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Dafna1 [17]
3 years ago
10

I need help with a conservation of energy question, 1. b) on the attachment :)

Physics
1 answer:
rosijanka [135]3 years ago
4 0

Explanation:

Dina tiktokneed help with a conservation of energy question, 1. b) on the attachment :)

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How can we produce energy by turning a turbine?​
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A pressure sensor inside of a mixing tank is designed to turn red when the pressure inside the tank exceeds 1.9 kPa. If the sens
spin [16.1K]

Answer:

19 N

Explanation:

From the question given above, the following data were obtained:

Pressure (P) = 1.9 kPa

Length (L) = 10 cm

Force (F) =?

Next, we shall convert 1.9 KPa to N/m². This can be obtained as follow:

1 KPa = 1000 N/m²

Therefore,

1.9 KPa = 1.9 KPa × 1000 N/m² / 1 KPa

1.9 KPa = 1900 N/m²

Thus, 1.9 KPa is equivalent to 1900 N/m².

Next, we shall convert 10 cm to m. This can be obtained as follow:

100 cm = 1 m

Therefore,

10 cm = 10 cm × 1 m / 100 cm

10 cm = 0.1 m

Thus, 10 cm is equivalent to 0.1 m

Next, we shall determine the area of the square. This can be obtained as follow:

Length (L) = 0.1 m

Area of square (A) =?

A = L²

A = 0.1²

A = 0.01 m²

Thus, the area of the square is 0.01 m².

Finally, we shall determine the force that must be exerted on the sensor in order for it to turn red. This can be obtained as follow:

Pressure (P) = 1900 N/m²

Area (A) = 0.01 m²

Force (F) =?

P = F/A

1900 = F / 0.01

Cross multiply

F = 1900 × 0.01

F = 19 N

Therefore, a force of 19 N must be exerted on the sensor in order for it to turn red.

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A marble at the front of a truck bed traveling at 20 m/s relative to the highway rolls toward the back of the truck bed with a s
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14 m/s in the direction of the truck
5 0
3 years ago
Kyle is flying a helicopter at 125 m/s on a heading of 325 o . If a wind is blowing at 25 m/s toward a direction of 240.0 o , wh
frosja888 [35]

Answer:

The resultant velocity of the helicopter is \vec v_{H} = \left(89.894\,\frac{m}{s}, -93.348\,\frac{m}{s}\right).

Explanation:

Physically speaking, the resulting velocity of the helicopter (\vec v_{H}), measured in meters per second, is equal to the absolute velocity of the wind (\vec v_{W}), measured in meters per second, plus the velocity of the helicopter relative to wind (\vec v_{H/W}), also call velocity at still air, measured in meters per second. That is:

\vec v_{H} = \vec v_{W}+\vec v_{H/W} (1)

In addition, vectors in rectangular form are defined by the following expression:

\vec v = \|\vec v\| \cdot (\cos \alpha, \sin \alpha) (2)

Where:

\|\vec v\| - Magnitude, measured in meters per second.

\alpha - Direction angle, measured in sexagesimal degrees.

Then, (1) is expanded by applying (2):

\vec v_{H} = \|\vec v_{W}\| \cdot (\cos \alpha_{W},\sin \alpha_{W}) +\|\vec v_{H/W}\| \cdot (\cos \alpha_{H/W},\sin \alpha_{H/W}) (3)

\vec v_{H} = \left(\|\vec v_{W}\|\cdot \cos \alpha_{W}+\|\vec v_{H/W}\|\cdot \cos \alpha_{H/W}, \|\vec v_{W}\|\cdot \sin \alpha_{W}+\|\vec v_{H/W}\|\cdot \sin \alpha_{H/W} \right)

If we know that \|\vec v_{W}\| = 25\,\frac{m}{s}, \|\vec v_{H/W}\| = 125\,\frac{m}{s}, \alpha_{W} = 240^{\circ} and \alpha_{H/W} = 325^{\circ}, then the resulting velocity of the helicopter is:

\vec v_{H} = \left(\left(25\,\frac{m}{s} \right)\cdot \cos 240^{\circ}+\left(125\,\frac{m}{s} \right)\cdot \cos 325^{\circ}, \left(25\,\frac{m}{s} \right)\cdot \sin 240^{\circ}+\left(125\,\frac{m}{s} \right)\cdot \sin 325^{\circ}\right)\vec v_{H} = \left(89.894\,\frac{m}{s}, -93.348\,\frac{m}{s}\right)

The resultant velocity of the helicopter is \vec v_{H} = \left(89.894\,\frac{m}{s}, -93.348\,\frac{m}{s}\right).

8 0
3 years ago
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