Doing a force balance on the car:
ma = Fr
ma = μmg
a = μg
a = 0.3(9.81)
a = 29.43 m/s2
Using the formula:
2ax = v2
2(29.43)(34) = v2
v = 44.74 m/s = 161.05 km/h
The car was going 44.74 m/s or 161.05 kph when the brakes were applied.
Positively charged protons in the nucleus of the gold atom .... rutherford scattering ???
Given that,
The acceleration of gravity is -9.8 m/s²
Initial velocity, u = 39.2 m/s
Time, t = 2 s
To find,
The final velocity of the shot.
Solution,
Let v is the final velocity of sling shot. Using first equation of motion to find it.
v = u +at
Here, a = -g
v = u-gt
v = (39.2)-(9.8)(2)
v = 19.6 m/s
So, its velocity after 2 seconds is 19.6 m/s.
Answer:
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