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
otez555 [7]
3 years ago
15

A coil of area 0.320 m2 is rotating at 100 rev/s with the axis of rotation perpendicular to a 0.430 T magnetic field. If the coi

l has 700 turns, what is the maximum emf generated in it
Physics
1 answer:
weqwewe [10]3 years ago
8 0

Answer:

The maximum  emf generated in the coil is 60527.49 V

Explanation:

Given;

area of coil, A = 0.320 m²

angular frequency, f = 100 rev/s

magnetic field, B = 0.43 T

number of turns, N = 700 turns

The maximum emf generated in the coil is calculated as,

E = NBAω

where;

ω is the angular speed = 2πf

E = NBA(2πf)

Substitute in the given values and solve for E

E = 700 x 0.43 x 0.32 x 2π x 100

E = 60527.49 V

Therefore, the maximum  emf generated in the coil is 60527.49 V

You might be interested in
A horizontal force of 750 N is needed to overcome the force of static friction between a level floor and a 250-kg crate. If g-9.
seraphim [82]

Answer:

0.31

Explanation:

horizontal force, F = 750 N

mass of crate, m = 250 kg

g = 9.8 m/s^2

The friction force becomes applied force = 750 N

According to the laws of friction,

Friction force = μ x Normal reaction of the surface

here, μ be the coefficient of friction

750 = μ x m g

750 = μ x 250 x 9.8

μ = 0.31

Thus, the coefficient of static friction is 0.31.

7 0
3 years ago
Bulky is working at a UPS loading dock. He boasts he can lift a 154 kg box 4 meters in 30
ivanzaharov [21]

Answer: Power is 200 W

Explanation: Power P = work done / time used.

P = W/t = mgh/t = 154 kg · 9.81 m/s²· 4 m / 30 s = 201 W

8 0
2 years ago
Place the following steps in order for motor control. 1. Upper motor neurons stimulate lower motor neurons. 2. Sensory informati
Alla [95]

Answer:

3, 5, 1, 4, 2

Explanation:

When the body needs to move, certain muscles need to be stimulated in order to perform a certain movement. There is a process that is followed that starts with the brain sending a signal, and ends with the movement being performed by the body.

3. This is the first step, because a pathway needs to be established before any electric impulses can be sent. Hence, the upper motor neurons in the premotor cortex <u>need to choose a pathway</u> (known as a motor program).

5. This is the second step, because once a pathway has been established, a <u>signal needs to be sent</u> down that pathway. Hence, the basal nuclei enable the thalamus to stimulate the upper motor neurons to send an electrical impulse down the pathway.

1. This is the third step, because once the upper motor neurons have been stimulated, they send an electrical impulse to stimulate the lower motor neurons.

4. This is the forth step, because once the electrical impulse reaches the lower motor neurons, they pass the impulse to a certain group of skeletal muscles, causing them to contract.

2. This is the final step, because once the skeletal muscles have contracted, the <u>body has completed its movement</u>. However, once the body has moved, sensory information is relayed to the cerebellum to <u>ensure that the correct movement is made</u>, and if it needs to be adjusted, then the whole process will start again in order to correct the movement.

4 0
3 years ago
MATHPHYS CAN U HELP ME PLEASE
ludmilkaskok [199]

Explanation:

(1) The heat added to warm the ice to 0°C is:

q = mCΔT = (0.041 kg) (2090 J/kg/°C) (0°C − (-11°C)) = 942.59 J

The heat added to melt the ice is:

q = mL = (0.041 kg) (3.33×10⁵ J/kg) = 13,653 J

The heat added to warm the water to 100°C is:

q = mCΔT = (0.041 kg) (4186 J/kg/°C) (100°C − 0°C) = 17,162.6 J

The heat added to evaporate the water is:

q = mL = (0.041 kg) (2.26×10⁶ J/kg) = 92,660 J

The heat added to warm the steam to 115°C is:

q = mCΔT = (0.041 kg) (2010 J/kg/°C) (115°C − 100°C) = 1236.15 J

The total heat needed is:

q = 942.59 J + 13,653 J + 17,162.6 J + 92,660 J + 1236.15 J

q = 125,654.34 J

(2) When the first two are mixed:

m C₁ (T₁ − T) + m C₂ (T₂ − T) = 0

C₁ (T₁ − T) + C₂ (T₂ − T) = 0

C₁ (6 − 11) + C₂ (25 − 11) = 0

-5 C₁ + 14 C₂ = 0

C₁ = 2.8 C₂

When the second and third are mixed:

m C₂ (T₂ − T) + m C₃ (T₃ − T) = 0

C₂ (T₂ − T) + C₃ (T₃ − T) = 0

C₂ (25 − 33) + C₃ (37 − 33) = 0

-8 C₂ + 4 C₃ = 0

C₂ = 0.5 C₃

Substituting:

C₁ = 2.8 (0.5 C₃)

C₁ = 1.4 C₃

When the first and third are mixed:

m C₁ (T₁ − T) + m C₃ (T₃ − T) = 0

C₁ (T₁ − T) + C₃ (T₃ − T) = 0

(1.4 C₃) (6 − T) + C₃ (37 − T) = 0

(1.4) (6 − T) + 37 − T = 0

8.4 − 1.4T + 37 − T = 0

2.4T = 45.4

T = 18.9°C

(3) Heat gained by the ice = heat lost by the tea

mL + mCΔT = -mCΔT

m (3.33×10⁵ J/kg) + m (2090 J/kg/°C) (30.8°C − 0°C) = -(0.176 kg) (4186 J/kg/°C) (30.8°C − 32.8°C)

m (397372 J/kg) = 1473.472 J

m = 0.004 kg

m = 4 g

4 grams of ice is melted and warmed to the final temperature, which leaves 128 grams unmelted.

(4) The heat added to warm the ice to 0°C is:

q = mCΔT = (0.028 kg) (2090 J/kg/°C) (0°C − (-67°C)) = 3920.84 J

The heat added to melt the ice is:

q = mL = (0.028 kg) (3.33×10⁵ J/kg) = 9324 J

The heat added to warm the melted ice to T is:

q = mCΔT = (0.028 kg) (4186 J/kg/°C) (T − 0°C) = (117.208 J/°C) T

The heat removed to cool the water to T is:

q = -mCΔT = -(0.505 kg) (4186 J/kg/°C) (T − 27°C)

q = (2113.93 J/°C) (27°C − T) = 57076.11 J − (2113.93 J/°C) T

The heat removed to cool the copper to T is:

q = -mCΔT = -(0.092 kg) (387 J/kg/°C) (T − 27°C)

q = (35.604 J/°C) (27°C − T) = 961.308 J − (35.604 J/°C) T

Therefore:

3920.84 J + 9324 J + (117.208 J/°C) T = 57076.11 J − (2113.93 J/°C) T + 961.308 J − (35.604 J/°C) T

13244.84 J + (117.208 J/°C) T = 58037.418 J − (2149.534 J/°C) T

(2266.742 J/°C) T = 44792.58 J

T = 19.8°C

(5) Kinetic energy of the hammer = heat absorbed by ice

KE = q

½ mv² = mL

½ (0.8 kg) (0.9 m/s)² = m (80 cal/g × 4.186 J/cal × 1000 g/kg)

m = 9.68×10⁻⁷ kg

m = 9.68×10⁻⁴ g

(6) Heat rate = thermal conductivity × area × temperature difference / thickness

q' = kAΔT / t

q' = (1.09 W/m/°C) (4.5 m × 9 m) (10°C − 4°C) / (0.09 m)

q' = 2943 W

After 10.7 hours, the amount of heat transferred is:

q = (2943 J/s) (10.7 h × 3600 s/h)

q = 1.13×10⁸ J

q = 113 MJ

6 0
3 years ago
What creates more pressure?
Evgesh-ka [11]
It is C, gasses with less kinetic energy, i did this and i think i remember it was C
8 0
2 years ago
Other questions:
  • A 0.25-m string, vibrating in its sixth harmonic, excites a 0.96-m pipe that is open at both ends into its second overtone reson
    5·1 answer
  • A particle is being accelerated through space by a 10 N force. Suddenly the particle encounters a second of 10 N force in the op
    6·1 answer
  • What is the power involved in lifting a 10-kg object 1.0 m in 0.50 s?
    9·1 answer
  • Col. Joe Kittinger of the United States Air Force crossed the Atlantic Ocean in nearly 86 hours. The distance he traveled was 57
    8·1 answer
  • Air pressure decreases as
    6·1 answer
  • a projectile is fired from the ground with an initial horizontal velocity of 8 meters per second. its initial vertical velocityi
    11·1 answer
  • How could the combustibility of a substance influence how the substances used
    11·1 answer
  • A car driver sees a rabbit on the road. The driver makes an emergency stop after he sees the rabbit. Figure given shows the spee
    9·2 answers
  • A car is moving with a velocity of45m/s. Is brought to rest in 5s.the distance travelled by car before it comes to rest is
    7·1 answer
  • A loop rests in the plane of a page of textbook while a magnetic field is directed into the page. A clockwise current is induced
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!