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denis-greek [22]
2 years ago
12

The two ends of a coiled wire are connected to the electrodes of a lightbulb.

Physics
1 answer:
barxatty [35]2 years ago
5 0

Answer:

a

Explanation:

The bar magnet moves downward with respect to the wire loop, so that the number of magnetic field lines going through the loop decreases with time. This causes an emf to be induced in the loop, creating an electric current.

in other words, the magnets motion creates a current in the loop

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vector is the answer of this blank

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At what speed do a bicycle and its rider, with a combined mass of 100 kg , have the same momentum as a 1600 kg car traveling at
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3 years ago
A 0.560 kg snowball is fired from a cliff 14.2 m high with an initial velocity of 13.3 m/s, directed 26.0° above the horizontal.
enot [183]

Answer:

a) v = 21.34 m/s

b) v = 21.34 m/s

c) v = 21.34 m/s

Explanation:

Mass of the snowball, m = 0.560 kg

Height of the cliff, h = 14.2 m

Initial velocity of the ball, u = 13.3 m/s

θ = 26°

The speed of the slow ball as it reaches the ground, v = ?

The initial Kinetic energy of the snow ball, KE_{0}  = 0.5 mu^{2}

Potential energy of the snow ball at the given height, PE = mgh

Final Kinetic energy of the ball as it reaches the ground, KE_{f} = 0.5mv^{2}

a) Using the principle of energy conservation,

KE_{0} + PE = KE_{f} \\0.5mu^{2} + mgh = 0.5mv^{2}\\v^{2} =2( 0.5u^{2} + gh)\\v^{2} =u^{2} + 2gh\\v = \sqrt{u^{2} + 2gh} \\v = \sqrt{13.3^{2} + 2*9.8*14.2}\\v = 21.34 m/s

b) The speed remains v = 21.34 m/s since it is not a function of the angle of launch

c)The principle of energy conservation used cancels out the mass of the object, therefore the speed is not dependent on mass

v = 21. 34 m/s

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3 years ago
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just guys

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and if not i need how old you are sorry just trying to be safe

7 0
3 years ago
A ball has a mass of 0.046kg. Calculate the change in gravitational potential energy when the ball is lifted through a vertical
loris [4]

Answer:

PE=0.92414J and KE=0.28175J

Explanation:

Gravitational potential energy=mass*gravity*height

PE=mgh

Data,

M=0.046kg

H=2.05m

g=9.8m/s^2

PE=0.046kg * 9.8m/s^2 * 2.05m

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KE=1/2mv^2

M=0.046kg

V=3.5m/s

KE=[(0.046kg)*(3.5m/s)^2]\2

KE=0.28175J

3 0
3 years ago
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