Answer:
17.6 N
Explanation:
The force exerted by the punter on the football is equal to the rate of change of momentum of the football:
where
is the change in momentum of the football
is the time elapsed
The change in momentum can be written as
where
m = 0.55 kg is the mass of the football
u = 0 is the initial velocity (the ball starts from rest)
v = 8.0 m/s is the final velocity
Combining the two equations and substituting the values, we find the force exerted on the ball:
A) 500 calories
B)5000 joules
C) 5 kilocalories
Answer is C
The have the knowledge of a monkey, they survive like monkey, they become one with monkey okay seriously tho they probably have survival knowledge
Answer:
"Crust" refers to a
terrestrial planet's outermost surface.
In general, the Earth's crust is divided into
older, thicker continental crust and younger, denser oceanic crust.
...
The thin, 40-kilometer (25-mile) deep crust of our planet — just
1 per cent of Earth's mass — contains all known universe existence.
Explanation:
Continental crust is thicker, 22 miles (35 km) on average and less dense than oceanic crust, which accounts for its mean surface elevation of about 3 miles (4.8 km) above that of the ocean floor (Archimedes’ principle). Continental crust is more complex than oceanic crust…
When the object is at rest, there is a zero net force due the cancellation of the object's weight <em>w</em> with the normal force <em>n</em> of the table pushing up on the object, so that by Newton's second law,
∑ <em>F</em> = <em>n</em> - <em>w</em> = 0 → <em>n</em> = <em>w</em> = <em>mg</em> = 112.5 N ≈ 113 N
where <em>m</em> = 12.5 kg and <em>g</em> = 9.80 m/s².
The minimum force <em>F</em> needed to overcome <u>maximum</u> static friction <em>f</em> and get the object moving is
<em>F</em> > <em>f</em> = 0.50 <em>n</em> = 61.25 N ≈ 61.3 N
which means a push of <em>F</em> = 15 N is not enough the get object moving and so it stays at rest in equilibrium. While the push is being done, the net force on the object is still zero, but now the horizontal push and static friction cancel each other.
So:
(a) Your free body diagram should show the object with 4 forces acting on it as described above. You have to draw it to scale, so whatever length you use for the normal force and weight vectors, the length of the push and static friction vectors should be about 61.3/112.5 ≈ 0.545 ≈ 54.5% as long.
(b) Friction has a magnitude of 15 N because it balances the pushing force.
(c) The object is in equilibrium and not moving, so the acceleration is zero.
