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MatroZZZ [7]
4 years ago
8

What are the two possible solutions to 5x^2−15=0?

Physics
1 answer:
Novosadov [1.4K]4 years ago
4 0

I have attached the solution as image.

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In a darkened room, a burning candle is placed 1.84 m from white wall. A lens is placed between candle and wall at a location th
KATRIN_1 [288]

Answer:

82.4 cm

Explanation:

The object and screen are kept fixed ie the distance between them is fixed and by displacing lens between them images are formed on the screen . In the first case let u be the object distance and v be the image distance

then ,

u + v = 184 cm

In the second case of image formation , v becomes u and u becomes v only then image formation in the second case is possible.

The difference between two object distance ie(  v - u ) is the distance by which lens is moved so

v - u = 82.4 cm

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3 years ago
The mass of earth 6*10^24kg and radius is 400kg Now find the value of acceleration due to gravity when a object is 3600km from t
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while test flying the new f-22 raptor fighter jet, test pilot bob put the plane in a horizontal loop while flying at the speed o
sveticcg [70]

Given

v = 343 m/s

ac = 5g

ac = 5*9.8 m/s^2

ac = 49 m/s^2

where,

v: velocity

ac = centripetal aceleration

Procedure

We call the acceleration of an object moving in uniform circular motion—resulting from a net external force—the centripetal acceleration ac; centripetal means “toward the center” or “center seeking”.

Formula

\begin{gathered} a_c=\frac{v^2}{r} \\ r=\frac{v^2}{a_c} \\ r=\frac{(343m/s)^2}{49m/s^2} \\ r=2401\text{ m} \end{gathered}

The minimum radius not to exceed the centripetal acceleration is 2401 m.

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1 year ago
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