Given:
u = 10⁵ m/s, the entrance velocity
v = 2.5 x 10⁶ m/s, the exit velocity
s = 1.6 cm = 0.016 m, distance traveled
Let a = the acceleration.
Then
u² + 2as = v²
(10⁵ m/s)² + 2*(a m/s²)*(0.016 m) = (2.5 x 10⁶ m/s)²
0.032a = 6.25 x 10¹² - 10¹⁰ = 6.24 x 10¹²
a = 1.95 x 10¹⁴ m/s²
Answer: 1.95 x 10¹⁴ m/s²
F=ma, so a=F/m. Thus, the more massive object accelerates more slowly, that is, the 21 kg object is three times slower than the 7 kg one (2nd variant, 7kg is 3 times faster than 21kg)
Answer:
306J
Explanation:
The formula for kinetic energy is given as ½mv².
Using this, we get our answer by substituting the values into the formula.
½ × 0.145kg × 65m/s
= 306.31J
can be rounded off to 306J
Answer:
Explanation:
I assume you are asking for the acceleration of the ball, not the velocity (in fact, the velocity is already given).
The acceleration is:
where in this case we have:
v = 24 m/s is the final velocity
u = 0 is the initial velocity
t = 8 s is the time interval for the velocity to change
Solving the equation,
Answer: 1,224.6km
S=r*theta
r=d/2=1.30m/2=.650m
Theta=(2pi/rev)rev
Theta=300,000rev(2pi/rev)=1,884,000rads
S=0.650m*1,844,000rad=1,224,600m
S=1,224,600m(1km/1000m)=1,224.6km
