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photoshop1234 [79]
3 years ago
13

Water flows along a streamline down a river of constant width. Over a short distance, the water slows from speed v to v/3. Which

of the following can you correctly conclude about the river's depth?
a. It became deeper by a factor of 3.
b. It became shallower by a factor of 3.
c. It became deeper by a factor of 3^2
d. It became shallower by a factor of 3^2
Physics
1 answer:
kvasek [131]3 years ago
8 0

Answer:

a. It became deeper by a factor of 3.

Explanation:

What we have is water flowing down a river with constant width. The water slows from speed v to v3 over a shirt distance

Using the equation of continuity

A1V1 = A2V2 ----1

A1 is the area of rectangle

V1 is the velocity of water

Area of rectangle = length x width

We rewrite equation 1 as

λ1w1v1 = λ2w2v2

We have w1 = w2

λ1v1 = λ2v2

λ1*v1 = λ2*v/3

λ1 = λ2/3

So it becomes deeper by a factor of 3

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What is the mass of a dog running at a speed of 5 m/s and a momentum of 120.5 kgm/s?
viktelen [127]

Given:

Momentum of the dog (p) = 120.5 kg m/s

Speed of the dog (v) = 5 m/s

To Find:

Mass of the dog (m)

Concept/Theory:

\underline{\underline{ \bf{\Large{Momentum}}}}

  • It is defined as the quantity of motion contained in a body.
  • It is measured as the product of mass of the body and it's speed.
  • It is represented by p.
  • It's SI unit is kg m/s
  • Mathematical Representation/Equation of Momentum: \boxed{ \bf{p = mv}}

Answer:

By using equation of momentum, we get:

\rm \longrightarrow m =  \dfrac{p}{v}  \\  \\  \rm \longrightarrow m =  \dfrac{120.5}{5}  \\  \\  \rm \longrightarrow m = 24.1 \: kg

\therefore Mass of the dog (m) = 24.1 kg

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At the same moment, one rock is dropped and one is thrown downward with an initial velocity of 29m/s from the top of a building
Inessa [10]

Answer:

The thrown rock strike 2.42 seconds earlier.

Explanation:

This is an uniformly accelerated motion problem, so in order to find the arrival time we will use the following formula:

x=vo*t+\frac{1}{2} a*t^2\\where\\x=distance\\vo=initial velocity\\a=acceleration

So now we have an equation and unkown value.

for the thrown rock

\frac{1}{2}(9.8)*t^2+29*t-300=0

for the dropped rock

\frac{1}{2}(9.8)*t^2+0*t-300=0

solving both equation with the quadratic formula:

\frac{-b\±\sqrt{b^2-4*a*c} }{2*a}

we have:

the thrown rock arrives on t=5.4 sec

the dropped rock arrives on t=7.82 sec

so the thrown rock arrives 2.42 seconds earlier (7.82-5.4=2.42)

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nevsk [136]
That they sometimes explode?
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The entropy of an isolated system must be conserved, so it never changes.a. Trueb. Fasle
Snowcat [4.5K]

Answer:

B: False

Explanation:

The second law of thermodynamics states that: the entropy of an isolated system will never decrease because isolated systems always tend to evolve towards thermodynamic equilibrium which is a state with maximum entropy.

Thus, it means that the entropy change will always be positive.

Therefore, the given statement in the question is false.

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