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Ulleksa [173]
2 years ago
5

A cricket can travel approximately 8 m/s. How many meters could a cricket travel in 75 s?

Physics
1 answer:
nika2105 [10]2 years ago
7 0

Answer:

Your answer will be 600meters

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The closest star to our solar system is Alpha Centauri, which is 4.12 × 10^16 m away. How long would it take light from Alpha Ce
bonufazy [111]

Time = (distance) / (speed)

Time = (4.12x10^16 m) / (3 x10^8 m/s)

Time = 1.37 x 10^8 seconds

Divide the seconds by 86,400 to get days. Then divide the days by 365 to get years.

Time = about 4.35 years

4 0
3 years ago
The escape velocity is defined to be the minimum speed with which an object of mass m must move to escape from the gravitational
s344n2d4d5 [400]

Answer:

v = √2G M_{earth} / R

Explanation:

For this problem we use energy conservation, the energy initiated is potential and kinetic and the final energy is only potential (infinite r)

        Eo = K + U = ½ m1 v² - G m1 m2 / r1

        Ef = - G m1 m2 / r2

When the body is at a distance R> Re, for the furthest point (r2) let's call it Rinf

       Eo = Ef

       ½ m1v² - G m1 M_{earth} / R = - G m1 M_{earth} / R

      v² = 2G M_{earth} (1 / R - 1 / Rinf)

If we do Rinf = infinity     1 / Rinf = 0

       v = √2G M_{earth} / R

      Ef = = - G m1 m2 / R

The mechanical energy is conserved  

 

      Em = -G m1  M_{earth} / R

      Em = - G m1  M_{earth} / R

     R = int        ⇒  Em = 0

6 0
2 years ago
In water what happens to a base?
Ratling [72]
Your answer would be B. It forms hydrogen ions (H+)
4 0
3 years ago
Read 2 more answers
The igneous rocks which were deposited on the surface and then cooled are known as __________.
mr Goodwill [35]
The igneous rocks which were deposited on the surface and then cooled are known as extrusive. These rocks are a result of a magma reaching the surface of the Earth which cools the magma quickly. Examples are rhyolite, basalt, obsidian and andesite.
6 0
3 years ago
You attach a meter stick to an oak tree, such that the top of the meter stick is 2.27 meters above the ground. later, an acorn f
Alexandra [31]

The acorn was at a height of <u>4.15 m</u> from the ground before it drops.

The acorn takes a time t to fall through a distance h₁, which is the length of the scale. When the acorn reaches the top of the scale, its velocity is u.

Calculate the speed of the acorn at the top of the scale, using the equation of motion,

s=ut+ \frac{1}{2} at^2

Since the acorn falls freely under gravity, its acceleration is equal to the acceleration due to gravity g.

Substitute 2.27 m for s (=h₁), 0.301 s for t and 9.8 m/s² for a (=g).

s=ut+ \frac{1}{2} at^2\\ (2.27 m)=u(0.301s)+\frac{1}{2}(9.8m/s^2)(0.301s)^2\\ u=\frac{1.8261m}{0.301s} =6.067m/s

If the acorn starts from rest and reaches a speed of 6.067 m/s at the top of the scale, it would have fallen a distance h₂ to achieve this speed.

Use the equation of motion,

v^2=u^2+2as

Substitute 6.067 m/s for v, 0 m/s for u, 9.8 m/s² for a (=g) and h₂ for s.

v^2=u^2+2as\\ (6.067m/s)^2=(0m/s)^2+2(9.8m/s^2)h_2\\ h_2=\frac{(6.067m/s)^2}{2(9.8m/s^2)} =1.878 m

The height h above the ground at which the acorn was is given by,

h=h_1+h_2=(2.27 m)+(1.878 m)=4.148 m

The acorn was at a height <u>4.15m</u> from the ground before dropping down.

3 0
3 years ago
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