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A 6.0-cm-diameter horizontal pipe gradually narrows to 4.0 cm. When water flows through this pipe at a certain rate, the gauge pressure in these two sections is 32.0 kPa and 24.0 kPa, respectively. What is the volume rate of flow?

Answers

Answer 1

Answer:

Answer:

a n c

Explanation:

Answer 2

Answer:

Final answer:

The volume rate of flow can be determined using the equation Q = Av, where Q is the volume rate of flow, A is the cross-sectional area of the pipe, and v is the average speed of the water. Given the diameter of the wider section of the pipe is 6.0 cm and the gauge pressure is 32.0 kPa, we can calculate the volume rate of flow using the provided information. The volume rate of flow is found to be 0.0018 m³/s.

Explanation:

The volume rate of flow can be determined using the equation Q = Av, where Q is the volume rate of flow, A is the cross-sectional area of the pipe, and v is the average speed of the water.

Given that the diameter of the wider section of the pipe is 6.0 cm, the radius is 3.0 cm and the gauge pressure is 32.0 kPa. Similarly, for the narrower section with a diameter of 4.0 cm, the radius is 2.0 cm and the gauge pressure is 24.0 kPa.

Using the equation Q = Av and the fact that the flow rate must be the same at all points along the pipe, we can set up the equation A₁v₁ = A₂v₂. Solving for v₂, we have v₂ = A₁v₁ / A₂ = πr₁²v₁ / πr₂², where r₁ is the radius of the wider section and r₂ is the radius of the narrower section.

Substituting the values, we get v₂ = (3.14)(3.0 cm)²(32.0 kPa) / [(3.14)(2.0 cm)²] = 18.0 cm/s. Since v = d/t, we can convert cm/s to m³/s by multiplying by 0.0001, so the volume rate of flow is 0.0018 m³/s.

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How does simple machines make work easier?

Answers

They make it so you would exert less force and make things easier to move

The simple machines can help make work easier by working quicker than the people and making other peoples jobs easier.

Explanation:

I hope this helped.

Find the net downward force on the tank's flat bottom, of area 1.60 m2 , exerted by the water and air inside the tank and the air outside the tank. Assume that the density of water is 1.00 g/cm3

Answers

Answer:

The net downward force on the tank is $1.85*10^(5)\ N$

Explanation:

Given that,

Area = 1.60 m²

Suppose the design of a cylindrical, pressurized water tank for a future colony on Mars, where the acceleration due to gravity is 3.71 meters per second per second. The pressure at the surface of the water will be 150 K Pa , and the depth of the water will be 14.4 m . The pressure of the air in the building outside the tank will be 88.0 K Pa.

We need to calculate the net downward force on the tank

Using formula of formula

$F=(P+\rho* g* h-P_(out))A$

Where, P = pressure

g = gravity at mars

h = height

A = area

Put the value into the formula

$F=(150*10^3+1.00*10^3*3.71*14.4-88.0*10^(3))*1.60$

$F=1.85*10^(5)\ N$

Hence, The net downward force on the tank is $1.85*10^(5)\ N$

Final answer:

The net downward force on the tank's flat bottom can be found by calculating the pressure at the bottom of the container.

Explanation:

Since the density is constant, the weight can be calculated using the density:

w = mg = pVg = pAhg.

The pressure at the bottom of the container is therefore equal to atmospheric pressure added to the weight of the fluid divided by the area.

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A car travels 13 km in a southeast direction and then 16 km 40 degrees north of east. What is the car's resultant direction?

Answers

Answer:

21.48 km 2.92° north of east

Explanation:

To find the resultant direction, we need to calculate a sum of vectors.

The first vector has module = 13 and angle = 315° (south = 270° and east = 360°, so southeast = (360+270)/2 = 315°)

The second vector has module 16 and angle = 40°

Now we need to decompose both vectors in their horizontal and vertical component:

horizontal component of first vector: 13 * cos(315) = 9.1924

vertical component of first vector: 13 * sin(315) = -9.1924

horizontal component of second vector: 16 * cos(40) = 12.2567

vertical component of second vector: 16 * sin(40) = 10.2846

Now we need to sum the horizontal components and the vertical components:

horizontal component of resultant vector: 9.1924 + 12.2567 = 21.4491

vertical component of resultant vector: -9.1924 + 10.2846 = 1.0922

Going back to the polar form, we have:

$module = √(horizontal^2 + vertical^2)$

$module = √(460.0639 + 1.1929)$

$module = 21.4769$

$angle = arc\ tangent(vertical/horizontal)$

$angle = arc\ tangent(1.0922/21.4491)$

$angle = 2.915\°$

So the resultant direction is 21.48 km 2.92° north of east.

Match each planet to an accurate characteristic

Answers

Answer:

venus - 2

earth - 3

mars - 4

mercury - 1

A model airplane with a mass of 0.741kg is tethered by a wire so that it flies in a circle 30.9 m in radius. The airplane engine provides anet thrust of 0.795 N perpendicular tothe tethering wire.(a) Find the torque the net thrust producesabout the center of the circle.
N·m

(b) Find the angular acceleration of the airplane when it is inlevel flight.
rad/s2

(c) Find the linear acceleration of the airplane tangent to itsflight path.
m/s2

Answers

(a) 24.6 Nm

The torque produced by the net thrust about the center of the circle is given by:

$\tau = F r$

where

F is the magnitude of the thrust

r is the radius of the wire

Here we have

F = 0.795 N

r = 30.9 m

Therefore, the torque produced is

$\tau = (0.795 N)(30.9 m)=24.6 N m$

(b) $0.035 rad/s^2$

The equivalent of Newton's second law for a rotational motion is

$\tau = I \alpha$

where

$\tau$ is the torque

I is the moment of inertia

$\alpha$ is the angular acceleration

If we consider the airplane as a point mass with mass m = 0.741 kg, then its moment of inertia is

$I=mr^2 = (0.741 kg)(30.9 m)^2=707.5 kg m^2$

And so we can solve the previous equation to find the angular acceleration:

$\alpha = (\tau)/(I)=(24.6 Nm)/(707.5 kg m^2)=0.035 rad/s^2$

(c) $1.08 m/s^2$

The linear acceleration (tangential acceleration) in a rotational motion is given by

$a=\alpha r$

where in this problem we have

$\alpha = 0.035 rad/s^2$ is the angular acceleration

r = 30.9 m is the radius

Substituting the values, we find

$a=(0.035 rad/s^2)(30.9 m)=1.08 m/s^2$

A car travels along a straight line at a constant speed of 53.0 mi/h for a distance d and then another distance d in the same direction at another constant speed. the average velocity for the entire trip is 26.5 mi/h. (a) what is the constant speed with which the car moved during the second distance d?

Answers

A distance of d is covered with 53 mile/hr initially.Time taken to cover this distance t1 = d/53 hourNext distance of d is covered with x mile hours.Time taken to cover this distance t2 = d/x hours.We have average speed = 26.5 mile / hour

= Total distance traveled/ total time taken

= $(2d)/((d)/(53)+(d)/(x)) = (2)/((1)/(53)+(1)/(x) ) = (106x)/(x+53)$

$26.5 = (106x)/(x+53) \n \n 79.5 x = 1404.5\n \n x = 17.67 miles/hour$

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