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3 years ago

Explain why the light bulb isn't lighting up in the circuit pictured on the right.

Physics

1 answer:

FromTheMoon [43]3 years ago

8 0

The wires should be attached to each screw on the light not only one

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a car travels from station A to B at 30kmph during its return trip A it travels 30 km pH for the first half distance and it 70 k

8_murik_8 [283]

Explanation:

If you cannot visualize it, just assume that the distance from station A to B is 420km. Each half is 210km.

When the car travels from A to B, it takes 420/30 = 14 hours.

When the car travels from B to the halfway point, it takes 210/30 = 7 hours.

When the car travels from the halfway point to A, it takes 210/70 = 3 hours.

Total time taken = 14 + 7 + 3 = 24 hours.

Total distance = 420km * 2 = 840km.

Hence, the average speed of the car is 840/24 = 35km/h.

3 0

3 years ago

Consider what will happen when a bar magnet is pushed toward the coil. when the coil "feels" the changing magnetic field caused

notka56 [123]

The question just basically explained what happens

4 0

4 years ago

34. A train, starting from rest, accelerates along the platform at a uniform rate of 0.6 m/s2. A passenger standing on the platf

ira [324]

Answer:

4.08 s

Explanation:

Let the passenger took "t" time to catch the train

so in this case the total distance moved by the train + 5 m = total distance moved by the passenger

so we will have

distance moved by train is given as

$d_1 = \frac{1}{2}(0.6) t^2$

also the distance moved by passenger

$d_2 = \frac{1}{2}(1.2) t^2$

so we will have

$d_1 + 5 = d_2$

$0.3 t^2 + 5 = 0.6 t^2$

$0.3 t^2 = 5$

t = 4.08 s

3 0

3 years ago

The barometric pressure in breckenridge, colorado (elevation 9600 feet) is 580 mm hg. how many atmospheres is this?

Maksim231197 [3]

1 atmospheric pressure = 760.0 mm Hg

Thus 580 mm Hg = (580 mm Hg/(760 mm Hg/atm))

= 0.763 atm

7 0

3 years ago

Equations E = 1 2πε0 qd z3 and E = 1 2πε0 P z3 are approximations of the magnitude of the electric field of an electric dipole,

tamaranim1 [39]

Answer:

The ratio of $E_{app}$ and $E_{act}$ is 0.9754

Explanation:

Given that,

Distance z = 4.50 d

First equation is

$E_{act}=\dfrac{qd}{2\pi\epsilon_{0}\times z^3}$

$E_{act}=\dfrac{Pz}{2\pi\epsilon_{0}\times (z^2-\dfrac{d^2}{4})^2}$

Second equation is

$E_{app}=\dfrac{P}{2\pi\epsilon_{0}\times z^3}$

We need to calculate the ratio of $E_{act}$ and $E_{app}$

Using formula

$\dfrac{E_{app}}{E_{act}}=\dfrac{\dfrac{P}{2\pi\epsilon_{0}\times z^3}}{\dfrac{Pz}{2\pi\epsilon_{0}\times (z^2-\dfrac{d^2}{4})^2}}$

$\dfrac{E_{app}}{E_{act}}=\dfrac{(z^2-\dfrac{d^2}{4})^2}{z^3(z)}$

Put the value into the formula

$\dfrac{E_{app}}{E_{act}}=\dfrac{((4.50d)^2-\dfrac{d^2}{4})^2}{(4.50d)^3\times4.50d}$

$\dfrac{E_{app}}{E_{act}}=0.9754$

Hence, The ratio of $E_{app}$ and $E_{act}$ is 0.9754

8 0

3 years ago