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aleksklad [387]
3 years ago
6

A water heater that has the shape of a right cylindrical tank with a radius of 1 foot and a height of 4 feet is being drained. h

ow fast is the water draining out of the tank in cubic feet​ / minute if the water level is dropping at 6​ inches/min?
Physics
1 answer:
Natali5045456 [20]3 years ago
5 0

<span> </span>For any prism-shaped geometry, the volume (V) is assumed by the product of cross-sectional area (A) and height (h). 

<span> V = Ah </span>

<span>
Distinguishing with respect to time gives the relationship between the rates. 
dV/dt = A*dh/dt</span>

<span> in the meantime the area is not altering </span>

<span>
dV/dt = π*(1 ft)^2*(-0.5 ft/min) </span>

<span>
dV/dt = -π/2 ft^3/min ≈ -1.571 ft^3/min 

Water is draining from the tank at the rate of π/2 ft^3/min.</span>

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3 years ago
You are heading toward an island in your speed boat when you see a friend standing on the shore of the island. You sound the boa
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Greater than

Explanation:

The Wavelength will be higher than what will be heard without any motion on the boat due to the Doppler Effect, which is the change in the frequency of a sound wave whenever there's a relative motion between the source of the wave and observer. The amount of shift in frequency depends on the speed of the source towards the observer; the higher the velocity of the source, the higher the shift.

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3 years ago
A person of mass 55 kg swings on a rope length 4 m from rest (when the rope makes an angle of 30 degrees with the vertical) and
vovangra [49]

Answer:

θ = 19.66°

Explanation:

To determine the angle that the rope makes with the vertical for the two people, you first take into account the potential energy of the first person before he swings on the rope:

U=mgh

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g: gravitational acceleration = 9.8m/s^2

m: mass of the first person = 55 kg

In the image attache below you can notice that the height h is:

h=4-4cos(30\°)=0.53m

Then, the potential energy is:

U=(55kg)(9.8m/s^2)(0.53m)=285.67J

When the first person picks up the second person (when the rope is exactly vertical), all the potential energy becomes kinetic energy. Next, when both people reaches the maximum height h' the energy must be equal to the initial potential energy of the first person:

U'=(m_1+m_2)gh'=285.67\ J

From the previous equation you can get h':

h'=\frac{285.67J}{(55kg+70kg)(9.8m/s^2)}=0.2332m

Finally, you obtain the angle between the rope at the height h,' and the vertical, by calculating the following:

h'=4-4cos(\theta)\\\\\theta=cos^{-1}(\frac{4-h'}{4})=cos^{-1}(\frac{4-0.2332}{4})=19.66\°

hence, the angle between the rope and the vertical, when the two people are in the rope is 19.66°

8 0
3 years ago
two horses are pulling a 325 kg wagon, initially at rest. The horses exert 250 N and 178 N forward forces, respectively. Ignorin
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Answer:

AFter 3.5 s, the wagon is moving at:   4.62\,\,\frac{m}{s}

Explanation:

Let's start by finding first the net force on the wagon, and from there the wagon's acceleration (using Newton's 2nd Law):

Net force = 250 N + 178 N = 428 N

Therefore, the acceleration from Newton's 2nd Law is:

F=m\,*\,a\\a = \frac{F}{m} \\a= \frac{428}{325}\, \frac{m}{s^2} \\a\approx 1.32 \,\,\frac{m}{s^2}

So now we apply this acceleration to the kinematic expression for velocity in an object moving under constant acceleration:

v_f=v_i+a\,*\,t\\v_f=0+1.32\,*\,3.5\,\,\frac{m}{s} \\v_f=4.62\,\,\frac{m}{s}

7 0
3 years ago
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