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Oxana [17]
2 years ago
8

Anybody help me, please? Describe one periodic trend that you learned about.

Physics
1 answer:
Korolek [52]2 years ago
8 0

Answer:

the renegade

Explanation: charklie dfamielo

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wave absorption results in some of the wave's energy being converted into thermal energy. Describe an example of a time you've e
FinnZ [79.3K]

Answer:

,,,,,,

Explanation:

m jkg yrfyuih iopoj kpoohjkbhj huiph i

5 0
2 years ago
A boulder of mass 10 kg rolls over a cliff and reaches the beach below with a velocity of 15 m/s. Find:
Kryger [21]

Answer:

<h3>formular kinetic energy=1/2 ×v</h3>
3 0
2 years ago
What is the wavelength associated with 0.113kg ball traveling with velocity of 43 m/s?
maks197457 [2]

Answer:

Wavelength = 1.36 * 10^{-34} meters

Explanation:

Given the following data;

Mass = 0.113 kg

Velocity = 43 m/s

To find the wavelength, we would use the De Broglie's wave equation.

Mathematically, it is given by the formula;

Wavelength = \frac {h}{mv}

Where;

h represents Planck’s constant.

m represents the mass of the particle.

v represents the velocity of the particle.

We know that Planck’s constant = 6.6262 * 10^{-34} Js

Substituting into the formula, we have;

Wavelength = \frac {6.6262 * 10^{-34}}{0.113*43}

Wavelength = \frac {6.6262 * 10^{-34}}{4.859}

Wavelength = 1.36 * 10^{-34} meters

7 0
2 years ago
Find the acceleration of a car with the mass of 1,200 kg and a force of
Mashutka [201]

Answer:9.17 m/s^2

Explanation:

mass=1200kg

Force=11 x 10^3 N

Acceleration=force ➗ mass

Acceleration=11 x 10^3 ➗ 1200

Acceleration=9.17

Acceleration=9.17 m/s^2

3 0
3 years ago
Determine the image distance and image height for a 5.00 cm tall object placed 30.0 cm from a double convex lens with a focal le
vlabodo [156]

Answer:

The image distance is 30 cm

image height = - 5 cm

Explanation:

The formula for calculating the image distance is expressed as

1/f = 1/u + 1/v

where

f is the focal length

u is the object distance

v is the image distance

From the information given,

u = 30

f = 15

By substituting these values into the formula,

1/15 = 1/30 + 1/v

1/v = 1/15 - 1/30 = (2 - 1)/30 = 1/30

Taking the reciprocal of both sides,

v = 30

The image distance is 30 cm

magnification = image height/object height = - v/u

Given that object height = 5 cm, then

image height/5 = - 30/30 = - 1

image height = - 5 * 1

image height = - 5 cm

8 0
1 year ago
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