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BlackZzzverrR [31]
3 years ago
8

If you are holding a book and you let go of it, why does it fall? Explain in terms of forces, potential energy, and kinetic ener

gy.
Physics
1 answer:
inessss [21]3 years ago
3 0

Answer:

Let the mass of the book be "m", acceleration due to gravity be "g", velocity be "v" and height be "h".

Now if we are holding a book at a certain height (h), <em><u>the potential energy will be maximum which is equal to mass× acceleration due to gravity× height (= mgh)</u>.</em>

(Remember: kinetic energy =0)

Now we consider that the book is dropped, in this case a force will act downward towards the centre of the earth, <em><u>Force= mass× acceleration due to gravity (F=mg)</u></em>. It is equal to the weight of the book.

While the book is falling, the potential energy stored in the book converts into kinetic energy and strikes the floor with <em><u>the maximum kinetic energy= (1/2)×mass×velocity² (=1/2mv²)</u>.</em>

(Remember: kinetic energy=0)

Due to this process the whole energy is conserved.

As the potential energy decreases kinetic energy increases.

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Please help me with both of them
8_murik_8 [283]

Answer:

101011010101010100101000101010100010011010100010100000101041204105210241012021012021021012022221222122122345788981633333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333311111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111114565656565+4+652147

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546213171549895465621324547998995656565656565656565722426579898541321447985331321

Explanation:

4 0
4 years ago
How is a net electric charge produced​
Vlad [161]

If an object has more protons than electrons, then the net charge on the object is positive. If there are more electrons than protons, then the net charge on the object is negative. If there are equal numbers of protons and electrons, then the object is electrically neutral.

Source: https://www.khanacademy.org/science/ap-physics-1/ap-electric-charge-electric-force-and-voltage/electric-charge-ap/a/electric-charge-ap1

Hope this helps :)

4 0
3 years ago
A wire has resistivity of 0.12 homes per metre​.
gulaghasi [49]

Answer:

(a) The resistance of 25m of wire is 3 ohms

(b) the length of this wire that has resistance 22 ohms is 183.33 m

Explanation:

Given;

resistivity of the wire, ρ = 0.12 ohms per meter

(a) The resistance of 25m of wire is calculated as follows;

R = \rho L\\\\R = 0.12 \ \frac{ohms}{m} \times 25\ m\\\\R = 3 \ ohms

(b) the length of this wire that has resistance 22 ohms is calculated as;

L = \frac{R}{\rho} \\\\L = \frac{22 \ ohms }{0.12 \ ohms/m} = 183.33 \ m

3 0
3 years ago
011 10.0 points
Ulleksa [173]

Answer:

2.47 m

Explanation:

Let's calculate first the time it takes for the ball to cover the horizontal distance that separates the starting point from the crossbar of d = 52 m.

The horizontal velocity of the ball is constant:

v_x = v cos \theta = (25)(cos 35.9^{\circ})=20.3 m/s

and the time taken to cover the horizontal distance d is

t=\frac{d}{v_x}=\frac{52}{20.3}=2.56 s

So this is the time the ball takes to reach the horizontal position of the crossbar.

The vertical position of the ball at time t is given by

y=u_y t - \frac{1}{2}gt^2

where

u_y = v sin \theta =(25)(sin 35.9^{\circ})=14.7 m/s is the initial vertical velocity

g = 9.8 m/s^2 is the acceleration of gravity

And substituting t = 2.56 s, we find the vertical position of the ball when it is above the crossbar:

y=(14.7)(2.56) - \frac{1}{2}(9.8)(2.56)^2=5.52 m

The height of the crossbar is h = 3.05 m, so the ball passes

h' = 5.52- 3.05 = 2.47 m

above the crossbar.

8 0
3 years ago
An object is placed in front of a diverging lens, such that the object-to-image distance is 71 cm.
Pachacha [2.7K]

Explanation:

Given that,

Object-to-image distance d= 71 cm

Image distance = 26 cm

We need to calculate the object distance

u -v= d

u=71+26=97\ cm

We need to calculate the focal length

Using formula of lens

\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}

put the value into the formula

\dfrac{1}{f}=\dfrac{1}{-26}+\dfrac{1}{97}

\dfrac{1}{f}=-\dfrac{71}{2522}

f=-35.52\ cm

The focal length of the lens is 35.52.

(B). Given that,

Object distance = 95 cm

Focal length = 29 cm

We need to calculate the distance of the image

Using formula of lens

\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}

Put the value in to the formula

\dfrac{1}{-29}=\dfrac{1}{v}-\dfrac{1}{95}

\dfrac{1}{v}=\dfrac{1}{-29}-\dfrac{1}{95}

\dfrac{1}{v}=-\dfrac{124}{2755}

v=-22.21\ cm

We need to calculate the magnification

Using formula of magnification

m=\dfrac{v}{u}

m=\dfrac{22.21}{95}

m=0.233

The magnification is 0.233.

The image is virtual.

Hence, This is the required solution.

4 0
3 years ago
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