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LenKa [72]
3 years ago
10

A constant magnetic field passes through a single rectangular loop whose dimensions are 0.35m x 0.55m. The magnetic field has a

magnitude of 2.1T and is inclined at an angle of 65 degrees with respect to the normal to the plane of the loops. (a) If the magnetic field decreases to zero in a time of 0.45s, what is the magnitude of the average emf induced in the loop.(b) If the magnetic field remains constant at its initail calue of 2.1T, what is the magnitude of the rate ΔA/Δt at which the area should change so that the average emf has the same magnitude as in part (a)?

Physics
1 answer:
Cloud [144]3 years ago
4 0

Answer:

E = 0.38V

Explanation:

See attachment below.

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Answer:

the correct answer is practicing refusal strategies can help students stay sober

Explanation:

7 0
3 years ago
You are traveling on an airplane. The velocity of the plane with respect to the air is 110.0 m/s due east. The velocity of the a
Mice21 [21]
1. Vpa = 180m/s. @ 0 deg.
  Vag = 40m/s @ 120 deg,CCW.


<span> Vpg = Vpa + Vag,
 Vpg = (180 + 40cos120) + i40sin120,
  Vpg = 160 + i34.64,
 Vpg=sqrt((160)^2 + (34.64)^2)=163.7m/s.
</span>
<span>2. tanA = Y / X = 34.64 / 160 = 0.2165,
  A = 12.2 deg,CCW. = 12.2deg. North of East. </span>

3.  1 hr = 3600s. <span>d = Vt = 163.7m/s * 3600s = 589,320m.

hope this helps</span>
8 0
3 years ago
40 POINTS EASY
11111nata11111 [884]
We know, Mechanical Energy = K.E. + P.E.
As ball is at ground, P.E. would be zero. But as it is in motion, it must have some K.E. and that is:

K.E. = 1/2 mv²
K.E. = 1/2 * 1 * 2²
K.E. = 4/2
K.E. = 2 J

In short, Your Answer would be Option B

Hope this helps!
7 0
3 years ago
A 5.30kg block hangs from a spring with a spring constant 1700 N/m. The block is pulled down 4.50cm from the equilibrium positio
Andru [333]

To solve this problem it is necessary to apply the concepts related to the frequency in a spring, the conservation of energy and the total mechanical energy in the body (kinetic or potential as the case may be)

PART A) By definition the frequency in a spring is given by the equation

f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}

Where,

m = mass

k = spring constant

Our values are,

k=1700N/m

m=5.3 kg

Replacing,

f = \frac{1}{2\pi} \sqrt{\frac{1700}{5.3}}

f=2.85 Hz

PART B) To solve this section it is necessary to apply the concepts related to the conservation of energy both potential (simple harmonic) and kinetic in the spring.

\frac{1}{2}kA^2 = \frac{1}{2}mv^2 + \frac{1}{2} kY^2

Where,

k = Spring constant

m = mass

y = Vertical compression

v = Velocity

This expression is equivalent to,

kA^2 =mV^2 +ky^2

Our values are given as,

k=1700 N/m

V=1.70 m/s

y=0.045m

m=5.3 kg

Replacing we have,

1700*A^2=5.3*1.7^2 +1700*(0.045)^2

Solving for A,

A^2 = \frac{5.3*1.7^2 +1700*(0.045)^2}{1700}

A ^2 = 0.011035

A=0.105 m \approx 10.5 cm

PART C) Finally, the total mechanical energy is given by the equation

E = \frac{1}{2}kA^2

E=\frac{1}{2}1700*(0.105)^2

E= 9.3712 J

3 0
3 years ago
Electromagnetic waves are ________ waves.
n200080 [17]
D. Transverse waves
3 0
3 years ago
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