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3 years ago

A man walks backward on a bus. To a stationary observer, the bus moves

Physics

1 answer:

NNADVOKAT [17]3 years ago

6 0

Explanation:

V(m,b)= V(m,e) - V(b,e)

2.3m/s (west) = V(m,e) - 15m/s (east)

to solve this change the direction of slower speed. so it will become negative value

-2.3m/s ( east) = V (m,e) - 15m/s (east)

V(m,e) = 15m/s (east) - 2.3m/s (east)

V(m,e) = 12.7 m/s (east)

velocity of the man in frame of reference of stationary observer is 12.7 m/s

it's slightly less than the velocity of bus so stationary observer see the man move slower than the one inside the bus

ps:- V(m,b) - velocity of man relative to bus

V(b,e) - velocity of bus relative to earth

V(m,e) - velocity of man relative to earth

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During destructive interference, two waves moving through the same medium will

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They will amplify eachother.

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3 years ago

Read 2 more answers

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

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 $3.05^{2}$ = 61.908 J

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

3 years ago

Par 1/2

BaLLatris [955]

part 1

mass = ρ x V

mass = 1739 kg/m³ x 3.8 km³ = 6608.2 kg

PE (potential energy)= mgh

PE = 6608.2 kg x 9.81 x 403

PE = 2.61 x 10⁷ J

part 2

megaton of TNT (Mt) =4.2 x 10¹⁵ J

convert PE to Mt:

2.61 x 10⁷ J : 4.2 x 10¹⁵ J = 6.21 x 10⁻⁹ Mt

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2 years ago

Un móvil se desplaza con movimiento uniforme con una rapidez de 36 Km/h ¿Cuál es la distancia recorrida al cabo de 0,5 horas?

Diano4ka-milaya [45]

Answer:

Distancia = 17,5 kilómetros.

Explanation:

Dados los siguientes datos;

Velocidad = 36 km/h

Tiempo = 0.5 horas

Para encontrar la distancia recorrida;

Distancia = velocidad * tiempo

Distancia = 35 * 0.5

Distancia = 17,5 kilómetros.

Por tanto, la distancia recorrida por el automóvil es de 17,5 kilómetros.

4 0

3 years ago

scientific endeavor should only be pursued after known tradeoffs associated with the research have been considered. True or Fals

AfilCa [17]

This is false. There were many times in history when people discovered something that they didn't even know was possible or didn't even plan to discover it. Knowing tradeoffs doesn't mean that something won't surprise you or that all will go according to plan.

8 0

3 years ago