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2 years ago

11

If our year were twice as long (that is, if Earth took twice as many days to complete each orbit around the Sun), but Earth's ro

tation period and axis tilt were unchanged, then ____

1 answer:

Flauer [41]2 years ago

6 0

Answer:

the four seasons would each be twice as long as they are now

Explanation:

A year is the result of the Earths revolution around the sun i.e., the Earth takes 365 days to complete one revolution around the sun. A day is the result of the Earth's rotation on its axis i.e., it takes 24 hours for Earth to complete one rotation on its axis.

The Earth is titled by about 23.5 degrees, this results in the seasons of Earth. The distance from the sun has nothing to do with the seasons. During the northern hemisphere winter the Earth is actually closer to the sun than in the summer. So, if our year was twice as long then the seasons would be twice as long.

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Consider a magnetic force acting on an electric charge in a uniform magnetic field. Which of the following statements are true?

7nadin3 [17]

Answer: 5) "The direction of the magnetic force acting on a moving charge in the magnetic field is perpendicular to the direction of the magnetic field"

3) "An electric charge moving perpendicular to the magnetic field experiences a magnetic force"

Explanation:

When a charge of magnitude q, moving with a velocity v is placed in a magnetic field of strength B, it experiences a force, the magnitude of this force F is given mathematically as

F =qvB sinx

Where x is the angle between the magnetic field and the velocity of the charge.

From this equation, we can see that the force is zero when magnetic field strength B is parallel to velocity (x=0) or when velocity v is zero.

Also the force F is maximum when the angle between magnetic field strength B and velocity is 90 ( that's they are perpendicular).

By the right hand rule, the force, velocity and strength of magnetic field are perpendicular to each other.

These points have made statement 3 and 5 of the questions to be correct.

7 0

3 years ago

If the work function of a material is such that red light of wavelength 700 nm just barely initiates the photoelectric effect, w

lawyer [7]

Answer: $2.13(10)^{-19} J$

Explanation:

The photoelectric effect consists of the emission of electrons (electric current) that occurs when light falls on a metal surface under certain conditions.

If the light is a stream of photons and each of them has energy, this energy is able to pull an electron out of the crystalline lattice of the metal and communicate, in addition, a kinetic energy.

<u>This is what Einstein proposed: </u>

Light behaves like a stream of particles called photons with an energy $E$

$E=h.f$ (1)

Where:

$h=6.63(10)^{-34}J.s$ is the Planck constant

$f$ is the frequency

Now, the frequency has an inverse relation with the wavelength $\lambda$ :

$f=\frac{c}{\lambda}$ (2)

Where $c=3(10)^{8}m/s$ is the speed of light in vacuum and $\lambda=400nm=400(10)^{-9}m$ is the wavelength of the absorbed photons in the photoelectric effect.

Substituting (2) in (1):

$E=\frac{h.c}{\lambda}$ (3)

So, the energy $E$ of the incident photon must be equal to the sum of the Work function $\Phi$ of the metal and the maximum kinetic energy $K_{max}$ of the photoelectron:

$E=\Phi+K_{max}$ (4)

Rewriting to find $K_{max}$ :

$K_{max}=E-\Phi$ (5)

Where $\Phi$ is the minimum amount of energy required to induce the photoemission of electrons from the surface of a metal, and its value depends on the metal:

$\Phi=h.f_{o}=\frac{h.c}{\lambda_{o}}$ (6)

Being $\lambda_{o}=700nm=700(10)^{-9}m$ the threshold wavelength (the minimum wavelength needed to initiate the photoelectric effect)

Substituting (3) and (6) in (5):

$K_{max}=\frac{h.c}{\lambda}-\frac{h.c}{\lambda_{o}}$

$K_{max}=h.c(\frac{1}{\lambda}-\frac{1}{\lambda_{o}})$ (7)

Substituting the known values:

$K_{max}=(6.63(10)^{-34}J.s)(3(10)^{8}m/s)(\frac{1}{400(10)^{-9}m}-\frac{1}{700(10)^{-9}m})$

$K_{max}=2.13(10)^{-19} J$ >>>>>This is the maximum kinetic energy that ejected electrons must have when violet light illuminates the material

7 0

3 years ago

1. While driving to Palm Desert you notice that some of the windmills near the freeway are spinning 20 times per minute. The len

igomit [66]

Answer:

a) $\alpha = -0.233 rad/s^{2}$

b) the rotational acceleration will remain the same, $\alpha = -0.233 rad/s^{2}$

c) When r = 36 m, $a_{c} = 157.25 m/s^{2}$

When r = 18 m, $a_{c} = 78.63 m/s^{2}$

d) When r = 36 m, $a_{c} = 39.31 m/s^{2}$

When r = 36 m, $a_{c} = 19.66 m/s^{2}$

Explanation:

The windmills are spinning 20 times per minute, the number of spins in 1 second = 20/60 = 1/3 spins/sec

frequency, f = 1/3 spins/sec

there are 1/3 spins in 1 second,

there will be 50 spins in 50/ (1/3) seconds = 150 seconds

i.e time taken to make 50 spins = 150 seconds

Time interval between 20 and 50 spins, Δt = 150 - 60 = 90

Δt = 90

Angular frequency at 20 spins:

w₁ = 2 π f

w₁ = 2 π * 1/3 = 2π/3 = 2.09 rad/s

Since the blade stops at 50 spins, angular frequency at 50 spins, w₂ = 0

The rotational acceleration,

$\alpha = \frac{\triangle w}{\triangle t} \\\alpha = \frac{w_{2} - w_{1} }{\triangle t} \\\alpha = \frac{0 - 2.09 }{90} \\\alpha = -0.233 rad/s^{2}$

b) The magnitude of the rotational acceleration does not depend on the radius, r. It depends on the angular frequency, therefore the magnitude of the rotational acceleration doe not change.

c) The centripetal acceleration is given by the formula, $a_{c} = w^{2} r$

While the windmill spins at 20 times per minute, the angular speed gotten, w = 2.09 rad/s

When r = 36 m

$a_{c} = 2.09^{2} * 36\\a_{c} = 157.25 m/s^{2}$

At the halfway point, r = 18 m

$a_{c} = 2.09^{2} * 18\\a_{c} = 78.63 m/s^{2}$

d) If the angular velocity of the blades were cut in halves:

w = 2.09/2

w = 1.045 rad/s

When r = 36 m

$a_{c} = 1.045^{2} * 36\\a_{c} = 39.31 m/s^{2}$

When r = 18 m

$a_{c} = 1.045^{2} * 18\\a_{c} = 19.66 m/s^{2}$

8 0

3 years ago

• 500 waves pass by in 2 second. These waves have a wavelength of 6

tamaranim1 [39]

Answer:

6/2times500=1500

Explanation:

s = d/t which means speed equals distance divided by time.

7 0

3 years ago

A girl weighing 600 N steps on a bathroom scale that contains a stiff spring. In equilibrium, the spring is compressed 1.0 cm un

8_murik_8 [283]

Answer:

The spring constant is 60,000 N

The total work done on it during the compression is 3 J

Explanation:

Given;

weight of the girl, W = 600 N

compression of the spring, x = 1 cm = 0.01 m

To determine the spring constant, we apply hook's law;

F = kx

where;

F is applied force or weight on the spring

k is the spring constant

x is the compression of the spring

k = F / x

k = 600 / 0.01

k = 60,000 N

The total work done on the spring = elastic potential energy of the spring, U;

U = ¹/₂kx²

U = ¹/₂(60000)(0.01)²

U = 3 J

Thus, the total work done on it during the compression is 3 J

3 0

3 years ago