S A voltage ΔV is applied to a series configuration of n resistors, each of resistance R. The circuit components are reconnected
1 answer:
The power of is series combination is Vn^2 times that of a parallel combination.
For series combination :
Req = R + R + R + ............... n times = nR
I = Δv/nr
Power = (Δv/nr)^2 × nr = Δv^2/nr
For parallel combination
1/req = 1/R + 1/R + 1/R +................(n times) = n/R
Req = R/n
Power = Δv/(R/n) = nΔv^2/R
Ratio = Δv^2/nr/n·Δv^2/R = 1/n^2
Hence, power of is series combination is Vn^2 times that of a parallel.
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Answer:
The last two bearings are
49.50° and 104.02°
Explanation:
Applying the Law of cosine (refer to the figure attached):
we have
x² = y² + z² - 2yz × cosX
here,
x, y and z represents the lengths of sides opposite to the angels X,Y and Z.
Thus we have,
or
substituting the values in the equation we get,
or
or
X = 26.47°
similarly,
or
or
Y = 49.50°
Consequently, the angel Z = 180° - 49.50 - 26.47 = 104.02°
The bearing of 2 last legs of race are angels Y and Z.
Answer:
A. The bomb will take <em>17.5 seconds </em>to hit the ground
B. The bomb will land <em>12040 meters </em>on the ground ahead from where they released it
Explanation:
Maverick and Goose are flying at an initial height of , and their speed is v=688 m/s
When they release the bomb, it will initially have the same height and speed as the plane. Then it will describe a free fall horizontal movement
The equation for the height y with respect to ground in a horizontal movement (no friction) is
[1]
With g equal to the acceleration of gravity of our planet and t the time measured with respect to the moment the bomb was released
The height will be zero when the bomb lands on ground, so if we set y=0 we can find the flight time
The range (horizontal displacement) of the bomb x is
[2]
Since the bomb won't have any friction, its horizontal component of the speed won't change. We need to find t from the equation [1] and replace it in equation [2]:
Setting y=0 and isolating t we get
Since we have
Replacing in [2]
A. The bomb will take 17.5 seconds to hit the ground
B. The bomb will land 12040 meters on the ground ahead from where they released it