Answer:
the ball travelled approximately 60 m towards north before stopping
Explanation:
Given the data in the question;
First course : 
= 0.75 m/s², 
= 20 m, 
= 10 m/s
now, form the third equation of motion;
v² = u² + 2as
we substitute
² = (10)² + (2 × 0.75 × 20)
² = 100 + 30
² = 130
= √130
= 11.4 m/s
for the Second Course:
= 11.4 m/s, 
= -1.15 m/s², 
= 0
Also, form the third equation of motion;
v² = u² + 2as
we substitute
0² = (11.4)² + (2 × (-1.15) × 
)
0 = 129.96 - 2.3
2.3 = 129.96
= 129.96 / 2.3
= 56.5 m
so;
|d| = √( ² + ² )
we substitute
|d| = √( (20)² + (56.5)² )
|d| = √( 400 + 3192.25 )
|d| = √( 3592.25 )
|d| = 59.9 m ≈ 60 m
Therefore, the ball travelled approximately 60 m towards north before stopping
Answer:
18m/s^2
Explanation:
Vf = Vi + at
t = distance/ average velocity
(120 + 0)/2 = 60 (average velocity)
400m/60m/s = 20/3 s
insert into first equation:
120 = 0 + a(20/3)
360 = 20a
18 = a
HOPE THIS HELPS!!!
Answer:
a)
b)
Explanation:
a) Let's use the constant velocity equation:
- v is the speed of the muon. 0.9*c
- c is the speed of light 3*10⁸ m/s
b) Here we need to use Lorentz factor because the speed of the muon is relativistic. Hence the time in the rest frame is the product of the Lorentz factor times the time in the inertial frame.
v is the speed of muon (0.9c)
Therefore the time in the rest frame will be:
No we use the value of Δt calculated in a)
I hope it helps you!
Answer:
k = 1,250 N/m
Explanation:
Use the formula F=kx, with F=5N and x=0.04m
Then the spring constant (k) is 5/0.04
Possibilities . . .
-- nuclear reaction
-- nuclear fission
-- nuclear fusion
-- radioactive decay.
Any of these makes it a true statement.
