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3 years ago

If the frequency of an FM wave is 8.85 × 10¥ hertz, what is the period of the FM wave?

2 answers:

7 0

The value of the period of the wave is the reciprocal of its frequency in hertz (1/s). From the given above, the value of the period is calculated below,
t = 1 / f = 1 / (8.85 x 10^7 1/s)
Thus, the value of period (t) is equal to 1.13 x 10^-8 s. The answer is letter B.

natka813 [3]3 years ago

6 0

for plato users it is 1.10 X 10^-8 just took the test got 100 percent so ik its right

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4. Compare how thermal and electrical energy transfer between objects and give an example of how you can

Leokris [45]

Answer:

4) we take a can of cold soda there is a transfer from hand to can. a current passes through a light bulb, it becomes incandescent and gives ligh

5) the law of energy conservation states that energy is neither created nor destroyed, therefore what happens is that you use energy and transform it into another type of energy,

Explanation:

4) This exercise is asked to give some examples of thermal and electrical energy transfer

the transfer of thermal energy occurs when we grasp a hot or cold body due to the difference in temperature between the body and us there is an exchange of heat, for example when drinking a cup of hot coffee there is a transfer of energy from the cup to the hand .

If we take a can of cold soda there is a transfer from hand to can.

Electrical transfer occurs, for example, when a current passes through a light bulb, it becomes incandescent and gives light.

In all modern electrical appliances there is transformation of electrical energy

5) the law of energy conservation states that energy is neither created nor destroyed, therefore what happens is that you use energy and transform it into another type of energy,

For example, when you lift a box, the potential energy of the box increases with height, but the energy of the person decreases because it does work that is negative.

In amusement park rides, the energy accumulated in one part of the game generally with height is transformed into energy of movement in another part of the game, but the total energy remains constant.

5 0

3 years ago

I’ll give brainliest if it’s correct ;-;z

BlackZzzverrR [31]

Explanation:

what is the question? could you pls provide it

6 0

3 years ago

A 10cm long wire is pulled along a U-shaped conducting rail in a perpendicular magnetic field. The total resistance of the wire

soldi70 [24.7K]

Answer:

a) $v=4.0m/s$

b) $B=2.958T$

Explanation:

From the question we are told that:

Wire Length $l=10cm=>0.10m$

Resistance $R=0.35$

Force $F=1.0N$

Power $P=4.0W$

Generally the equation for Power is mathematically given by

$P=Fv$

Therefore

$v=\frac{P}{F}$

$v=\frac{4.0}{1.0}$

$v=4.0m/s$

Generally the equation for Magnetic Field is mathematically given by

$B=\frac{\sqrt{PR}}{vl}$

$B=\frac{\sqrt{4*0.35}}{4*0.10}$

$B=2.958T$

5 0

3 years ago

A 0.877 mol sample of N2(g) initially at 298 K and 1.00 atm is held at constant volume while enough heat is applied to raise the

MissTica

Answer : The value of $q\text{ and }\Delta U$ is 286.2 J and 286.2 J respectively.

Explanation : Given,

Moles of sample = 0.877 mol

Change in temperature = 15.7 K

First we have to calculate the heat absorbed by the system.

Formula used :

$q=n\times c_v\times \Delta T$

where,

q = heat absorbed by the system = ?

n = moles of sample = 0.877 mol

$\Delta T$ = Change in temperature = 15.7 K

$c_v$ = heat capacity at constant volume of $N_2$ (diatomic molecule) = $\frac{5}{2}R$

R = gas constant = 8.314 J/mol.K

Now put all the given value in the above formula, we get:

$q=0.877mol\times \frac{5}{2}\times 8.314J/mol.K\times 15.7K$

$q=286.2J$

Now we have to calculate the change in internal energy of the system.

$\Delta U=q+w$

As we know that, work done is zero at constant volume. So,

$\Delta U=q=286.2J$

Therefore, the value of $q\text{ and }\Delta U$ is 286.2 J and 286.2 J respectively.

8 0

4 years ago

Read 2 more answers

Certain neutron stars (extremely dense stars) are believed to be rotating at about 10 rev/s. If such a star has a radius of 18 k

Aleks [24]

Answer:

mass of the neutron star =3.45185×10^26 Kg

Explanation:

When the neutron star rotates rapidly, a material on its surface to remain in place, the magnitude of the gravitational acceleration on the central material must be equal to magnitude of the centripetal acc. of the rotating star.

That is

$\frac{GM_{ns}}{R^2}= \omega^2 R$

M_ns = mass odf the netron star.

G= gravitational constant = 6.67×10^{-11}

R= radius of the star = 18×10^3 m

ω = 10 rev/sec = 20π rads/sec

therefore,

$M_{ns}= \frac{\omega^2R^3}{G} = \frac{4\pi^2\times(18\times10^3)^3}{6.67\times10^{-11}}$

= 3.45185... E26 Kg

= 3.45185×10^26 Kg

4 0

3 years ago