The magnitude of the resultant force on the balloon is 374.13 N.
The given forces from the image;
- <em>Upward force = 514 N</em>
- <em>Downward force = 267 N</em>
- <em>Eastward force = 678 N</em>
- <em>Westward force = 397 N</em>
The net vertical force on the balloon is calculated as follows;
The net horizontal force on the balloon is calculated as follows;
The magnitude of the resultant force on the balloon is calculated as follows;
Thus, the magnitude of the resultant force on the balloon is 374.13 N.
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Answer:
13.7m
Explanation:
Since there's no external force acting on the astronaut or the satellite, the momentum must be conserved before and after the push. Since both are at rest before, momentum is 0.
After the push
Where 
is the mass of the astronaut, 
is the mass of the satellite, 
is the speed of the satellite. We can calculate the speed 
of the astronaut:
So the astronaut has a opposite direction with the satellite motion, which is further away from the shuttle. Since it takes 7.5 s for the astronaut to make contact with the shuttle, the distance would be
d = vt = 1.83 * 7.5 = 13.7 m
Newtons First Law of Motion:
An object at rest stays at rest and an object in motion<span> stays in </span>motion <span>with the same speed and in the same direction unless acted upon by an unbalanced force.</span>
Therefore, the relationship between force and motion is that it takes force to change the speed or direction of any object in motion.
Answer:-6800J
Explanation: 8.0m x 850N = 6800
Rewritten as a negative when brake/stop -6800
Q: The small piston of a hydraulic lift has a cross-sectional of 3.00 cm2 and its large piston has a cross-sectional area of 200 cm2. What downward force of magnitude must be applied to the small piston for the lift to raise a load whose weight is Fg = 15.0 kN?
Answer:
225 N
Explanation:
From Pascal's principle,
F/A = f/a ...................... Equation 1
Where F = Force exerted on the larger piston, f = force applied to the smaller piston, A = cross sectional area of the larger piston, a = cross sectional area of the smaller piston.
Making f the subject of the equation,
f = F(a)/A ..................... Equation 2
Given: F = 15.0 kN = 15000 N, A = 200 cm², a = 3.00 cm².
Substituting into equation 2
f = 15000(3/200)
f = 225 N.
Hence the downward force that must be applied to small piston = 225 N
