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Nutka1998 [239]
1 year ago
15

You toss a conductive open ring of diameter d = 1.75 cm up in the air. The ring is flipping around a horizontal axis at a rate o

f 6.25 flips per second. One flip is a full rotation. At your location, the earth's magnetic field is B e = 30.5 μT. What is the maximum emf induced in the ring?
Physics
1 answer:
Mama L [17]1 year ago
8 0

Answer:

The maximum emf induced in the ring

= (2.882 × 10⁻⁷) V

Explanation:

According to the law of electromagnetic induction, the emf induced in the ring is given by

E = N BA w sin wt

The maximum emf induced is

E = N BA w

B = 30.5 μT = (30.5 × 10⁻⁶) T

A = (πD²/4)

D = 1.75 cm = 0.0175 m

A = (π×0.0175²/4) = 0.000240625 m²

Nw = 2π × 6.25 = 39.29 rad/s

E = 30.5 × 10⁻⁶ × 0.000240625 × 39.29

E = (2.882 × 10⁻⁷) V

Hope this Helps!!!

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Consider a bird that flies at an average speed of 10.7 m/sm/s and releases energy from its body fat reserves at an average rate
Wittaler [7]

Answer:

455165.278 m

Explanation:

P = Power = 3.7 W

v = Velocity = 10.7 m/s

Amount of fat = 4 g

1 gram of fat provides about 9.40 (food) Calories

Energy given by 4 g of fat

E=4\times 9.4\times 4186\\\Rightarrow E=157393.6\ J

Time required to burn the fat

t=\dfrac{E}{P}\\\Rightarrow t=\dfrac{157393.6}{3.7}\\\Rightarrow t=42538.811\ s

Distance traveled by the bird

s=vt\\\Rightarrow s=10.7\times 42538.811\\\Rightarrow s=455165.2777\ m

The bird will fly 455165.278 m

4 0
1 year ago
Read 2 more answers
A pitcher throws a 0.15 kg baseball so that it crosses home plate horizontally with a speed of 10 m/s. It is hit straight back a
Maru [420]

Answer:

-5.1 kg m/s

Explanation:

Impulse is the change in momentum.

Change in momentum= final momentum - initial momentum=mv_{2} +mv_{1}

Plugging in the values= -0.15*24 - (0.15*10) (The motion towards the pitcher is negative as the initial motion is considered to be positive)

Impulse=-5.1 kg m/s (-ve means that it is the impulse towards the pitcher)

4 0
1 year ago
A 3.0-kg cart moving to the right with a speed of 1.0 m/s has a head-on collision with a 5.0-kg cart that is initially moving to
Maurinko [17]

Answer:

v = 0.8 m/s towards left

Explanation:

As we know that there is no external force on the system of two cart so total momentum of the system is conserved

so we will say

m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}

now plug in all data into the above equation

3(1) + 5(-2) = 3(-1) + 5 v

here we assumed that left direction of motion is negative while right direction is positive

so we can solve it for speed v now

3 - 10 = - 3 + 5 v

5 v = -4

v = -0.8 m/s

3 0
2 years ago
A block of mass m begins at rest at the top of a ramp at elevation h with whatever PE is associated with that height. The block
melomori [17]

This question is incomplete, the complete question is;

A block of mass m begins at rest at the top of a ramp at elevation h with whatever PE is associated with that height. The block slides down the ramp over a distance d until it reaches the bottom of the ramp.

How much of its original total energy (in J) survives as KE when it reaches the ground? m = 9.9 kg h = 4.9 m d = 5 m μ = 0.3 θ = 36.87°

Answer:

the amount of its original total energy (in J) that survives as KE when it reaches the ground will is 358.975 J

Explanation:

Given that;

m = 9.9 kg

h = 4.9 m

d = 5 m

μ = 0.3

θ = 36.87°

Now from conservation of energy, the energy is;

Et = mgh

we substitute

Et = 9.9 × 9.8 × 4.9

= 475.398 J

Also the loss of energy i

E_loss = (umg cosθ) d

we substitute

E_loss  = 0.3 × 9.9 × 9.8 × cos36.87°  × 5

= 116.423 J

so the amount of its original total energy (in J) that survives as KE when it reaches the ground will be

E = Et - E_loss

E = 475.398 J - 116.423 J

E = 358.975 J

5 0
1 year ago
What common laboratory measuring device will we likely always use in experiments that measure enthalpy changes?
adelina 88 [10]
The most common measuring device to be used in measuring enthalpy changes is the thermometer. 
5 0
1 year ago
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