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2 years ago

Derive formula for pressure in any liquid

Physics

1 answer:

mestny [16]2 years ago

7 0

Answer:

Besides, we often calculate pressure for gases and fluids. In those conditions, the pressure of liquid or gas is equal to the density of that fluid multiplied by the acceleration due to the gravity and the height (depth) of the fluid above a certain point. Pressure = density of fluid × acceleration due to gravity × height of the fluid column

Explanation:

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A sled of mass 50 kg is pulled along a snow-covered, flat ground. The static friction coefficient is 0.3 and the kinetic frictio

Diano4ka-milaya [45]

Answer:

a) We kindly invite you to see below the Free Body Diagram of the forces acting on the sled.

b) The weight of the sled is 490.35 newtons.

c) A force of 147.105 newtons is needed to start the sled moving.

d) A force of 49.035 newtons is needed to keep the sled moving at a constant velocity.

Explanation:

a) We kindly invite you to see below the Free Body Diagram of the forces acting on the sled. All forces are listed:

$F$ - External force exerted on the sled, measured in newtons.

$f$ - Friction force, measured in newtons.

$N$ - Normal force from the ground on the mass, measured in newtons.

$W$ - Weight, measured in newtons.

b) The weight of the sled is determined by the following formula:

$W = m\cdot g$ (1)

Where:

$m$ - Mass, measured in kilograms.

$g$ - Gravitational acceleration, measured in meters per square second.

If we know that $m = 50\,kg$ and $g = 9.807\,\frac{m}{s^{2}}$ , the weight of the sled is:

$W = (50\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)$

$W = 490.35\,N$

The weight of the sled is 490.35 newtons.

c) The minimum force needed to start the sled moving on the horizontal ground is:

$F_{min,s} = \mu_{s}\cdot W$ (2)

Where:

$\mu_{s}$ - Static coefficient of friction, dimensionless.

$W$ - Weight of the sled, measured in newtons.

If we know that $\mu_{s} = 0.3$ and $W = 490.35\,N$ , then the force needed to start the sled moving is:

$F_{min,s} = 0.3\cdot (490.35\,N)$

$F_{min,s} = 147.105\,N$

A force of 147.105 newtons is needed to start the sled moving.

d) The minimum force needed to keep the sled moving at constant velocity is:

$F_{min,k} = \mu_{k}\cdot W$ (3)

Where $\mu_{k}$ is the kinetic coefficient of friction, dimensionless.

If we know that $\mu_{k} = 0.1$ and $W = 490.35\,N$ , then the force needed to keep the sled moving at a constant velocity is:

$F_{min,k} = 0.1\cdot (490.35\,N)$

$F_{min,k} = 49.035\,N$

A force of 49.035 newtons is needed to keep the sled moving at a constant velocity.

8 0

3 years ago

A wire 3.22 m long and 7.32 mm in diameter has a resistance of 11.9 mΩ. A potential difference of 33.7 V is applied between the

Scorpion4ik [409]

Answer:

(a) Current is 2831.93 A

(b) $8.40A/m^2$

Explanation:

Length of wire l = 3.22 m

Diameter of wire d = 7.32 mm = 0.00732 m

Cross sectional area of wire

$A=\pi r^2=3.14\times 0.00366^2=4.20\times 10^{-5}m^2$

Resistance $R=11.9mohm=11.9\times 10^{-3}ohm$

Potential difference V = 33.7 volt

(A) current is equal to

$i=\frac{V}{R}=\frac{33.7}{11.9\times 10^{-3}}=2831.93A$

(B) Current density is equal to

$J=\frac{i}{A}$

$J=\frac{2831.93}{4.20\times 10^{-5}}=8.40A/m^2$

$R=\frac{\rho l}{A}$

$11.9\times 10^{-3}=\frac{\rho \times 3.22}{4.20\times 10^{-5}}$

$\rho =15.52\times 10^{-9}ohm-m$

4 0

3 years ago

If a 1-megaton nuclear weapon is exploded

Elenna [48]

Answer:

1,780,000 N

Explanation:

0.2 atm × (1.013×10⁵ Pa/atm) = 20,260 Pa

Force = pressure × area

F = 20,260 Pa × (3.89 m × 22.6 m)

F = 1,780,000 N

4 0

2 years ago

A big wheel at a theme park has a diameter of 14m and people on the ride complete one revolution in 24s. calculate the distance

just olya [345]

Explanation:

We'll call the radius r and the diameter d:

We also assume that the riders are at a distance r = d/2 = 7m from the center of the wheel.

The period of the wheel is 24s. The tangent velocity of the wheel (and the riders) will be: (2pi/T)*r = 0.8 m/s (circa).

It means that in 3 minutes (180 seconds) they'll run 0.8 m/s * 180s = 144m.

Hopefully I understood the question. If yes, that's the answer.

7 0

2 years ago

What is 1.0 x 10^9 in standard form?

jasenka [17]

1.0 x 10^9= 1000000000

3 0

3 years ago