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o-na [289]
3 years ago
13

What is the velocity of a wave that has a wavelength of 20 meters and frequency of 0.5 Hz?

Physics
2 answers:
KiRa [710]3 years ago
4 0
Answer:

10 m/s

I hope this helps :)
k0ka [10]3 years ago
3 0

Answer:

v = 10 m/s

Explanation:

recall that velocity is related to wavelength and frequency by the formula

v = fλ

where v = velocity, f = frequency and λ= wavelength

Simply substitute these into the formula:

v = fλ

v = (0.5)(20)

v = 10 m/s

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What is the acceleration of a 1000kg car subject to a 550N net force?
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Answer:

a=550÷1000

a=0.55m/s²

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2 years ago
A block of amber is placed in water and a laser beam travels from the water through the amber. The angle of incidence is 35 degr
7nadin3 [17]
Answer: 1.88

Explanation

Applying Snell’s Law, sin(1)/sin(2) = n(2)/n(1), where n is the index of refraction and sin 1 and 2 being of incidence and refracted respectively.

1) sin35/sin24 = n(2)/1.33
2) 1.41 = n(2)/1.33
3) n(2) = 1.41 x 1.33
4) n(2) = 1.88

Hope this helps :)
7 0
3 years ago
An object is most likely to sink in water if
LiRa [457]

Answer:

High density D answers to your questions

8 0
2 years ago
Read 2 more answers
One differnce between magnetic poles and elcyrical charges is that​
Sphinxa [80]

Answer:

In the electric field, the like charges repel each other, and the unlike charges attract each other, whereas in a magnetic field the like poles repel each other and the unlike poles attract each other.

Explanation:

8 0
3 years ago
Find the final temperature of 375 grams of tea (c = 4.184 J/g°C) if its initial temperature is 95°C just before it is placed in
Minchanka [31]

Answer:

the final temperature of the tea is 7.39⁰C.

Explanation:

Given;

mass of the tea, m = 375 g

specific heat capacity of the tea, C = 4.184 JJ/g°C

initial temperature of the tea, t₁ = 95°C

the final temperature of the tea, t₂ = ?

Energy lost by the refrigerator, Q = 137,460 J

The energy lost by the refrigerator is given by the following formula;

-Q = mc(t₂ - t₁)

-137,460 =375 x 4.184(t₂ - 95°C)

-137,460 = 1569(t₂ - 95°C)

t_2-95 = \frac{-137,460}{1569} \\\\t_2-95 = -87.61\\\\t_2 = -87.61 + 95\\\\t_2 = 7.39 \ ^0 C

Therefore, the final temperature of the tea is 7.39⁰C.

4 0
3 years ago
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