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3 years ago

Which of the following is equal to impulse? A. Ft B. m/s C. pt D. mv

Physics

1 answer:

tankabanditka [31]3 years ago

5 0

Answer: A (Ft)

Explanation: The impulse experienced by the object equals the change in momentum of the object. In equation form, F • t = m • Δ v

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Answer:

it is accelerating 14 m/s

Explanation:

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3 years ago

A 1.20 g sample of an unknown has a volume of 1.73 cm what is the density of the unknown

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1.73 divide by 1.20=1.4416

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3 years ago

Gravity is an attractive force between two or more objects. The amount of force is directly proportional of the _______ of the o

Monica [59]

Blank 1: mass
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3 years ago

How many lenses with different focal lengths can be obtained by combining two surfaces whose radii of curvature are 4.00 cm and

Dovator [93]

Answer:

The lenses with different focal length are four.

Explanation:

Given that,

Radius of curvature R₁= 4

Radius of curvature R₂ = 8

We know ,

Refractive index of glass = 1.6

When, R₁= 4, R₂ = 8

We need to calculate the focal length of the lens

Using formula of focal length

$\dfrac{1}{f}=(n-1)(\dfrac{1}{R_{1}}+\dfrac{1}{R_{2}})$

Put the value into the formula

$\dfrac{1}{f}=(1.6-1)(\dfrac{1}{4}+\dfrac{1}{8})$

$\dfrac{1}{f}=\dfrac{9}{40}$

$f=4.44\ cm$

When , R₁= -4, R₂ = 8

Put the value into the formula

$\dfrac{1}{f}=(1.6-1)(\dfrac{1}{-4}+\dfrac{1}{8})$

$\dfrac{1}{f}=-\dfrac{3}{40}$

$f=-13.33\ cm$

When , R₁= 4, R₂ = -8

Put the value into the formula

$\dfrac{1}{f}=(1.6-1)(\dfrac{1}{4}-\dfrac{1}{8})$

$\dfrac{1}{f}=\dfrac{3}{40}$

$f=13.33\ cm$

When , R₁= -4, R₂ = -8

Put the value into the formula

$\dfrac{1}{f}=(1.6-1)(\dfrac{1}{-4}-\dfrac{1}{8})$

$\dfrac{1}{f}=-\dfrac{9}{40}$

$f=-4.44\ cm$

Hence, The lenses with different focal length are four.

8 0

2 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

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

• But we know that, v/u is M

${ \tt{f + fM = v}} \\ { \tt{f(1 +M) = v }} \\ { \tt{1 +M = \frac{v}{f} }} \\ \\ { \boxed{ \mathfrak{formular : } \: { \tt{ M = \frac{v}{f} - 1 }}}}$

v is image distance, v is 10 cm
f is focal length, f is 5 cm
M is magnification.

${ \tt{M = \frac{10}{5} - 1 }} \\ \\ { \tt{M = 5 - 1}} \\ \\ { \underline{ \underline{ \pmb{ \red{ \: magnification \: is \: 4}}}}}$

 » Nature of Image :

Image is magnified
Image is erect or upright
Image is inverted
Image distance is identical to object distance.

4 0

2 years ago