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Nezavi [6.7K]
3 years ago
14

A Circular loop in the plane of a paper lies in a 0.65 TT magnetic field pointing into the paper. The loop's diameter changes fr

om 17.5 cmcm to 6.6 cmcm in 0.48 ss .
a) Determine the direction of the induced current and justify your answer.b) Determine the magnitude of the average induced emf.c) If the coil resistance is 2.5 Ω, what is the average induced current?
Physics
1 answer:
Kobotan [32]3 years ago
6 0

Answer:

Explanation:

magnetic filed, B = 0.65 T

initial diameter, d = 17.5 cm

final diameter, d' = 6.6 cm

time, t = 0.48 s

(a) According to Lenz's law, the direction of induced current is clockwise.

(b) Let e is the induced emf.

initial area, A = π r² = 3.14 x 0.0875 x 0.0875 = 0.024 m²

final area, A' = π r'² = 3.14 x 0.033 x 0.033 = 0.00342 m²

change in area, ΔA = A - A' = 0.024 - 0.00342 = 0.02058 m²

The magnitude of induced emf is given by

e=\frac{d\phi }{dt}

e=B\frac{\Delta A }{dt}

e = 0.65 x 0.02058 / 0.48

e = 0.028 V

(c) R = 2.5 ohm

i = e / R

i = 0.028 / 2.5

i = 0.011 A

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Ratling [72]

Answer:

a). 1.218 m/s

b). R=2.8^{-3}

Explanation:

m_{bullet}=6.99g*\frac{1kg}{1000g}=6.99x10^{-3}kg

v_{bullet}=341\frac{m}{s}

Momentum of the motion the first part of the motion have a momentum that is:

P_{1}=m_{bullet}*v_{bullet}

P_{1}=6.99x10^{-3}kg*341\frac{m}{s} \\P_{1}=2.3529

The final momentum is the motion before the action so:

a).

P_{2}=m_{b1}*v_{fbullet}+(m_{b2}+m_{bullet})*v_{f}}

P_{2}=1.202 kg*0.554\frac{m}{s}+(1.523kg+6.99x10^{-3}kg)*v_{f}

P_{1}=P_{2}

2.529=0.665+(1.5299)*v_{f}\\v_{f}=\frac{1.864}{1.5299}\\v_{f}=1.218 \frac{m}{s}

b).

kinetic energy

K=\frac{1}{2}*m*(v)^{2}

Kinetic energy after

Ka=\frac{1}{2}*1.202*(0.554)^{2}+\frac{1}{2}*1.523*(1.218)^{2}\\Ka=1.142 J

Kinetic energy before

Kb=\frac{1}{2}*mb*(vf)^{2}\\Kb=\frac{1}{2}*6.99x10^{-3}kg*(341)^{2}\\Kb=406.4J

Ratio =\frac{Ka}{Kb}

R=\frac{1.14}{406.4}\\R=2.8x10^{-3}

3 0
3 years ago
Pls how to solve this problem. help would be appreciated ​
Bingel [31]

Answer:

80 m/s

Explanation:

Given:

a = -5 m/s²

v = 0 m/s

Δx = 640 m

Find: v₀

v² = v₀² + 2a(x − x₀)

(0 m/s)² = v₀² + 2(-5 m/s²) (640 m)

v₀ = 80 m/s

7 0
3 years ago
A small, 300 g cart is moving at 1.20 m/s on an air track when it collides with a larger, 2.00 kg cart at rest?
stiv31 [10]

Answer:

The speed of the large cart after collision is 0.301 m/s.

Explanation:

Given that,

Mass of the cart, m_1 = 300\ g = 0.3\ kg

Initial speed of the cart, u_1=1.2\ m/s

Mass of the larger cart, m_2 = 2\ kg

Initial speed of the larger cart, u_2=0

After the collision,

Final speed of the smaller cart, v_1=-0.81\ m/s (as its recolis)

To find,

The speed of the large cart after collision.

Solution,

Let v_2 is the speed of the large cart after collision. It can be calculated using conservation of momentum as :

m_1u_1+m_2u_2=m_1v_1+m_2v_2

m_1u_1+m_2u_2-m_1v_1=m_2v_2

v_2=\dfrac{m_1u_1+m_2u_2-m_1v_1}{m_2}

v_2=\dfrac{0.3\times 1.2+0-0.3\times (-0.81)}{2}

v_2=0.301\ m/s

So, the speed of the large cart after collision is 0.301 m/s.

4 0
3 years ago
Mt. Asama, Japan, is an active volcano. In 2009, an eruption threw solid volcanic rocks that landed 1 km horizontally from the c
Nata [24]

Answer:

a) 69.3 m/s

b) 18.84 s

Explanation:

Let the initial velocity = u

The vertical and horizontal components of the velocity is given by uᵧ and uₓ respectively

uᵧ = u sin 40° = 0.6428 u

uₓ = u cos 40° = 0.766 u

We're given that the horizontal distance travelled by the projectile rock (Range) = 1 km = 1000 m

The range of a projectile motion is given as

R = uₓt

where t = total time of flight

1000 = 0.766 ut

ut = 1305.5

The vertical distance travelled by the projectile rocks,

y = uᵧ t - (1/2)gt²

y = - 900 m (900 m below the crater's level)

-900 = 0.6428 ut - 4.9t²

Recall, ut = 1305.5

-900 = 0.6428(1305.5) - 4.9 t²

4.9t² = 839.1754 + 900

4.9t² = 1739.1754

t = 18.84 s

Recall again, ut = 1305.5

u = 1305.5/18.84 = 69.3 m/s

7 0
3 years ago
A floating ice block is pushed through a displacement d = (13.5 m)i + (-14.3 m)j along a straight embankment by rushing water, w
fredd [130]

To solve this problem, we are going to use the formula for work which is Fd where x and y are measured separately.

 

X direction: W = 13.5 x 230 = 3105 Joules

Y direction: W = -14.3 x -165 = 2360 Joules

So the total work is getting the sum of the two: 3105 + 2360 = 5465 Joules

7 0
3 years ago
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