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2 years ago

Which refers to the number of wavelengths that pass a fixed point in a second?

Physics

1 answer:

pav-90 [236]2 years ago

7 0

Answer:

The word that refers to the number of wavelengths that pass a fixed point in a second is:
C.) Frequency

Frequency is also known as hertz.

Explanation:

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What changes occurred when Zoroastrianism became a state religion during the Sasanian Empire?

statuscvo [17]

Zoroastrianism affected the way the Persians governed their subjects by allowing the territories under their rule to worship their own religion.

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2 years ago

a car travelling at 50km/h from rest covers a distance of 10km in 40minutes. Calculate the acceleration

Margarita [4]

Answer:

$9.67\cdot 10^{-3}m/s^2$

Explanation:

We can solve the problem by using the following suvat equation:

$v^2-u^2=2as$

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the displacement

For the car in this problem:

u = 0 (it starts from rest)

$v=50 km/h \cdot \frac{1000}{3600}=13.9 m/s$ is the final velocity

s = 10 km = 10 000 m is the displacement

Solving for a, we find:

$a=\frac{v^2-u^2}{2s}=\frac{13.9^2}{2(10000)}=9.67\cdot 10^{-3}m/s^2$

7 0

2 years ago

Example 2.1

tatiyna

Answer:

5 min 20 km --------

Explanation:

5 min 20 km --------

8 0

2 years ago

At take off, a plane flies 100 km north before turning to fly 200 km east. How far is its destination from where the plane took

Snowcat [4.5K]

The answer is 300 km

To the destination

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2 years ago

If the distance between the Earth and Moon were half what it is now, by what factor would the force of gravity between them be c

hichkok12 [17]

Answer:

Explanation:

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

$m_1$ = Mass of Earth

$m_2$ = Mass of Moon

r = Distance between Earth and Moon

Old gravitational force

$F_o=\dfrac{Gm_1m_2}{r^2}$

New gravitational force

$F_n=\dfrac{Gm_1m_2}{(\dfrac{1}{2}r)^2}$

Dividing the equations

$\dfrac{F_n}{F_o}=\dfrac{\dfrac{Gm_1m_2}{(\dfrac{1}{2}r)^2}}{\dfrac{Gm_1m_2}{r^2}}\\\Rightarrow \dfrac{F_n}{F_o}=\dfrac{\dfrac{Gm_1m_2}{\dfrac{1}{4}r^2}}{\dfrac{Gm_1m_2}{r^2}}\\\Rightarrow \dfrac{F_n}{F_o}=4$

The ratio is $\dfrac{F_n}{F_o}=4$

The new force would be 4 times the old force

7 0

2 years ago