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Radda [10]
2 years ago
15

A U-shaped tube, open to the air on both ends, contains mercury. Water is poured into the left arm until the water column is 10.

0 cm deep. How far upward from its initial position does the mercury in the right arm rise?
Physics
1 answer:
Masja [62]2 years ago
8 0

Answer:

0.368 cm

Explanation:

x = distance by which the mercury rise

d = depth of the water = 10 cm = 0.10 m

ρ = density of water = 1000 kgm⁻³

ρ' = density of mercury = 13600 kgm⁻³

P₀ = atmospheric pressure

Using equilibrium of pressure on both side

P₀ + ρ g d = P₀ + ρ' g (2x)

(1000) (0.10) = (13600) (2x)

x = 0.00368 m

x = 0.368 cm

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A person throws a ball with a force of 140 N that accelerates at 15 m/s? What is the mass of the
mr_godi [17]

Newton's 2nd law

\tt \sum F=m.a

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2 years ago
An empty parallel plate capacitor is connected between the terminals of a 18.8-V battery and charges up. The capacitor is then d
Basile [38]

Answer:

p.d' = 37.6 V

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New Capacitor C_1=C_2/2

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C=\frac{eA}{d}

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 p.d' = 37.6 V

7 0
3 years ago
What is the equivalent resistance of a circuit that contains three 10.0 12
lana66690 [7]

Answer:

Explanation:

The equivalent resistance for three resistors connected in parallel is given as

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now we.need to.insert the value of 3 resistances but only 2 are given in the question.

3 0
2 years ago
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The moment of inertia of a uniform-density disk rotating about an axle through its center can be shown to be . This result is ob
Naddik [55]

(a) 0.2888 kg m^2

The moment of inertia of a uniform-density disk is given by

I=\frac{1}{2}MR^2

where

M is the mass of the disk

R is its radius

In this problem,

M = 16 kg is the mass of the disk

R = 0.19 m is the radius

Substituting into the equation, we find

I=\frac{1}{2}(16 kg)(0.19 m)^2=0.2888 kg m^2

(b) 142.5 J

The rotational kinetic energy of the disk is given by

K=\frac{1}{2}I\omega^2

where

I is the moment of inertia

\omega is the angular velocity

We know that the disk makes one complete rotation in T=0.2 s (so, this is the period). Therefore, its angular velocity is

\omega=\frac{2\pi}{T}=\frac{2\pi}{0.2 s}=31.4 rad/s

And so, the rotational kinetic energy is

K=\frac{1}{2}(0.2888 kg m^2)(31.4 rad/s)^2=142.5 J

(c) 9.07 kg m^2 /s

The rotational angular momentum of the disk is given by

L=I\omega

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Substituting the values found in the previous parts of the problem, we find

L=(0.2888 kg m^2)(31.4 rad/s)=9.07 kg m^2 /s

8 0
3 years ago
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