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liberstina [14]
3 years ago
11

Newton's first law of motion is sometimes called the law of _________.

Physics
1 answer:
Gre4nikov [31]3 years ago
3 0
Newtons first law of motion is also known as the law of inertia
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A fisherman notices that his boat is moving up and down periodically without any horizontal motion, owing to waves on the surfac
Y_Kistochka [10]

Answer:

See Explanation

Explanation:

a) We know that;

v = λf

Where;

λ = wavelength of the wave

f = frequency of the wave

v = velocity of the wave

So;

T = 2 * 2.10 s = 4.2 s

Hence f = 1/4.2 s

f = 0.24 Hz

The wavelength =  6.5 m

Hence;

v = 6.5 m * 0.24 Hz

v = 1.56 m/s

b)The amplitude of the wave is;

A =  0.600 m/2 = 0.300 m

c) Since the wave speed does not depend on the amplitude of the wave then the answer in (a) above remains the same

Where d = 0.30 m

A = 0.30 m/2 = 0.15 m

6 0
3 years ago
Which is more reliable—using a manual stop watch or using light gates?<br> Explain.
Kitty [74]
Light gates are more reliable. When using a manual stop watch, it is difficult to stop it at an exact time. A light gate is able to detect when an object passes through a 'gate' with the infrared transmitter and receiver. 
7 0
4 years ago
What is the attractive force between all matter in the universe?
kvv77 [185]
The attractive force between all matter in the universe is gravity.
4 0
3 years ago
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How do Air molecules enable sound to travel from a radios speaker to your ears?
MatroZZZ [7]
The radio frequencies push one air molecule that then bumps into a different air molecule.....which then hits another and another causing a line of crashing molecules that lead inside your ear and hits your ear drum causing it to vibrate which causes the sounds.
6 0
3 years ago
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A certain spring stretches 3 cm when a load of 15 n is suspended from it. how much will the spring stretch if 30 n is suspended
Alik [6]
Initially, the spring stretches by 3 cm under a force of 15 N. From these data, we can find the value of the spring constant, given by Hook's law:
k= \frac{F}{\Delta x}
where F is the force applied, and \Delta x is the stretch of the spring with respect to its equilibrium position. Using the data, we find
k= \frac{15 N}{3.0 cm}=5.0 N/cm

Now a force of 30 N is applied to the same spring, with constant k=5.0 N/cm. Using again Hook's law, we can find the new stretch of the spring:
\Delta x =  \frac{F}{k}= \frac{30 N}{5.0 N/cm}=6 cm
4 0
3 years ago
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