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3 years ago

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in a Mercury thermometer the level of Mercury Rises when its bulb comes in contact with a hot object what is the reason for this

rise in the level of Mercury

1 answer:

olasank [31]3 years ago

3 0

Answer:

because thermometric liquid readily expands on heating or contracts on cooling even for a small difference in the temperature of the body.

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A disk with radius R and uniform positive charge density s lies horizontally on a tabletop. A small plastic sphere with mass M a

Yanka [14]

Answer:

a. F = Qs/2ε₀[1 - z/√(z² + R²)] b. h = (1 - 2mgε₀/Qs)R/√[1 - (1 - 2mgε₀/Qs)²]

Explanation:

a. What is the magnitude of the net upward force on the sphere as a function of the height z above the disk?

The electric field due to a charged disk with surface charge density s and radius R at a distance z above the center of the disk is given by

E = s/2ε₀[1 - z/√(z² + R²)]

So, the net force on the small plastic sphere of mass M and charge Q is

F = QE

F = Qs/2ε₀[1 - z/√(z² + R²)]

b. At what height h does the sphere hover?

The sphere hovers at height z = h when the electric force equals the weight of the sphere.

So, F = mg

Qs/2ε₀[1 - z/√(z² + R²)] = mg

when z = h, we have

Qs/2ε₀[1 - h/√(h² + R²)] = mg

[1 - h/√(h² + R²)] = 2mgε₀/Qs

h/√(h² + R²) = 1 - 2mgε₀/Qs

squaring both sides, we have

[h/√(h² + R²)]² = (1 - 2mgε₀/Qs)²

h²/(h² + R²) = (1 - 2mgε₀/Qs)²

cross-multiplying, we have

h² = (1 - 2mgε₀/Qs)²(h² + R²)

expanding the bracket, we have

h² = (1 - 2mgε₀/Qs)²h² + (1 - 2mgε₀/Qs)²R²

collecting like terms, we have

h² - (1 - 2mgε₀/Qs)²h² = (1 - 2mgε₀/Qs)²R²

Factorizing, we have

[1 - (1 - 2mgε₀/Qs)²]h² = (1 - 2mgε₀/Qs)²R²

So, h² = (1 - 2mgε₀/Qs)²R²/[1 - (1 - 2mgε₀/Qs)²]

taking square-root of both sides, we have

√h² = √[(1 - 2mgε₀/Qs)²R²/[1 - (1 - 2mgε₀/Qs)²]]

h = (1 - 2mgε₀/Qs)R/√[1 - (1 - 2mgε₀/Qs)²]

4 0

3 years ago

a watermelon is balanced by a dog, a pumpkin, a flowerpot, and a baseball as shown below. Is the weight of the watermelon equal

alukav5142 [94]

Answer:

pumpkin

Explanation:

watermelon and pumpkins are close to shape and size

6 0

3 years ago

Read 2 more answers

Another name for the skin as a whole.

sergejj [24]

Organ, integument, dermis

5 0

3 years ago

a car moving at 11 m/s crashes into an obstacle and stops in 0.26s. compute the Force that a seatbelt exerts on a 21-kg child to

Natalka [10]

Answer:

890 N

Explanation:

Acceleration is change in velocity over change in time.

a = Δv / Δt

a = (11 m/s − 0 m/s) / 0.26 s

a = 42.3 m/s²

Force is mass times acceleration.

F = ma

F = (21 kg) (42.3 m/s²)

F ≈ 890 N

4 0

3 years ago

A boy throws a ball of mass 0.22 kg straight upward with an initial speed of 29 m/s. When the ball returns to the boy, its speed

maksim [4K]

Answer:

The work is -67.76 J

Explanation:

The law of conservation of energy is considered one of one of the fundamental laws of physics and states that the total energy of an isolated system remains constant. except when it is transformed into other types of energy.

This is summed up in the principle that energy can neither be created nor destroyed in the universe, only transformed into other forms of energy.

In this case you must calculate the loss of kinetic energy. This loss is actually the work done against the resistive force in the air. Friction is the only force other than gravity that acts on the ball.

So, the loss of kinetic energy is $\frac{1}{2} *m*(vf^{2} -vi^{2} )$

You know:

mass=m=0.22 kg
Initial velocity of the ball: $vi= 29 \frac{m}{s}$

Final velocity of the ball: $vf= 15 \frac{m}{s}$

Replacing:

$\frac{1}{2} *0.22 kg*(15^{2} -29^{2} )$ = -67.76 J

Friction work is always negative because friction is always against displacement.

<u><em>The work is -67.76 J</em></u>

5 0

3 years ago