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Sati [7]
2 years ago
5

Why does a Parallel Circuit still continue to function after a switch turns off one bulb but continues to pass through the secon

d bulb? (please be as specific as possible)​
Physics
1 answer:
sertanlavr [38]2 years ago
6 0

Explanation:

In short, when we switch off a branch, the remaining branch still forms a complete circuit with the battery, so electricity still pass through the remaining branch.

Technically speaking, the voltage in each branch is the same, and the resistance of the light bulb remains unchanged (neglect effect of temp.) by V=IR, the current should remain unchanged after switching off the other branch.

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Please help Need good grade
ivann1987 [24]

Answer:

The other angle is 120°.

Explanation:

Given that,

Angle = 60

Speed = 5.0

We need to calculate the  range

Using formula of range

R=\dfrac{v^2\sin(2\theta)}{g}...(I)

The range for the other angle is

R=\dfrac{v^2\sin(2(\alpha-\theta))}{g}....(II)

Here, distance and speed are same

On comparing both range

\dfrac{v^2\sin(2\theta)}{g}=\dfrac{v^2\sin(2(\alpha-\theta))}{g}

\sin(2\theta)=\sin(2\times(\alpha-\theta))

\sin120=\sin2(\alpha-60)

120=2\alpha-120

\alpha=\dfrac{120+120}{2}

\alpha=120^{\circ}

Hence, The other angle is 120°

6 0
3 years ago
How far does a runner run if he runs for 60 seconds at 5 m/s
mylen [45]

Answer: 20 miles

Explanation:

4 0
3 years ago
Astronomers have found water on mars.
vodomira [7]

The thin atmosphere of Mars is thought to be due to the planet's lack of a magnetic field, which has allowed the Solar wind to blow away much of the gas the planet once had. Venus, despite still having a thick atmosphere of CO2, surprisingly has a similar problem

4 0
1 year ago
Taking test, need physics help<br> ASAP please!
Anuta_ua [19.1K]

Answer:

im pretty sure its c the third answer i got that one right

Explanation:

your welcome :)

8 0
2 years ago
Pls help answer embed <br>​
Savatey [412]

Answer:

C = 1.01

Explanation:

Given that,

Mass, m = 75 kg

The terminal velocity of the mass, v_t=60\ m/s

Area of cross section, A=0.33\ m^2

We need to find the drag coefficient. At terminal velocity, the weight is balanced by the drag on the object. So,

Weight of the object = drag force

R = W

or

\dfrac{1}{2}\rho CAv_t^2=mg

Where

\rho is the density of air = 1.225 kg/m³

C is drag coefficient

So,

C=\dfrac{2mg}{\rho Av_t^2}\\\\C=\dfrac{2\times 75\times 9.8}{1.225\times 0.33\times (60)^2}\\\\C=1.01

So, the drag coefficient is 1.01.

5 0
3 years ago
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