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dusya [7]
3 years ago
7

If a certain mass of mercury has a volume of 0.002 m3 at a temperature of 20°C, what will be the volume at 50°C?

Physics
1 answer:
Klio2033 [76]3 years ago
5 0
From tables, the density of mercury is
13545 kg/m^3 at 20°C,
13472 kg/m^3 at 50°C.

Because mass = density * volume, the mass of mercury at 20°C is
m = (13545 kg/m^3)*(0.002 m^3) = 27.09 kg

Let V  = volume of mercury at 50°C.
Because the mass of mercury does not change, therefore at 50°C,
(13472 kg/m^3)*(V m^3) = 27.09
V = 27.09/13472 = 0.0020108 m^3

Answer:  B. 0.002010812 m³
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     The force that prevents motion when the surfaces of two objects come into contact is known as friction. Friction decreases a machine's mechanical advantage, or, to put it another way, reduces the output to input ratio.

<h3>How can I figure out the frictional force?</h3>

        The resistive force of friction (Fr) divided by the normal or perpendicular force (N) pushing the objects together yields the coefficient of friction (fr), which is a numerical value.

The formula fr = Fr/N serves as a representation of it.

Therefore, 100N of force is needed to move an item with a mass of 50 kg.

It will accelerate by 10 m/s2.

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8 0
1 year ago
Light rays travel from one medium into another and refract away from the boundary. What changes about the light to cause this re
vaieri [72.5K]

Answer:

a

Explanation:

I think its a its speed increases.....

8 0
3 years ago
Read 2 more answers
An incompressible fluid (water) is flowing through a pipe of diameter 20 cm with
sergey [27]

Answer:

115 kPa

Explanation:

Use Bernoulli equation:

P₁ + ½ ρ v₁² + ρgh₁ = P₂ + ½ ρ v₂² + ρgh₂

Assuming no elevation change, h₁ = h₂.

P₁ + ½ ρ v₁² = P₂ + ½ ρ v₂²

Plugging in values:

(582,000 Pa) + ½ (1000 kg/m³) (1.28 m/s)² = P + ½ (1000 kg/m³) (30.6 m/s)²

P = 115,000 Pa

P = 115 kPa

3 0
3 years ago
What is the energy (in evev) of a photon of visible light that has a wavelength of 500 nmnm?
lisabon 2012 [21]
<h3>Answer:</h3>
  • E≈2,5 eV
<h3>Explanation:</h3>

_______________

λ=500 nm = 500·10⁻⁹ m

c=3·10⁸ m/s

h=6,63·10⁻³⁴ J·s = 4,14·10⁻¹⁵ eV·s

_______________

E - ?

_______________

\displaystyle \boldsymbol{E}=h\nu =h \frac{c}{\lambda} =4,14\cdot 10^{-15} \; eV\cdot s\cdot \; \frac{3\cdot 10^8\; m/s}{500\cdot 10^{-9}\; m} =2,484\;  eV\approx \boldsymbol{2,5\; eV}

6 0
2 years ago
Calculate the work done by a force of 30 N in lifting a load of 2 kg to a height of 10 m (g = 10 m/s2)
Mars2501 [29]

Answer:

<u>300 J</u>

Explanation:

Given :

  1. Applied force = 30 N
  2. Mass = 2 kg
  3. Height = 10 m
  4. g = 10 m/s²

Work done by the force :

  • Work done = Force x Displacement
  • Work done = mg x h
  • Work done = 30 N x 10 m
  • Work done = <u>300 J</u>

<u></u>

<u><em>Note</em></u> :

  • What you have calculated is the work done by gravitational force on the object (that too, incorrectly)
  • But in the end, it asks for work done by the force of 30N
  • Hence, the given answer ~
3 0
2 years ago
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