Answer:
Explanation:
Given that
Mass of bowling ball M1=7.2kg
The radius of bowling ball r1=0.11m
Mass of billiard ball M2=0.38kg
The radius of the Billiard ball r2=0.028m
Gravitational constant
G=6.67×10^-11Nm²/kg²
The magnitude of their distance apart is given as
r=r1+r2
r=0.028+0.11
r=0.138m
Then, gravitational force is given as
F=GM1M2/r²
F=6.67×10^-11×7.2×0.38/0.138²
F=9.58×10^-9N
The force of attraction between the two balls is
F=9.58×10^-9N
This is the same question that I just answered.
Have present the definition of acceleration:
a = Δv / Δt, this is change in velocity per unit of time.
a and v are in bold to mean that they are vectors.
1) a body traveling in a straight line and increasing in speed: CORRECT:
Acceleration is the change in velocity, either magnitude or direction or both. So, a body increasing in speed is accelerated.
2) a body traveling in a straight line and decreasing in speed: CORRECT
A decrease in speed is a change in velocity, so it means acceleration.
3) a body traveling in a straight line at constant speed: FALSE.
That body is not changing either direction or speed so its motion is not accelerated but uniform.
4) a body standing still : FALSE.
That body is not changind either direction or speed.
5) a body traveling at a constant speed and changing direction: CORRECT.
The change in direction means that the body is accelerated. The acceleration due to change in direction is named centripetal acceleration.
Answer:
X=92.49 m
Explanation:
Given that
u= 21 m/s
h= 97 m
Time taken to cover vertical distance h
h= 1/2 g t²
By putting the values
97 = 1/2 x 10 x t² ( g = 10 m/s²)
t= 4.4 s
The horizontal distance
X= u .t
X= 21 x 4.4
X=92.49 m
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>
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Answer:
you can measure by scale beacause we dont no sorry i cant help u but u can ask me some other Q
