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kirill [66]
3 years ago
7

g The bottom end of a long vertical tube filled with liquid is opened in a basin exposed to air having pressure 100.8 kilo-Pasca

ls. The top end of the tube remains closed. The maximum height of liquid in the tube that this air pressure can support is 10.5 meters. Determine the density of the liquid in units of bevelled fraction numerator k g over denominator m cubed end fraction.
Physics
1 answer:
Masja [62]3 years ago
3 0

Answer:

979.6 kg/m³

Explanation:

We know pressure P = hρg where h = height of liquid = 10.5 m, ρ = density of liquid and g = acceleration due to gravity = 9.8 m/s²

So, density ρ = P/hg

Since P = 100.8 kPa = 100.8 × 10³ Pa

substituting the values of the variables into the equation for ρ, we have

ρ = P/hg

= 100.8 × 10³ Pa ÷ (10.5 m × 9.8 m/s²)

= 100.8 × 10³ Pa ÷ 102.9 m²/s²

= 0.9796 × 10³ kg/m³

= 979.6 kg/m³

So, the density of the liquid is 979.6 kg/m³

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A boy reaches out of a window and tosses a ball straight up with a speed of 10 m/s. The ball is 20 m above the ground as he rele
kirill115 [55]

Explanation:

It is given that,

Speed of the ball, v = 10 m/s

Initial position of ball above ground, h = 20 m

(a) Let H is the maximum height reached by the ball. It can be calculated using the conservation of energy as :

\dfrac{1}{2}mv^2=mgh'            

h'=\dfrac{v^2}{2g}

h'=\dfrac{(10)^2}{2\times 9.8}

h' = 5.1 m

The maximum height above ground,

H = 5.1 + 20

H = 25.1 meters

So, the maximum height reached by the ball is 25.1 meters.

(b) The ball's speed as it passes the window on its way down is same as the initial speed i.e. 10 m/s.

Hence, this is the required solution.                              

6 0
3 years ago
Chegg Given that the mean radius of the Moon’s orbit is 3.84 x 108 m and its period is 2.36 x 106 sec, at what altitude above th
alukav5142 [94]

Answer:

The altitude of geostationary satellite is 3.58\times10^{7}\ m

Explanation:

Given that,

Radius of moon's orbit r=3.84\times10^{8}\ m

Time period T=2.36\times10^{6}\ sec

We need to calculate the orbital radius of geostationary satellite is

Using formula of time period

T=\sqrt{\dfrac{4\pi^2}{GM}a^3}

a=((\dfrac{GM}{4\pi^2})T^2)^{\dfrac{1}{3}}

Where, G = gravitational constant

M = Mass of earth

T = time period of geostationary satellite orbit

Put the value in to the formula

a=((\dfrac{6.67\times10^{-11}\times5.97\times10^{24}}{4\times\pi^2})\times(86160)^2)^{\dfrac{1}{3}}

a=4.217\times10^{7}\ m

We need to calculate the altitude of geostationary satellite

Using formula of altitude

h = a-R_{e}

Where, R = radius of earth

a = radius of geostationary satellite

Put the value into the formula

h =4.217\times10^{7}-6.38\times10^{6}

h =35790000\ m

h=3.58\times10^{7}\ m

Hence, The altitude of geostationary satellite is 3.58\times10^{7}\ m

4 0
4 years ago
Closing Summary Questions: Phases of Matter
snow_tiger [21]

Answer:

Explanation:

Particles in all states of matter are in constant motion and this is very rapid at room temperature. A rise in temperature increases the kinetic energy and speed of particles; it does not weaken the forces between them. The particles in solids vibrate about fixed positions; even at very low temperatures.

Even with all of these state changes, it is important to remember that the substance stays the same—it is still water, which consists of two hydrogen atoms and one oxygen atom. Changing states of matter are only physical changes; the chemical properties of the matter stays the same regardless of its physical state!

5 0
3 years ago
A 6000 kg roller coaster goes around a loop of radius 30 m at 6 m/s. What is the centripetal acceleration?
Margarita [4]

Answer:

The answer to the question is 7200

7 0
3 years ago
A train travels at a speed of 30 m/s. The train starts at an initial position of 1000 meters and travels for 30 seconds. What is
pychu [463]
1000 + 30x30 = 1900. Hope that helps
6 0
3 years ago
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