22. a - (vf^2 - vi^2)/(2d)
a = (0 - 23^2)/(170)
a = -3.1 m/s^2
23. Find the time (t) to reach 33 m/s at 3 m/s^2
33-0/t = 3
33 = 3t
t = 11 sec to reach 33 m/s^2
Find the av velocuty: 33+0/2 = 16.5 m/s
Dist = 16.5 * 11 = 181.5 meters to each 33m/s speed. Runway has to be at least this long.
24. The sprinter starts from rest. The average acceleration is found from:
(Vf)^2 = (Vi)^2 = 2as ---> a = (Vf)^2 - (Vi)^2/2s = (11.5m/s)^2-0/2(15.0m) = 4.408m/s^2 estimated: 4.41m/s^2
The elapsed time is found by solving
Vf = Vi + at ----> t = vf-vi/a = 11.5m/s-0/4.408m/s^2 = 2.61s
25. Acceleration of car = v-u/t = 0ms^-1-21.0ms^-1/6.00s = -3.50ms^-2
S = v^2 - u^2/2a = (0ms^-1)^2-(21.0ms^-1)^2/2*-3.50ms^-2 = 63.0m
26. Assuming a constant deceleration of 7.00 m/s^2
final velocity, v = 0m/s
acceleration, a = -7.00m/s^2
displacement, s - 92m
Using v^2 = u^2 - 2as
0^2 - u^2 + 2 (-7.00) (92)
initial velocity, u = sqrt (1288) = 35.9 m/s
This is the speed pf the car just bore braking.
I hope this helps!!
Answer:
The frictional force acting on the block is 14.8 N.
Explanation:
Given that,
Weight of block = 37 N
Coefficients of static = 0.8
Kinetic friction = 0.4
Tension = 24 N
We need to calculate the maximum friction force
Using formula of friction force
Put the value into the formula
So, the tension must exceeds 29.6 N for the block to move
We need to calculate the frictional force acting on the block
Using formula of frictional force
Put the value in to the formula
Hence, The frictional force acting on the block is 14.8 N.
If 1 ken is 1.97 meter, then 1 square ken is 3.8809 square meters, and one cubic ken is 7.645373. As for the cylindrical tank, the volume of it would be 10.835 times the radius of the cylinder time 1.97^2 times pi. As you didn't specify the radius, I can't give the exact answer but that would be how to get it.
Answer:
b) Betelgeuse would be 
times brighter than Sirius
c) Since Betelgeuse brightness from Earth compared to the Sun is 
the statement saying that it would be like a second Sun is incorrect
Explanation:
The start brightness is related to it luminosity thought the following equation:
(1)
where 
is the brightness, 
is the star luminosity and 
, the distance from the star to the point where the brightness is calculated (measured). Thus:
b) 
and 
where 
is the Sun luminosity (
) but we don't need to know this value for solving the problem. 
is light years.
Finding the ratio between the two brightness we get:
c) we can do the same as in b) but we need to know the distance from the Sun to the Earth, which is 
. Then
Notice that since the star luminosities are given with respect to the Sun luminosity we don't need to use any value a simple states the Sun luminosity as the unit, i.e 1. From this result, it is clear that when Betelgeuse explodes it won't be like having a second Sun, it brightness will be 5 orders of magnitude smaller that our Sun brightness.
Answer:
false
Explanation:
discovered colours of the rainbow
