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3 years ago

How can we relate density and pressure in liquids ??

Physics

2 answers:

Masteriza [31]3 years ago

6 0

Pressure increases with increasing depth. h2=2hh

Otrada [13]3 years ago

5 0

<u>How can we relate density and pressure in liquids ??</u>

$\huge{ \texttt{ \red{answers}}}$

It is the pressure at each point of a static fluid. When we immerse ourselves in a liquid, the pressure increases with depth.
In the case of a liquid (constant density) the pressure increases linearly with depth.

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$\bold{BRAINLY MENTALMENTE}$

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The cricket ball has a mass of 0.16kg and it hits the bat with a speed of 25 m/s. After being in contact with the bat for 0.0013

zavuch27 [327]

change in velocity is 3m/s,the force exterted by bat on ball is 4.8N and the acceleration is 30m/s².

<h3>Give some difference between velocity and speed.</h3>

1) Speed is scalar quantity but velocity is vector quantity.

2) Speed is distance followed by an individual with respect to time and velocity is displacement with respect to time .

1) final velocity- intial velocity

25-22= 3m/s

2) The force is

F= ma

. 0.16×30= 4.8N

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7 0

1 year ago

PLEASE help me with these questions as soon as possible i need to turn it in!!!!

Sloan [31]

A. I believe this might be the answer but
6 J/60 J= .1
.1 x 100= 10%

b. I'm not sure of this question

19. I'm not sure if I did this correct way or if there is a better way (I'm rusty)
300 J/.50 %= 600
600/100= 6 (answer)

20. (200 J)(.75%) = 150 (answer)
Check:
Efficiency: output/input x 100
150 J/200 J= .75
.75 x 100= 75%

3 0

3 years ago

By means of a rope whose mass is negligible, two blocks are suspended over a pulley, as the drawing shows. The pulley can be tre

tankabanditka [31]

Answer:

Mass of the pull is 77 kg

Explanation:

Here we have for

Since the rope moves along with pulley, we have

For the first block we have

T₁ - m₁g = -m₁a = -m₁g/4

T₁ = 3/4(m₁g) = 323.4 N

Similarly, as the acceleration of the second block is the same as the first block but in opposite direction, we have

T₂ - m₂g = m₂a = m₂g/4

T₂ = 5/4(m₂g) = 134.75 N

T₂r - T₁r = I·∝ = 0.5·M·r²(-α/r)

∴ $M = -\frac{2}{a} (T_2-T_1)$

$M = -\frac{2}{2.45} (134.75-323.4) = 77 \, kg$

Mass of the pull = 77 kg.

5 0

3 years ago

A bicycile traves with an average velocty 15km/h north for 20min what is his displasment

Mila [183]

Answer:

5000 m equivalent to 5 Km

Explanation:

Average velocity = $\frac{Displacement}{Time}$

so, Displacement =Average velocity × time

We should convert Km/h to m/s so Km/h ⇒ $\frac{1000}{60*60}$ m/s, also convert time to second so, 20min ⇒(20* 60)seconds

Displacement = (15 × $\frac{1000}{60*60}$ ) × (20×60) =5000m OR 5Km

6 0

2 years ago

Three 20.0 ohm resistors are

V125BC [204]

Answer:

6.67 ohm

Explanation:

From the question given above, the following data were obtained:

Resistor 1 (R₁) =20 ohm

Resistor 2 (R₂) = 20 ohm

Resistor 3 (R₃) = 20 ohm

Equivalent Resistance (R) =?

Since the resistors are arranged in parallel connection, the equivalent resistance can be obtained as follow:

1/R = 1/R₁ + 1/R₂ + 1/R₃

1/R = 1/20 + 1/20 + 1/20

1/R = (1 + 1 + 1) / 20

1/R = 3/20

Invert

R = 20/3

R = 6.67 ohm

Therefore, the equivalent resistance is 6.67 ohm.

5 0

3 years ago