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NemiM [27]
3 years ago
9

While in a stream 39 cm deep, they look down into the water and see a craw fish at the bottom. How deep does the stream appear t

o the student? (nwater = 1.33)
Physics
1 answer:
Helga [31]3 years ago
7 0

Answer:

The  depth of stream to the student is  d_1  =  0.2932 \  m

Explanation:

From the question we are told that

   The actual  depth of the stream is d =  39 \ cm  =  0.39 \ m  

    The  refractive index of the water is  n =  1.33

Generally the apparent depth of the stream is mathematically represented as

         d_1  =  \frac{d}{1.33}

substituting values

        d_1  =  \frac{ 0.39}{1.33}

        d_1  =  0.2932 \  m

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What’s the unit for velocity?
torisob [31]

Answer:

meter per second

Explanation:

It could be any other unit such as yard or feet, put it will be whatever measure per second or whatever time.

Examples

feet per second

miles per hour

4 0
3 years ago
What is Tension variables?
Soloha48 [4]

Answer:

The tension on an object is equal to the mass of the object x gravitational force plus/minus the mass x acceleration. T = mg + ma.

Explanation:

6 0
2 years ago
A nautical mile is 6076 feet, and 1 knot is a unit of speed equal to 1 nautical mile/hour. How fast is a boat going 8 knots goin
mezya [45]

Answer:

The speed of the boat is equal to 13.50 ft/s.

Explanation:

given,

1  nautical mile = 6076 ft

1 knot = 1 nautical mile /hour

1 knot = 6076 ft/hr

speed of boat = 8 knots

 8 knots = 8 nautical mile /hour

               =8 \times \dfrac{6076\ ft}{1\nautical\ mile}\times \dfrac{1\ hour}{60\times 60\ s}

               = 13.50 ft/s

The speed of the boat is equal to 13.50 ft/s.

5 0
3 years ago
A ball with a mass of 2000 g is floating on the surface of a pool of water. What is the minimum volume that the ball could have
Doss [256]

Answer:

2000\; {\rm cm^{3}}.

Explanation:

When the ball is placed in this pool of water, part of the ball would be beneath the surface of the pool. The volume of the water that this ball displaced is equal to the volume of the ball that is beneath the water surface.

The buoyancy force on this ball would be equal in magnitude to the weight of water that this ball has displaced.

Let m(\text{ball}) denote the mass of this ball. Let m(\text{water}) denote the mass of water that this ball has displaced.

Let g denote the gravitational field strength. The weight of this ball would be m(\text{ball}) \, g. Likewise, the weight of water displaced would be m(\text{water})\, g.

For this ball to stay afloat, the buoyancy force on this ball should be greater than or equal to the weight of this ball. In other words:

\text{buoyancy} \ge m(\text{ball})\, g.

At the same time, buoyancy is equal in magnitude the the weight of water displaced. Thus:

\text{buoyancy} = m(\text{water}) \, g.

Therefore:

m(\text{water})\, g = \text{buoyancy} \ge m(\text{ball})\, g.

m(\text{water}) \ge m(\text{ball}).

In other words, the mass of water that this ball displaced should be greater than or equal to the mass of of the ball. Let \rho(\text{water}) denote the density of water. The volume of water that this ball should displace would be:

\begin{aligned}V(\text{water}) &= \frac{m(\text{water})}{\rho(\text{water})} \\ &\ge \frac{m(\text{ball}))}{\rho(\text{water})}  \end{aligned}.

Given that m(\text{ball}) = 2000\; {\rm g} while \rho = 1.00\; {\rm g\cdot cm^{-3}}:

\begin{aligned}V(\text{water}) &\ge \frac{m(\text{ball}))}{\rho(\text{water})}  \\ &= \frac{2000\; {\rm g}}{1.00\; {\rm g\cdot cm^{-3}}} \\ &= 2000\; {\rm cm^{3}}\end{aligned}.

In other words, for this ball to stay afloat, at least 2000\; {\rm cm^{3}} of the volume of this ball should be under water. Therefore, the volume of this ball should be at least 2000\; {\rm cm^{3}}\!.

3 0
1 year ago
Name and describe two forces that act on your body as you walk down the hallway.
defon
Gravity & Resistance ?
3 0
3 years ago
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