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Fed [463]
3 years ago
10

Why is mercury not suitable as the liquid in the U-tube? (the relationship between density of the liquid and the pressure in the

liquid)​
Physics
1 answer:
adoni [48]3 years ago
8 0
Mercury Is more suitable than water cause it is denser than water.

Relation between density and pressure

Pressure = force/area

Replace force by mass * acceleration
Pressure = mass * acceleration/ area

We can replace mass by density * volume since density = mass/ volume
Pressure = density * volume * acceleration/ area

Volume is length^3 whereas area is length ^2. So volume / area = length

Pressure = density * length * acceleration

Gravity is form of acceleration so

Pressure = density * length * gravity
Length can be height or depth

Finally pressure = density * height * gravity

P = ρ g h => pressure of liquid
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If two balls have the same volume,
Lena [83]

Here, we are required to find the relationship between balls of different mass(a measure of weight) and different volumes.

  • 1. Ball A will have the greater density
  • 2. Ball C and Ball D have the same density.
  • 3. Ball Q will have the greater density.
  • 4. Ball X and Y will have the same density

The density of an object is given as its mass per unit volume of the object.

Mathematically;.

  • Density = Mass/Volume.

For Case 1:

  • Va = Vb and Ma = 2Mb
  • D(b) = (Mb)/(Vb) and D(a) = 2(Mb)/Vb
  • Therefore, the density of ball A,
  • D(a) = 2D(b).
  • Therefore, ball A has the greater density.

For Case 2:

  • Vc = 3Vd,

  • Vd = (1/3)Vc

  • Md = (1/3)Mc

  • D(c) = (Mc)/(Vc) and D(d) = (1/3)Md/(1/3)Vd

  • D(c) = D(d).

  • Therefore, ball C and D have the same density

For Case 3:

  • Vp = 2Vq and Mp = Mq
  • D(p) = (Mq)/2(Vq) and D(q) = (Mq)/Vq
  • Therefore, the density of ball P is half the density of ball Q
  • Therefore, ball Q has the greater density.

For case 4:

  • Mx = (1/2)My
  • Vx = Vy

Therefore, Ball X and Ball Y have the same density.

Read more:

brainly.com/question/18110802

8 0
2 years ago
The center of the Hubble space telescope is 6940 km from Earth’s center. If the gravitational force between Earth and the telesc
Law Incorporation [45]
The gravitational force between two objects is given by:
F=G \frac{m_1 m_2}{r^2}
where
G is the gravitational constant
m1 and m2 are the masses of the two objects
r is the separation between the two objects

The distance of the telescope from the Earth's center is r=6940 km=6.94 \cdot 10^6 m, the gravitational force is F=9.21 \cdot 10^4 N and the mass of the Earth is m_1=5.98 \cdot 10^{24} kg, therefore we can rearrange the previous equation to find m2, the mass of the telescope:
m_2 =  \frac{Fr^2}{Gm_1}= \frac{(9.21 \cdot 10^4 N)(6.94\cdot 10^6)^2}{(6.67\cdot 10^{-11})(5.98\cdot 10^{24})} =11121 kg
6 0
3 years ago
Read 2 more answers
What is the direction of the force for a negative charge moving downward in a magnetic field pointing to the left?
velikii [3]

Answer: A

Out of the screen

Explanation:

Using right hand rule, the magnetic force is perpendicular to the plane form by the magnetic field of a charged particle and its speed. Which will be into the screen.

But the negative charged particle moves in the opposite direction of the positive charged particle. Therefore, the magnetic force direction will be out of the screen

5 0
3 years ago
The current through a 10 ohm resistor connected to a 120 volt power supply is
Leno4ka [110]

Answer:I=12 A

Explanation:

Given

Resistance R=10 \Omega

Voltage V=120 V

According to ohm's law current through a conductor is directly proportional to the voltage applied.

V\propto I

V=IR

where V=Voltage

I=Current

R=resistance

I=\frac{V}{R}

I=\frac{120}{10}

I=12 A

4 0
2 years ago
On a straight, level, two-lane road, two cars moving in opposite directions approach and pass each other. Car A is in the eastbo
ludmilkaskok [199]

Answer:

a) 42 m/s, positive direction (to the east), b) 42 m/s, negative direction (to the west).

Explanation:

a) Let consider that Car A is moving at positive direction. Then, the relative velocity of Car A as seen by the driver of Car B is:

\vec v_{A/B} = \vec v_{A} - \vec v_{B}\\\vec v_{A/B} = 11 \frac{m}{s} \cdot i + 31 \frac{m}{s} \cdot i\\\vec v_{A/B} = 42 \frac{m}{s} \cdot i

42 m/s, positive direction (to the east).

b) The relative velocity of Car B as seen by the drive of Car A is:

\vec v_{B/A} = \vec v_{B} - \vec v_{A}\\\vec v_{B/A} = -31 \frac{m}{s} \cdot i - 11 \frac{m}{s} \cdot i\\\vec v_{B/A} = - 42 \frac{m}{s} \cdot i

42 m/s, negative direction (to the west).

5 0
3 years ago
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