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Ivanshal [37]
3 years ago
5

Maisie drags a 1kg mass along a table with a Newton meter so that it accelerates at 0.25 m/s. The Newton-meter reads 0.4N, calcu

late the force of friction between the mass and the table
Physics
1 answer:
Temka [501]3 years ago
7 0
Applying one newton of force to an object of 1 kg accelerates it by 1 meter per second

maisie applied .4 newtons to the block, so it should be accelerating at .4 m/s
but it actually accelerated at .25 m/s or .25 newtons reached the block

the difference in force applied and force received is .15, which is the force of friction because that's how much force it stopped from reaching the block
force of friction is .15 newtons
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If the Sun subtends a solid angle Ω on the sky, and the flux from the Sun just above the Earth’s atmosphere, integrated over all
Arada [10]

Answer:

A)Ω = 7.8 × 10^−5 steradians.

B) TE = 5800K

C) fλ(λ1) = (π ^2 ) /ΩBλ(T)

Explanation:

A) First of all, if we assume that the Sun emits isotropically at a luminosity (L⊙) , the flux at a given distance R from the sun would be f(d) = L⊙/ (4πd^2)

The ratio of flux at the solar photosphere to the flux at the Earth’s atmosphere would be: F⊙/{f(d⊙)} = (R⊙)^2 / (d⊙)^2

Now if we think of this relationship of the flux and the earth as a conical pattern, we'll deduce that the solid angle subtended by the sun at Earth’s surface to be;

Ω = π[(R⊙)^2 / (d⊙)^2]

Combining this with the ratio earlier gotten, well arrive at;

F⊙ = {f(d⊙ )π} /Ω

Now let's express The radius of the sun (R) in terms of its angular diameter (2α) and this gives;

R⊙ ≈ αd⊙

Now combining this with the equation for Ω earlier, we get;

Ω ≈ πα^2

So, = π((0.57/2π) /180)^2 = 7.8 × 10^−5 steradians.

B) from Stefan-Boltzmann Law,

F⊙ = σ(TE)^4

From the beginning, we know that;

F⊙ = {f(d⊙ )π} /Ω

And so replacing that in the stephan boltzmann law, we get ;

{f(d⊙ )π} /Ωσ = (TE)^4

So, (TE)^4 = {π (1.4 kWm^(−2))} / [(7.8 × 10^(−5 ) steradians x (5.66961 × 10^(−8))]

In stephan boltzmann law, σ = 5.66961 × 10^(−8)

And so, TE is approximately 5800K.

C) In order to relate fλ(λ1) with T, let's assume the sun’s surface to be an isotropically emitting blackbody, i.e its specific intensity is Iλ = Bλ(T). Hence, the flux at Sun’s surface for a given wavelength would be;

Fλ(λ1) = πBλ(T)

Now, if we combine this with the expression of F⊙ gotten earlier, well get the relation;

fλ(λ1) = (π ^2 ) /ΩBλ(T)

7 0
3 years ago
As a clinical chemist in charge of the testing of blood samples. You are required to mix a solution of sodium chloride with wate
Natalka [10]
<span>7.8x102 x 1.95x10<span>3 this is the answer mate
</span></span>
6 0
3 years ago
Examine the graph. Select the statement that best describes the energy change in the particles of a substance during melting.
serg [7]

Answer:

B)  Absorbed energy results in the change in potential energy.

Explanation:

i got it wrong on usatestpreo

3 0
3 years ago
Read 2 more answers
I appreciate any help I get on this, thank you in advanced :)
Reika [66]
The picture isn’t clear so I can’t read the dimensions of the box but I can try my best to guide u through the question.

For part a u need to find the volume of the box as that will equal the volume of sand that can be filled inside.
For this u multiply the height, width and length of the box.

For part b the mass of sand alone will be
=Mass of box + sand - Mass of empty box
=216 - 40
=176 grams

For part c the density of sand can be calculated by the formula
Density= Mass/Volume
So the mass (176g) / volume from part a

For part d u need to know that something will float if it has a lower density than what it is floating in. If the final density of sand that was found in part c is less than the density of gold (19.3 g/cm^3) it will float. Otherwise it will sink.

Hope this helped!
3 0
3 years ago
In the mobile m1=0.42 kg and m2=0.47 kg. What must the unknown distance to the nearest tenth of a cm be if the masses are to be
LuckyWell [14K]

Complete Question

The complete question is shown on the first uploaded image

Answer:

Explanation:

From he question we are told that

    The first mass is   m_1 = 0.42kg

      The second mass is  m_2 = 0.47kg

From the question we can see that at equilibrium the moment about the point where the  string  holding the bar (where m_1 \ and \ m_2 are hanged ) is attached is zero  

   Therefore we can say that

               m_1 * 15cm  = m_2 * xcm

Making x the subject of the formula  

                x = \frac{m_1 * 15}{m_2}

                    = \frac{0.42 * 15}{0.47}

                     x = 13.4 cm

Looking at the diagram we can see that the tension T  on the string holding the bar where m_1  \  and   \ m_2 are hanged  is as a result of the masses (m_1 + m_2)

     Also at equilibrium the moment about the point where the string holding the bar (where (m_1 +m_2)  and  m_3 are hanged ) is attached is  zero

   So basically

          (m_1 + m_2 ) * 20  = m_3 * 30

          (0.42 + 0.47)  * 20 = 30 * m_3

 Making m_3 subject

          m_3 = \frac{(0.42 + 0.47) * 20 }{30 }

                m_3 = 0.59 kg

3 0
3 years ago
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