Please help me with my question

The frequency of a wave is 1) A) measured in cycles per second. B) measured in hertz (Hz). C) the number of peaks passing by any

7nadin3 [17]

Answer: E) all of the above

Explanation:

Frequency is the number of vibrations in one second. It is also defined as the number of crests that pass a point in a given time.

The frequency and the wavelength has an inverse relationship.

Frequency is measured in cycles per second or Hertz(Hz).

The relationship between wavelength and frequency of the wave follows the equation:

$\nu=\frac{c}{\lambda}$

where,

$\nu$ = frequency of the wave

c = speed of light

$\lambda$ = wavelength of the wave

Thus all the given statements are true.

6 0

3 years ago

What is the force that opposes the movement of an object through water??

anastassius [24]

In air, you'd call it air resistance, pilots call it drag. In water, some call it
water resistance, or just plain drag.

Whatever it's called, it's the friction between the object and the fluid, plus
the force needed to push the fluid out of the way to let the object get through.

5 0

3 years ago

Read 2 more answers

Four identical masses m are evenly spaced on a frictionless 1D track. The first mass is sent at speed v toward the other three.

SpyIntel [72]

Answer:

The speed decreases 75%.

Explanation:

Since no friction present, assuming no external forces acting during the three collisions, total momentum must be conserved.
For the first collission, only mass 1 is moving before it, so we can write the following equation:

$p_{i} = p_{f} = m*v_{o} (1)$

Since both masses are identical, and they stick together after the collision, we can express the final momentum as follows:

$p_{f1} = 2*m*v_{1} (2)$

From (1) and (2) we get:
v₁ = v₀/2 (3)
Since the masses are moving on a frictionless 1D track, the speed of the set of mass 1 and 2 combined together before colliding with mass 3 is just v₁, so the initial momentum prior the second collision (p₁) can be expressed as follows:

$p_{1} = 2*m*v_{1} = 2*m*\frac{v_{o} }{2} = m*v (4)$

Since after the collision the three masses stick together, we can express this final momentum (p₂) as follows:

$p_{2} = 3*m*v_{2} (5)$

From (4) and (5) we get:
v₂ = v₀/3 (6)

Since the masses are moving on a frictionless 1D track, the speed of the set of mass 1, 2 and 3 combined together before colliding with mass 4 is just v₂, so the initial momentum prior the third collision (p₂) can be expressed as follows:

$p_{2} = 3*m*v_{2} = 3*m*\frac{v_{o} }{3} = m*v (7)$

Since after the collision the four masses stick together, we can express this final momentum (p₃) as follows:

$p_{3} = 4*m*v_{3} (8)$

From (7) and (8) we get:
v₃ = v₀/4
This means that after the last collision, the speed will have been reduced to a 25% of the initial value, so it will have been reduced in a 75% regarding the initial value of v₀.

5 0

3 years ago

An object with a mass of 5kg accelerates at 2m/s2. How much force in Newtons(N) is needed to cause this to happen??

Reptile [31]

Answer: The formula of Newtons second law of motion is F=MA so therefore it would be written like this Force = Mass X Acceleration

F = 5 x 2

F = 10 N

6 0

3 years ago

Microscopes are often equipped with blue filters. give a reason for using blue light in a microscope

timama [110]

The blue light removes red parts of the color spectrum giving it a cooler color

8 0

4 years ago