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3 years ago

An object is located 25.0 cm from a convex mirror. The image distance is -50.0 cm. What is the magnification?

Physics

1 answer:

Lynna [10]3 years ago

4 0

Answer:

$\boxed{\sf Magnification \ (m) = 2}$

Given:

Object distance (u) = 25.0 cm

Image distance (v) = -50.0 cm

To Find:

Magnification (m)

Explanation:

$\boxed{\bold{\sf Magnification \: (m) = - \frac{Image \: distance \: (v)}{Object \: distance \: (u)}}}$

Substituting values of Image distance(v) & Object distance (u) in the equation:

$\sf \implies m = - \frac{( - 50)}{25}$

-(-50) = 50:

$\sf \implies m = \frac{50}{25}$

$\sf \implies m = \frac{2 \times \cancel{25}}{ \cancel{25}}$

$\sf \implies m = 2$

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What are the horizontal and vertical components of a lizard’s displacement if it has climbed 7m directly up a tree?

madam [21]

Answer:

The horizontal component is zero.

The vertical component is $7\sin\theta$

Explanation:

Given that,

The lizard climb 7m directly up on a tree.

We know that,

The horizontal component is

$x=\cos\theta$

The vertical component is

$y=\sin\theta$

If the lizard climb 7m directly up on a tree then,

We need to find the components

Using given data

The horizontal component of lizard is

$x=0$

The vertical component is

$y=7\sin\theta$

Hence, The horizontal component is zero.

The vertical component is $7\sin\theta$

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3 years ago

An ideal spring hangs from the ceiling. A 2.15 kg mass is hung from the spring, stretching the spring a distance d = 0.0895 m fr

Igoryamba

Answer:

The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.

Explanation:

Given that,

Mass = 2.15 kg

Distance = 0.0895 m

Amplitude = 0.0235 m

We need to calculate the spring constant

Using newton's second law

$F= mg$

Where, f = restoring force

$kx=mg$

$k=\dfrac{mg}{x}$

Put the value into the formula

$k=\dfrac{2.15\times9.8}{0.0895}$

$k=235.41\ N/m$

We need to calculate the kinetic energy of the mass

Using formula of kinetic energy

$K.E=\dfrac{1}{2}mv^2$

Here, $v = A\omega$

$K.E=\dfrac{1}{2}m\times(A\omega)^2$

Here, $\omega=\sqrt{\dfrac{k}{m}}^2$

$K.E=\dfrac{1}{2}m\times A^2\sqrt{\dfrac{k}{m}}^2$

$K.E=\dfrac{1}{2}kA^2$

Put the value into the formula

$K.E=\dfrac{1}{2}\times235.41\times(0.0235)^2$

$K.E=0.06500\ J$

Hence, The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.

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3 years ago

As temperature increase what energy increases

garik1379 [7]

Kinetic energy would increase sir.

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4 years ago

A ball is thrown upward from the top of a building at an angle of 30.0° to the horizontal and with

Ray Of Light [21]

Answer:

See below

Explanation:

Vertical position = 45 + 20 sin (30) t - 4.9 t^2

when it hits ground this = 0

0 = -4.9t^2 + 20 sin (30 ) t + 45

0 = -4.9t^2 + 10 t +45 = 0 solve for t =4.22 sec

max height is at t= - b/2a = 10/9.8 =1.02

use this value of 't' in the equation to calculate max height = 50.1 m

it has 4.22 - 1.02 to free fall = 3.2 seconds free fall

v = at = 9.81 * 3.2 = 31.39 m/s VERTICAL

it will <u>also</u> still have horizontal velocity = 20 cos 30 = 17.32 m/s

total velocity will be sqrt ( 31.39^2 + 17.32^2) = 35.85 m/s

Horizontal range = 20 cos 30 * t = 20 * cos 30 * 4.22 = 73.1 m

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2 years ago

Which graph shows an object that is dropped?

Olegator [25]

Answer:

I'm pretty sure it's the third one where velocity goes from positive to negative

Explanation:

the positive velocity is before the object hits the ground and the negative is after

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3 years ago