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Elenna [48]
3 years ago
10

A 100-turn coil has a radius of 7.50 cm and a resistance of 50.0 W. At what rate must a perpendicular magnetic field change to p

roduce a current of 5.00 A in the coil
Physics
1 answer:
Furkat [3]3 years ago
3 0

Answer:

dB / dt = -141.47 T / s

Explanation:

For this exercise let's use Faraday's law to calculate the induced electromotive force

          E = - N dФ / dt

the magnetic flux is

         Ф = B. A

in this case the direction of the field and the normal to area are parallel so the scalar product is reduced to the algebraic product

         Ф = B A

  we substitute

        E = - N A dB / dt

the area of ​​the loop is

        A = π r²

we substitute

          E = - N π r² dB / dt

in the exercise indicate that the resistance of the coil is R = 50.0 Ω,

           E = i R

we substitute

           i R = -N π r² dB / dt

           dB / dT = - i R / N π r²

let's calculate

           dB / dt = - 5.00 50.0 / (100 π 0.075²)

           dB / dt = -141.47 T / s

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A 200-g block is attached to a horizontal spring and executes simple harmonic motion with a period of 0.250 s. The total energy
m_a_m_a [10]

Answer:

(A) Spring constant will be 126.58 N/m

(B) Amplitude will be equal to 0.177 m

Explanation:

We have given mass of the block m = 200 gram = 0.2 kg

Time period T = 0.250 sec

Total energy is given TE = 2 J

(A) For mass spring system time period is equal to T=2\pi \sqrt{\frac{m}{K}}

So 0.250=2\times 3.14 \sqrt{\frac{0.2}{K}}

0.0398=\sqrt{\frac{0.2}{K}}

Now squaring both side

0.00158=\frac{0.2}{K}

K = 126.58 N/m

So the spring constant of the spring will be 126.58 N/m

(B) Total energy is equal to TE=\frac{1}{2}KA^2, here K is spring constant and A is amplitude

So 2=\frac{1}{2}\times 126.58\times A^2

A^2=0.0316

A = 0.177 m

So the amplitude of the wave will be equal to 0.177 m

6 0
3 years ago
"The International Space Station (ISS) orbits at a distance of 350 km above the surface of the Earth. (a) Determine the gravitat
vagabundo [1.1K]

Answer:

(a) g = 8.82158145m/s^2.

(b) 7699.990192m/s.

(c)5484.3301s = 1.5234 hours.(extremely fast).

Explanation:

(a) Strength of gravitational field 'g' by definition is

g = \frac{M_{(earth)} }{r^2} G , here G is Gravitational Constant, and r is distance from center of earth, all the values will remain same except r which will be radius of earth + altitude at which ISS is in orbit.

r = 6721,000 meters, putting this value in above equation gives g = 8.82158145m/s^2.

(b) We have to essentially calculate centripetal acceleration that equals new 'g'.

a_{centripetal}=\frac{V^2}{r} =g here g is known, r is known and v is unknown.

plugging in r and g in above and solving for unknown gives V = 7699.990192m/s.

(c)  S = vT,  here T is time period or time required to complete one full revolution.

S =  earth's circumfrence , V is calculated in (B) T is unknown.

solving for unknown gives T = 5484.3301s = 1.5234hours.

3 0
3 years ago
Calculate the acceleration of a 1000 kg car if the motor provides a small thrust of 1000 N and the static and dynamic friction c
grin007 [14]

Explanation :

It is given that,

Mass of the car, m = 1000 kg              

Force applied by the motor, F_A=1000\ N

The static and dynamic friction coefficient is, \mu=0.5

Let a is the acceleration of the car. Since, the car is in motion, the coefficient of sliding friction can be used. At equilibrium,

F_A-\mu mg=ma

\dfrac{F_A-\mu mg}{m}=a

a=\dfrac{1000-0.5(1000)(9.81)}{1000}

a=-3.905\ m/s^2

So, the acceleration of the car is -3.905\ m/s^2. Hence, this is the required solution.

6 0
3 years ago
A proton accelerates from rest in a uniform electric field of 610 N/C. At some later time, its speed is 1.4 106 m/s.(a) Find the
nata0808 [166]

Answer:

a) 5.851× 10¹⁰m/s²

b) 2.411×10⁻¹¹s

c)  1.70×10⁻¹¹m

d) 1.661×10⁻²⁷KJ

Explanation:

A proton in the field experience a downward force of magnitude,

F = eE. The force of gravity on the proton will be negligible compared to the electric force

F = eE

a= eE/m

= 1.602×10⁻¹⁹ × 610/1.67×10⁻²⁷

= 5.851× 10¹⁰m/s²

b)

V = u + at

u= 0

v= 1.4106m/s

v= (0)t + at

t= v/a

= 1.4106m/s/5.851 ×10¹⁰

= 2.411×10⁻¹¹s

c)

S = ut + at²

= (o)t + 5.851×10¹⁰×(2.411×10⁻¹¹)²

= 1.70×10⁻¹¹m

d)

Ke = 1/2mv²

    = (1.67×10⁻²⁷×)(1.4106)²/2

 =  1.661×10⁻²⁷KJ

5 0
3 years ago
How is force related to math
dybincka [34]

Answer:

Newton's second law of motion describes the relationship between force and acceleration. They are directly proportional. If you increase the force applied to an object, the acceleration of that object increases by the same factor. In short, force equals mass times acceleration.

Explanation:

8 0
2 years ago
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