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Nesterboy [21]
3 years ago
6

Suppose a drinking straw is in a glass of water. Someone blows air across the top of the straw. The water level in the straw ___

__.
drops

rises

stays the same
Physics
2 answers:
DIA [1.3K]3 years ago
8 0

The correct answer is rises. I hope this helps.

Mark Brainiest if this was helpful.

enyata [817]3 years ago
7 0
The answer is A. drops! Suppose a drinking straw is in a glass of water. Someone blows air across the top of the straw. The water level in the straw will be drops. 

Hope it helped!
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If a 400-mm diameter pipe with a pipe roughness coefficient of 100 flows full of pressurized water with a head loss of 0.4 ft pe
RoseWind [281]

Answer:

Q = 913.9 gpm

Explanation:

The Hazen Williams equation can be written as follows:

P = \frac{4.52\ Q^{1.85}}{C^{1.85}d^{4.87}}

where,

P = Friction Loss per foot of pipe = \frac{0.4}{1000\ ft} = 4 x 10⁻⁴

Q = Flow Rate in gallon/min (gpm) = ?

d = pipe diameter in inches = (400 mm)(0.0393701 in/1 mm) = 15.75 in

C = roughness coefficient = 100

Therefore,

4\ x \ 10^{-4} = \frac{4.52\ Q^{1.85}}{(100)^{1.85}(15.75)^{4.87}}\\\\Q^{1.85} = \frac{4\ x \ 10^{-4}}{1.33\ x\ 10^{-9}} \\\\Q = (300384.75)^\frac{1}{1.85}

<u>Q = 913.9 gpm</u>

5 0
3 years ago
I need the solution to this
posledela

Answer:

He could jump 2.6 meters high.

Explanation:

Jumping a height of 1.3m requires a certain initial velocity v_0. It turns out that this scenario can be turned into an equivalent: if a person is dropped from a height of 1.3m in free fall, his velocity right before landing on the ground will be v_0. To answer this equivalent question, we use the kinematic equation:

v_0 = \sqrt{2gh}=\sqrt{2\cdot 9.8\frac{m}{s^2}\cdot 1.3m}=5.0\frac{m}{s}

With this result, we turn back to the original question on Earth: the person needs an initial velocity of 5 m/s to jump 1.3m high, on the Earth.

Now let's go to the other planet. It's smaller, half the radius, and its meadows are distinctly greener. Since its density is the same as one of the Earth, only its radius is half, we can argue that the gravitational acceleration g will be <em>half</em> of that of the Earth (you can verify this is true by writing down the Newton's formula for gravity, use volume of the sphere times density instead of the mass of the Earth, then see what happens to g when halving the radius). So, the question now becomes: from which height should the person be dropped in free fall so that his landing speed is 5 m/s ? Again, the kinematic equation comes in handy:

v_0^2 = 2g_{1/2}h\implies \\h = \frac{v_0^2}{2g_{1/2}}=\frac{25\frac{m^2}{s^2}}{2\cdot 4.9\frac{m}{s^2}}=2.6m

This results tells you, that on the planet X, which just half the radius of the Earth, a person will jump up to the height of 2.6 meters with same effort as on the Earth. This is exactly twice the height he jumps on Earth. It now all makes sense.

6 0
3 years ago
if you run around a circle at 4.5 m/s and the circle has a radius of 7.7 m, what is your centripetal acceleration?
madreJ [45]

Answer:

Centripetal acceleration,

a_{c} =2.63\ m/s^{2} }

Explanation:

Centripetal acceleration:

Centripetal acceleration is the idea that any object moving in a circle, in something called circular motion, will have an acceleration vector pointed towards the center of that circle.

Centripetal means towards the center.

Examples of centripetal acceleration (acceleration pointing towards the center of rotation) include such situations as cars moving on the cicular part of the road.

An acceleration is a change in velocity.

Formula for Centripetal acceleration:

a_{c} =\frac{(velocity)^{2} }{radius}

Given here,

Velocity = 4.5 m/s

radius = 7.7 m

To Find :

a_{c} = ?

Solution:

We have,

a_{c} =\frac{(velocity)^{2} }{radius}

Substituting  given value in it we get

a_{c} =\frac{(4.5)^{2}}{7.7} \\\\a_{c} =\frac{20.25}{7.7}\\\\a_{c} =2.629\ m/s^{2} \\\\\therefore a_{c} =2.63\ m/s^{2

Centripetal acceleration,

a_{c} =2.63\ m/s^{2} }

7 0
3 years ago
Which of the following statements accurately describes oceanic electric fields?
vichka [17]

Answer:

i sure it is D

Explanation:

3 0
3 years ago
Read 2 more answers
A thin, uniform rod is bent into a square of side length a. If the total mass is M, find the moment of inertia about an axis thr
Papessa [141]

Answer:

The  moment of inertia about an axis through the center and perpendicular to the plane of the square is

    I_s =  \frac{Ma^2}{3}

Explanation:

From the question we are told that

   The length of one side of the square is  a

   The total mass of the square is  M

Generally the mass of one size of the square is mathematically evaluated as

    m_1 = \frac{M}{4}

Generally the moment of inertia of one side of the square is mathematically represented as

        I_g =  \frac{1}{12}  *  m_1 * a^2

Generally given that m_1 = m_2 = m_3 = m_4 = m it means that this moment inertia evaluated above apply to every side of the square  

Now substituting for  m_1

  So

       I _g=  \frac{1}{12}  *  \frac{M}{4} * a^2

Now according to  parallel-axis theorem the moment of inertia of one side of the square about an axis through the center and perpendicular to the plane of the square is mathematically represented as

      I_a =  I_g + m [\frac{q}{2} ]^2

=>    I_a =  I_g + {\frac{M}{4} }* [\frac{q}{2} ]^2

substituting for I_g

=>    I_a =  \frac{1}{12}  *  \frac{M}{4} * a^2 + {\frac{M}{4} }* [\frac{q}{2} ]^2

=>    I_a = \frac{Ma^2}{48} + \frac{Ma^2}{16}

=>    I_a = \frac{Ma^2}{12}

Generally the moment of inertia of the square about an axis through the center and perpendicular to the plane of the square is mathematically represented as

      I_s = 4 * I_a

=>   I_s = 4 * \frac{Ma^2}{12}

=>   I_s =  \frac{Ma^2}{3}

8 0
3 years ago
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