Explanation:
The distance that a car travels down the interstate can be calculated with the following formula:
Distance = Speed x Time
(A) Speed of the car, v = 70 miles per hour = 31.29 m/s
Time, d = 6 hours = 21600 s
Distance = Speed x Time
D = 31.29 m/s × 21600 s
D = 675864 meters
or
(b) Time, d = 10 hours = 36000 s
Distance = Speed x Time
D = 31.29 m/s × 36000 s
D = 1126440 meters
or
(c) Time, d = 15 hours = 54000 s
Distance = Speed x Time
D = 31.29 m/s × 54000 s
D = 1689660 meters
or
Hence, this is the required solution.
30 Grams would be your answer (I took the test and got it right)
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'.
Answer:
speed of the charge electric is v = - (Eo q/m) cos t
Explanation:
The electric charge has a very small mass so it follows the oscillations of the electric field. We force ourselves on the load,
F = q Eo sint
a) To find the velocity of the particle, let's use Newton's second law to find the acceleration and of this by integration the velocity
F = ma
q Eo sint = ma
a = Eo q / m sint
a = dv / dt
dv = adt
∫ dv = ∫ a dt
v-vo = I (Eoq / m) sin t dt
v- vo = Eo q / m (-cos t)
We evaluate the integral from the initial point, as the particle starts from rest Vo = 0, for t = 0
v = - (Eo q / m) cos t
b) Kinetic energy
K = ½ m v2
K = ½ m (Eoq / m)²2 (sint)²
K = ¹/₂ Eo² q² / m sin² t
c) The average kinetic energy over a period
K = ½ m v2
<v2> = (Eoq / m) 2 <cos2 t>
The average of cos2 t = ½, substitute and calculate
K = ½ m (Eoq / m)² ½
K = ¼ Eo² q² / m
Answer:
I honestly don't know(sorry)
