Physics, 29.11.2021 21:50 rozzy5132

In a game of pool, the cue ball rolls toward the stationary eight ball with a velocity of 1.35 m/s and they collide. The mass of the cue ball is 170grams and the mass of the eight ball is 156grams. The velocity of the cue ball immediately after the collision is 0.060 m/s

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Answer from: Quest

answer: the rock sample is a sedimentary rock.

explanation:

it is given that:

mass of rock sample, m = 2.7 g

volume of rock sample, $v = 1.1 cm^{3}$

so, density of rock sample will be

$density = \frac{mass}{volume}$

$\rho =\frac{2.7}{1.1}g/cm^{3}\\\\\rho =2.45 g/cm^{3}$

since, the density of rock sample is $2.45 g/cm^{3}$ . hence, the rock sample is a sedimentary rock, whose densities varies between $2g/cm^{3} - 2.6 g/cm^{3}$ .

Answer from: Quest

Protons have a positive charge and, collectively, give an element its atomic number.

Answer from: Quest

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.

Answer from: Quest

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

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