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Anastaziya [24]
3 years ago
14

1. Why should you use scientific notation when writing big numbers?

Physics
1 answer:
Fynjy0 [20]3 years ago
6 0

Answer:

It helps the answer look clean. It also makes the work easier to work with.

Explanation:

Instead of writing a lot of zeros, all you have to do is add exponents to the number to show how much the decimal moved.

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Consider a blackbody that radiates with an intensity I1I1I_1 at a room temperature of 300K300K. At what intensity I2I2I_2 will t
kap26 [50]

Answer:

Explanation:

We shall apply Stefan's formula

E = AσT⁴

When T = 300

I₁ = Aσ x 300⁴

When T = 400K

I₂ = Aσ x 400⁴

I₂ / I₁ = 400⁴ / 300⁴

= 256 / 81

= 3.16

I₂ = 3.16 I₁ .

5 0
3 years ago
which characteristic of northern renaissance artworks do albrecht dürer and pieter bruegel the elder share? a. they rarely inclu
VladimirAG [237]
According to my assumptions,the answer closest to the perfection is :

c. they show royalty instead of peasants.
7 0
3 years ago
Which of the following expressions will have units of kg⋅m/s2? Select all that apply, where x is position, v is velocity, m is m
netineya [11]

Answer: m \frac{d}{dt}v_{(t)}

Explanation:

In the image  attached with this answer are shown the given options from which only one is correct.

The correct expression is:

m \frac{d}{dt}v_{(t)}

Because, if we derive velocity v_{t} with respect to time t we will have acceleration a, hence:

m \frac{d}{dt}v_{(t)}=m.a

Where m is the mass with units of kilograms (kg) and a with units of meter per square seconds \frac{m}{s}^{2}, having as a result kg\frac{m}{s}^{2}

The other expressions are incorrect, let’s prove it:

\frac{m}{2} \frac{d}{dx}{(v_{(x)})}^{2}=\frac{m}{2} 2v_{(x)}^{2-1}=mv_{(x)} This result has units of kg\frac{m}{s}

m\frac{d}{dt}a_{(t)}=ma_{(t)}^{1-1}=m This result has units of kg

m\int x_{(t)} dt= m \frac{{(x_{(t)})}^{1+1}}{1+1}+C=m\frac{{(x_{(t)})}^{2}}{2}+C This result has units of kgm^{2} and C is a constant

m\frac{d}{dt}x_{(t)}=mx_{(t)}^{1-1}=m This result has units of kg

m\frac{d}{dt}v_{(t)}=mv_{(t)}^{1-1}=m This result has units of kg

\frac{m}{2}\int {(v_{(t)})}^{2} dt= \frac{m}{2} \frac{{(v_{(t)})}^{2+1}}{2+1}+C=\frac{m}{6} {(v_{(t)})}^{3}+C This result has units of kg \frac{m^{3}}{s^{3}} and C is a constant

m\int a_{(t)} dt= \frac{m {a_{(t)}}^{2}}{2}+C This result has units of kg \frac{m^{2}}{s^{4}} and C is a constant

\frac{m}{2} \frac{d}{dt}{(v_{(x)})}^{2}=0 because v_{(x)} is a constant in this derivation respect to t

m\int v_{(t)} dt= \frac{m {v_{(t)}}^{2}}{2}+C This result has units of kg \frac{m^{2}}{s^{2}} and C is a constant

6 0
3 years ago
Two loudspeakers emit identical sound waves along the x-axis. The sound at a point on the axis has maximum intensity when the sp
EleoNora [17]

The concept required to solve this problem is related to the wavelength.

The wavelength can be defined as the distance between two positive crests of a wave.

The waves are in phase, then the first distance is

\Delta x_1 = 20cm

And out of the phase when

\Delta x_2 = 30cm

Thus the wavelength is

\Delta x_2-\Delta x_1 = \frac{\lambda}{2}

Here,

\lambda =  Wavelength

If we rearrange the equation to find it, we will have

\lambda = 2 (\Delta x_2-\Delta x_1 )

\lambda= 2(30-20)

\lambda = 20cm

Therefore the wavelength of the sound is 20cm.

5 0
3 years ago
I need to know the answer to the question
galben [10]
I would go with A. An Observation. Because he's looking at visual differences, for example: "based on smell, color change, and release of bubbles.
4 0
3 years ago
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