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2 years ago

8

A book is pushed with an initial horizontal velocity of 5.0 meters per second off the top of a 1.19 meter high desk. How far awa

y does the book land?

1 answer:

kipiarov [429]2 years ago

8 0

Answer:

2.45 m

Explanation:

First of all, we have to calculate the time of flight of the book, by using the equation for the vertical motion:

$h=\frac{1}{2}gt^2$

where

h = 1.19 m is the vertical distance covered by the book

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

t is the time of flight

Solving for t,

$t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2(1.19)}{9.8}}=0.49 s$

Now we can find the horizontal distance covered by the book, which is given by

$d=v_x t$

where

$v_x = 5.0 m/s$ is the horizontal velocity

t = 0.49 s is the time of flight

Substituting,

$d=(5.0)(0.49)=2.45 m$

So the book lands 2.45 m away.

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Explanation:

It is given that,

Magnetic field, B = 0.1 T

Acceleration, $a=6\times 10^{15}\ m/s^2$

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Mass of electron, $m=9.1\times 10^{-31}\ kg$

(a) The force acting on the electron when it is accelerated is, F = ma

The force acting on the electron when it is in magnetic field, $F=qvB\ sin\theta$

Here, $\theta=90$

So, $ma=qvB$

Where

v is the velocity of the electron

B is the magnetic field

$v=\dfrac{ma}{qB}$

$v=\dfrac{9.1\times 10^{-31}\ kg\times 6\times 10^{15}\ m/s^2}{1.6\times 10^{-19}\ C\times 0.1\ T}$

v = 341250 m/s

or

$v=3.41\times 10^5\ m/s$

So, the speed of the electron is $3.41\times 10^5\ m/s$

(b) In 1 ns, the speed of the electron remains the same as the force is perpendicular to the cross product of velocity and the magnetic field.

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2 years ago

The water level in a tank is 20 m above the ground. a hose is connected to the bottom of the tank, and the nozzle at the end of

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Answer:

P_(pump) = 98,000 Pa

Explanation:

We are given;

h2 = 30m

h1 = 20m

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First of all, we know that the sum of the pressures in the tank and the pump is equal to that of the Nozzle,

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P_(tank)+ P_(pump) = P_(nozzle)

Now, the pressure would be given by;

P = ρgh

So,

ρgh_1 + P_(pump) = ρgh_2

Thus,

P_(pump) = ρg(h_2 - h_1)

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P_(pump) = 1000•9.8(30 - 20)

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1 year ago

Two oppositely charged but otherwise identical conducting plates of area 2.50 square centimeters are separated by a dielectric 1

7nadin3 [17]

Answer:

A). σ = 3.823 x $10^{-5}$ $C^{2}$ /N- $m^{2}$

B). $\sigma ^{'}=2.76\times 10^{-5}$ C/ $m^{2}$

C). $U=10.322$ J

Explanation:

A). We know magnitude of charge per unit area for a conducting plate is given by

$\sigma =k.\varepsilon _{0}.E$

where, E is resultant electric field = 1.2 x $10^{6}$ V/m

$\varepsilon _{0}$ is permittivity of free space = 8.85 x $10^{-12}$ $C^{2}$ /N- $m^{2}$

k is dielectric constant = 3.6

∴ $\sigma =k.\varepsilon _{0}.E$

= 3.6 x 8.85 x $10^{-12}$ x 1.2 x $10^{6}$

= 3.823 x $10^{-5}$ $C^{2}$ /N- $m^{2}$

B).Now we know that the magnitude of charge per unit area on the surface of the dielectric plate is given by

$\sigma ^{'}=\sigma\left ( 1-\frac{1}{k} \right )$

$\sigma ^{'}=3.823\times 10^{-5}\left ( 1-\frac{1}{3.6} \right )$

$\sigma ^{'}=2.76\times 10^{-5}$ C/ $m^{2}$

C).

Area of the plate, A = 2.5 $cm^{2}$

= 2.5 x $10^{-4}$ $m^{2}$

diameter of the plate, d = 1.8 mm

= 1800 m

∴ Total energy stored in the capacitor

$U=\frac{1}{2}k\varepsilon _{0}E^{2}Ad$

$U=\frac{1}{2}\times 3.6\times8.85 \times10^{-12}\times\left ( 1.2\times 10^{6} \right ) ^{2}\times 2.5\times 10^{-4}\times 1800$

$U=10.322$ J

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Answer:

Explanation:

Volume of block A = 10 x 6 x 1 = 60 cm³

Mass of block A = 630 g

density of mass A = mass / density

= 630 / 60 = 10.5g / cm³

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Mass of block A = 604 g

density of mass A = mass / density

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astraxan [27]

Answer:

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Explanation:

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Substitute into the formula as shown;

Q = 0.025×2.256×10^6

Q = 56400Joules

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1 year ago