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3 years ago

(6.9 * 10-6)(770 * 102)

Physics

1 answer:

meriva3 years ago

5 0

Answer:

4948020

Explanation:

(6.9*10-6)(770*102)

Multiply 6.9 by 10 .

( 69 − 6 ) ( 770 ⋅ 102 )

Subtract 6 from 69 .

63 ( 770 ⋅ 102 )

Multiply 770 by 102 .

63 ⋅ 78540

Multiply 63 by 78540 .

4948020

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a hammer falls off a roof and strikes the ground with a certain kinetic energy. If it fell from a roof twice as tall, how ouwld

S_A_V [24]

If it starts from twice the height, then it has twice the potential energy up there. This means it'll have twice the kinetic energy when it hits the ground.

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3 years ago

A package is dropped from a hovering helicopter.

aivan3 [116]

Answer:

The height of the package when it was released was 1102.5 meters

Explanation:

A package is dropped from a hovering helicopter

Dropped means the initial velocity of the package is zero

The package takes 15 seconds to strike the ground

Ignore the air resistance

We need to know the height of the package when it was released

$h=ut+\frac{1}{2}gt^{2}$ , where h is the height of the package

from the ground, u is the initial velocity, g is the acceleration of gravity

and t is the time

u = 0 , t = 15 seconds , g = 9.8 m/s²

$h=(0)(15)+\frac{1}{2}(9.8)(15)^{2}$

$h=0+1102.5$

h = 1102.5 meters

The height of the package when it was released was 1102.5 meters

4 0

3 years ago

A certain metal with work function of Φ = 1.7 eV is illuminated in vacuum by 1.4 x 10-6 W of light with a wavelength of λ = 600

Makovka662 [10]

Answer:

1) n = 4.47*10^12 photons

2) K = 0.25 eV

Explanation:

This is a problem about the photoelectric effect.

1) In order to calculate the number of photons that impact the metal, you take into account the power of the light, which is given by:

$P=\frac{E}{t}=1.4*10^{-6}\frac{J}{s}$ (1)

Furthermore you calculate the energy of a photon with a wavelength of 600nm, by using the following formula:

$E_p=h\frac{c}{\lambda}$ (2)

c: speed of light = 3*10^8 m/s

h: Planck's constant = 6.626*10^-34 Js

λ: wavelength = 600*10^-9 m

You replace the values of the parameters in the equation (2):

$E_p=(6.626*10^{-34}Js)\frac{3*10^8m/s}{600*10^{-9}m}=3.131*10^{-19}J$

Next, you calculate the quotient between the power of the light (equation (1)) and the energy of the photon:

$n=\frac{P}{E_p}=\frac{1.4*10^{-6}J/s}{3.131*10^{-19}J}=4.47*10^{12}photons$

The number of photons is 4.47*10^12

2) The kinetic energy of the electrons emitted by the metal is given by the following formula:

$K=E_p-\Phi$ (3)

Ep: energy of the photons

Φ: work function of the metal = 1.7 eV

You first convert the energy of the photons to eV:

$E_p=3.131*10^{-19}J*\frac{6.242*10^{18}eV}{1J}=1.954eV$

You replace in the equation (3):

$K=1.95eV-1.7eV=0.25eV$

The kinetic energy of the electrons emitted by the metal is 0.25 eV

6 0

4 years ago

Two uniformly charged conducting plates are parallel to each other. They each have area A. Plate #1 has a positive charge Q whil

Marrrta [24]

Answer:5

Explanation:

Given

First Plate has a charge of +Q and area A

Second Plate has a charge of -3 Q and area A

We Know electric Field due to sheet charge is given by

$E=\frac{Q}{2A\epsilon }$

Where Q=charge over the Plate

A=Area of plate

$\epsilon$ = Permittivity of free space

Electric Field Due to Positive charge will always be away from it while for negative charge it is towards it.

Net Electric Field at a point between between the Plates is the superimposition of two electric with direction

$E_{net}=\frac{Q}{2A\epsilon }+\frac{3Q}{2A\epsilon }$

$E_{net}=\frac{2Q}{A\epsilon }$

Net electric Field is towards the negative charged plate

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4 years ago

Large sections of troposphere with same temperature and humidity is called

77julia77 [94]

Answer:

air mass

Explanation:

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polar
tropical
continental and
maritime.

7 0

3 years ago