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Sonbull [250]
4 years ago
12

WILL GIVE BRAINLYEST!!!!!!!!!!!!

Physics
2 answers:
lakkis [162]4 years ago
6 0

I would say it would be the weight read.

Only because the Control is the type of liquid and the constants are Size of object and amount of liquid

Eva8 [605]4 years ago
4 0

Answer:

the amount of liquid in which the object was submerged

Explanation:

As we know that spring balance will read the spring force that is exerted by spring on the object

Now when the object is suspended by the spring then in that case the spring force is balanced by the weight so spring will read the correct mass of the object.

Now the object is suspended by the spring and then submerged into the liquid so in that case we can say

F_{spring} + F_b = mg

F_{spring} = mg - \rho_{liquid}Vg

so here the reading will be less and it depends on volume of object and density of liquid

so independent variable will be

the amount of liquid in which the object was submerged

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9) Of all the types of light the Sun gives off, it emits the greatest amount of light at visible wavelengths of light. If the Su
erastova [34]

Answer:

*  most of the emission would be in the infrared part, the visible radiation would be very small.

*total intensity of the semition decreases that the intensity depends on the fourth power of the temperature

Explanation:

The radiation emitted by the Sun is approximately the radiation of a black body, if the Sun were to cool, the maximum emission wavelength changes

          λ T = 2,898 10⁻³

          λ = 2,898 10⁻³ / T

if the temperature decreases the maximum wavelength the greater values ​​are moved, that is to say towards the infrared. Therefore the emission curve also moves, in this case most of the emission would be in the infrared part, the visible radiation would be very small.

Furthermore, the total intensity of the semition decreases that the intensity depends on the fourth power of the temperature according to Stefan's law

           P = σ A eT⁴

7 0
3 years ago
Two homogeneous bodies of the same volume
lord [1]

Answer:

No, it is not necessary for them to have same mass.

Explanation:

Let both bodies have a density d1 and d2 respectively.

Since their volumes are equal V1 = V2

we know that, https://tex.z-dn.net/?f=%5Cfrac%7Bmass%7D%7Bvolume%7D

Hence, d1 =  and d2 =    

Taking the ratio of densities,we get

This implies that unless the bodies have same densities, the mass of the two bodies will not be same.

8 0
3 years ago
A capacitor is connected across an ac source. Suppose the frequency of the source is doubled. What happens to the capacitive rea
fgiga [73]

The capacitive reactance is reduced by a factor of 2.

<h3>Calculation:</h3>

We know the capacitive reactance is given as,

Xc = \frac{1}{2\pi fC}

where,

Xc\\ = capacitive reactance

f = frequency

C = capacitance

It is given that frequency is doubled, i.e.,

f' = 2f

To find,

Xc =?

Xc' = \frac{1}{2\pi f'C}

      = \frac{1}{2\pi 2f C}

      = \frac{1}{2} (\frac{1}{2\pi fC} )\\

Xc' = \frac{1}{2} Xc

Therefore, the capacitive reactance is reduced by a factor of 2.

I understand the question you are looking for is this:

A capacitor is connected across an AC source. Suppose the frequency of the source is doubled. What happens to the capacitive reactant of the inductor?

  1. The capacitive reactance is doubled.
  2. The capacitive reactance is traduced by a factor of 4.
  3. The capacitive reactance remains constant.
  4. The capacitive reactance is quadrupled.
  5. The capacitive reactance is reduced by a factor of 2.

Learn more about capacitive reactance here:

brainly.com/question/23427243

#SPJ4

3 0
2 years ago
Why forces are balanced and unbalanced? need help with this the lesson is tommorow
Naya [18.7K]
"Balanced" means that if there's something pulling one way, then there's also
something else pulling the other way. 

-- If there's a kid sitting on one end of a see-saw, and another one with the
same weight sitting on the other end, then the see-saw is balanced, and
neither end goes up or down.  It's just as if there's nobody sitting on it.

-- If there's a tug-of-war going on, and there are 300 freshmen pulling on one
end of a rope, and another 300 freshmen pulling in the opposite direction on
the other end of the rope, then the hanky hanging from the middle of the rope
doesn't move.  The pulls on the rope are balanced, and it's just as if nobody
is pulling on it at all.

-- If a lady in the supermarket is pushing her shopping cart up the aisle, and her
two little kids are in front of the cart pushing it in the other direction, backwards,
toward her.  If the kids are strong enough, then the forces on the cart can be
balanced. Then the cart doesn't move at all, and it's just as if nobody is pushing
on it at all.

From these examples, you can see a few things:

-- There's no such thing as "a balanced force" or "an unbalanced force".
It's a <em><u>group</u> of forces</em> that is either balanced or unbalanced.

-- The group of forces is balanced if their strengths and directions are
just right so that each force is canceled out by one or more of the others.

-- When the group of forces on an object is balanced, then the effect on the
object is just as if there were no force on it at all.
4 0
3 years ago
Read 2 more answers
The kinetic energy of a proton and that of an a-particle are 4 eV and 1 eV,
madam [21]

Answer:

  • <u><em>(b) 1:1</em></u>

Explanation:

<u></u>

<u>1. Formulae:</u>

  • E = hf  
  • E = h.v/λ
  • λ = h/(mv)
  • E = (1/2)mv²

Where:

  • E = kinetic energy of the particle
  • λ = de-Broglie wavelength
  • m = mass of the particle
  • v = speed of the particle
  • h = Planck constant

<u><em>2. Reasoning</em></u>

An alha particle contains 2 neutrons and 2 protons, thus its mass number is 4.

A proton has mass number 1.

Thus, the relative masses of an alpha particle and a proton are:

       \dfrac{m_\alpha}{m_p}=4

For the kinetic energies you find:

          \dfrac{E_\alpha}{E_p}=\dfrac{m_\alpha \times v_\alpha^2}{m_p\times v_p^2}

            \dfrac{1eV}{4eV}=\dfrac{4\times v_\alpha^2}{1\times v_p^2}\\\\\\\dfrac{v_p^2}{v_\alpha^2}=16\\\\\\\dfrac{v_p}{v_\alpha}=4

Thus:

           \dfrac{m_\alpha}{m_p}=4=\dfrac{v_p}{v__\alpha}

          m_\alpha v_\alpha=m_pv_p

From de-Broglie equation, λ = h/(mv)  

       \dfrac{\lambda_p}{\lambda_\alpha}=\dfrac{m_\lambda v_\lambda}{m_pv_p}=\dfrac{1}{1}=1:1

5 0
3 years ago
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