KE = 1/2 x 80 x 60^2
KE = 144000
Answer:
If the temperature of the air in the balloon is less than the temperature of the air surrounding the balloon then the balloon will appear slightly deflated because of the difference in temperature.
As the temperature of the air in the balloon reaches the surrounding air temperature, then the balloon will appear to be fully inflated because the temperature of the air in the balloon is the same as the surrounding air temperature.
Answer: a) 7.1 * 10^3 N; b) -880 N directed out of the curve.
Explanation: In order to solve this problem we have to use the Newton laws, then we have the following:
Pcos 15°-N=0
Psin15°-f= m*ac
from the first we obtain N, the normal force
N=750Kg*9.8* cos (15°)= 7.1 *10^3 N
Then to calculate the frictional force (f) we can use the second equation
f=P sin (15°) -m*ac where ac is the centripetal acceletarion which is equal to v^2/r
f= 750 *9.8 sin(15°)-750*(85*1000/3600)^2/150= -880 N
Answer:
4.91 x 10⁻⁷ m
Explanation:
the applicable formula is
v = fλ
where
v = velocity (i.e speed) = given as 3.0 x 10⁸ m/s
f = frequency = given asw 6.11 x 10¹⁴
λ = wavelength
if we rearrange the equation and substitute the values given above,
v = fλ
λ = v/f
= 3.0 x 10⁸ / 6.11 x 10¹⁴
= 4.91 x 10⁻⁷ m