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Nataly [62]
2 years ago
15

Submarines need to be extremely strong to withstand the extremely high pressure of water pushing down on them. An experimental r

esearch submarine with a volume of 1,510 liters has an internal pressure of 1.20atm. If the pressure of the ocean breaks the submarine forming a bubble with a pressure of 253atm pushing on it, how big will that bubble be? You must show your work to receive full credit.
Physics
1 answer:
Brrunno [24]2 years ago
8 0
1500 I think so but I not sure
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3. A model rocket is launched straight upward at 58.8 m/s.
SIZIF [17.4K]

Let us assume that rocket only runs in initial energy and not using its own to flying.

Also , let upward direction is +ve and downward direction is -ve .

Initial velocity , u = 58.8 m/s .

Acceleration due to gravity , g=-9.8\ m/s^2 .

Final velocity , v - = 0 m/s .

We know , by equation of motion .

v^2-u^2=2gh\\\\2gh_{max}=0^2-58.8^2\\\\h_{max}=\dfrac{0^2-58.8^2}{-2\times 9.8}\\\\h_{max}= 176.4\ m

Hence, this is the required solution .

8 0
3 years ago
Unas niñas en el receso estaban jugando a derramar flatulencias en un frasco de forma cilíndrica, cuyo radio tenía 5" y una altu
hram777 [196]

Answer:

Los gases siempre se adaptaran al volumen del contenedor en el que estén.

Entonces, si baja la presión del gas, la temperatura se mantendrá constante, lo que implica que tiene que aumentar la temperatura, pues para un gas ideal tenemos:

P*V = n*k*T

donde si el frasco esta cerrado, tenemos que n, k y V son constantes, entonces:

P = (n*k/V)*T

P = constante*T

(Notar que si el frasco no estuviera cerrado, entonces el numero n variaría de forma incontrolable, y este problema no se podría resolver de forma sencilla)

Entonces, si la presión baja, también baja la temperatura, pero el volumen se mantendrá constante, eso es lo importante.

Como el volumen se mantiene constante, el volumen que tomara el gas va a ser igual al volumen del frasco, sabemos que el volumen de un cilindro es:

V = (pi*r^2*h)

donde:

r = radio

pi = 3.14

h = altura.

en este caso, r = 5'' = 5 in

                       h = 7 in

Entonces el volumen sera:

V = 3.14*(5in)^2*7in = 549.5 in^2

3 0
3 years ago
A scientist examines a large pot of boiling water and a small cup of boiling water. The scientist determines that the large pot
saul85 [17]

Answer:

d. The large pot of water and small cup of water have the same temperature, but the large pot of water has higher thermal energy.

Explanation:

Temperature is a measure of the average kinetic energy of individual molecules. While internal energy refers to the total kinetic energy of the molecules within the object. Since in this case we have the same amount of average kinetic energy, then the large pot of water and small cup of water have the same temperature. While the large pot of water has higher thermal energy, since has more water particles than the small cup.

7 0
2 years ago
Read 2 more answers
If an object is thrown upward at 128 feet per second from a height of 76 feet, its height S after t seconds is given by the foll
faltersainse [42]
<h2>a) Average velocity in first 4 seconds is 64 ft/s upward</h2><h2>b) Average velocity in second 4 seconds is 63.5 ft/s downward</h2>

Explanation:

a) Given S(t) = 76 + 128t − 16t²

    s(0) = 76 + 128 x 0 − 16 x 0² = 76 ft

    s(4) = 76 + 128 x 4 − 16 x 4² = 332 ft

    Displacement in 4 seconds = 332 - 76 = 256 ft

    Time = 4 - 0 = 4 s

    \texttt{Velocity = }\frac{256}{4}=64ft/s

    Average velocity in first 4 seconds is 64 ft/s upward

a) Given S(t) = 76 + 128t − 16t²

    s(4) = 76 + 128 x 4 − 16 x 4² = 332 ft

    s(8) = 76 + 128 x 8 − 16 x 8² = 78 ft

    Displacement in 4 seconds = 78 - 332 = -254 ft

    Time = 4 - 0 = 4 s

    \texttt{Velocity = }\frac{-254}{4}=-63.5ft/s

    Average velocity in second 4 seconds is 63.5 ft/s downward

3 0
2 years ago
Energy is to be stored in a flywheel in the shape of a uniform solid disk with a radius of R = 1.22 m and a mass of 67.0 kg . To
eduard

Answer: 71.7 KJ

Explanation:

The rotational kinetic energy of a rotating body can be written as follows:

Krot = ½ I ω2

Now, any point on the rim of the flywheel, is acted by a centripetal force, according to Newton’s 2nd Law, as follows:

Fc = m. ac

It can be showed that the centripetal acceleration, is related with the angular velocity and the radius, as follows:

ac =  ω2 r

We know that this acceleration has a limit value, so , we can take this limit to obtain a maximum value for the angular velocity also.

As the flywheel is a solid disk, the rotational inertia I is just ½ m r2.

Replacing in the expression for the Krot, we have:

Krot= ½ (1/2 mr2.ac/r) = ¼ mr ac = ¼ 67.0 Kg. 1.22 m . 3,510 m/s2 = 71. 7 KJ

5 0
2 years ago
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