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Tema [17]
4 years ago
14

A rectangular loop of wire of width 10 cm and length 20 cm has a current of 2.5 A flowing through it. Two sides of the loop are

oriented parallel to a uniform magnetic field of strength 0.037 T, the other two sides being perpendicular to the magnetic field.
A)What is the magnitude of the magnetic moment of the loop?
B)What torque does the magnetic field exert on the loop?
Physics
1 answer:
Dahasolnce [82]4 years ago
4 0

Answer:

(a) 0.05 Am^2

(b) 1.85 x 10^-3 Nm

Explanation:

width, w = 10 cm = 0.1 m

length, l = 20 cm = 0.2 m

Current, i = 2.5 A

Magnetic field, B = 0.037 T

(A) Magnetic moment, M = i x A

Where, A be the area of loop

M = 2.5 x 0.1 x 0.2 = 0.05 Am^2

(B) Torque, τ = M x B x Sin 90

τ = 0.05 x 0.037 x 1

τ = 1.85 x 10^-3 Nm

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Why can't heat be transferred between two objects that have the same temperature
Inessa [10]

Answer: Because temperature is a measure of the average kinetic energy of the atoms or molecules in the system. The zeroth law of thermodynamics says that no heat is transferred between two objects in thermal equilibrium; therefore, they are the same temperature.

Explanation:9 (- _ -)

3 0
3 years ago
A football quarterback throws a football for a long pass. While in the motion of throwing, the quarterback moves the ball , star
sergejj [24]

This question is incomplete, the complete question is;

A football quarterback throws a 0.408 kg football for a long pass. While in the motion of throwing, the quarterback moves the ball 1.909 m, starting from rest, and completes the motion in 0.439 s. Assuming the acceleration is constant, what force does the quarterback apply to the ball during the pass ;

a) F_throw = 8.083 N

b) F_throw = 9.181 N

c) F_throw = 2.284 N

d) F_throw = 16.014 N

e) None of these is correct

Answer:

the quarterback applied a force of 8.083 N to the ball during the pass

so Option a) F_throw = 8.083 N is the correct answer

Explanation:

Given that;

m = 0.408 kg

d = 1.909 m

u = 0 { from rest}

t = 0.439 s

Now using Kinetic equation

d = ut + 1/2 at²

we substitute

1.909 = (0 × 0.439) + 1/2 a(0.439)²

1.909 = 0 + 0.09636a

1.909 = 0.09636a

a = 1.909 / 0.09636

a = 19.8111 m/s²

Now force applied will be;

F = ma

we substitute

F = 0.408 ×  19.8111

F = 8.0828 ≈ 8.083 N

Therefore the quarterback applied a force of 8.083 N to the ball during the pass

so Option a) F_throw = 8.083 N is the correct answer

5 0
3 years ago
Once the merry-go-round travels at this new angular speed, with what force does the person need to hold on?.
natita [175]

For a merry go round with a radius of R=1.8 m and moment of inertia I=184 kg-m^2 is spinning with an initial angular speed of w=1.48 rad/s   is mathematically given as

F= 618.9 N

<h3>What is the centripetal force?</h3>

Generally, the equation for the angular speed  is mathematically given as

w = v/R

Therefore

w= 4.7/1.8

w= 2.611 rad/s

Where total momentum

Tm= 642.96 + 272.32

Tm= 915.28

and total inertia

Ti= 184 + 246.24

Ti= 430.24

In conclusion, centripetal force

F= mrw^2

F = m*R*w2^2

F = 76*1.8*2.127^2

F= 618.9 N

Read more about mass

brainly.com/question/15959704

CQ

Flag

a merry go round with a radius of R=1.8 m and moment of inertia I=184 kg-m^2 is spinning with an initial angular speed of w=1.48 rad/s in the counter clockwise direction when viewed from above a person with mass m=76 kg and velocity v=4.7 m/s runs on a path tangent to the merry go round once at the merry go round the person jumps on and holds on to the rim of the merry go round angular speed of the merry go round after the person jumps on 2.127 rad/s Once the merry go round travels at this new angular speed with what force does the person need to hold on?

3 0
3 years ago
5. A 5.0 kg object accelerates uniformly from rest for 5.0 s and reaches a final velocity of 20.0 m/s. At 3.0 s, what is the obj
irina [24]

Answer:

jack JACKKK

Explanation:

OK YAHHHHH SO DO THIS

3 0
4 years ago
A train going 14m/s moves 250 m while accelerating to a stop. What is the train’s deceleration?
Elanso [62]

Answer:

-0.056 is the deceleration

6 0
3 years ago
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