Answer:
What is the difference between the electrons in the nucleus and the ones in orbit around the nucleus? ... The Strong Nuclear Force is an attractive force between protons and neutrons that keep the nucleus together and the Weak Nuclear Force is responsible for the radioactive decay of certain nuclei
Explanation:
Use the exact values you enter to make later calculations. Jack and Jill are on two different floors of their high rise office building and looking out of their respective windows. Jack sees a flower pot go past his window ledge and Jill sees the same pot go past her window ledge a little while later. The time between the two observed events was 4.0 s. Assume air resistance is negligible. (a) If the speed of the pot as it passes Jill's window is 60.0 m/s, what was its speed when Jack saw it go by
The part B of the question is missing and it is;
b) What is the height between the two window ledges?
Answer:
A) 20.76 m/s
B) 161.52 m
Explanation:
A) To calculate the initial speed we use the formula from Newton's first law of motion:
v = u + at
Making u the subject gives;
u = v - at
Where;
v is the final velocity which is the speed when Jill sees the pot = 60 m/s
u is the initial velocity which is the speed when Jack sees the pot go by
t is the time between the two observed events = 4 s
a in this question is acceleration due to gravity = 9.81 m/s².
Plugging in the relevant values into the initial velocity equation gives;
u = 60 - (9.81 × 4)
u = 20.76 m/s
B) To get the height difference, we will use the formula;
(y1 - y0) = ut + ½at²
Thus, plugging in the relevant values, we have;
y1 - y0 = (20.76 × 4) + (½ × 9.81 × 4²)
(y1 - y0) = 161.52 m
Describing the Motion of an Object
SARE
Use the information from the graph to answer the
question.
What is the acceleration of the object?
m/s2
Velocity vs. Time
Velocity (m/s)
45
40
35
30
25
20
10
5
0
0
2
10
12
4. 6 8
Time (s)
Answer:
i actually dont know im so sorry but i will keep trying to find the answerr
Suppose a fast-pitch softball player does a windmill pitch, moving her hand through a circular arc with her arm straight. She releases the ball at a speed of 21.3 m/s21.3 m/s (about 47.6 mph47.6 mph ). Just before the ball leaves her hand, the ball's radial acceleration is 871 m/s2871 m/s2 . What is the length of her arm from the pivot point at her shoulder?
Answer:
R = 52.08 cm
Explanation:
given data
tangential velocity v = 21.3 m/s
radial acceleration aR = 871 m/s²
solution
we will get here length of her arm from the pivot point at her shoulder so
here we know aR = [tex]\frac{v^2}{R}[/tex] ........................1
so put here value and we get
871 = [tex]\frac{21.3^2}{R}[/tex]
R = 0.5208840 m
R = 52.08 cm
Answer:
Explanation:
linear speed of ball v = 21.3 m /s
radial acceleration = 871 m /s²
radial acceleration = v² / R
where R is length of arm which acts as radius of circular path of the ball .
Putting the values
21.3² / R = 871
R = .52 m.