Physics, 05.12.2019 18:31, Annamea

Asubmarine releases an air bubble with a volume of 1.5 cm^3 from 200 m below the surface of the ocean. how large is the bubble when it reaches the surface? assume the temperature changes from 22 degree c to 24 degree c between 200 m deep and the surface. (the density of water is 1000 kg m^3.) a. 51 cm^3 b. 34 cm^3 c. 31 cm^3 d. 21 cm^3 e. 8 cm^3 1f. 6cm^3 g. 1.5 cm^3

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answered: Guest

answer: the metamorphic rocks are formed when they are exposed to high pressure and high temperatures. this happens when one of the earth's plates comes in contact with the other plate by traveling in opposite directions, straight into each other or slipping under one another. the friction that is caused by this activity results in melting of rocks and when it cools they are known as metamorphic rocks. this process needs a controlled amount of pressure and temperature otherwise it may turn into magma.

answered: Guest

force is based upon mass and acceleration.the statement is not always true.

it can be explained from newton's second law of motion.

as per newton's second law of motion the rate of change of momentum is directly proportional to the applied force and takes place along the direction of force.

mathematically it can be written as -

$f\alpha \frac{dp}{dt}$

⇒ $f= k \frac{dp}{dt}$

here k is the proportionality constant whose value is 1.

$f=\frac{dp}{dt}$

$f= \frac{d[mv]}{dt}$

$f= m\frac{dv}{dt} +v\frac{dm}{dt}$

$f=ma+v\frac{dm}{dt}$ [a is the acceleration=dv/dt]

hence we see that force not only depends on mass and acceleration but also on velocity and rate of change of mass. force f will depend on mass and acceleration only when mass is constant.

hence the statement that force is based upon mass and acceleration is false.

answered: Guest

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

$a = 32 ft/s^2$

so as we know that

$a = \frac{v^2}{r}$

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

$a_c = 32 ft/s^2$

$a_t = 0$

now radius of curvature is given as

$r = \frac{v^2}{a_c}$

$r = \frac{150^2}{32}$

$r = 703.125ft$

so the radius of curvature just after it is dropped is 703.125 ft

answered: Guest

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Asubmarine releases an air bubble with a volume of 1.5 cm^3 from 200 m below the surface of the ocea...