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asambeis [7]
3 years ago
9

If a coil is rotated in a magnetic field at twice the speed -- ie twice the frequency -- the amplitude of the EMF induced in the

loop will...
A. remain unchaged

B. reduce by half

C. double
Physics
1 answer:
valkas [14]3 years ago
6 0

Answer:

C. double

Explanation:

The voltage of a EMF depends largely on the amount of spins that the coild is able to do, in this case as yuo are doubling the speed at which the coil is rotating you are also doubling the number of spins that the coil is doing in a certain amount of time, so the amplitude of the EMF induce on the loop will also double.

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A 10.0 kg ball weighs 98.0 N in air and weighs 65.0 N when submerged in water. The volume of the ball is:_________.A) 0.00245 m3
Kamila [148]

Answer: B) 0.00337 m3.

Explanation:

Given data:

Mass of the ball = 10kg

Weight of the ball in air = 98N

Weight of the ball in water = 65N

Solution:

To get the Volume of the ball when submerged in water, we divide the weight of the ball in water with the difference in apparent weight by 9.8m/s^2.

= 98 - 65 / 9.8

= 33 / 9.8

= 3.37kg

The volume of the ball is 3.37kg

The density of water is 1kg per Liter.

So 3.37 kg of water would have a volume of 3.37 Liters.

Therefore the ball would have a volume of 3.37 Liters (or 0.00337 cubic meters).

7 0
2 years ago
Will two separate 50db sounds together constitute a 100db sound explain mathematical
vfiekz [6]
Putting together two distinct 50 dB sound, do not create a 100 dB sound. Since decibels are logarithm of energy, creating two sounds together only makes the energy increase but the logarithm only goes up by somehow little. So increasing the sound by 10 dB, only makes it 10000 times louder because each 10 dB increase in sound makes the sound 10 times louder.

Twice as loud is an increase of 10Log (2) = 3.01 dB. So, 53,01 dB is twice as loud as 50dB.
5 0
3 years ago
The density of gold is 19 300kg/m cube. what is the mass of gold cube with the length 0.2015m?
Sergeeva-Olga [200]

Answer:

157.9 kg

Explanation:

Density: This can be defined as the ratio of the mass of a body and it's volume.

The S.I unit of density is kg/m³.

From the question,

Density = Mass/volume

D = m/v............................ Equation 1

Where D = Density of gold, m = mass of gold, v = volume of gold.

make m the subject of the equation

m = Dv.................... Equation 2

Since the gold is a cube,

v = l³................... Equation 3

Where l = length of the gold cube.

Substitute equation 3 into equation 2

m = Dl³............... Equation 4

Given: D = 19300 kg/m³, l = 0.2015 m

Substitute into equation 4

m = 19300(0.2015)³

m = 157.9 kg.

4 0
3 years ago
A child whose weight is 235 N slides down a 4.90 m playground slide that makes an angle of 37.0° with the horizontal. The coeffi
Basile [38]

Answer:

Explanation:

a) First, let's calculate the value of the Friction force, which is given by the formula:

Ff = u*W

As the Friction force has an X component, it would be:

Ff = u*m*g*cosФ

Where m*g is the weight of the child.

Solving for Ff:

Ff = 0.051 * 235cos37 = 9.57 N

Now, to get the energy transferred to thermal energy (or heat) we need to get the Work done, so:

Wf = Ff * d

Wf = 9.57 * 4.9 = 46.9 J

b) We need to get the downslope component of the child weight, which is:

Wy = 235sin37 = 141.43 N

As you can imagine, the gravity also does work, so:

Wg = Wy * d

Wg = 141.43 * 4.9 = 693 J

Now, let's get the kinetic energy, which can be obtained by this expression:

ΔKe = Wg - Wf

ΔKe = 693 - 46.9 = 646.1 J

The formula for kinetic energy is:

Ke = 1/2 m*V^2

We have the innitial speed which is 0.355 m/s, and the mass can be obtained by m*g so:

Fw= m*g ----> m = Fw/g

m = 235 / 9.8 = 23.97 kg

so the innitial energy is:

Ke = 1/2 * 23.97 * (0.355)^2

Ke = 1.51 J

This could be Ke1, so to get Ke2:

ΔKe = Ke2 - Ke1

Ke2 = Δke + Ke1

Ke2 = 646.1 + 1.51 = 647,61 J

Finally, the speed at the bottom would be:

v = √2Ke/m

v = √2*647.61/23.97

v = 7.35 m/s

7 0
3 years ago
A 60 mAs results in an exposure of 85 mGya, with all factors remaining the same, what would the new exposure be if 120 mAs is us
Cloud [144]

Answer: d₂ = 170 mGya

Explanation:

the relationship between absonbed 'd' and exposure 'E' is given as;

D(Gv) = F . x (AS/xB)

F is a conversion coefficient depending on medium

so we can simply write

d₁/d₂ = x₁/x₂

Given that;

our x₁ = 60 mAs, x₂ = 120 mAs,  d₁ = 85 mGya, d₂ = ?

from the given formula,

d₂ = (x₂d₁ / x₁)

now we substitute

d₂ = (120 × 85) / 60

d₂ = 170 mGya

∴ if 120 mAa is used,  the new exposure will be 170 mGya

8 0
3 years ago
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