Answer:
the car with the hay should slow to 16m/s if the bale of hay is dropped into it.
<span>Px = 0
Py = 2mV
second, Px = mVcosφ
Py = –mVsinφ
add the components
Rx = mVcosφ
Ry = 2mV – mVsinφ
Magnitude of R = âš(Rx² + Ry²) = âš((mVcosφ)² + (2mV – mVsinφ)²)
and speed is R/3m = (1/3m)âš((mVcosφ)² + (2mV – mVsinφ)²)
simplifying
Vf = (1/3m)âš((mVcosφ)² + (2mV – mVsinφ)²)
Vf = (1/3)âš((Vcosφ)² + (2V – Vsinφ)²)
Vf = (V/3)âš((cosφ)² + (2 – sinφ)²)
Vf = (V/3)âš((cos²φ) + (4 – 2sinφ + sin²φ))
Vf = (V/3)âš(cos²φ) + (4 – 2sinφ + sin²φ))
using the identity sin²(Ď)+cos²(Ď) = 1
Vf = (V/3)âš1 + 4 – 2sinφ)
Vf = (V/3)âš(5 – 2sinφ)</span>
The acceleration of gravity on Earth is 9.8 m/s² .
The speed of a falling object keeps increasing smoothly,
in such a way that the speed is always 9.8 m/s faster than
it was one second earlier.
If you 'drop' the penny, then it starts out with zero speed.
If you also start the clock at the same instant, then
After 1.10 sec, Speed = (1.10 x 9.8) = 10.78 meters/sec
After 1.85 sec, Speed = (1.85 x 9.8) = 18.13 meters/sec
But you want this second one given in a different unit of speed.
OK then:
= (18.13 meter/sec) x (3,600 sec/hr) x (1 mile/1609.344 meter)
= (18.13 x 3,600 / 1609.344) (mile/hr) = 40.56 mph (rounded)
We did notice that in an apparent effort to make the question
sound more erudite and sophisticated, you decided to phrase
it in terms of 'velocity'. We can answer it in those terms, if we
ASSUME that there is no wind, and the penny therefore doesn't
acquire any horizontal component of motion on its way down.
With that assumption in force, we are able to state unequivocally
and without fear of contradiction that each 'speed' described above ...
with the word 'downward' appended to it ... does become a 'velocity'.
<h3>You forgot to add question...Add questions before asking so we can help</h3>
Answer:
"look for a large stripe or a minus sign (or both) on one side of the capacitor."
Explanation:
i hope it helped :)
