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3 years ago

Look at the circuit diagram.

Physics

2 answers:

Nikitich [7]3 years ago

8 0

Answer: The correct option is B.

Explanation:

In the given circuit, the circuit is closed as the battery is connected to the circuit and the current is flowing.

The two bulbs connected in the circuit are in series combination as the same amount of current is flowing in the bulbs.

In the parallel combination, the amount of current flowing in the circuit across resistance is different.

Therefore, the provided circuit is closed series circuit.

Fiesta28 [93]3 years ago

6 0

B. closed series circuit because the resistors are one after the other and there is no open switch.

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Some lenses are shaped with one flat side and one spherically-shaped side. This shape is designed to focus parallel light rays o

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Answer:

Some lenses are used to focus light to a pre-defined point based on the amount of curvature of their surfaces.

In a piano design convex, some surfaces are flat while others has positive lenses (biconvex)

Explanation:

Solution

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For that of a plano-convex design, one surface has a positive curve and for biconvex lenses, both surfaces are positively curved while the other remains flat.

when used practically, plano-convex lenses are most commonly used where the object being imaged is far apart from lens.

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3 years ago

How much tension must a cable withstand if it is used to accelerate a 1800-kg car vertically upward at 0.80 m/s2 ?

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Force =tension =mass x acceleration

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3 years ago

Read 2 more answers

Take 100 points look the picture

Marina86 [1]

Answer:

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3 years ago

On the way to the moon, the Apollo astronauts reach a point where the Moon’s gravitational pull is stronger than that of Earth’s

Drupady [299]

Answer:

rm = 38280860.6[m]

Explanation:

We can solve this problem by using Newton's universal gravitation law.

In the attached image we can find a schematic of the locations of the Earth and the moon and that the sum of the distances re plus rm will be equal to the distance given as initial data in the problem rt = 3.84 × 108 m

$r_{e} = distance earth to the astronaut [m].\\r_{m} = distance moon to the astronaut [m]\\r_{t} = total distance = 3.84*10^8[m]$

Now the key to solving this problem is to establish a point of equalisation of both forces, i.e. the point where the Earth pulls the astronaut with the same force as the moon pulls the astronaut.

Mathematically this equals:

$F_{e} = F_{m}\\F_{e} =G*\frac{m_{e} *m_{a}}{r_{e}^{2} } \\$

$F_{m} =G*\frac{m_{m}*m_{a} }{r_{m} ^{2} } \\where:\\G = gravity constant = 6.67*10^{-11}[\frac{N*m^{2} }{kg^{2} } ] \\m_{e}= earth's mass = 5.98*10^{24}[kg]\\ m_{a}= astronaut mass = 100[kg]\\m_{m}= moon's mass = 7.36*10^{22}[kg]$

When we match these equations the masses cancel out as the universal gravitational constant

$G*\frac{m_{e} *m_{a} }{r_{e}^{2} } = G*\frac{m_{m} *m_{a} }{r_{m}^{2} }\\\frac{m_{e} }{r_{e}^{2} } = \frac{m_{m} }{r_{m}^{2} }$

To solve this equation we have to replace the first equation of related with the distances.

$\frac{m_{e} }{r_{e}^{2} } = \frac{m_{m} }{r_{m}^{2} } \\\frac{5.98*10^{24} }{(3.84*10^{8}-r_{m} )^{2} } = \frac{7.36*10^{22} }{r_{m}^{2} }\\81.25*r_{m}^{2}=r_{m}^{2}-768*10^{6}* r_{m}+1.47*10^{17} \\80.25*r_{m}^{2}+768*10^{6}* r_{m}-1.47*10^{17} =0$

Now, we have a second-degree equation, the only way to solve it is by using the formula of the quadratic equation.

$r_{m1,2}=\frac{-b+- \sqrt{b^{2}-4*a*c } }{2*a}\\ where:\\a=80.25\\b=768*10^{6} \\c = -1.47*10^{17} \\replacing:\\r_{m1,2}=\frac{-768*10^{6}+- \sqrt{(768*10^{6})^{2}-4*80.25*(-1.47*10^{17}) } }{2*80.25}\\\\r_{m1}= 38280860.6[m] \\r_{m2}=-2.97*10^{17} [m]$

We work with positive value

rm = 38280860.6[m] = 38280.86[km]

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4 years ago

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