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Over [174]
3 years ago
12

What and where is the asteroid belt? Please ANSWER THIS

Physics
2 answers:
Mrrafil [7]3 years ago
7 0

Answer:

The asteroid belt (sometimes referred to as the main asteroid belt) orbits between Mars and Jupiter. It consists of asteroids and minor planets forming a disk around the sun. It also serves as a sort of dividing line between the inner rocky planets and outer gas giants.

Explanation:

Hope this helps! :)

Rudiy273 years ago
3 0

Answer:

The asteroid belt is a region of our solar system, between the orbits of Mars and Jupiter, in which many small bodies orbit our sun.

Explanation:

Hope this helps!

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A bird flies from the South Pole to the North Pole. Part of the journey is 1000 miles that takes 2 weeks. What is the bird’s vel
viktelen [127]
1000 miles = 1610km = 1.61x10^6m
2 weeks = 14 days = 14x24x1440

V=d/t = 1.61x10^6/14x24x1440
= 3.33m/s
6 0
3 years ago
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During World War I, the Germans had a gun called Big Bertha that was used to shell Paris. The shell had an initial speed of 2.61
bonufazy [111]

Answer:

The shell hit at a distance of 1.9 x 10² km

The time of flight of the shell was 5.3 x 10² s

Explanation:

The position of the shell is given by the vector "r":

r  = (x0 + v0 * t * cos α ; y0 + v0 * t * sin α + 1/2 g t²)

where:

x0 = initial horizontal position

v0 = magnitude of the initial velocity

t = time

α = launching angle

y0 = initial vertical position

g = acceleration of gravity

When the shell hit, the vertical component (ry) of the vector position r is 0. See figure.

Then:

ry = 0 =  y0 + v0 * t * sin α + 1/2 g t²

Since the gun is at the center of our system of reference, y0 and x0 = 0

0 = t (v0 sin α + 1/2 g t)

t= 0 is discarded as solution

v0 sin α + 1/2 g t = 0

t = -2v0 sin α / g

t = (-2 * 2610 m/s * sin 81.9°)/ (-9.8 m/s²) = 5.3 x 10² s. This is the time of flight of the shell until it hit.

Then, the distance at which the shell hit is:

Distance = Module of r = ( x0 + v0 * t * cos α; 0) = x0 + v0 * t * cos α  

Distance = 2.61 km/s * 5.3 x 10² s * cos 81.9 = 1.9 x 10² km

7 0
3 years ago
A microcomputer that is smaller, lighter, and less powerful than a notebook, and that has a touch sensitive screen, is called a
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<span>A microcomputer that is smaller, lighter, and less powerful than a notebook, and that has a touch sensitive screen, is called a tablet. Tablets are used similarly to computers in the way that information can be stored, viewed and edited on them.</span>
8 0
3 years ago
a stone is projected vertically up from the top of a tower 73.5m with velocity 24.5 m/s . find the time taken by the stone to re
hoa [83]

The stone's altitude at time t is given by

y=73.5\,\mathrm m+\left(24.5\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2

where g=9.80\dfrac{\rm m}{\mathrm s^2} is the acceleration due to gravity. The stone reaches the ground when y=0:

0=73.5\,\mathrm m+\left(24.5\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2\implies t=7.11\,\rm s

6 0
3 years ago
A runner begins from rest at the starting line and travles for 6.5 seconds, a runner reaches a speed of 13.4 m/s what is the run
Butoxors [25]

The acceleration of the runner in the given time is 2.06m/s².

Given the data in the question;

Since the runner begins from rest,

  • Initial velocity; u = 0
  • Final velocity; v = 13.4m/s
  • Time elapsed; t = 6.5s

Acceleration of the runner; a = \ ?

<h3>Velocity and Acceleration</h3>

Velocity is the speed at which an object moves in a particular direction.

Acceleration is simply the rate of change of the velocity of a particle or object with respect to time. Now, we can see the relationship from the First Equation of Motion

v = u + at

Where v is final velocity, u is initial velocity, a is acceleration and t is time elapsed.

To determine the acceleration of the runner, we substitute our given values into the equation above.

v = u + at\\\\13.4m/s = 0 + (a * 6.5s)\\\\13.4m/s = a * 6.5s\\\\a = \frac{13.4m/s}{6.5s}\\ \\a = 2.06m/s^2

Therefore, the acceleration of the runner in the given time is 2.06m/s².

Learn more about Equations of Motion: brainly.com/question/18486505

3 0
2 years ago
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