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Nezavi [6.7K]
2 years ago
10

00

Physics
1 answer:
zhuklara [117]2 years ago
4 0

Answer:

In a column of fluid, pressure increases with depth as a result of the weight of the overlying fluid. Thus a column of fluid, or an object submerged in the fluid, experiences greater pressure at the bottom of the column than at the top. This difference in pressure results in a net force that tends to accelerate an object upwards.

The pressure at a depth in a fluid of constant density is equal to the pressure of the atmosphere plus the pressure due to the weight of the fluid, or p = p 0 + ρ h g , p = p 0 + ρ h g , 14.4

Granite: 2.70 × 10 32.70 × 10 3

Lead: 1.13 × 10 41.13 × 10 4

Iron: 7.86 × 10 37.86 × 10 3

Oak: 7.10 × 10 27.10 × 10 2

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Why aren't humans as evolved as we think we are?
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3 years ago
A parallel-plate air capacitor is made from two plates 0.210 m square, spaced 0.815 cm apart. it is connected to a 120 v battery
GuDViN [60]

Answer:

at the beginning: 2.3\cdot 10^{-10} F

when the plates are pulled apart: 1.1\cdot 10^{-10} F

Explanation:

The capacitance of a parallel-plate capacitor is given by

C=k \epsilon_0 \frac{A}{d}

where

k is the relative permittivity of the medium (for air, k=1, so we can omit it)

\epsilon_0 = 8.85\cdot 10^{-12} F/m is the permittivity of free space

A is the area of the plates of the capacitor

d is the separation between the plates

In this problem, we have:

A=0.210 m^2 is the area of the plates

d=0.815 cm=8.15\cdot 10^{-3} m is the separation between the plates at the beginning

Substituting into the formula, we find

C=(1)(8.85\cdot 10^{-12}F/m)\frac{0.210 m^2}{8.15\cdot 10^{-3} m}=2.3\cdot 10^{-10} F

Later, the plates are pulled apart to d=1.63 cm=0.0163 m, so the capacitance becomes

C=(1)(8.85\cdot 10^{-12}F/m)\frac{0.210 m^2}{0.0163 m}=1.1\cdot 10^{-10} F

4 0
3 years ago
A 0.500-kg block slides up a plane inclined at a 30° angle. If it slides 1.50 m before coming to rest while encountering a frict
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B it’s Intail velocity
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A car is 200 m from a stop sign and traveling toward the sign at 40.0 m/s. At this time, the driver suddenly realizes that she m
victus00 [196]

Answer:

The acceleration of the car will be a=9600m/sec^

Explanation:

We have given that distance from stop sign s = 200 m

Time t = 0.2 sec

We have to find the constant acceleration

Now from second equation of motion s=ut+\frac{1}{2}at^2

200=40\times 0.2+\frac{1}{2}\times a\times 0.2^2

a=9600m/sec^

So the acceleration of the car will be a=9600m/sec^

6 0
3 years ago
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