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grandymaker [24]
3 years ago
15

Which shows a decrease in fluid pressure? A. A fan is turned from high speed to low speed. B. Oxygen is compressed as it is put

into a pressurized tank. C. Water in a creek slows down as it enters a slow-moving river. D. A river speeds up when it moves into a narrower channel.
Physics
2 answers:
Volgvan3 years ago
8 0

Answer:

Option A.

A fan is turned from high speed to low speed.

Explanation:

It is important to note that air is also a fluid.

In a system, static pressure of air increases with the speed of rotation of the fan. This is because when the speed of the fan is increased, the force with which it is pushing the air molecules is increased. Since pressure is a relationship between force and area, the pressure of the air molecules will be increased.

Conversely, when the speed of the fan is reduced, the priming force on the air molecules will be reduced, hence the pressure of the air will drop.

This makes option A the correct option

Olegator [25]3 years ago
4 0

Answer:

A. A fan is turned from high speed to low speed.

Explanation:

Liquid and gas are fluids.

In the case of the fan, it requires an amount of pressure to blow air particles. The higher the pressure, the faster the rotation and the more the air particles will be blown .

This means the fan being turned from high to low speed indicates a reduction in the fluid pressure and amount of air particles blown. This validates option A.

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What force is necessary to keep a mass of 0.8 kg revolving in a horizontal circle of radius 0.7 m with a period of 0.5 s? What i
inn [45]

Answer:

88.34 N directed towards the center of the circle

Explanation:

Applying,

F = mv²/r................... Equation 1

F = Force needed to keep the mass in a circle, m = mass of the mass, v = velocity of the mass, r = radius of the circle.

But,

v = 2πr/t................... Equation 2

Where t = time, π = pie

Substitute equation 2 into equation 1

F = m(2πr/t)²/r

F = 4π²r²m/t²r

F = 4π²rm/t²............. Equation 3

From the question,

Given: m = 0.8 kg, r = 0.7 m, t = 0.5 s

Constant: π = 3.14

Substitute these values into equation 3

F = 4(3.14²)(0.7)(0.8)/0.5²

F = 88.34 N directed towards the center of the circle

8 0
3 years ago
Seasonal changes bring about scenes like this one. Plant cells respond to changes in _________________ and as a result, photosyn
jekas [21]
Light intensity & temp change 
4 0
3 years ago
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The linear density of a string is 1.9 × 10-4 kg/m. A transverse wave on the string is described by the equation y = (0.034 m) si
garri49 [273]

Answer:

a. v = 13.572 m / s

b. T = 2.578 x 10 ⁻³ N

Explanation:

μ = 1.9 x 10 ⁻⁴ kg / m

y = y ₙ * sin ( kx + wt )

a.

y = 0.034 m * sin ( 2.8 m⁻¹) x + (38 s⁻¹)t

R = 2.8 m⁻¹

W = 38 s⁻¹

To determine speed of the string

v = W / R = 38 / 2.8

v = 13.572 m / s

b.

v = √ T / μ

v ² = T / μ

To determine the tension on the string

T = v ² * μ

T = 13.572 m/s * 1.9 x 10 ⁻⁴ kg / m

T = 2.578 x 10 ⁻³ N

8 0
4 years ago
An eleven quart container of ice cream is to be made in the form of a cube. calculate the length of a side of the cube. 1 gallon
tatyana61 [14]

The length of a side of the cube is 21.83 cm.

We need to know about cube volume to solve this problem. Cube volume can be calculated by multiplying the length of the cube. It can be written as

V = L³

where V is cube volume and L is cube length.

From the question above, we know that

V = 11 quart = 2.75 gallon

1 gallon = 3.786 liters

Convert volume to liters

V = 2.75 gallon

V = 2.75 x 3.786 liters

V = 10.41 liters

V = 10.41 dm³

Find the length

V = L³

10.41 = L³

L = ³√10.41

L = 2.18 dm

Convert length to cm

L = 2.183 dm

L = 21.83 cm

Find more on cube volume at: brainly.com/question/1972490

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7 0
2 years ago
No matter what values of m and k you used for the spring, what is the ratio of the period to k m.
Leto [7]

The relationship between the period of an oscillating spring and the attached mass determines the ratio of the period to \sqrt{\dfrac{m}{k} }.

Response:

  • The ratio of the period to  \sqrt{\dfrac{m}{k} } is always approximately<u> 2·π : 1</u>

<u />

<h3>How is the value of the ratio of the period to \sqrt{\dfrac{m}{k} } calculated?</h3>

Given:

The relationship between the period, <em>T</em>, the spring constant <em>k</em>, and the

mass attached to the spring <em>m</em> is presented as follows;

T =  \mathbf{2 \cdot \pi \cdot \sqrt{\dfrac{m}{k} }}

Therefore, the fraction of of the period to \sqrt{\dfrac{m}{k} }, is given as follows;

\mathbf{\dfrac{T}{ \sqrt{\dfrac{m}{k} }}} = 2 \cdot \pi

2·π ≈ 6.23

Therefore;

T :{ \sqrt{\dfrac{m}{k} }} = 2 \cdot \pi : 1

Which gives;

  • The ratio of the period to  \sqrt{\dfrac{m}{k} } is always approximately<u> 2·π : 1</u>

Learn more about the oscillations in spring here:

brainly.com/question/14510622

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