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3 years ago

Day and night are caused by

Physics

1 answer:

JulijaS [17]3 years ago

4 0

Answer:

Explanation:

How is the rotation of the earth responsible for causing day and night?

We get day and night because the Earth spins (or rotates) on an imaginary line called its axis and different parts of the planet are facing towards the Sun or away from it. It takes 24 hours for the world to turn all the way around, and we call this a day.

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Could someone help me with some info for an essay?I WILL REPOST FOR 100 POINTS IF YOU CAN.

polet [3.4K]

Answer:

Self-motivation is the surest way to stay focused. Nearly all the most significant tasks in life are tests for our motivation. Facing challenges and achieving success requires focus; this means we need to be responsible for our motivation. People who are motivated towards achieving their goals are focused on success. The following ideas will help you stay focused and motivated in your work and studies.

Explanation:

3 0

3 years ago

A simple model of a hydrogen atom is a positive point charge +e (representing the proton) at the center of a ring of radius a wi

Norma-Jean [14]

Answer:

Now e is due to the ring at a

We say

1/4πEo(ea/ a²+a²)^3/2

= 1/4πEo ea/2√2a³

So here E is faced towards the ring

Next is E due to a point at the centre

E² = 1/4πEo ( e/a²)

Finally we get the total

Et= E²-E

= e/4πEo(2√2-1/2√2)

So the direction here is away from the ring

8 0

4 years ago

Consider the motion of a 4.00-kg particle that moves with potential energy given by U(x) = + a) Suppose the particle is moving w

gtnhenbr [62]

Correct question:

Consider the motion of a 4.00-kg particle that moves with potential energy given by

$U(x) = \frac{(2.0 Jm)}{x}+ \frac{(4.0 Jm^2)}{x^2}$

a) Suppose the particle is moving with a speed of 3.00 m/s when it is located at x = 1.00 m. What is the speed of the object when it is located at x = 5.00 m?

b) What is the magnitude of the force on the 4.00-kg particle when it is located at x = 5.00 m?

Answer:

a) 3.33 m/s

b) 0.016 N

Explanation:

a) given:

V = 3.00 m/s

x1 = 1.00 m

x = 5.00

$u(x) = \frac{-2}{x} + \frac{4}{x^2}$

At x = 1.00 m

$u(1) = \frac{-2}{1} + \frac{4}{1^2}$

= 4J

Kinetic energy = (1/2)mv²

$= \frac{1}{2} * 4(3)^2$

= 18J

Total energy will be =

4J + 18J = 22J

At x = 5

$u(5) = \frac{-2}{5} + \frac{4}{5^2}$

$= \frac{4-10}{25} = \frac{-6}{25} J$

= -0.24J

Kinetic energy =

$\frac{1}{2} * 4Vf^2$

= 2Vf²

Total energy =

2Vf² - 0.024

Using conservation of energy,

Initial total energy = final total energy

22 = 2Vf² - 0.24

Vf² = (22+0.24) / 2

$Vf = \sqrt{frac{22.4}{2}$

= 3.33 m/s

b) magnitude of force when x = 5.0m

$u(x) = \frac{-2}{x} + \frac{4}{x^2}$

$\frac{-du(x)}{dx} = \frac{-d}{dx} [\frac{-2}{x}+ \frac{4}{x^2}$

$= \frac{2}{x^2} - \frac{8}{x^3}$

At x = 5.0 m

$\frac{2}{5^2} - \frac{8}{5^3}$

$F = \frac{2}{25} - \frac{8}{125}$

= 0.016N

8 0

4 years ago

A hockey puck slides off the edge of a horizontal platform with an initial velocity of 28.0 m/shorizontally in a city where the

kozerog [31]

Answer:

θ = 12.60°

Explanation:

In order to calculate the angle below the horizontal for the velocity of the hockey puck, you need to calculate both x and y component of the velocity of the puck, and also you need to use the following formula:

$\theta=tan^{-1}(\frac{v_y}{v_x})$ (1)