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.
Answer:
d. Two soccer balls that are touching each other
Explanation:
Let
be the mass of a tennis ball,
is the mass of a soccer ball.
As the mass of a soccer ball is more than the mass of a tennis ball, so

Let
be the distance between the centers of both the balls near each other and
be the distance between the centers of both the balls touching each other.
So, 
The gravitational force, F, between the two objects having masses M and m and separated by distance d is

Where G is the universal gravitational constant.
As, the gravitational force is directly proportional to the product of both the masses and inversely proportional to the square of the distance between them, so selecting the larger mass (
, soccer ball) separated by a lesser distance (
, touching) to get more gravitational force.
Therefore, there will be a larger gravitational force between them when two soccer balls touching each other.
Hence, option (d) is correct.
I honestly think it is B, if not I'm sorry for the incorrect answer, but B seems to be the only one to make sense
Answer:
0.3625
Explanation:
From the given information:
Consider the equilibrium conditions;
On the ladder, net torque= 0
Thus,
; and

However, by rearrangement;

![\mu= \dfrac{ (323 \ N) ( 10.8 \ m) \ cos 56^0 + (734 \ N) (7.46 \ m) \ cos \ 66.46^0}{\Big [(323 \ N)+(734 \ N) \Big] (10.8 \ m)}](https://tex.z-dn.net/?f=%5Cmu%3D%20%5Cdfrac%7B%20%28323%20%20%5C%20N%29%20%28%2010.8%20%5C%20m%29%20%5C%20cos%2056%5E0%20%2B%20%28734%20%5C%20N%29%20%287.46%20%5C%20m%29%20%5C%20cos%20%20%5C%20%2066.46%5E0%7D%7B%5CBig%20%5B%28323%20%5C%20N%29%2B%28734%20%5C%20N%29%20%5CBig%5D%20%2810.8%20%5C%20m%29%7D)

Transition
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