Answer:
F = 4.47 10⁻⁶ N
Explanation:
The expression they give for the strength of the tide is
F = 2 G m M a / r³
Where G has a value of 6.67 10⁻¹¹ N m² / kg² and M which is the mass of the Earth is worth 5.98 10²⁴ kg
They ask us to perform the calculation
F = 2 6.67 10⁻¹¹ 135 5.98 10²⁴ 13 / (6.79 10⁶)³
F = 4.47 10⁻⁶ N
This force is directed in the single line at the astronaut's mass centers and the space station
Answer:
184 feets
Explanation:
Given the data:
time (sec) __ velocity (ft/sec)
0 __________30
1 __________ 54
2 __________56
3 __________34
4 __________ 8
5 __________ 2
6 __________22
Using left end approximation:
(0,1) ___ f(0) = 30
(1,2) ___ f(1) = 54
(2,3) ___f(2) = 56
(3,4) ___f(3) = 34
(4,5) ___f(4) = 8
(5,6) __ f(5) = 2
Hence, the Total distance traveled during the 6 second interval is:
Change ; dT = 1
1 * (30 + 54 + 56 + 34 + 8 + 2) = 184
Answer:
(a) The spring constant is 59.23 N/m
(b) The total energy involved in the motion is 0.06 J
Explanation:
Given;
mass, m = 240 g = 0.24 kg
frequency, f = 2.5 Hz
amplitude of the oscillation, A = 4.5 cm = 0.045 m
The angular speed is calculated as;
ω = 2πf
ω = 2 x π x 2.5
ω = 15.71 rad/s
(a) The spring constant is calculated as;
(b) The total energy involved in the motion;
E = ¹/₂kA²
E = (0.5) x (59.23) x (0.045)²
E = 0.06 J
B, air blowing from across the field is as a bullet fired from a rifle
