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Schach [20]
3 years ago
14

How do asteroids and comets differ?

Physics
1 answer:
aalyn [17]3 years ago
4 0

Answer:

Explained

Explanation:

Asteroids: Asteroids are generally made up of rocky material, metals and their size are large in  comparison to comets. Asteroid belt is found between Jupiter and Mars.

Comets: Comets are made up of ice, dust and rocky material. Generally smaller in size and are far from the Sun. With each orbit nearing the sun ice of the comet vaporize in form of blazing tail.

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How long is the photons journey from the milky way to earth
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Well i’d love to answer your question but its missing some information. firstly the earth is within the milky way galaxy so something cannot travel “from” but if you said “through” the answer would be 299,792,458 meters a second
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True or False. A projectile is an object that once set in motion, continues in motion by its own inertia.
bazaltina [42]
The answer is true.
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The small ball of mass m = 0.5 kg is attached to point A via string and is moving at constant speed in a horizontal circle of ra
babymother [125]

Answer:

d = 2.45 meters

Explanation:

Mass of the ball, m = 0.5 kg

Radius of the circle, r = 0.16 m

The angular speed of the ball around the circle is, \omega=2\ rad/s

The attached figure shows the whole scenario. Let F_t is the force acting on the ball in tangential direction. The forces will balanced each other at equilibrium.

In horizontal direction,

T\ sin\theta=F_t=mr\omega^2................(1)

In vertical direction,

T\ cos\theta=mg...............(2)

From equation (1) and (2) :

tan\theta=\dfrac{r\omega^2}{g}

Also,

tan\theta=\dfrac{r}{d}

d=\dfrac{g}{\omega^2}d=\dfrac{9.8}{2^2}

d = 2.45 meters

So, the value of d is 2.45 meters. Hence, this is the required solution.  

3 0
3 years ago
Two cars are facing each other. Car A is at rest while car B is moving toward car A with a constant velocity of 20 m/s. When car
lapo4ka [179]

Answer:

Let's define t = 0s (the initial time) as the moment when Car A starts moving.

Let's find the movement equations of each car.

A:

We know that Car A accelerations with a constant acceleration of 5m/s^2

Then the acceleration equation is:

A_a(t)  = 5m/s^2

To get the velocity, we integrate over time:

V_a(t) = (5m/s^2)*t + V_0

Where V₀ is the initial velocity of Car A, we know that it starts at rest, so V₀ = 0m/s, the velocity equation is then:

V_a(t) = (5m/s^2)*t

To get the position equation we integrate again over time:

P_a(t) = 0.5*(5m/s^2)*t^2 + P_0

Where P₀ is the initial position of the Car A, we can define P₀ = 0m, then the position equation is:

P_a(t) = 0.5*(5m/s^2)*t^2

Now let's find the equations for car B.

We know that Car B does not accelerate, then it has a constant velocity given by:

V_b(t) =20m/s

To get the position equation, we can integrate:

P_b(t) = (20m/s)*t + P_0

This time P₀ is the initial position of Car B, we know that it starts 100m ahead from car A, then P₀ = 100m, the position equation is:

P_b(t) = (20m/s)*t + 100m

Now we can answer this:

1) The two cars will meet when their position equations are equal, so we must have:

P_a(t) = P_b(t)

We can solve this for t.

0.5*(5m/s^2)*t^2 = (20m/s)*t + 100m\\(2.5 m/s^2)*t^2 - (20m/s)*t - 100m = 0

This is a quadratic equation, the solutions are given by the Bhaskara's formula:

t = \frac{-(-20m/s) \pm \sqrt{(-20m/s)^2 - 4*(2.5m/s^2)*(-100m)}  }{2*2.5m/s^2} = \frac{20m/s \pm 37.42 m/s}{5m/s^2}

We only care for the positive solution, which is:

t = \frac{20m/s + 37.42 m/s}{5m/s^2} = 11.48 s

Car A reaches Car B after 11.48 seconds.

2) How far does car A travel before the two cars meet?

Here we only need to evaluate the position equation for Car A in t = 11.48s:

P_a(11.48s) = 0.5*(5m/s^2)*(11.48s)^2 = 329.48 m

3) What is the velocity of car B when the two cars meet?

Car B is not accelerating, so its velocity does not change, then the velocity of Car B when the two cars meet is 20m/s

4)  What is the velocity of car A when the two cars meet?

Here we need to evaluate the velocity equation for Car A at t = 11.48s

V_a(t) = (5m/s^2)*11.48s = 57.4 m/s

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