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nikdorinn [45]
3 years ago
5

The pressure, p, of water (in pounds per square foot) at a depth of d feet below the surface is given by the formula p=15+15/33d

. If a diver is 100 feet below surface, what is the pressure of the water on the diver?
a.15.0 pounds per square feet
b.38.63 pounds per square feet
c.1545.45 pounds per square feet
d.60.45 pounds per square feet
Physics
2 answers:
Viefleur [7K]3 years ago
8 0
The pressure of the water on the diver is given in an expression written as:

<span>p=15+15/33d

where p is the pressure and d is the distance of the diver </span><span>below the surface.

The pressure is calculated as follows:

</span>p=15+15/33(100) = 15.00  pounds per square feet

Therefore, the correct answer is option A.
Nadusha1986 [10]3 years ago
6 0

Answer:

CCCCCCCCCCC

Explanation:

The rate of change for the diver is

8.6 − 4.3

20 − 10

= 0.43. Therefore, the rate of change for the submarine is 4(0.43) = 1.72 and p = 1.72d

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Answer:

satisfaction, enjoyment and fair play

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3 years ago
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What is the acceleration of a 1000 kg car subject to a 550 N net force
Xelga [282]

Answer:

The acceleration of a 1000 kg car subject to a 550 N net force = 0.55 m/s^2

Explanation:

Given:

F = 550 N

m = 1000 kg

To Find:

a = ?

Solution:

So by the equation by Newton's 2nd Law of Motion,

F = m x a

550 N = 1000 kg x a

a = 550 N/ 1000 kg

a = 0.55 m/s^2

Therefore,

The acceleration of a 1000 kg car subject to a 550 N net force = 0.55 m/s^2

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4 0
3 years ago
An orange light (f = 5.2 * 10'4Hz) is
KonstantinChe [14]

Answer:

2.145×10^-10 V or 0.2145nV

Explanation:

From hf=eV

h= Plank's constant = 6.6×10^-34JS

f= frequency of the electromagnetic wave = 5.2×10^4 Hz

e= electronic charge= 1.6×10^-19 C

V= voltage

V= hf/e

V= 6.6×10^-34JS × 5.2×10^4 Hz/ 1.6×10^-19 C

V= 2.145×10^-10 V or 0.2145nV

Therefore the voltage created is 2.145×10^-10 V or 0.2145nV

7 0
3 years ago
T=2pi square root 1/g solve for g.<br> Explanation would be really helpful.
Natalija [7]

I added individual steps for clarity. Note that g must be positive if the solution is to be real.

T=2\pi \sqrt{\frac{1}{g}}=2\pi g^{-\frac{1}{2}}\\g^{-\frac{1}{2}} = \frac{T}{2\pi}\\(g^{-\frac{1}{2}})^{-2} = (\frac{T}{2\pi})^{-2}\\g = \frac{4\pi^2}{T^2}\,\,\,, g>0}

Let me know if you have any questions.

7 0
3 years ago
An object is 39 cm away from a concave mirrors surface along the principles axis. If the mirrors focal length is 9.50 cm, how fa
Tatiana [17]

Answer:

12.6 cm

Explanation:

We can use the mirror equation to find the distance of the image from the mirror:

\frac{1}{f}=\frac{1}{p}+\frac{1}{q}

where here we have

f = 9.50 cm is the focal length

p = 39 cm is the distance of the object from the mirror

Solving the equation for q, we find:

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{9.50 cm}-\frac{1}{39 cm}=0.080 cm^{-1}\\q = \frac{1}{0.080 cm}=12.6 cm

5 0
3 years ago
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