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11Alexandr11 [23.1K]
3 years ago
14

An ideal incompressible fluid flows at 12 m/s in a horizontal pipe. If the pipe widens to twice its original radius, what is the

flow speed in the wider section?
a) 4 m/sb) 6 m/sc) 12 m/sd) 3 m/s
Physics
1 answer:
grandymaker [24]3 years ago
5 0

Answer:

d) 3 m/s

Explanation:

As we know by equation of continuity

A_1v_1 = A_2v_2

here we have

A_1 = \pi r^2

A_2 = \pi (2r)^2

A_2 = 4\pi r^2

now from above equation we have

\pi r^2 v_1 = 4\pi r^2 v_2

v_2 = \frac{v_1}{4}

here we have

v_1 = 12 m/s

v_2 = 3 m/s

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5 0
3 years ago
A simple pendulum consisting of a bob of mass m attached to a string of length L swings with a period T. If the pendulum is take
Juli2301 [7.4K]

To solve this problem we will use the definition of the period in a simple pendulum, which warns that it is dependent on its length and gravity as follows:

T =2\pi \sqrt{\frac{L}{g}}

Here,

L = Length

g = Acceleration due to gravity

We can realize that 2 \pi is a constant so it is proportional to the square root of its length over its gravity,

T \propto \sqrt{\frac{L}{g}}

Since the body is in constant free fall, that is, a point where gravity tends to be zero:

g \rightarrow 0 \Rightarrow T \rightarrow \infty

The value of the period will tend to infinity. This indicates that the pendulum will no longer oscillate because both the pendulum and the point to which it is attached are in free fall.

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3 years ago
How much heat is lost by 2.0 grams of water if the temperature drops from 31 °C to 29 °C? The specific heat of water is 4.184 J/
Elanso [62]

Given :

Mass of water, m = 2 grams.

The temperature of water drops from 31 °C to 29 °C .

The specific heat of water is 4.184 J/(g • °C).

To Find :

Amount of heat lost in this process.

Solution :

We know, heat lost is given by :

Heat\ lost,H = ms( T_f - T_i)\\\\H = 2\times 4.184 \times ( 31 - 29 )\ J\\\\H = 16.736\ J

Therefore, amount of heat lost in this process is 16.736 J.

4 0
3 years ago
What does the term mccarthyism describe? To whom does it refer?
kolbaska11 [484]
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6 0
3 years ago
Two bicycle tires are set rolling with the same initial speed of 4.0 m/s along a long, straight road, and the distance each trav
umka2103 [35]

Answer:

The coefficient of rolling friction will be "0.011".

Explanation:

The given values are:

Initial speed,

v_i = 4.0 \ m/s

then,

v_f=\frac{4.0}{2}

    =2.0 \ m/s

Distance,

s = 18.2 m

The acceleration of a bicycle will be:

⇒ a=\frac{v_f^2-v_i^2}{2s}

On substituting the given values, we get

⇒    =\frac{(2.0)^2-(4.0)^2}{2\times 18.2}

⇒    =\frac{4-8}{37}

⇒    =\frac{-4}{37}

⇒    =0.108 \ m/s^2

As we know,

⇒  f=ma

and,

⇒  \mu_rmg=ma

⇒       \mu_r=\frac{a}{g}

On substituting the values, we get

⇒       =\frac{0.108}{9.8}

⇒       =0.011

7 0
3 years ago
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