Answer: T= 715 N
Explanation:
The only external force (neglecting gravity) acting on the swinging mass, is the centripetal force, which. in this case, is represented by the tension in the string, so we can say:
T = mv² / r
At the moment that the mass be released, it wil continue moving in a straight line at the same tangential speed that it had just an instant before, which is the same speed included in the centripetal force expression.
So the kinetic energy will be the following:
K = 1/2 m v² = 15. 0 J
Solving for v², and replacing in the expression for T:
T = 1.9 Kg (3.97)² m²/s² / 0.042 m = 715 N
Answer:
Earth's mass is 5.98 x 1024 kg and has a radius of 6.37 x 106 m. 8.06 N/kg ... what would be the weight of a 70.0-kg student on the surface of Jupiter?Explanation:
Explanation:
Given
Velocity v = 23.0m/s
Distance S = 3.45m
Required
Time it will take the skier to reach the ground;
Using the equation of motion;
S = ut + 1/2gt²
3.45 = 23t + 1/2(9.8)t²
3.45 = 23t + 4.9t²
4.9t²+23t-3.45 = 0
Factorize;
t = -23 ±√23²-4(4.9)(-3.45)/2(4.9)
t = -23 ±√529+67.62/9.8
t = -23±√596.62/9.8
t = -23±24.43/9.8
t = 1.43/9.8
t = 0.146 secs
Hence take the skier 0.146 secs to reach the ground.
b) Horizontal distance covered is the range;
Range = U√2H/g
Range = 23√2(3.45)/9.8
Range = 23√6.9/9.8
Range = 23√0.7041
Range = 23(0.8391)
Range = 19.29m
Hence the horizontal distance travelled in air is 19.29m
Answer:
m v^2 / R = G M m / R^2 gravitational attraction = centripetal force
M = v^2 R / G solving for M
period = 6 h 25 min = (6 * 3600 + 25 * 60) sec = 23,100 sec = T
v = 2 pi R / T
M = 4 pi^2 R^3 / (G T^2)
M = 39.5 * (8.6E7)^3 / (6.67E-11 * 2.31E4^2)
M = 39.5 * 636 / (6.67 * 5.34) * 10^24
M = 7.05 * 10^26 kg