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hodyreva [135]
3 years ago
12

Different between AM and FM

Physics
2 answers:
Hatshy [7]3 years ago
7 0
AM is morning and fmis evening or night till 12
lidiya [134]3 years ago
5 0
The difference is in how the carrier wave is modulated, or altered. With AM radio, the amplitude, or overall strength, of the signal is varied to incorporate the sound information. With FM, the frequency (the number of times each second that the current changes direction) of the carrier signal is varied.
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A car starts from rest and after 7 seconds it is moving at 42 m/s. What is the car’s average acceleration? A. 0.17 m/s2 B. 1.67
aliya0001 [1]
Acceleration =velocity /time
=42/7
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3 years ago
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The picture shows an object resting on a balance.
Maslowich

Answer:

4.90kgm^-2

Explanation:

4 0
2 years ago
An 89 kg man drops from rest on a diving board −3.1 m above the surface of the water and comes to rest 0.5 s after reaching the
OLga [1]

To solve this problem we will use the linear motion kinematic equations, for which the change of speed squared with the acceleration and the change of position. The acceleration in this case will be the same given by gravity, so our values would be given as,

m= 89 kg\\x = 3.1 m\\t = 0.5s\\a = g = 9.8m/s^2

Through the aforementioned formula we will have to

v_f^2-v_i^2 = 2ax

The particulate part of the rest, so the final speed would be

v_f^2 = 2gx

v_f=\sqrt{2(9.8)(3.1)}

v_f = 7.79m/s

Now from Newton's second law we know that

F = ma

Here,

m = mass

a = acceleration, which can also be written as a function of velocity and time, then

F = m\frac{dv}{dt}

Replacing we have that,

F = (89)\frac{7.79}{0.5}

F = 1386.62N

Therefore the force that the water exert on the man is 1386.62

3 0
3 years ago
The__ of friction is a number that represents the resistance to sliding
Tresset [83]

Answer:

B. coefficient

Explanation:

i dont have to explain right?

3 0
3 years ago
To aid in the prevention of tooth decay, it is recommended that drinking water contain 0.800 ppm fluoride, F−. How many grams of
Leviafan [203]

Answer:

2.23 × 10^6 g of F- must be added to the cylindrical reservoir in order to obtain a drinking water with a concentration of 0.8ppm of F-

Explanation:

Here are the steps of how to arrive at the answer:

The volume of a cylinder = ((pi)D²/4) × H

Where D = diameter of the cylindrical reservoir = 2.02 × 10^2m

H = Height of the reservoir = 87.32m

Therefore volume of cylindrical reservoir = (3.142×202²/4)m² × 87.32m = 2798740.647m³

1ppm = 1g/m³

0.8ppm = 0.8 × 1g/m³

= 0.8g/m³

Therefore to obtain drinking water of concentration 0.8g/m³ in a reservoir of volume 2798740.647m³, F- of mass = 0.8g/m³ × 2798740.647m³ = 2.23 × 10^6 g must be added to the tank.

Thank you for reading.

4 0
3 years ago
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