Answer:
When a right-handed golfer’s tee shot curves to far to the left, the shot is described as a “hook.” This shot typically occurs when the club head moves across the ball from left to right, imparting side-spin on the ball
Answer:
assume nitrogen is an ideal gas with cv=5R/2
assume argon is an ideal gas with cv=3R/2
n1=4moles
n2=2.5 moles
t1=75°C <em>in kelvin</em> t1=75+273
t1=348K
T2=130°C <em>in kelvin</em> t2=130+273
t2=403K
u=пCVΔT
U(N₂)+U(Argon)=0
<em>putting values:</em>
=>4x(5R/2)x(Tfinal-348)=2.5x(3R/2)x(T final-403)
<em>by simplifying:</em>
Tfinal=363K
Answer:
226.6 N
Explanation:
The work done by the push is given by:
where
p is the magnitude of the push
d is the displacement of the crate
is the angle between the direction of the push and the displacement of the crate
On the contrary, the work done by the frictional force is:
where
is the frictional force, with m=1.00x102 kg being the mass of the crate ang g=9.81 m/s^2 the gravitational acceleration
d is the displacement of the crate
the negative sign is due to the fact that the frictional force acts in the opposite direction to the displacement of the crate.
The net work must be zero, so:
Substituting the previous equations into this one, we find:
which we can solve to find p: