"<span>a layer in the earth's stratosphere at an altitude of about 6.2
miles (10 km) containing a high concentration of ozone, which absorbs
most of the ultraviolet radiation reaching the earth from the sun."
Hope this helps!
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Answer:
Distance, d = 0.1 m
It is given that,
Initial velocity of meson,
Finally, the meson is coming to rest v = 0
Acceleration of the meson, (opposite to initial velocity)
Using third equation of motion as :
s is the distance the meson travelled before coming to rest.
So,
s = 0.1 m
The meson will cover the distance of 0.1 m before coming to rest. Hence, this is the required solution.
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The orbiting velocity of the satellite is 4.2km/s.
To find the answer, we need to know about the orbital velocity of a satellite.
<h3>What's the expression of orbital velocity of a satellite?</h3>
- Mathematically, orbital velocity= √(GM/r)
- r = radius of the orbital, M = mass of earth
<h3>What's the orbital velocity of a satellite orbiting earth with a radius 3.57 times the earth radius?</h3>
- M= 5.98×10²⁴ kg, r= 3.57× 6.37×10³ km = 22.7×10⁶m
- Orbital velocity= √(6.67×10^(-11)×5.98×10²⁴/22.7×10⁶)
=4.2km/s
Thus, we can conclude that the orbiting velocity of the satellite is 4.2km/s.
Learn more about the orbital velocity here:
brainly.com/question/22247460
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Answer:
The acceleration of the cart is 1.0 m\s^2 in the negative direction.
Explanation:
Using the equation of motion:
Vf^2 = Vi^2 + 2*a*x
2*a*x = Vf^2 - Vi^2
a = (Vf^2 - Vi^2)/ 2*x
Where Vf is the final velocity of the cart, Vi is the initial velocity of the cart, a the acceleration of the cart and x the displacement of the cart.
Let x = Xf -Xi
Where Xf is the final position of the cart and Xi the initial position of the cart.
x = 12.5 - 0
x = 12.5
The cart comes to a stop before changing direction
Vf = 0 m/s
a = (0^2 - 5^2)/ 2*12.5
a = - 1 m/s^2
The cart is decelerating
Therefore the acceleration of the cart is 1.0 m\s^2 in the negative direction.