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3 years ago

The time required for one cycle, a complete motion that returns to its starting point, is called the _____.

Physics

1 answer:

GenaCL600 [577]3 years ago

8 0

Answer: the Time that is required is called a period

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All BUT one statement applies to electromagnetic radiation. That is

koban [17]

A) Radio waves are a form of electromagnetic radiation.

TRUE

It is the type of electromagnetic wave with least energy range and longest wavelength range among all

B) Electromagnetic radiation does not require a medium to travel.

TRUE

This radiation can travel through vacuum also and its speed is maximum in vacuum

C) In a vacuum, electromagnetic radiation travels at the speed of light.

TRUE

speed of all electromagnetic radiation is same as speed of light in vacuum

D) Electromagnetic radiation travels in the form of longitudinal waves.

FALSE

All electromagnetic radiation is always in form of transverse waves

So correct options are

4 0

3 years ago

A car accelerates from zero to a speed of 110

Verizon [17]

The car's rate of acceleration : a = 2.04 m/s²

<h3>Further explanation</h3>

Given

speed = 110 km/hr

time = 15 s

Required

The acceleration

Solution

110 km/hr⇒30.56 m/s

Acceleration is the change in velocity over time

a = Δv : Δt

Input the value :

a = 30.56 m/s : 15 s

a = 2.04 m/s²

3 0

3 years ago

A laser beam is incident at an angle of 30.0° from the vertical onto a solution of corn syrup in water. The beam is refracted to

dimaraw [331]

Answer with Explanation:

We are given that

Angle of incidence, $i=30^{\circ}$

Angle of refraction, $r=19.24^{\circ}$

a.Refractive index of air, $n_1=1$

We know that

$n_2sinr=n_1sini$

$n_2=\frac{n_1sin i}{sin r}=\frac{sin30}{sin19.24}=1.517$

b.Wavelength of red light in vacuum, $\lambda=632.8nm=632.8\times 10^{-9} m$

$1nm=10^{-9} m$

Wavelength in the solution, $\lambda'=\frac{\lambda}{n_2}$

$\lambda'=\frac{632.8}{1.517}=417nm$

c.Frequency does not change .It remains same in vacuum and solution.

Frequency, $\nu=\frac{c}{\lamda}=\frac{3\times 10^8}{632.8\times 10^{-9}}$

Where $c=3\times 10^8 m/s$

Frequency, $\nu=4.74\times 10^{14}Hz$

d.Speed in the solution, $v=\frac{c}{n_2}$

$v=\frac{3\times 10^8}{1.517}=1.98\times 10^8m/s$

5 0

3 years ago

A batter hits a foul ball. The 0.140-kg baseball that was approaching him at 40.0 m/s leaves the bat at 30.0 m/s in a direction

lara31 [8.8K]

<h3>Answer</h3><h3>7 Ns</h3><h3>Explanation</h3>

Given in the question,

mass of foul ball = 0.140 kg

initial speed with which ball was hit with the bat = 30 m/s

final speed = 40 m/s

According to the scenario the whole scene is making a right angle triangle

So, to the solve the question we will use pythagorus theorem

<h3>Hypotenuse² = base² + height²</h3>

Here,

Hypotenuse= Magnitude of impulse

Base = 1st change of momentum

height = 2nd change of momentum

1st impulse (1st change of momentum)

p = m(1)v(1) = (0.14 kg)(40.0 m/s) = 5.6 kg m / s = 5.6 N s

2nd impulse (2nd change of momentum)

p = m(2)v(2) = (0.14 kg)(30.0 m/s) = 4.2 kg m / s = 4.2 N s

Magnitude of impulse (hypotenuse of triangle)

impulse² = (5.6)² + (4.2)²

impulse² = 31.36 + 17.64

impulse² = 49

impulse² = √49

impulse = 7.0 N s

7 0

3 years ago

During science class, Ava and Justin were discussing the rotation of the Sun and debating over which model would best illustrate

agasfer [191]

Answer:

it would b B

Explanation:

3 0

3 years ago