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Grace [21]
2 years ago
9

A scuba diver stays underwater for 30 minutes. How is this possible when the air tank is so small?​

Physics
1 answer:
irina1246 [14]2 years ago
6 0

Answer:

Here's the answer:

Explanation:

An average open-water certified diver using a standard aluminium 80-cubic-foot tank on a 40-foot dive will be able to stay down for about 45 to 60 minutes before surfacing with a safe reserve of air still in the tank.

Three Factors That Determine How Long a Diver's Air Will Last:

1. Tank Volume

2. Depth

3. Air Consumption Rate

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1. An object in free fall will have an initial velocity equal to zero when: a. It is thrown vertically down
Ne4ueva [31]

Answer:

b. It is dropped

Explanation:

If the initial velocity is zero, the object move from rest. That happens if the object is dropped

6 0
3 years ago
a ball rolls along the floor with a constant velocity of 3 m/s. How far will it have gone after 234 seconds?
sergij07 [2.7K]

Answer:

The ball travel for 702m

Explanation:

distance = speed × time

speed = 3m/s

time = 234s

distance = 3 × 234

= 702m

7 0
2 years ago
Read 2 more answers
Predict the deformation or elongation of a spring that has a constant of elasticity of 400 N/m when a force of 75 N is applied i
morpeh [17]

Answer:

Explanation:

Give that,

Spring constant (k)=40N/m

Force applied =75N

Since the force is applied to the right, we don't know if it is compressing or stretching the spring

So let assume it compress

Using hooke's law

F=-ke

e=-F/k

Then, e=-75/40

e=-1.875m

The deformation is 1.875m.

Let assume it stretch

Using hooke's law

-F=-ke

e=F/k

Then, e=75/40

e=1.875m

The elongation is 1.875m

3 0
3 years ago
A force of 660 n stretches a certain spring a distance of 0.300 m. what is the potential energy of the spring when a 70.0 kg mas
kkurt [141]
A force of 660 n stretches a certain spring a distance of 0.300 m. what is the potential energy of the spring when a 70.0 kg mass hangs vertically from it?
5 0
3 years ago
If you could shine a very powerful flashlight beam toward the Moon, estimate the diameter of the beam when it reaches the Moon.
grin007 [14]

To develop this problem it is necessary to apply the Rayleigh Criterion (Angular resolution)criterion. This conceptos describes the ability of any image-forming device such as an optical or radio telescope, a microscope, a camera, or an eye, to distinguish small details of an object, thereby making it a major determinant of image resolution.  By definition is defined as:

\theta = 1.22\frac{\lambda}{d}

Where,

\lambda= Wavelength

d = Width of the slit

\theta= Angular resolution

Through the arc length we can find the radius, which would be given according to the length and angle previously described.

The radius of the beam on the moon is

r = l\theta

Relacing \theta

r = l(\frac{1.22\lambda}{d})

r = 1.22\frac{l\lambda}{d}

Replacing with our values we have that,

r = 1.22*(\frac{(384*10^3km)(\frac{1000m}{1km})(550*10^{-9}m)}{7*10^{{-2}}})

r = 3680.91m

Therefore the diameter of the beam on the moon is

d = 2r

d = 2 * (3690.91)

d = 7361.8285m

Hence, the diameter of the beam when it reaches the moon is 7361.82m

8 0
3 years ago
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