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svet-max [94.6K]
3 years ago
7

A large magnetic flux change through a coil must induce a greater emf in the coil than a small flux change. A) True B) False

Physics
1 answer:
bixtya [17]3 years ago
3 0

Answer:

False

Explanation:

Faraday's law gives the relationship between the induced emf and the rate of change of magnetic flux i.e.

\epsilon=\dfrac{-d\phi}{dt}

The given statement "A large magnetic flux change through a coil must induce a greater emf in the coil than a small flux change" is false. The reason is that if the rate of change of magnetic flux is greater, then its will induce more emf. It would mean it does not say about emf.

Hence, it is false.

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Two truckers are traveling directly away from each other at the same speed. If one trucker sounds her horn at a frequency of 221
8_murik_8 [283]

Answer:

v = 8.8 m /s

Explanation:

For listener and source going away from each other the formula of Doppler effect is as follows

\frac{f}{f_0} = \frac{V-v}{V+v}

V is velocity of sound , v is velocity of listner and source of sound

f₀ is apparent frequency and f is real frequency

V = 343 , v = ? ,f = 210 , f₀ = 221

Put these value in the relation above

[tex]\frac{210}{221} = \frac{343-v}{343+v}[/tex]

v = 8.8 m /s

6 0
3 years ago
Substance A has a heat capacity that is much greater than that of substance B. If 10.0 g of substance A initially at 30.0 ∘C is
astra-53 [7]

Explanation:

In a heat exchange, the temperature change is inversely proportional to the specific heat capacity. Since substance A has a heat capacity that is much greater than that of substance B, the temperature change of substance A will be less than the temperature change of substance B. Therefore, the final temperature is closer to that of 30^\circ C than 80^\circ C.

8 0
3 years ago
Stop lamps must be visible within____ to the rear
Charra [1.4K]
The answer is 300 feet. The stop lamp or lamps on the rear of a vehicle must show a red light that is set in motion upon application of the service or foot brake and, in a vehicle manufactured or assembled on or after January 1, 1964, must be visible from a distance of not less than 300 feet to the rear in normal sunlight. Take note, if the vehicle is manufactured or assembled January 1, 1964, the stop lamp or lamps must be visible from a distance of not less than 100 feet. Also, the stop lamp may be combined with one or more other rear lamps. 
3 0
4 years ago
Read 2 more answers
Find the absolute value of the change of the gravitational potential energy (GPE) of the puck-Earth system from the moment the p
jekas [21]

Answer:

0.16joules

Explanation:

Using the relation for The gravitational potential energy

E= Mgh

Where,

E= Potential energy

h = Vertical Height

M = mass

g = Gravitational Field Strength

To find the vertical component of angle of launch Where the angle is 22°

h= sin theta

So E = mghsintheta

= 0.18 x 0.98 x 0.253 sin22

=0.16joules

Explanation:

7 0
3 years ago
Two tugboats pull a disabled supertanker. Each tug exerts a constant force of 1.8×106N , one an angle 11degrees west of north an
sp2606 [1]

Answer: 2,9 ×10¹⁰J.

Explanation:

1. The <em><u>work </u></em>is a measure of energy and is calculated as the product of the displacement times the parallel force to such displacement.

That means that the only components of the force that contribute to work are those that result parallel to the displacement.

2. Since it is given that the <em>two tugboats "pull the tanker a distance 0.83km toward the north"</em>, that is the displacement, and you have to calculate the net force toward the north.

3. <u>Tugboat #1</u>.

a) Force magnitude: F₁ = 1.8×10⁶N

b) Angle: α = 11° West of North

c) North component of the force F₁: Fy₁ = F₁cos(α) =  1.8×10⁶N  × cos(11°) = 1.77×10⁶N

4. <u>Tugboat #2</u>:

a) Force magnitude: F₂ = 1.8×10⁶N

b) Angle:  = 11° East of North

c) North component of the force F₂: Fy₂ = F₂cos(β) =  1.8×10⁶N  × cos(11°) = 1.77×10⁶N =

5. <u>Total net force, Fn</u>:

Fn = Fy₁ + Fy₂ = 1.77×10⁶N + 1.77×10⁶N = 3.54×10⁶N

6. <u>Work, W</u>:

Displacement, d = 0.83 km = 8,300 m

W = Fn×d = 3.54×10⁶N×8,300m = 29,000 ×10⁶J = 2,9 ×10¹⁰J

The answer is rounded to two significant figures because both data, Force and displacement, have two significant figures.

7 0
3 years ago
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