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3 years ago

AM radio waves have longer wavelength than FM radio waves? true or false?

Physics

1 answer:

Mars2501 [29]3 years ago

8 0

Answer:

true

Explanation:

the wavelength of AM radio is 10 m whereas FM has 1 m

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A 1-kg rock is suspended by a massless string from one end of a

maxonik [38]

Answer:

The weight of measuring stick is 9.8 N

Explanation:

given information:

the mass of the rock, $m_{r}$ = 1 kg

measuring stick, x =1 m

d = 0.25 m

to find the weight of measuring stick, we can use the following equation:

τ = Fd

τ = 0

$F_{r} d$ - $F_{s}d$ = 0

F_{r} = the force of the rock

F_{s} = the force of measuring stick

$F_{s} =F_{r}$

= m g

= 1 kg x 9.8 m/s

= 9.8 N

thus, the weight of measuring stick is 9.8 N

6 0

3 years ago

Charge: A piece of plastic has a net charge of +2.00 μC. How many more protons than electrons does this piece of plastic have? (

snow_lady [41]

Answer

given,

net charge = +2.00 μC

we know,

1 coulomb charge = 6.28 x 10¹⁸electrons

1 micro coulomb charge = 6.28 x 10¹⁸ x 10⁻⁶ electron

= 6.28 x 10¹² electrons

2.00 μC = 2 x 6.28 x 10¹² electrons

= 1.256 x 10¹³ electrons

since net charge is positive.

The number of protons should be 1.256 x 10¹³ more than electrons.

hence, +2.00 μC have 1.256 x 10¹³ more protons than electrons.

6 0

3 years ago

______ interference occurs when two waves overlap and the resulting wave has a larger amplitude.

ratelena [41]

Constructive is correct

8 0

3 years ago

Read 2 more answers

The 9-inch-long elephant nose fish in the Congo River generates a weak electric field around its body using an organ in its tail

Alik [6]

Answer:

$1.34\cdot 10^{-16} C$

Explanation:

The strength of the electric field produced by a charge Q is given by

$E=k\frac{Q}{r^2}$

where

Q is the charge

r is the distance from the charge

k is the Coulomb's constant

In this problem, the electric field that can be detected by the fish is

$E=3.00 \mu N/C = 3.00\cdot 10^{-6}N/C$

and the fish can detect the electric field at a distance of

$r=63.5 cm = 0.635 m$

Substituting these numbers into the equation and solving for Q, we find the amount of charge needed:

$Q=\frac{Er^2}{k}=\frac{(3.00\cdot 10^{-6} N/C)(0.635 m)^2}{9\cdot 10^9 Nm^2 C^{-2}}=1.34\cdot 10^{-16} C$

4 0

3 years ago

Melissa heated a few blue crystals in a test tube. After a few minutes, she observed that the crystals had turned white and a fi

seraphim [82]

I think it will be developed theory

4 0

3 years ago