Ten seconds after an electric fan is turned on, the fan rotates at 300 rev/min. its average angular acceleration is

A large container, 120 cm deep is filled with water. If a small hole is punched in its side 77.0 cm from the top, at what initia

prisoha [69]

Answer:

The water will flow at a speed of 3,884 m/s

Explanation:

Torricelli's equation

v = $\sqrt{2gh}$

*v = liquid velocity at the exit of the hole

g = gravity acceleration

h = distance from the surface of the liquid to the center of the hole.

v = $\sqrt{2*9,8m/s^2*0,77m}$ = 3,884 m/s

6 0

2 years ago

A father demonstrates projectile motion to his children by placing a pea on his fork's handle and rapidly depressing the curved

MariettaO [177]

Answer:

4.17 m/s

Explanation:

To solve this problem, let's start by analyzing the vertical motion of the pea.

The initial vertical velocity of the pea is

$u_y = u sin \theta = (7.39)(sin 69.0^{\circ})=6.90 m/s$

Now we can solve the problem by applying the suvat equation:

$v_y^2-u_y^2=2as$

where

$v_y$ is the vertical velocity when the pea hits the ceiling

$a=g=-9.8 m/s^2$ is the acceleration of gravity

s = 1.90 is the distance from the ceiling

Solving for $v_y$ ,

$v_y = \sqrt{u_y^2+2as}=\sqrt{(6.90)^2+2(-9.8)(1.90)}=3.22 m/s$

Instead, the horizontal velocity remains constant during the whole motion, and it is given by

$v_x = u cos \theta = (7.39)(cos 69.0^{\circ})=2.65 m/s$

Therefore, the speed of the pea when it hits the ceiling is

$v=\sqrt{v_x^2+v_y^2}=\sqrt{2.65^2+3.22^2}=4.17 m/s$

5 0

2 years ago

A flywheel is a mechanical device used to store rotational kinetic energy for later use. Consider a flywheel in the form of a un

Kamila [148]

Answer:

a) 6738.27 J

b) 61.908 J

c) $\frac{4492.18}{v_{car} ^{2} }$

Explanation:

The complete question is

A flywheel is a mechanical device used to store rotational kinetic energy for later use. Consider a flywheel in the form of a uniform solid cylinder rotating around its axis, with moment of inertia I = 1/2 mr2.

Part (a) If such a flywheel of radius r1 = 1.1 m and mass m1 = 11 kg can spin at a maximum speed of v = 35 m/s at its rim, calculate the maximum amount of energy, in joules, that this flywheel can store?

Part (b) Consider a scenario in which the flywheel described in part (a) (r1 = 1.1 m, mass m1 = 11 kg, v = 35 m/s at the rim) is spinning freely at its maximum speed, when a second flywheel of radius r2 = 2.8 m and mass m2 = 16 kg is coaxially dropped from rest onto it and sticks to it, so that they then rotate together as a single body. Calculate the energy, in joules, that is now stored in the wheel?

Part (c) Return now to the flywheel of part (a), with mass m1, radius r1, and speed v at its rim. Imagine the flywheel delivers one third of its stored kinetic energy to car, initially at rest, leaving it with a speed vcar. Enter an expression for the mass of the car, in terms of the quantities defined here.

moment of inertia is given as

$I$ = $\frac{1}{2}$ $mr^{2}$

where m is the mass of the flywheel,

and r is the radius of the flywheel

for the flywheel with radius 1.1 m

and mass 11 kg

moment of inertia will be

$I$ = $\frac{1}{2}$ $*11*1.1^{2}$ = 6.655 kg-m^2

The maximum speed of the flywheel = 35 m/s

we know that v = ωr

where v is the linear speed = 35 m/s

ω = angular speed

r = radius

therefore,

ω = v/r = 35/1.1 = 31.82 rad/s

maximum rotational energy of the flywheel will be

E = $Iw^{2}$ = 6.655 x $31.82^{2}$ = 6738.27 J

b) second flywheel has

radius = 2.8 m

mass = 16 kg

moment of inertia is

$I$ = $\frac{1}{2}$ $mr^{2}$ = $\frac{1}{2}$ $*16*2.8^{2}$ = 62.72 kg-m^2

According to conservation of angular momentum, the total initial angular momentum of the first flywheel, must be equal to the total final angular momentum of the combination two flywheels

for the first flywheel, rotational momentum = $Iw$ = 6.655 x 31.82 = 211.76 kg-m^2-rad/s