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2 years ago

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8) A flat circular loop having one turn and radius 5.0 cm is positioned with its plane perpendicular to a uniform 0.60-T magneti

c field. The area of the loop is suddenly reduced to essentially zero in 0.50 ms. What emf is induced in the loop

1 answer:

Marrrta [24]2 years ago

7 0

Answer:

EMF induced in the loop is 9.4 V

Explanation:

As we know that initial magnetic flux of the loop is given as

$\phi_1 = B.A$

$\phi_1 = (0.60)(\pi (0.05)^2)$

$\phi_1 = 4.7 \times 10^{-3} Wb$

As soon as the area of the loop becomes zero the final magnetic flux of the loop is ZERO

Now as per faraday's law of electromagnetic induction the EMF is induced due to rate of change in magnetic flux

so we will have

$EMF = \frac{\Delta \phi}{\Delta t}$

so we will have

$EMF = \frac{4.7 \times 10^{-3} - 0}{0.50 \times 10^{-3}}$

$EMF = 9.4 V$

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A firecracker breaks up into several pieces, one of which has a mass of 200 g and flies off along the x-axis with a speed of 82.

MakcuM [25]

Answer:

The magnitude of the total momentum is 21.2 kg m/s and its direction is 39.5° from the x-axis.

Explanation:

Hi there!

The total momentum is calculated as the sum of the momenta of the pieces.

The momentum of each piece is calculated as follows:

p = m · v

Where:

p = momentum.

m = mass.

v = velocity.

The momentum is a vector. The 200 g-piece flies along the x-axis then, its momentum will be:

p = (m · v, 0)

p = (0.200 kg · 82.0 m/s, 0)

p = (16.4 kg m/s, 0)

The 300 g-piece flies along the y-axis. Its momentum vector will be:

p =(0, m · v)

p = (0, 0.300 kg · 45.0 m/s)

p = (0, 13.5 kg m/s)

The total momentum is the sum of each momentum:

Total momentum = (16.4 kg m/s, 0) + (0, 13.5 kg m/s)

Total momentum = (16.4 kg m/s + 0, 0 + 13.5 kg m/s)

Total momentum = (16.4 kg m/s, 13.5 kg m/s)

The magnitude of the total momentum is calculated as follows:

$|p| = \sqrt{(16.4 kgm/s)^2+(13.5 kg m/s)^2}= 21.2 kg m/s$

The direction of the momentum vector is calculated using trigonometry:

cos θ = px/p

Where px is the horizontal component of the total momentum and p is the magnitude of the total momentum.

cos θ = 16.4 kg m/s / 21.2 kg m/s

θ = 39.3 (39.5° if we do not round the magnitude of the total momentum)

Then, the magnitude of the total momentum is 21.2 kg m/s and its direction is 39.5° from the x-axis.

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Which statement describes one way in which global winds affect weather and climate? A. Polar easterlies move warm air to the mid

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The answer your looking for is "D".

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An astronomer observes that the wavelength of light from a distant star is shifted toward the red part of the visible spectrum.

balandron [24]

Answer:

The distance between the earth and the star is increasing.

Explanation:

When we observe an object and its electromagnetic radiation has been displaced to blue, it means that it is getting closer to us, causing the light waves it emits to get closer together and its wavelength to decrease towards blue, this is knowm as blueshift.

On the contrary, when an object is rapidly moving away from us, the light waves or electromagnetic radiation it emits have been stretched from their normal wavelength to a longer wavelength, towards the red part of the spectrum. This is known as redshift.

This phenomenon of changes in wavelength and frequency due to movement (whether the source approaches or moves away) is described by the Doppler effect.

So for this case because the light we perceive from the star has moved to the red part of the visible spectrum, we can conclude that it is moving away from the earth, and that the distance between the star and the earth is increasing.

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2 years ago

Sachi wants to throw a water balloon to knock over a target and win a prize. The target will only fall over if it is hit with a

posledela

Answer:

3.1 m/s²

Explanation:

Given:

Mass of the balloon (m) = 11.4 g = 0.0114 kg ( 1 kg = 1000 g)

Force acting on the balloon (F) = 0.035 N

Acceleration with which the balloon must be hit (a) = ?

Now, we know that, from Newton's second law, net force acting on an object is equal to the product of its mass and acceleration.

Therefore, framing in equation form, we have:

$F=ma$

Rewriting in terms of acceleration 'a', we get:

$a=\frac{F}{m}$

Now, substitute the given values and solve for 'a'. This gives,

$a=\frac{0.035\ N}{0.0114\ kg}\\\\a=3.07\approx 3.1\ m/s^2(Nearest\ tenth)$

Therefore, the acceleration of the water balloon to reach the target must be equal to 3.1 m/s².

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A 5.00μF parallel-plate capacitor is connected to a 12.0 V battery. After the capacitor is fully charged, the battery is disconn

EastWind [94]

(a) 12.0 V

In this problem, the capacitor is connected to the 12.0 V, until it is fully charged. Considering the capacity of the capacitor, $C=5.00 \mu F$ , the charged stored on the capacitor at the end of the process is

$Q=CV=(5.00 \mu F)(12.0 V)=60 \mu C$

When the battery is disconnected, the charge on the capacitor remains unchanged. But the capacitance, C, also remains unchanged, since it only depends on the properties of the capacitor (area and distance between the plates), which do not change. Therefore, given the relationship

$V=\frac{Q}{C}$

and since neither Q nor C change, the voltage V remains the same, 12.0 V.

(b) (i) 24.0 V

In this case, the plate separation is doubled. Let's remind the formula for the capacitance of a parallel-plate capacitor:

$C=\frac{\epsilon_0 \epsilon_r A}{d}$

where:

$\epsilon_0$ is the permittivity of free space

$\epsilon_r$ is the relative permittivity of the material inside the capacitor

A is the area of the plates

d is the separation between the plates

As we said, in this case the plate separation is doubled: d'=2d. This means that the capacitance is halved: $C'=\frac{C}{2}$ . The new voltage across the plate is given by

$V'=\frac{Q}{C'}$

and since Q (the charge) does not change (the capacitor is now isolated, so the charge cannot flow anywhere), the new voltage is

$V'=\frac{Q}{C'}=\frac{Q}{C/2}=2 \frac{Q}{C}=2V$

So, the new voltage is

$V'=2 (12.0 V)=24.0 V$

(c) (ii) 3.0 V

The area of each plate of the capacitor is given by:

$A=\pi r^2$

where r is the radius of the plate. In this case, the radius is doubled: r'=2r. Therefore, the new area will be

$A'=\pi (2r)^2 = 4 \pi r^2 = 4A$

While the separation between the plate was unchanged (d); so, the new capacitance will be

$C'=\frac{\epsilon_0 \epsilon_r A'}{d}=4\frac{\epsilon_0 \epsilon_r A}{d}=4C$

So, the capacitance has increased by a factor 4; therefore, the new voltage is

$V'=\frac{Q}{C'}=\frac{Q}{4C}=\frac{1}{4} \frac{Q}{C}=\frac{V}{4}$

which means

$V'=\frac{12.0 V}{4}=3.0 V$

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2 years ago