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3 years ago

I.Name two commonly used thermometric liquids.

Physics

2 answers:

PtichkaEL [24]3 years ago

6 0

Answer:

mercury and alcohol

ii) used to test temperatures

ivann1987 [24]3 years ago

6 0

Mercury and Alcohol
Mercury:

i) It is a good conductor of heat and therefore the whole liquid reaches the temperature of the surroundings quickly.

ii) It does not wet (cling to the sides of) the tube.

Alcohol:

i) Alcohol has greater value of temperature coefficient of expansion than mercury.

ii) it's freezing point is below –100°C.

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A glass ball of radius 3.74 cm sits at the bottom of a container of milk that has a density of 1.04 g/cm3. The normal force on t

Gelneren [198K]

Answer:

The mass of the ball is 0.23 kg

Explanation:

Given that

radius ,r= 3.74 cm

Density of the milk ,ρ = 1.04 g/cm³ = 1.04 x 10⁻³ kg/cm³

Normal force ,N= 9.03 x 10⁻² N

The volume of the ball V

$V=\dfrac{4}{3}\pi r^3$

$V=\dfrac{4}{3}\times \pi \times 3.74^3\ cm^3$

V= 219.13 cm³

The bouncy force on the ball = Fb

Fb = ρ V g

Fb + N = m g

m=Mass of the ball = Density x volume

m = γ V , γ =Density of the Ball

ρ V g + N = γ V g ( take g= 10 m/s²)

$\gamma =\dfrac{N+\rho V g}{V g}$

$\gamma =\dfrac{9.03\times 10^{-2}+1.04\times 10^{-3}\times 219.13\times 10}{219.13\times 10}$

γ = 0.00108 kg/cm³

m = γ V

m = 0.00108 x 219.13

m= 0.23 kg

The mass of the ball is 0.23 kg

5 0

3 years ago

???whats the answers??

OverLord2011 [107]

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3 years ago

IN WHAT CONDITION DO SOUND ECHO

DerKrebs [107]

Answer:

The conditions necessary for hearing the echo. The distance between the sound source and the reflecting surface must not be less than 17 metres where the time period between hearing the original sound and its echo should not be less than 0.1 of a second.

6 0

2 years ago

What fraction of the water must evaporate to remove precisely enough energy to keep the temperature constant? water at 37°c has

mart [117]

The fraction of the water must evaporate to remove precisely enough energy to keep the temperature constant when water at 37°c has a latent heat of vaporization of lv = 580 kcal/kg is 2.58 times 10 to the minus 3.

Vaporization is the process by which a substance is transformed from its liquid or solid state into its gaseous (vapour) state. Boiling is the term for the vaporization process when conditions permit the creation of vapour bubbles within a liquid. Sublimation is the process of directly converting a solid to a liquid.

Boiling and evaporation are the two processes that cause vaporization. Evaporation is the process by which a liquid body's surface changes from a liquid to a gas, as in the case of a drop of water on hot concrete evaporating into a gas. A liquid is said to be boiling when it is heated to the point at which it begins to give off steam, as when you boil water on a stove. The process of converting a substance from its liquid or solid state into its gaseous (vapour) state is known as vaporization.

To learn more about vaporization please visit - brainly.com/question/12625048
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5 0

2 years ago

A block, M1=10kg, slides down a smooth, curved incline of height 5m. It collides elastically with another block, M2=5kg, which i

erma4kov [3.2K]

Answer:

2.86 m

Explanation:

Given:

M₁ = 10 kg

M₂ = 5 kg

$\mu_k$ = 0.5

height, h = 5 m

distance traveled, s = 2 m

spring constant, k = 250 N/m

now,

the initial velocity of the first block as it approaches the second block

u₁ = √(2 × g × h)

u₁ = √(2 × 9.8 × 5)

u₁ = 9.89 m/s

let the velocity of second ball be v₂

now from the conservation of momentum, we have

M₁ × u₁ = M₂ × v₂

on substituting the values, we get

10 × 9.89 = 5 × v₂

v₂ = 19.79 m/s

now,

let the velocity of mass 2 when it reaches the spring be v₃

from the work energy theorem, we have

Work done by the friction force = change in kinetic energy of the mass 2

$0.5\times5\times9.8\times2 = \frac{1}{2}\times5\times( v_3^2-19.79^2)$

v₃ = 20.27 m/s

now, let the spring is compressed by the distance 'x'

therefore, from the conservation of energy

we have

Energy of the spring = Kinetic energy of the mass 2

$\frac{1}{2}kx^2=\frac{1}{2}mv_3^2$

on substituting the values, we get

$\frac{1}{2}\times250\times x^2=\frac{1}{2}\times5\times20.27^2$

x = 2.86 m

8 0

3 years ago