Answer:

Explanation:
Acceleration is given by

where
u is the initial velocity
v is the final velocity
t is the time interval
In this problem:
is the initial velocity
is the final velocity
t = 2 s is the time
Substituting, we find the acceleration:

Answer:
3.185×10^-29 kgm/s
Explanation:
Momentum(p)=mass×velocity
=9.1×10^-31×3.5×10
=3.185×10^-29 kgm/s
Answer:
a) The velocity is 2.94m/s
b) 0.441
Explanation:
a) Assume gravity is 9.8m/s^2
Use the equation below to solve for the velocity at 0.30 seconds
,
vf =unknown velocity vi= initial velocity vi=0m/s a= 9.8m/s^2 t=0.30seconds
Step 1: Substitute the variables with the knowns

Step 2: Solve

b)
Use the equation below to solve for the displacement at 0.30 seconds

Step 1: Substitute the same variables with the knowns

Note that vi*t=0 as vi=0m/s
Step 2: Solve
x=0.441m
Answer:
Data:-vi=om/s (b/c as in question penny is dropped from building means before coming to ground its initial state or velocity was considered as zero ) now distance or height h=380m and now we have to find the final velocity vf=? and the time t=?
Explanation:
So applying second eq of motion s=vit+1/2×gt² (here we have taken a gravity b/c when ever body is in vertical position then acceleration due to gravity is applied ) s=0×t+1/2×gt² , s=0+1/2×9.8×t² ,380=4.9t² we have to find t so 4.9t²=380 , t²=380÷4.9 , t²=77.55 now sq root on b/s

so t=8.806s and now apply 1st eq o²f motion to find out vf so vf=vi+gt , vf=0+9.8×8.806 ,vf=86.298 and if you want to verify that either this is answer is correct or not so put the value of t in second eq of motion and if you got distance same as give in the question so your value of t is considered as correct likewise s=vit+1/2gt² , s=0+1/2×9.8(8.806)²,s=4.9×77.55 ,s=380m (proved) I hope it would be helpfull
Answer:
The magnitude is:
The direction of E is in the negative x-direction.
Explanation:
The electric field equation is:

Where:
- Q is the charge (we can choose the electron or the proton)
- r is the distance (in our case is at the midpoint 973/2 nm)
- k is the Coulomb constant (
)
Using the electron charge (
)
The magnitude is:
The direction of E is in the negative x-direction.
I hope it helps you!