-- Accelerating at the rate of 8 m/s², Andy's speed
after 30 seconds is
(8 m/s²) x (30.0 s) = 240 m/s .
-- His average speed during that time is
(1/2) (0 + 240 m/s) = 120 m/s .
-- In 30 sec at an average speed of 120 m/s,
Andy will travel a distance of
(120 m/s) x (30 sec) = 3,600 m
= 3.6 km .
"But how ? ! ?", you ask.
How in the world can Andy leave a stop light and then
cover 3.6 km = 2.24 miles in the next 30 seconds ?
The answer is: His acceleration of 8 m/s², or about 0.82 G
is what does it for him.
At that rate of acceleration ...
-- Andy achieves "Zero to 60 mph" in 3.35 seconds,
and then he keeps accelerating.
-- He hits 100 mph in 5.59 seconds after jumping the light ...
and then he keeps accelerating.
-- He hits 200 mph in 11.2 seconds after jumping the light ...
and then he keeps accelerating.
-- After accelerating at 8 m/s² for 30 seconds, Andy and his
car are moving at 537 miles per hour !
We really don't know whether he keeps accelerating,
but we kind of doubt it.
A couple of observations in conclusion:
-- We can't actually calculate his displacement with the information given.
Displacement is the distance and direction between the starting- and
ending-points, and we're not told whether Andy maintains a straight line
during this tense period, or is all over the road, adding great distance
but not a lot of displacement.
-- It's also likely that sometime during this performance, he is pulled
over to the side by an alert cop in a traffic-control helicopter, and
never actually succeeds in accomplishing the given description.
Answer:100joules
Explanation:
Mass(m)=5kg
Acceleration due to gravity(g)=10m/s^2
Height(h)=2 meters
Work=m x g x h
Work=5 x 10 x 2
Work=100joules
Answer and Explanation:
curents i = 2.9 A
i ' = 4.4 A
the magnitude (in T.m) of the path integral of B.dl around the window frame = μo * current enclosed
= μo* ( i '- i )
Since from Ampere's law
where μ o = permeability of free space = 4π * 10 ^-7 H / m
plug the values we get the magnitude (in T.m) of the path integral of B.dl = ( 4π*10^-7 ) (2.9+4.4)
= 1.884 * 10^-6 Tm
Answer:
A. 243 N
Explanation:
Friction is the force that opposes the relative motion between systems that are in contact.
This friction force that opposes the motion of the oak chest across the oak surface will be equal and opposite to that exerted by the woman.
First find the normal force which is the force that would point directly upwards to support weight of the block.
Normal force, N= mg where m is the mass of the chest and g is the acceleration due to gravity.
Given m=40 kg and g=9.80 m/s²
N force=40×9.80 =392N
Then find the force of friction which is given by the formula;
<em>F=μN where μ is friction coefficient for the oak chest and N is the normal force on the chest</em>
Given <em>μ</em>=0.620 and N force = 392 N then it will be;
F=0.620× 392 =243.04 N
Answer : 243 N