F=M×A
F=3 ×15
F=45N
so the force is 45 N
I believe the correct answer from the choices listed above is the first option. Decreasing a telescope's eyepiece focal length will increase magnification. <span>The magnification of the </span>telescope<span> image is (</span>focal length<span> of the objective) divided by (</span>focal length <span>of the </span>eyepiece<span>). Hope this answers the question.</span>
Answer:
Spiral Galaxy
Explanation:
Spiral galaxies are named by their spiral structures that extend from the center into the galactic disc. The spiral arms are sites of ongoing star formation and are brighter than the surrounding disc because of the young, hot OB stars that inhabit them.
How fast a car goes is known as its speed.
Speed = (distance covered) divided by (time to cover the distance)
It has nothing to do with the direction the car is going.
______________________________________
The car's velocity is its speed AND the direction it's going.
30 miles per hour . . . speed
40 miles per hour north . . . velocity
20 miles per hour south
20 miles per hour west . . . . . same speed, different velocity
-- 'Velocity' is NOT a big word that you use when you mean
'speed' but you want to sound smarter. It's a different thing.
-- If you don't know anything about the direction the car is going,
then you can't say anything about its velocity.
-- If the car is going around a curve, then its velocity is constantly
changing, even if its speed is constant.
Answer:
a) The maximum height the ball will achieve above the launch point is 0.2 m.
b) The minimum velocity with which the ball must be launched is 4.43 m/s or 0.174 in/ms.
Explanation:
a)
For the height reached, we use 3rd equation of motion:
2gh = Vf² - Vo²
Here,
Vo = 3.75 m/s
Vf = 0m/s, since ball stops at the highest point
g = -9.8 m/s² (negative sign for upward motion)
h = maximum height reached by ball
therefore, eqn becomes:
2(-9.8m/s²)(h) = (0 m/s)² - (3.75 m/s²)²
<u>h = 0.2 m</u>
b)
To find out the initial speed to reach the hoop at height of 3.5 m, we again use 3rd eqn. of motion with h= 3.5 m - 2.5m = 1 m (taking launch point as reference), and Vo as unknown:
2(-9.8m/s²)(1 m) = (0 m/s)² - (Vo)²
(Vo)² = 19.6 m²/s²
Vo = √19.6 m²/s²
<u>Vo = 4.43 m/s</u>
Vo = (4.43 m/s)(1 s/1000 ms)(39.37 in/1 m)
<u>Vo = 0.174 in/ms</u>
<u />
