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2 years ago

What is 6,210 bucks converted to kilobucks?

2 answers:

8 0

6,210/1000=6.21

Kilo means thousand so each kilobuck has a thousand bucks. Divide 6210 (number of bucks) by 1000 (number of bucks in a kilobuck) to find your answer.

Vinvika [58]2 years ago

4 0

When 6210 bucks is converted to kilobucks, the result obtained is 6.21 kilobucks

<h3>Conversion scale </h3>

1000 bucks = 1 Kilobuck

Using the above convesion scale, we can express 6210 bucks in kilobucks

<h3>How to express bucks in kilobucks</h3>

1000 bucks = 1 Kilobuck

Therefore,

6210 bucks = (6210 × 1) / 1000

6210 bucks = 6.21 kilobucks

Thus, 6210 bucks is equivalent to 6.21 kilobucks

Learn more about conversion:

brainly.com/question/2139943

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A steel ball and a wooden ball of the same diameter are released together from the top of a tower. In the absence of air resista

ella [17]

Answer:

False

Explanation:

The steel ball and the wooden ball do not have the same force acting on them because their masses are different. But, they have the same acceleration which is the acceleration due to gravity g = 9.8 m/s².

Using the equation of motion under freefall, s = ut +1/2gt². Since u = 0,

s = 1/2gt² ⇒ t = √(2s/g)

Since. s = height is the same for both objects, they land at the same time neglecting air resistance.

8 0

3 years ago

Explain the universal law of gravity and it's calculate from as well as the value of g

horrorfan [7]

Answer:

According to the Newton's law of gravitational every object in the universe attracts every other objects with a force which is called gravitational force.This gravitational force is (i) directly proportional to the product of their masses and (ii) inversely proportional to the square of the distance between their centres.

Explanation:

Newton's law of gravitational is called the universal law because it is applicable to all the bodies either terrestrial or celestial having any shape,size,mass or at any distance apart with any medium between them,at any time(past,present or future).

3 0

3 years ago

2. A solid plastic cube of side 0.2 m is submerged in a liquid of density 0.8 hgm calculate the

kotegsom [21]

Answer:

vpg = 0.064 N

Explanation:

Upthrust = Volume of fluid displaced

upthrust liquid on the cube g=10ms−2

vpg =0.2 x 0.2 x 0.2 x0.8 x 10= 0.064N

vpg = 0.064 N

hope it helps.

3 0

3 years ago

A transparent oil with index of refraction 1.28 spills on the surface of water (index of refraction 1.33), producing a maximum o

Ad libitum [116K]

Answer:

The thickness of the oil slick is $1.95\times10^{-7}\ m$

Explanation:

Given that,

Index of refraction = 1.28

Wave length = 500 nm

Order m = 1

We need to calculate the thickness of oil slick

Using formula of thickness

$2nt= m\lambda$

Where, n = Index of refraction

t = thickness

$\lambda$ = wavelength

Put the value into the formula

$2\times1.28 \times t=1\times\times500\times10^{-9}$

$t = \dfrac{1\times\times500\times10^{-9}}{2\times1.28 }$

$t=1.95\times10^{-7}\ m$

Hence, The thickness of the oil slick is $1.95\times10^{-7}\ m$

4 0

3 years ago

Compare the wavelengths of an electron (mass = 9.11 × 10−31 kg) and a proton (mass = 1.67 × 10−27 kg), each having (a) a speed o

Ad libitum [116K]

Answer:

Part A:

The proton has a smaller wavelength than the electron.

$\lambda_{proton} = 6.05x10^{-14}m$ < $\lambda_{electron} = 1.10x10^{-10}m$

Part B:

The proton has a smaller wavelength than the electron.

$\lambda_{proton} = 1.29x10^{-13}m$ < $\lambda_{electron} = 5.525x10^{-12}m$

Explanation:

The wavelength of each particle can be determined by means of the De Broglie equation.

$\lambda = \frac{h}{p}$ (1)

Where h is the Planck's constant and p is the momentum.

$\lambda = \frac{h}{mv}$ (2)

Part A

Case for the electron:

$\lambda = \frac{6.624x10^{-34} J.s}{(9.11x10^{-31}Kg)(6.55x10^{6}m/s)}$

But $J = Kg.m^{2}/s^{2}$

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(9.11x10^{-31}Kg)(6.55x10^{6}m/s)}$

$\lambda = 1.10x10^{-10}m$

Case for the proton:

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(1.67x10^{-27}Kg)(6.55x10^{6}m/s)}$

$\lambda = 6.05x10^{-14}m$

Hence, the proton has a smaller wavelength than the electron.

<em>Part B </em>

For part b, the wavelength of the electron and proton for that energy will be determined.

First, it is necessary to find the velocity associated to that kinetic energy:

$KE = \frac{1}{2}mv^{2}$

$2KE = mv^{2}$

$v^{2} = \frac{2KE}{m}$

$v = \sqrt{\frac{2KE}{m}}$ (3)

Case for the electron:

$v = \sqrt{\frac{2(7.89x10^{-15}J)}{9.11x10^{-31}Kg}}$

but $1J = kg \cdot m^{2}/s^{2}$

$v = \sqrt{\frac{2(7.89x10^{-15}kg \cdot m^{2}/s^{2})}{9.11x10^{-31}Kg}}$

$v = 1.316x10^{8}m/s$

Then, equation 2 can be used:

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(9.11x10^{-31}Kg)(1.316x10^{8}m/s)}$

$\lambda = 5.525x10^{-12}m$

Case for the proton :

$v = \sqrt{\frac{2(7.89x10^{-15}J)}{1.67x10^{-27}Kg}}$

But $1J = kg \cdot m^{2}/s^{2}$

$v = \sqrt{\frac{2(7.89x10^{-15}kg \cdot m^{2}/s^{2})}{1.67x10^{-27}Kg}}$

$v = 3.07x10^{6}m/s$

Then, equation 2 can be used:

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(1.67x10^{-27}Kg)(3.07x10^{6}m/s)}$

$\lambda = 1.29x10^{-13}m$

Hence, the proton has a smaller wavelength than the electron.

7 0

3 years ago