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2 years ago

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A 15.0 kg load of bricks hangs from one end of a rope that passes over a small, frictionles pulley. A 28.0 kg counterweight is s

uspended from the other end of the rope . the system is released from rest. A0 draw two free-body diagrams, one for the load od bricks and one for the counterweight. B) What is the magnitude of the upward acceleration of the load of bricks

1 answer:

Talja [164]2 years ago

3 0

Answer:

A) The free body diagrams for both the load of bricks and the counterweight are attached.

B) a = 2.96 m/s²

Explanation:

A)

The free body diagrams for both the load of bricks and the counterweight are attached.

B)

The acceleration of upward acceleration of the load of bricks is given by the following formula:

a = g(m₁ - m₂)/(m₁ + m₂)

where,

a = upward acceleration of load of bricks = ?

g = 9.8 m/s²

m₁ = heavier mass = mass of counterweight = 28 kg

m₂ = lighter mass = mass of load of bricks = 15 kg

Therefore, using these values in equation, we get:

a = (9.8 m/s²)(28 kg - 15 kg)/(28 kg + 15 kg)

<u>a = 2.96 m/s²</u>

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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:

<em>a) 6738.27 J</em>

<em>b) 61.908 J</em>

<em>c) </em> $\frac{4492.18}{v_{car} ^{2} }$

<em></em>

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 $31.82^{2}$ = <em>6738.27 J</em>

<em></em>

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

for their combination, the rotational momentum is

$(I_{1} +I_{2} )w$

where the subscripts 1 and 2 indicates the values first and second flywheels

$(I_{1} +I_{2} )w$ = (6.655 + 62.72)ω

where ω here is their final angular momentum together

==> 69.375ω

Equating the two rotational momenta, we have

211.76 = 69.375ω

ω = 211.76/69.375 = 3.05 rad/s

Therefore, the energy stored in the first flywheel in this situation is

E = $Iw^{2}$ = 6.655 x $3.05^{2}$ = <em>61.908 J</em>

<em></em>

<em></em>

c) one third of the initial energy of the flywheel is

6738.27/3 = 2246.09 J

For the car, the kinetic energy = $\frac{1}{2}mv_{car} ^{2}$

where m is the mass of the car

$v_{car}$ is the velocity of the car

Equating the energy

2246.09 = $\frac{1}{2}mv_{car} ^{2}$

making m the subject of the formula

mass of the car m = $\frac{4492.18}{v_{car} ^{2} }$

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2 years ago

Which of the following statements is FALSE?

Levart [38]

A thrust fault is a reverse fault with an extremely high dip (close to 90°). This is the false statement.

Answer: Option D

<u>Explanation:</u>

Faults are the fracture or fracture zone occurring on the rocks. These fractures can travel through the rocks leading to massive destruction. So, depending upon the direction of their travel, the faults can be classified as normal, reverse and strike slip fault. Also, the angle of dip along the fault is one of the important criteria for determining the type of faults.

There is dip-slip fault which has its movement along the vertical fault plane while the strike slip fault will be in horizontal direction. Similarly, an oblique fault will be acting in both vertical and the horizontal direction. So, the fourth statement related to thrust fault is false as in reverse fault or thrust fault the dip will be shallow and not high.

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2 years ago

An electric eel (Electrophorus electricus) can produce a shock of up to 600 V and a current of 1 A for a duration of 2 ms, which

Irina-Kira [14]

Answer:

$2\times 10^{-3}\ C$

6000

1.2 J

$3.33\times 10^{-6}\ F$

Explanation:

I = Current = 1 A

t = Time = 2 ms

n = Number of electrocyte

V = Voltage = 100 mV

Charge is given by

$Q=It\\\Rightarrow Q=1\times 2\times 10^{-3}\\\Rightarrow Q=2\times 10^{-3}\ C$

The charge flowing through the electrocytes in that amount of time is $2\times 10^{-3}\ C$

The maximum potential is given by

$V_m=nV\\\Rightarrow n=\dfrac{V_m}{V}\\\Rightarrow n=\dfrac{600}{100\times 10^{-3}}\\\Rightarrow n=6000$

The number of electrolytes is 6000

Energy is given by

$E=Pt\\\Rightarrow E=V_mIt\\\Rightarrow E=600\times 1\times 2\times 10^{-3}\\\Rightarrow E=1.2\ J$

The energy released when the electric eel delivers a shock is 1.2 J

Equivalent capacitance is given by

$C_e=\dfrac{Q}{V_m}\\\Rightarrow C_e=\dfrac{2\times 10^{-3}}{600}\\\Rightarrow C_e=3.33\times 10^{-6}\ F$

The equivalent capacitance of all the electrocyte cells in the electric eel is $3.33\times 10^{-6}\ F$

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2 years ago

After soccer practice, Coach Miller goes to the roof of the school to retrieve the errant soccer balls. The height of the school

blsea [12.9K]

With gravitational acceleration at 9.8, initial height at 3.5m and distance at 22m the initial horizontal velocity is 26.03 ms and the flight time is .845 seconds

3 0

2 years ago

7. A stream of water strikes a stationary turbine blade horizontally, as the drawing illustrates. The oncoming water stream has

NNADVOKAT [17]

Answer:

The magnitude of the average force exerted on the water by the blade is 960 N.

Explanation:

Given that,

The mass of water per second that strikes the blade is, $\dfrac{m}{t}=30\ kg/s$

Initial speed of the oncoming stream, u = 16 m/s

Final speed of the outgoing water stream, v = -16 m/s

We need to find the magnitude of the average force exerted on the water by the blade. It can be calculated using second law of motion as :

$F=\dfrac{\Delta P}{\Delta t}$

$F=\dfrac{m(v-u)}{\Delta t}$

$F=30\ kg/s\times (-16-16)\ m/s$

F = -960 N

So, the magnitude of the average force exerted on the water by the blade is 960 N. Hence, this is the required solution.

6 0

2 years ago