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2 years ago

A sports car skids off of a wet mountain road at 48.5 m/s and lands in the river, 110

Physics

1 answer:

Oxana [17]2 years ago

3 0

Answer:

about 4.74 seconds

Explanation:

The time to fall distance d from height h is given by ...

t = √(2d/g)

t = √(2·110 m/(9.8 m/s^2)) ≈ 4.74 s

It will take the car about 4.74 seconds to fall 110 meters to the river.

We assume the car's speed is horizontal, so does not add or subtract anything to/from the time to fall from the height.

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What is the speed of light? What is the speed of sound?

Rasek [7]

Answer:

speed of light simulating traveling at the speed of light. Speed of light, speed at which light waves propagate through different materials. In particular, the value for the speed of light in a vacuum is now defined as exactly 299,792,458 metres per second

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3 years ago

A leaf is floating on the Surface of the water,what will happen to its movement? Explain

docker41 [41]

Answer:

MRCORRECT has answered the question

Explanation:

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3 years ago

A block lies on a horizontal frictionless surface and

zhenek [66]

Answer:

0.1 m

Explanation:

F = Force exerted on spring = 3 N

k = Spring constant = 60 N/m

x = Displacement of the block

As the energy of the system is conserved we have

$Fx=\dfrac{1}{2}kx^2$

$\\\Rightarrow x=\dfrac{2F}{k}$

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$\\\Rightarrow x=0.1\ m$

The position of the block is 0.1 from the initial position.

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3 years ago

2. A 20 cm object is placed 10cm in front of a convex lens of focal length 5cm. Calculate

adoni [48]

Answer:

 » Image distance :

${ \tt{ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} }} \\$

v is image distance
u is object distance, u is 10 cm
f is focal length, f is 5 cm

${ \tt{ \frac{1}{v} + \frac{1}{10} = \frac{1}{5} }} \\ \\ { \tt{ \frac{1}{v} = \frac{1}{10} }} \\ \\ { \tt{v = 10}} \\ \\ { \underline{ \underline{ \pmb{ \red{ \: image \: distance \: is \: 10 \: cm \: \: }}}}}$

 » Magnification :

• Let's derive this formula from the lens formula:

${ \tt{ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} }} \\$

» Multiply throughout by fv

${ \tt{fv( \frac{1}{v} + \frac{1}{u} ) = fv( \frac{1}{f} )}} \\ \\ { \tt{ \frac{fv}{v} + \frac{fv}{u} = \frac{fv}{f} }} \\ \\ { \tt{f + f( \frac{v}{u} ) = v}}$