Explanation:
Given data
velocity v= 25m/s
The time it takes to put on brake t= 0.3s
the distance covered when the brake was put on is
v=s/t
s= v*t
s= 25*0.3s
s= 7.5m
hence the distance covered is 7.5m
Also the rate of decrease in aceleration is 5m/s^2
we can also calculate the distance covered at this rate
v^2=u^2+2as
25^2= 0+2*5*s
625=10s
divide both sides by 10
s=625/10
s= 62.5m
The total distance covered between putting on the brakes and decelareation is 7.5+62.5= 70m
Given that the tree is 75m ahead, the car would not hit the tree
The speed of light is......299 792 458 m / s
<span>They are used to measure and map effluent and pollution discharges from factories and sewerage plants, and the movement of sand around harbours, rivers and bays. Radioactive materials used for such purposes have short half-lives and decay to background levels within days.</span>
Answer:
E = 31.329 N/C.
Explanation:
The differential electric field 
at the center of curvature of the arc is
<em>(we have a cosine because vertical components cancel, leaving only horizontal cosine components of E. )</em>
where 
is the radius of curvature.
Now
,
where 
is the charge per unit length, and it has the value
Thus, the electric field at the center of the curvature of the arc is:
Now, we find 
and 
. To do this we ask ourselves what fraction is the arc length 3.0 of the circumference of the circle:
and this is
radians.
Therefore,
evaluating the integral, and putting in the numerical values we get:
