The answer is 4.7 m for the spàthan. this is a very question! I hope this helps my friend! :-)
Answer:
the net work is 0
Explanation:
in an isolated system, there is no energy entering or exiting the system, and since work is dependent on the change in energy (W = ΔE) , with no change in energy, there is no work done. the net work done on or by the system must be 0
Answer:
Option (3)
Explanation:
Formula used to calculate acceleration is,
F = ma
Where F = force exerted on a mass
m = mass
a = acceleration due to force exerted on the mass
Option (1),
When F = 100 N and m = 100 kg
100 = 100a
a = 1 m per sec²
Option (2)
For F = 1 N and m = 100 kg
1 = 100a
a = 
a = 0.01 m per sec²
Option (3)
For F = 100 N and m = 1 kg
100 = 1(a)
a = 100 m per sec²
Option (4)
For F = 1 N and m = 1 kg
1 = 1(a)
a = 1 m per sec²
Therefore. acceleration in Option (3) is the maximum.
Answer:
u = 3.35 m/s
Explanation:
given,
mass , m = 0.455 kg
R = 0.675 m
Height of Loop = 1.021 m
the speed required at the top of loop be v
equating the force vertically


v² = 6.622
v = 2.57 m/s
Let the initial speed of ball be u
using conservation of energy

where, 



0.7 u² = 7.85092
u² = 11.2156
u = 3.35 m/s
the initial speed is 3.35 m/s
In component form, the displacement vectors become
• 350 m [S] ==> (0, -350) m
• 400 m [E 20° N] ==> (400 cos(20°), 400 sin(20°)) m
(which I interpret to mean 20° north of east]
• 550 m [N 10° W] ==> (550 cos(100°), 550 sin(100°)) m
Then the student's total displacement is the sum of these:
(0 + 400 cos(20°) + 550 cos(100°), -350 + 400 sin(20°) + 550 sin(100°)) m
≈ (280.371, 328.452) m
which leaves the student a distance of about 431.8 m from their starting point in a direction of around arctan(328.452/280.371) ≈ 50° from the horizontal, i.e. approximately 431.8 m [E 50° N].