Which scenario requires the most power?

A light plane is headed due south with a speed of 200 km/h relative to still air. After 1.00 hour, the pilot notices that they h

Lilit [14]

Answer: the total velocity of the air is 67.69km/h to the north and 35.4km/h to the east.

Explanation: The initial velocity of the plane is 200km/h south (supose that south is our positive x-axis here and east is the positive y-axis)

In one hour, the plane is located 137km away from the initial position, and the position in X is equal to 137km*cos(15°) = 132.33, this means that the velocity in the x axis is equal to 132.33 km/h, knowing that the initial velocity of the plane was 200km in the x-axis, this means that the velocity of the air must be:

132.33km/h - 200km/h = -67.69km/h

km and the position in "y" is equal to 137km*sin(15°) = 35.4km

This means that the velocity of the air in the y-axis is 35.4km/h

So the total velocity of the air is 67.69km/h to the north and 35.4km/h to the east.

8 0

3 years ago

A block is pulled along a horizontal surface at a constant speed by a force (14.1i 0 j 5.1k). The direction k is perpendicular t

Pepsi [2]

Answer:

$W =84.6\ Nm$

Explanation:

given,

F = 14.1 i + 0 j + 5.1 k

displacement = 6 m

Assuming block is moving in x- direction

we know,

dW = F dx

$\int dW = F\int dx$

$W = F\int_0^6 dx$

$W = F[x]_0^6$

$W = 14.1 \times 6$

$W =84.6\ Nm$

hence, work done by the force is equal to $W =84.6\ Nm$

7 0

3 years ago

constant force F=5i+5j−1kF=5i+5j−1k is applied to an object that is moving along a straight line from the point (−5,−3,−4)(−5,−3

meriva

Answer:

W = 71J

Explanation:

Given force F = (5i+5j−1k)N

d = Δr

r1 = (−5,−3,−4)m

r2 = (2,5,0)m

Δr = r2 – r1 = (2-(-5), 5-(-3), 0-(-4))

Δr = (2+5, 5+3, 0+4) = (7i+ 8j +4k)m

W = F•d = (5i+5j−1k)•(7i+ 8j +4k)

W = 5×7 + 5×8 +-1×4 = 35 + 40 - 4

W = 71J

6 0

3 years ago

How high does the water rise in the bell after enough time has passed for the air to reach thermal equilibrium

Minchanka [31]

The height risen by water in the bell after enough time has passed for the air to reach thermal equilibrium is 3.8 m.

<h3>Pressure and temperature at equilibrium </h3>

The relationship between pressure and temperature can be used to determine the height risen by the water.

$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$

where;

V₁ = AL
V₂ = A(L - y)
P₁ = Pa
P₂ = Pa + ρgh
T₁ = 20⁰C = 293 K
T₂ = 10⁰ C = 283 k

$\frac{PaAL}{T_1} = \frac{(P_a + \rho gh)A(L-y)}{T_2} \\\\\frac{PaL}{T_1} = \frac{(P_a + \rho gh)(L-y)}{T_2} \\\\L-y = \frac{PaLT_2}{T_1(P_a + \rho gh)} \\\\y = L (1 - \frac{PaT_2}{T_1(P_a + \rho gh)})\\\\y = 4.2(1 - \frac{101325 \times 283}{293(101325\ +\ 1000 \times 9.8 \times 100)} )\\\\y = 3.8 \ m$

Thus, the height risen by water in the bell after enough time has passed for the air to reach thermal equilibrium is 3.8 m.

The complete question is below:

A diving bell is a 4.2 m -tall cylinder closed at the upper end but open at the lower end. The temperature of the air in the bell is 20 °C. The bell is lowered into the ocean until its lower end is 100 m deep. The temperature at that depth is 10°C. How high does the water rise in the bell after enough time has passed for the air to reach thermal equilibrium?

Learn more about thermal equilibrium here: brainly.com/question/9459470

#SPJ4

3 0

2 years ago

(a) Find the magnitude of an earthquake that has an intensity that is 37.25 (that is, the amplitude of the seismograph reading i

MAVERICK [17]

Answer:

The magnitude of an earthquake is 5.6.

Explanation:

The magnitude of an earthquake can be found as follows:

$M = log(\frac{I}{S})$

Where:

I: is the intensity of the earthquake = 37.25 cm

S: is the intensity of a standard earthquake = 10⁻⁴ cm

Hence, the magnitude is:

$M = log(\frac{I}{S}) = log(\frac{37.25}{10^{-4}}) = 5.6$

Therefore, the magnitude of an earthquake is 5.6.

I hope it helps you!

6 0

3 years ago