Answer:
The answer to your question is 5.4 cm
Explanation:
This problem refers to calculate the change in length in one dimension due to a change in temperature.
Data
α = 12 x 10⁻⁶
Lo = 150 meters
ΔT = 30 °C
Formula
ΔL/Lo = αΔT
solve for ΔL
ΔL = αLoΔT
Substitution
ΔL = (12 x 10⁻⁶)(150)(30)
Simplification
ΔL = 0054 m = 5.4 cm
Answer:
Is the equation for Ec=1/2 m(Dv)^2 where Dv is the difference between the angular speed & the areolar speed?
Answer:
time rising = 34 / 9.8 = 3.47 sec
total time in air = 2 * 3.47 sec = 6.94 sec
(time rising must equal time falling)
R = 17 m/s * 6.94 s = 118 m
Can also use range formula
R = v^2 sin (2 theta) / g
tan theta = 34 / 17 = 2
theta = 63.4 deg
2 theta = 126.9 deg
sin 126.9 = .8
v^2 = 17^2 + 34^2 = 1445 m^2/s^2
R = 1445 * .8 / 9.8 = 118 m agreeing with answer found above
Answer: 2.1 × 10^7 m/s
Explanation:
Please see the attachments below
Answer:
The height of the object is 5007.4 miles.
Explanation:
Given that,
Weight of object = 200 lb
We need to calculate the value of 
Using formula of gravitational force

Put the value into the formula



We need to calculate the height of the object
Using formula of gravitational force

Put the value into the formula





Hence. The height of the object is 5007.4 miles.