I’m not sure because a baseball player run 10 meters north but get out and he turn around and that a other 10 meters for turning around and 4 more meters to his dugout it’s could be d or a because it’s closer to 24
Answer:
1) 65 m
2) 40 m/s downward
Explanation:
Using for both questions the kinematic equation
v² = u² + 2as
and ignoring air resistance.
1) h = 60 + √(20²/(2(9.8))) = 64.51753...
2) v = √(20² + 2(9.8)(60)) = 39.69886...
Answer:
A chain
Explanation:
A series circuit is a circuit, where the resistors are arranged in a chain. They are aligned like this so the current only has one path to take. The current quantity is the same through each resistor.
If you ever have trouble remembering this you can think of a movie series which goes on and on and on. And remembering a parallel circuit is simpler because everything in the circuit will be parallel!
Answer:
V = 9.33 V
Explanation:
By definition, the capacitance of a capacitor, is the proportion between the charge on one of charged surfaces and the potential difference between these surfaces, as follows:
![C = \frac{Q}{V}](https://tex.z-dn.net/?f=C%20%3D%20%5Cfrac%7BQ%7D%7BV%7D)
for a parallel-plate type capacitor, it can be showed, applying Gauss'Law to the a gaussian surface with the shape of a pillbox parallel to the surface of one of the plates, half outside the surface, half inside it, that the capacitance can be expressed as follows:
![C =\frac{\epsilon*A}{d}](https://tex.z-dn.net/?f=C%20%3D%5Cfrac%7B%5Cepsilon%2AA%7D%7Bd%7D)
As we can see, if a slab of paraffin, with dielectric constant of 2.25, is inserted, the capacitance will be increased in the same factor:
![Cf = 2.25 * Co](https://tex.z-dn.net/?f=Cf%20%3D%202.25%20%2A%20Co)
Now, if the capacitor, once charged, is disconnected from the battery, the charge Q will remain constant.
So, if the capacitance will be increased 2.25 times, the only way to do this is that the voltage between plates be reduced in the same factor:
![Cf = 2.25 * Co = 2.25* \frac{Q}{Vo} = \frac{Q}{Vf} \\ Vf= \frac{Vo}{2.25} =\frac{21.0V}{2.25} = 9.33 V](https://tex.z-dn.net/?f=Cf%20%3D%202.25%20%2A%20Co%20%3D%202.25%2A%20%5Cfrac%7BQ%7D%7BVo%7D%20%3D%20%5Cfrac%7BQ%7D%7BVf%7D%20%5C%5C%20Vf%3D%20%5Cfrac%7BVo%7D%7B2.25%7D%20%3D%5Cfrac%7B21.0V%7D%7B2.25%7D%20%3D%209.33%20V)
Answer:
runway use is 3307.8 feet
Explanation:
given data
velocity = 140 kts = 140 × 0.5144 m/s = 72.016 m/s
time = 28 seconds
weight = 28000 lbs
to find out
How many feet of runway was used
solution
we will use here first equation of motion for find acceleration
v = u + at ..............1
here v is velocity given and u is initial velocity that is 0 and a is acceleration and t is time
put here value in equation 1
72.016 = 0 + a(28)
a = 2.572 m/s²
and
now apply third equation of motion
s = ut + 0.5×a×t² .......................2
here s is distance and u is initial speed and t is time and a is acceleration
put here all value in equation 2
s = 0 + 0.5×2.572×28²
s = 1008.24 m = 3307.8 ft
so runway use is 3307.8 feet