1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Vladimir [108]
3 years ago
10

Shrinking Loop. A circular loop of flexible iron wire has an initial circumference of 168 cm , but its circumference is decreasi

ng at a constant rate of 15.0 cm/s due to a tangential pull on the wire. The loop is in a constant uniform magnetic field of magnitude 0.900 T , which is oriented perpendicular to the plane of the loop. Assume that you are facing the loop and that the magnetic field points into the loop.
Required:
Find the magnitude of the emf EMF induced in the loop after exactly time 8.00s has passed since the circumference of the loop started to decrease.
Physics
1 answer:
DedPeter [7]3 years ago
6 0

Answer:

103.1 V

Explanation:

We are given that

Initial circumference=C=168 cm

\frac{dC}{dt}=-15cm/s

Magnetic field,B=0.9 T

We have to find the magnitude of the  emf induced in the loop after exactly time 8 s has passed since the circumference of the loop started to decrease.

Magnetic flux=\phi=BA=B(\pi r^2)

Circumference,C=2\pi r

r=\frac{C}{2\pi}

r=\frac{168}{2\pi} cm

\frac{dr}{dt}=\frac{1}{2\pi}\frac{dC}{dt}=\frac{1}{2\pi}(-15)=-\frac{15}{2\pi} cm/s

\int dr=-\int \frac{15}{2\pi}dt

r=-\frac{15}{2\pi}t+C

When t=0

r=\frac{168}{2\pi}

\frac{168}{2\pi}=C

r=-\frac{15}{2\pi}t+\frac{168}{2\pi}

E=-\frac{d\phi}{dt}=-\frac{d(B\pi r^2)}{dt}=-2\pi rB\frac{dr}{dt}

E=-2\pi(-\frac{5}{2\pi}t+\frac{168}{2\pi})B\times -\frac{15}{2\pi}

t=8 s

B=0.9

E=2\pi\times \frac{15}{2\pi}\times 0.9(-\frac{15}{2\pi}(8)+\frac{168}{2\pi})

E=103.1 V

You might be interested in
a baseball was thrown with an initial velocity of 14 m/s at an angle above the horizontal . It remained in the air for 2 s. Whic
liberstina [14]

Answer:

Explanation:acceleration

3 0
3 years ago
A 25kg ball is thrown into the air. when thrown it is going 10 m/s. calculate how high it travels?
Darina [25.2K]

Answer:

5m

Explanation:

V^2 = U^2 - 2gh

V=0 , U=10m/s, g=10m/s^2

0^2 = 10^2 - 2×10×h

0 = 100 - 20h

100=20h

H = 100/20 = 5m

3 0
3 years ago
A certain part of an aircraft engine has a volume of 1.4 ft^3. (a) Find the weight of the piece when it is made of lead. (b) If
nydimaria [60]

Answer:

(a) 4400.83 N

(b) 1051.7 N

(c) 3349.13 N

Explanation:

V = 1.4 ft^3 = 0.0396 m^3

density of lead = 11.34 x 10^3 kg/m^3

density of aluminium = 2710 kg/m^3

(a) When lead is used

Weight = mass x gravity = volume x density x gravity

            = 0.0396 x 11.34 x 1000 x 9.8 = 4400.83 N

(b) when aluminium is used

Weight = mass x gravity = volume x density x gravity

           = 0.0396 x 2710 x 9.8 = 1051.7 N

(c) Difference in weight = 4400.83 - 1051.7 = 3349.13 N

6 0
3 years ago
A dart in a toy dart gun is pressed 4 cm against a spring. if it has 1.5 J of stored potential energy what is the spring constan
Sphinxa [80]
U = 0.5k(x^2)
1.5 J = 0.5k(0.04m^2)
k = 1875 N/m

Check my math i might be wrong
8 0
2 years ago
A barbell is 1.5 m long. Three weights, each of mass 20 kg, are hung on the left and two weights of the same mass, on the right.
VMariaS [17]

Answer:

b. -11.6 cm

Explanation:

We have given parameters:

Length, l = 1.5 m = 150 cm

Mass of weight, m_1 = 20 kg

Width, x = 4 cm

Distance d = 4 cm

Mass of bar, m_{bar} = 5 kg

We are asked to find the center of mass from the mid-point, X_{CM} = ?

Since 3 weights are on the left and 2 weights are on the right, we know:

m_{left} = 3 * 20 = 60 kg

m_{right} = 2 * 20 = 40 kg

And also we know that, M = \frac{l}{2} = 150/2 = 75 cm

For the left side, center of mass is:

x_{left} = \frac{3 * 4}{2} = 6 cm

From the midpoint, the distance to the left is:

X_{left} = -(M - 4 - x_{left}) = -(75 - 4 -6) = -65 cm

For the right side, center of mass is:

x_{right} = \frac{2 * 4}{2} = 4 cm

From the midpoint, the distance to the right will be:

X_{right} = (M - 4 - x_{right}) = (75 - 4 - 4) = 67 cm

Hence,

X_{CM} = \frac{m_{right}*x_{right} + m_{left}*x_{left} }{m_{right} + m_{left} + m_{bar}} = \frac{40 * 67 - 60 * 65}{40 + 60 + 5} = -11.62 cm

4 0
3 years ago
Other questions:
  • A ball is released at the top of a ramp at t =0. which is the speed of the ball at t=4
    9·2 answers
  • How much horsepower does a camaro ss have?
    8·2 answers
  • If the acceleration was reduced 1/4 what will be the new force
    7·1 answer
  • A person holding a 30 kg box 1 m above the floor puts it down. How much work was done ? Remember , the force of an objects weigh
    6·2 answers
  • How does water deposit sediment
    13·1 answer
  • A plate moves 200 m in 10,000 years. What is its rat in cm/year
    5·1 answer
  • Explain what happens when a ferromagnetic material is made into a permanent magnet.
    8·1 answer
  • What is the function of law
    7·1 answer
  • How do exited atoms give off light
    14·1 answer
  • How will you show that white light is made of several colours?
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!