Answer:
b) -10 m/s
Explanation:
In perfectly elastic head on collisions of identical masses, the velocities are exchanged with one another.
Answer:
Explanation:
This is a circular motion questions
Where the oscillation is 27.3days
Given radius (r)=3.84×10^8m
Circular motion formulas
V=wr
a=v^2/r
w=θ/t
Now, the moon makes one complete oscillation for 27.3days
Then, one complete oscillation is 2πrad
Therefore, θ=2πrad
Then 27.3 days to secs
1day=24hrs
1hrs=3600sec
Therefore, 1day=24×3600secs
Now, 27.3days= 27.3×24×3600=2358720secs
t=2358720secs
Now,
w=θ/t
w=2π/2358720 rad/secs
Now,
V=wr
V=2π/2358720 ×3.84×10^8
V=1022.9m/s
Then,
a=v^2/r
a=1022.9^2/×3.84×10^8
a=0.0027m/s^2
Answer:
a = 2 [m/s²]
Explanation:
To be able to solve this problem we must make it clear that the starting point when the time is equal to zero, the velocity is 5 [m/s] and when three seconds have passed the velocity is 11 [m/s], this point is the final point or the final velocity.
We can use the following equation.
where:
Vf = final velocity = 11 [m/s]
Vo = initial velocity = 5 [m/s]
a = acceleration [m/s²]
t = time = 3 [s]
![11 = 5 + a*3\\6=3*a\\a= 2[m/s^{2} ]](https://tex.z-dn.net/?f=11%20%3D%205%20%2B%20a%2A3%5C%5C6%3D3%2Aa%5C%5Ca%3D%202%5Bm%2Fs%5E%7B2%7D%20%5D)
Answer:
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-- The area under a velocity/time graph, between two points in time, is the difference in displacement during that period of time.
-- The area under a speed/time graph, between two points in time, is the distance covered during that period of time.
