Answer:
s = 1.7 m
Explanation:
from the question we are given the following:
Mass of package (m) = 5 kg
mass of the asteriod (M) = 7.6 x 10^{20} kg
radius = 8 x 10^5 m
velocity of package (v) = 170 m/s
spring constant (k) = 2.8 N/m
compression (s) = ?
Assuming that no non conservative force is acting on the system here, the initial and final energies of the system will be the same. Therefore
• Ei = Ef
• Ei = energy in the spring + gravitational potential energy of the system
• Ei = \frac{1}{2}ks^{2} + \frac{GMm}{r}
• Ef = kinetic energy of the object
• Ef = \frac{1}{2}mv^{2}
• \frac{1}{2}ks^{2} + (-\frac{GMm}{r}) = \frac{1}{2}mv^{2}
• s = 
s = 
s = 1.7 m
Answer:
44.6 N
Explanation:
Draw a free body diagram of the block. There are four forces on the block:
Weight force mg pulling down,
Normal force N pushing up,
Friction force Nμ pushing left,
and applied force F pulling right 30° above horizontal.
Sum of forces in the y direction:
∑F = ma
N + F sin 30° − mg = 0
N = mg − F sin 30°
Sum of forces in the x direction:
∑F = ma
F cos 30° − Nμ = 0
F cos 30° = Nμ
N = F cos 30° / μ
Substitute:
mg − F sin 30° = F cos 30° / μ
mg = F sin 30° + (F cos 30° / μ)
Plug in values:
mg = 20 N sin 30° + (20 N cos 30° / 0.5)
mg = 44.6 N
Answer:
The acceleration of the both masses is 0.0244 m/s².
Explanation:
Given that,
Mass of one block = 602.0 g
Mass of other block = 717.0 g
Radius = 1.70 cm
Height = 60.6 cm
Time = 7.00 s
Suppose we find the magnitude of the acceleration of the 602.0-g block
We need to calculate the acceleration
Using equation of motion
Where, s = distance
t = time
a = acceleration
Put the value into the formula
Hence, The acceleration of the both masses is 0.0244 m/s².
