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Scilla [17]
3 years ago
15

After a surface interval my pressure group is D. I want to do a repetitive dive to 38 feet for 32 minutes. What will be my new p

ressure group when I reach the surface?
Physics
2 answers:
ExtremeBDS [4]3 years ago
5 0

Answer:

New pressure group is N

Explanation:

The Repetitive Dive table has to be checked, the residual nitrogen for the maximum depth of 38 feet as a "D" diver, then add that to the actual dive time to get the total nitrogen time. The total nitrogen time and the maximum depth of 38 feet would put you on a pressure group N.

gladu [14]3 years ago
3 0

Answer: N

Explanation: Finding the new pressure group would involve looking into the Repetitive Dive Table, where we will be interested in finding the Residual Nitrogen time(RNT). The Residual Nitrogen Time value is the intersection of the pressure group D at 12m which according to the table is 26 min. This value '26' will be added to the next dive which is ; 32 min + 26min =58.

The intersection of value '58' at 12m is located in the table to find the next value of the pressure group. Which according to the table is N.

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Determine whether the interference is constructive or destructive at each location indicated.
OlgaM077 [116]

Answer:

A. Constructive

B. Destructive

C. Destructive

D. Constructive

Explanation:

Constructive interference takes place at locations along the path of two superposed waves where the waves are in phase such that a high or low point of one of the waves corresponds with a high or low point of the other wave which gives a resulting wave amplitude which is the sum of the amplitudes of the individual waves

Destructive interference takes place at locations along the path of two superposed waves where one wave is out of phase with the other wave such that a high or low point of one of the waves coincides with a low or high point of the other wave respectively thereby cancelling the effect of the other wave and giving a resulting wave that has an amplitude which is the difference in the amplitudes of the individual waves

Therefore;

At point A, the peak of each wave partially coincides resulting in constructive interference

At point B, the peak of the blue wave and the trough of the red wave partially coincides resulting in destructive interference

At point C, the through of the blue wave and the peak of the back wave partially coincides resulting in destructive interference

At point D, the trough of each wave partially coincides resulting in constructive interference.

6 0
2 years ago
Write all the different ways you can think of that describe what it means to be healthy
My name is Ann [436]

Answer:

Eat more healthy foods.  Workout and build your immune system.

Explanation:

Eat  Healthy foods like Carrots, Apples, Bannas, Pears, and anything that deals with not much of any sugar. An example of unhealthy foods is Cakes, Chocolates, Candy, and more. Drink a lot of water.

3 0
3 years ago
The pressure of a box pushes down on the floor is 50 Pa if the box weighs 400 N what is the area of the base of the box
Wewaii [24]

Answer:8m^2

Explanation:

Area=force÷pressure

Area=400÷50

A=8m^2

5 0
3 years ago
Help ASAP please & thank you ​
jasenka [17]

Answer:

I think its A plz tell me if im right

7 0
2 years ago
One ring of radius a is uniformly charged with charge +Q and is placed so its axis is the x-axis. A second ring with charge –Q i
kati45 [8]

Answer:

The force exerted on an electron is 7.2\times10^{-18}\ N

Explanation:

Given that,

Charge = 3 μC

Radius a=1 m

Distance  = 5 m

We need to calculate the electric field at any point on the axis of a charged ring

Using formula of electric field

E=\dfrac{kqx}{(a^2+x^2)^{\frac{3}{2}}}\hat{x}

E_{1}=\dfrac{kqx}{(a^2+x^2)^{\frac{3}{2}}}\hat{x}

Put the value into the formula

E_{1}=\dfrac{9\times10^{9}\times3\times10^{-6}\times5}{(1^2+5^2)^{\frac{3}{2}}}

E_{1}=1.0183\times10^{3}\ N/C

Using formula of electric field again

E_{2}=\dfrac{kqx}{(a^2+x^2)^{\frac{3}{2}}}\hat{x}

Put the value into the formula

E_{2}=\dfrac{9\times10^{9}\times(-3\times10^{-6})\times5}{((0.5)^2+5^2)^{\frac{3}{2}}}

E_{2}=-1.064\times10^{3}\ N/C

We need to calculate the resultant electric field

Using formula of electric field

E=E_{1}+E_{2}

Put the value into the formula

E=1.0183\times10^{3}-1.064\times10^{3}

E=-0.045\times10^{3}\ N/C

We need to calculate the force exerted on an electron

Using formula of electric field

E = \dfrac{F}{q}

F=E\times q

Put the value into the formula

F=-0.045\times10^{3}\times(-1.6\times10^{-19})

F=7.2\times10^{-18}\ N

Hence, The force exerted on an electron is 7.2\times10^{-18}\ N

8 0
3 years ago
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