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vredina [299]
2 years ago
7

If a light bulb is missing or broken in a parallel circuit will the other bulb light?

Physics
1 answer:
Yuki888 [10]2 years ago
6 0
It should as long as it's connected to a power source.
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a current of 180 mini amphere passes through a conductor for 5minute calculate the quantity of electricity transported​
oksano4ka [1.4K]

Answer:

Explanation:

You can calculate the total electric charge that passes through the conductor as q=It=(180\times 10^{-3})(5\times 60)= 54 C. It means that the number of electron that passes through the conductor is:

n=\frac{q}{e}=\frac{54}{1.6\times 10^{-19}}=33.75\times 10^{19}

8 0
2 years ago
Acceleration is the magnitude of average velocity.​
lina2011 [118]

Answer:

false

Explanation:

just did the question on apex, true was wrong

4 0
3 years ago
Complete each statement with the word that makes it true.
Nuetrik [128]

Answer:

reactive

halogens

nonmetals

Explanation:

5 0
2 years ago
compare to visible light, the wavelength of x-rays is shorter longer or the same and the frequency is lower higher or the same
Doss [256]

Answer:shorter higher

Explanation:Compare to visible light, the wavelength of X-rays is shorter and then frequency is higher

7 0
2 years ago
For a given initial projectile speed Vo, calculate what launch angle A gives the longest range R. Show your work, don't just quo
pickupchik [31]
The optimal angle of 45° for maximum horizontal range is only valid when initial height is the same as final height. 

<span>In that particular situation, you can prove it like this: </span>

<span>initial velocity is Vo </span>
<span>launch angle is α </span>

<span>initial vertical velocity is </span>
<span>Vv = Vo×sin(α) </span>

<span>horizontal velocity is </span>
<span>Vh = Vo×cos(α) </span>

<span>total time in the air is the the time it needs to fall back to a height of 0 m, so </span>
<span>d = v×t + a×t²/2 </span>
<span>where </span>
<span>d = distance = 0 m </span>
<span>v = initial vertical velocity = Vv = Vo×sin(α) </span>
<span>t = time = ? </span>
<span>a = acceleration by gravity = g (= -9.8 m/s²) </span>
<span>so </span>
<span>0 = Vo×sin(α)×t + g×t²/2 </span>
<span>0 = (Vo×sin(α) + g×t/2)×t </span>
<span>t = 0 (obviously, the projectile is at height 0 m at time = 0s) </span>
<span>or </span>
<span>Vo×sin(α) + g×t/2 = 0 </span>
<span>t = -2×Vo×sin(α)/g </span>

<span>Now look at the horizontal range. </span>
<span>r = v × t </span>
<span>where </span>
<span>r = horizontal range = ? </span>
<span>v = horizontal velocity = Vh = Vo×cos(α) </span>
<span>t = time = -2×Vo×sin(α)/g </span>
<span>so </span>
<span>r = (Vo×cos(α)) × (-2×Vo×sin(α)/g) </span>
<span>r = -(Vo)²×sin(2α)/g </span>

<span>To find the extreme values of r (minimum or maximum) with variable α, you must find the first derivative of r with respect to α, and set it equal to 0. </span>

<span>dr/dα = d[-(Vo)²×sin(2α)/g] / dα </span>
<span>dr/dα = -(Vo)²/g × d[sin(2α)] / dα </span>
<span>dr/dα = -(Vo)²/g × cos(2α) × d(2α) / dα </span>
<span>dr/dα = -2 × (Vo)² × cos(2α) / g </span>

<span>Vo and g are constants ≠ 0, so the only way for dr/dα to become 0 is when </span>
<span>cos(2α) = 0 </span>
<span>2α = 90° </span>
<span>α = 45° </span>
4 0
2 years ago
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