Answer:
<h2> 27m/s</h2>
Explanation:
Given data
initital velocity u=15m/s
deceleration a=3m/s^2
time t= 4 seconds
final velocity v= ?
Applying the expression
v=u+at------1
substituting our data into the expression we have
v=15+3*4
v=15+12
v=27m/s
The velocity after 4 seconds is 27m/s
Dependent Variable=Responding Variable.
Answer:
98.33 %
Explanation:
On an elliptical orbit, angular momentum will be conserved .
Angular momentum = I ω = mvR
So mv₁R₁ = mv₂R₂
= v₁R₁ = v₂R₂
where v₁ is velocity and R₁ radius in low orbit (perigee)and v₂ and R₂ is velocity and radius in high orbit ( apogee ).
Here R₁ = Radius of the earth , R₂ is distance between moon and earth.
R₁ / R₂ = 1/60
v₁ /v₂ = R₂ / R₁ = 60
v₂ / v₁ = 1 / 60
1 - (v₂ / v₁ ) = 1 -( 1 / 60)
(v₁ -v₂)/v₁ = ( 60-1 )/60
(v₁ -v₂)/v₁ x 100 = 5900/60 = 98.33 %
The net force performs a total amount of work equal to
(45 N) (0.80 m) = 36 J
on the bullet, and this is in turn is equal to the change in the bullet's kinetic energy by the work-energy theorem. So we have
W = ∆K = 1/2 mv²
since the bullet starts at rest, where m = its mass and v = its final velocity.
Solve for v :
36 J = 1/2 (0.0050 kg) v² ⇒ v = 120 m/s