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2 years ago

6

Suppose an electrical wire is replaced with one having every linear dimension doubled (i.e., the length and radius have twice th

eir original values). How is the resistance of the wire affected?

1 answer:

Flauer [41]2 years ago

4 0

Answer:

The new resistance becomes half of the initial resistance.

Explanation:

The resistance of a wire is given by :

$R=\dfrac{\rho L}{A}$

$\rho$ = resistivity of material

L and A are linear dimension

If the electrical wire is replaced with one having every linear dimension doubled i.e. l' = 2l and r' = 2r

New resistance of wire is given by :

$R'=\dfrac{\rho L'}{A'}$

$R'=\dfrac{\rho (2L)}{\pi (2r)^2}$

$R'=\dfrac{1}{2}\dfrac{\rho L}{A}$

$R'=\dfrac{1}{2}R$

The new resistance becomes half of the initial resistance. Hence, this is the required solution.

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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 = $\frac{4492.18}{v_{car} ^{2} }$

3 0

2 years ago

If you wished to warm 100 kg of water by 15 degrees celsius for your bath, how much heat would be required? (give your answer in

Anit [1.1K]

For the answer to the question above,
<span>Q = amount of heat (kJ) </span>
<span>cp = specific heat capacity (kJ/kg.K) = 4.187 kJ/kgK </span>
<span>m = mass (kg) </span>
<span>dT = temperature difference between hot and cold side (K). Note: dt in °C = dt in Kelvin </span>

<span>Q = 100kg * (4.187 kJ/kgK) * 15 K </span>
<span>Q = 6,280.5 KJ = 6,280,500 J = 1,501,075.5 cal</span>

6 0

2 years ago

an elastic cord 61 cm long when a weight of 75N hangs from it, but 85cm when a weight of 210N hangs from it. what is the spring

pishuonlain [190]

Answer:

560 N/m

Explanation:

F = kx

75 N = k (0.61 m − L)

210 N = k (0.85 m − L)

Divide the equations:

2.8 = (0.85 − L) / (0.61 − L)

2.8 (0.61 − L) = 0.85 − L

1.708 − 2.8L = 0.85 − L

0.858 = 1.8L

L = 0.477

Plug into either equation and find k.

75 = k (0.61 − 0.477)

k = 562.5

Rounded to two significant figures, k = 560 N/m.

3 0

2 years ago

How many slices of bread did each climber have to eat to compensate for the increase of the gravitational potential energy of th

N76 [4]

Answer:

So No of slices to be consumed by each person = n = 65

Explanation:

Energy released by one slice = E1

$E1=10^6\ J$

h = 8850 m ; m = 79 kg ,η= 10.5%

We know that potential energy given as

u = m g h

u = 79 x 9.81 x 8850

$u=6.8\times 10^6\ J$

we know from the defination of efficiency that, η= E(out)/E(in)

Now amount of PE has to be compensated, In our case, E(out) =u

$0.105=\dfrac{E(out)}{E(in)}$

$0.105=\dfrac{6.8\times 10^6}{E(in)}$

$E(in)=64.76\times 10^6\ J$

Let n be the number of bread slices to be consumed.

n = E(in)/E1

$n=\dfrac{64.76\times 10^6}{10^6}$

n=64.76

So No of slices to be consumed by each person = n = 65

3 0

2 years ago

A roller coaster car drops a maximum vertical distance of 35.4 m. Determine the maximum speed of the car at the bottom of that d

marissa [1.9K]

Answer:

The maximum speed of the car at the bottom of that drop is 26.34 m/s.

Explanation:

Given that,

The maximum vertical distance covered by the roller coaster, h = 35.4 m

We need to find the maximum speed of the car at the bottom of that drop. It is a case of conservation of energy. The energy at bottom is equal to the energy at top such that :

$mgh=\dfrac{1}{2}mv^2$

$v=\sqrt{2gh}$

$v=\sqrt{2\times 9.8\times 35.4}$

v = 26.34 m/s

So, the maximum speed of the car at the bottom of that drop is 26.34 m/s. Hence, this is the required solution.

8 0

2 years ago