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Yuki888 [10]
4 years ago
7

N=-D(n2-n1)/(x2-x1) D is diffastion.What are the dimensions of D.

Physics
1 answer:
MrMuchimi4 years ago
8 0

We don't have much to go on.

The dimensions of D depend on the dimensions of N, n, and x, and we don't know what any of those stand for.

It might help if we had ever heard of 'diffastion', but we're striking out there too.

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Starting from one oasis, a camel walks 25 km in a direction 30° south of west and then walks 30 km toward the north to a second
shepuryov [24]

Answer:

distance between both oasis ( 1 and 2) is  27.83 Km

Explanation:

let d is the distance between oasis1 and oasis 2

from figure

OC  = 25cos 30

OE = 25sin30

OE = CD

Therefore BC =  30-25sin30

distance between both oasis ( 1 and 2) is calculated by using phytogoras theorem

in\Delta BCO

OB^2 = BC^2 + OC^2

PUTTING ALL VALUE IN ABOVE EQUATION

d^2 = 930-25sin30)^2 + (25cos30)^2

d^2 = 775

d = 27.83 Km

distance between both oasis ( 1 and 2) is  27.83 Km

3 0
3 years ago
An ideal spring hangs from the ceiling. A 1.25-kg mass is hung from the spring. After all vibrations have died away, the spring
ch4aika [34]

The kinetic energy of the mass at the instant it passes back through its equilibrium position is about 1.20 J

\texttt{ }

<h3>Further explanation</h3>

Let's recall Elastic Potential Energy formula as follows:

\boxed{E_p = \frac{1}{2}k x^2}

where:

<em>Ep = elastic potential energy ( J )</em>

<em>k = spring constant ( N/m )</em>

<em>x = spring extension ( compression ) ( m )</em>

Let us now tackle the problem!

\texttt{ }

<u>Given:</u>

mass of object = m = 1.25 kg

initial extension = x = 0.0275 m

final extension = x' = 0.0735 - 0.0275 = 0.0460 m

<u>Asked:</u>

kinetic energy = Ek = ?

<u>Solution:</u>

<em>Firstly , we will calculate the spring constant by using </em><em>Hooke's Law</em><em> as follows:</em>

F = k x

mg = k x

k = mg \div x

k = 1.25(9.8) \div 0.0275

k = 445 \frac{5}{11} \texttt{ N/m}

\texttt{ }

<em>Next , we will use </em><em>Conservation of Energy</em><em> formula to solve this problem:</em>

Ep_1 + Ek_1 = Ep_2 + Ek_2

\frac{1}{2}k (x')^2 + mgh + 0 = \frac{1}{2}k x^2 + Ek

Ek = \frac{1}{2}k (x')^2 + mgh - \frac{1}{2}k x^2

Ek = \frac{1}{2}k ( (x')^2 - x^2 ) + mgh

Ek = \frac{1}{2}(445 \frac{5}{11}) ( 0.0460^2 - 0.0275^2 ) + 1.25(9.8)(0.0735)

\boxed {Ek \approx 1.20 \texttt{ J}}

\texttt{ }

<h3>Learn more</h3>
  • Kinetic Energy : brainly.com/question/692781
  • Acceleration : brainly.com/question/2283922
  • The Speed of Car : brainly.com/question/568302
  • Young Modulus : brainly.com/question/9202964
  • Simple Harmonic Motion : brainly.com/question/12069840

\texttt{ }

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Elasticity

8 0
3 years ago
Read 2 more answers
Newtons law of gravitational force
JulijaS [17]
\large\begin{array}{I} \mathtt{ F= \dfrac{GMm}{R^{2}} } \end{array}
4 0
3 years ago
How many infrared photons of frequency 2.57 x 1013 Hz would need to be absorbed simultaneously by a tightly bound molecule to br
victus00 [196]

Answer:

94

Explanation:

f = 2.57 x 10^13 Hz

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Energy of each photon = h f

Where, h is Plank's constant

Energy of each photon = 6.63 x 10^-34 x 2.57 x 10^13 = 1.7 x 10^-20 J

Number of photons = Total energy / energy of one photon

N = (1.6 x 10^-18) / (1.7 x 10^-20) = 94.11 = 94

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3 years ago
What are some uses of the electromagnetic
den301095 [7]
If you mean electromagnetic waves
Radio waves can be used well in radio signals
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