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pishuonlain [190]
3 years ago
6

The deepest point of the pacific ocean is 11,033 m, in the mariana trench. what is the gauge pressure in the water at that point

? the density of seawater is 1025 kg/m3.
Physics
1 answer:
Liula [17]3 years ago
4 0

Given that,

Depth of seawater, h = 11,033 m

Density seawater, p (rho) = 1025 kg/m³

Gauge Pressure , P = ??

Since, we know that:

Pressure, P = pgh

Pressure = 1025 * 9.81 * 11033

Pressure = 1109395723.3 N/m²

or

Pressure = 1.1 x 10∧8 Pascal

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The rectangular plates in a parallel-plate capacitor are 0.063 m x 5.4 m. A distance of 3.5 x 10^-5m separate the plates. The pl
Aleks04 [339]

The capacitance of the capacitor is 1.8\cdot 10^{-9}F

Explanation:

The capacitance of a parallel-plate capacitor is given by the equation

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

where

k is the dielectric constant of the medium

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

A is the area of the plates

d is the separation between the plates

For the capacitor in this problem, we have:

k = 2.1 is the dielectric constant

d=3.5\cdot 10^{-5}m is the separation between the plates

A=0.063m \cdot 0.054m = 0.0034 m^2 (I assumed that 5.4 m is a typo, since it is not a realistic size for the side of the plate)

Therefore, the capacitance of the  capacitor is

C=\frac{(2.1)(8.85\cdot 10^{-12})(0.0034)}{3.5\cdot 10^{-5}}=1.8\cdot 10^{-9}F

Learn more about capacitors:

brainly.com/question/10427437

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5 0
3 years ago
What is the magnitude of the electric field at a distance of 89 cm from a 27 μC charge, in units of N/C?
GuDViN [60]

Answer:

306500 N/C

Explanation:

The magnitude of an electric field around a single charge is calculated with this equation:

E(r) = \frac{1}{4 \pi *\epsilon 0} \frac{q}{r^2}

With ε0 = 8.85*10^-12 C^2/(N*m^2)

Then:

E(0.89) = \frac{1}{4 \pi *8.85*10^-12} \frac{27*10^-6}{0.89^2}

E(0.89) = 306500 N/C

3 0
3 years ago
To avoid using the variable "a"
sdas [7]

Answer:

option c .

Explanation:

..............................

6 0
2 years ago
A plane cruising at 233 m/s accelerates at 17 m/s 2 for 4.8 s. What is its final velocity? Answer in units of m/s. 013 (part 2 o
Volgvan

Answer:

Final velocity will be 314.6 m/sec

Distance traveled = 1314.24 m

Explanation:

We have given initial velocity u = 233 m/sec

Acceleration a=17m/sec^2

Time t = 4.8 sec

From first equation of motion v=u+at, here v is final velocity, u is initial velocity and t is time

So v=233+17\times 4.8=314.6m/sec

Now we have to find distance traveled

From second equation of motion

S=ut+\frac{1}{2}at^2=233\times 4.8+\frac{1}{2}\times 17\times 4.8^2=1314.24m

So distance traveled in given time will be 1314.24 m

4 0
3 years ago
An unstrained horizontal spring has a length of 0.36 m and a spring constant of 320 N/m. Two small charged objects are attached
lakkis [162]

Answer:

1.been both -ve charged or both +be charged particles

2. 3.52mC

Explanation:

For the charge particle to cause an extension or movement of the string from its unrestrained position they would have been both -ve charged or both +be charged particles that's because like charges repel.

Now the Force sustain by the extended string is

F = Ke;

Where K is the force constant of the string, 320 N/m

e is the extension,0.033 m

F = 320 × 0.033 =10.56N

2.But according to columns law of charge;

F = kQ1 Q2

But Q1=Q2{ since the charge are of the same magnitude}.

Hence F = KQ^2

Where K is columns constant =9×10^9F/m

Hence Q=√F/K

Q= √10.56/9×10^9

=3.52×10^-3C

= 3.52mC

8 0
3 years ago
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