery is a source of potential difference; the potential difference between the terminals of the battery is always the same. A battery is a source of current; the current leaving the battery is always the same.

Answers

Answer:

All are true except the last point that says that a battery is a current source and the current at the outlet is always the same.

Explanation:

A battery is an electro-chemical device which converts the chemical energy into usable electrical energy thus it provides electrical energy.
Since, the battery maintains a a constant potential difference between its terminals, once connected.
Since, the movement of electric current requires energy, which is supplied by the electric potential energy stored in the battery.
The current in the battery flows as per the Ohm's law and we can not say that the current leaving will always remain constant.
As the current is the flow of electric charge, and charges are not stored in batteries unlike capacitors, thus the current at the leaving end will depend on Ohm's law and will vary accordingly.

A 50-gram ball is released from rest 80 m above the surface of the Earth. During the fall to the Earth, the total thermal energy of the ball and the air in the system increases by 15 J. Just before it hits the surface its speed is

Answers

Answer:

Speed of ball just before it hit the surface is 31.62 m/s .

Explanation:

Given :

Mass of ball , m = 50 g = 0.05 kg .

Height from which it falls , h = 80 m .

Thermal energy , E = 15 J .

Now , Initial energy of the system is :

$E_i=(mv^2)/(2)+mgh$

Here , initial velocity is zero .

Therefore , $E_i=mgh=40\ J$

Now , final energy of the system :

$E_f=(mv^2)/(2)+mg(0)+15\n\nE_f=(0.05* v^2)/(2)+15$

Since , no external force is applied .

Therefore , total energy of the system will be constant .

By conservation of energy :

$E_i=E_f\n40=(0.05v^2)/(2)+15\n\n25=(0.05v^2)/(2)\n\nv=31.62\ m/s$

Therefore , speed of ball just before it hit the surface is 31.62 m/s .

Final answer:

Using the principle of conservation of energy, the speed of the ball just before hitting the Earth's surface is found to be 79.2 m/s after accounting for the 15 J increase in thermal energy.

Explanation:

This question is concerned with the concept of conservation of energy, specifically the principles of potential and kinetic energy. When the ball is 80 meters above the Earth's surface, the total gravitational potential energy is m*g*h = 50g*9.8m/s²*80m = 39200 J (where m is mass, g is gravity, and h is height), and the kinetic energy is 0.

As the ball falls, its potential energy gets converted into kinetic energy, but we also know that the total thermal energy of the ball and the air in the system increases by 15 J. That means that not all the potential energy is converted into kinetic energy, 15 J is lost to thermal energy. So, the kinetic energy of the ball when it hits the Earth is 39200 J - 15 J = 39185 J.

Finally, we know that kinetic energy equals (1/2)*m*v², where v is the speed of the ball. Rearranging this formula to solve for v we get, v = sqrt((2*kinetic energy)/m) = sqrt((2*39185 J)/50g) = 79.2 m/s. So, just before the ball hits the surface, its speed is 79.2 m/s.

Learn more about Potential and Kinetic Energy here:

brainly.com/question/15764612

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What would the position of arrows on a target need to be to illustrate measurements that are neither accurate nor precise

Answers

Answer:

The position of the arrows will not be on the target i.e. outside the bull's eye, neither will they be close to one another (widely scattered).

Explanation:

Accuracy refers to the closeness of a measurement to an actual or accepted value while precision refers to the closeness of measurements to one another.

Using archery as an illustration of precision and accuracy, measurements (arrows) that are neither accurate not precise are those arrows that will be far away or outside the bull's eye region (target) of the board and also far apart from one another.

In a nutshell, the arrows will be distant from the bull's eye or target (not accurate) and also distant from one another (not precise).

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