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FromTheMoon [43]

1 year ago

7

Two loudspeakers 42.0 m apart and facing each other emit identical 115 Hz sinusoidal sound waves in a room where the sound speed

is 345 m/s. Susan is walking along a line between the speakers. As she walks, she finds herself moving through loud and quiet spots. ?

1 answer:

Tems11 [23]1 year ago

3 0

Answer:

She passes through a loud spot at x = 19.5m from the first speaker. Constructive interference occurs here.

At the quiet spot destructive interference occurs. The minimum distance for this to occur is 20.25 from the first speaker.

Explanation:

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A truck covered 2/7 of a journey at an average speed of 40

slavikrds [6]

Answer:8h

Explanation:

8 0

1 year ago

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José is pinned against the walls of the Rotor, a ride with a radius of 3.00 meters that spins so fast that the floor can be remo

zaharov [31]

r = radius of the circle of the ride = 3.00 meters

v = linear speed of the person during the ride = 17.0 m/s

m = mass of the person in angular motion in the ride

L = angular momentum of the person in the ride = 3570 kg m²/s

Angular momentum is given as

L = m v r

inserting the values

3570 kg m²/s = m (17 m/s) (3.00 m)

m = 3570 kg m²/s/(51 m²/s)

m = 7 kg

hence the mass comes out to be 7 kg

8 0

1 year ago

A small ball of mass 2.00 kilograms is moving at a velocity 1.50 meters/second. It hits a larger, stationary ball of mass 5.00 k

rewona [7]

The kinetic energy of the small ball before the collision is

   KE = (1/2) (mass) (speed)²

   = (1/2) (2 kg) (1.5 m/s)

   =    (1 kg) (2.25 m²/s²)

   =    2.25 joules.

Now is a good time to review the Law of Conservation of Energy:

   Energy is never created or destroyed.
   If it seems that some energy disappeared,
   it actually had to go somewhere.
   And if it seems like some energy magically appeared,
   it actually had to come from somewhere.

The small ball has 2.25 joules of kinetic energy before the collision.
If the small ball doesn't have a jet engine on it or a hamster inside,
and does not stop briefly to eat spinach, then there won't be any
more kinetic energy than that after the collision. The large ball
and the small ball will just have to share the same 2.25 joules.

3 0

2 years ago

A truck is traveling down a road with a 4-percent grade at a speed of 75 mi/h when its brakes are applied to slow it down to 22.

kvasek [131]

Answer:

3.964 s

Explanation:

Metric unit conversion:

1 miles = 1.6 km = 1600 m.

1 hour = 60 minutes = 3600 seconds

75 mph = 75 * 1600 / 3600 = 33.3 m/s

22.5 mph = 22.5 * 1600/3600 = 10 m/s

Let g = 9.81 m/s2

Friction is the product of coefficient and normal force, which equals to the gravity

$F_f = \mu N = \mu mg$

The deceleration caused by friction is friction divided by mass according to Newton 2nd law.

$a = F_f / m = \mu mg / m = \mu g = 0.6 *9.81 = 5.886 m/s^2$

So the time required to decelerate from 33.3 m/s to 10 m/s so the wheels don't slide, with the rate of 5.886 m/s2 is

$t = \frac{\Delta v}{a} = \frac{33.3 - 10}{5.886} = 3.964 s$

3 0

1 year ago

Consider a star that is a sphere with a radius of 6.32 108 m and an average surface temperature of 5350 K. Determine the amount

Mariulka [41]

Answer:

The value is $\Delta s = 8.537 *10^{25 } \ J/K$

Explanation:

From the we are told that

The radius of the sphere is $r = 6.32 *10^{8} \ m$

The temperature is $T_x = 5350 \ K$

The average temperature of the rest of the universe is $T_r = 2.73 \ K$

Generally the change in entropy of the entire universe per second is mathematically represented as

$\Delta s = s_r - s_x$

Here $s_r$ is the entropy of the rest of the universe which is mathematically represented as

$s_r = \frac{Q}{T_r}$

Here Q is the quantity of heat radiated by the star which is mathematically represented as

$Q = 4 \pi * r^2 * \sigma * T^4_x$

Here $\sigma$ is the Stefan-Boltzmann constant with value

$\sigma = 5.67 * 10^{-8 }W\cdot m^{-2} \cdot K^{-4}.$

=> $Q = 4 \pi * (6.32*10^{8})^2 * 5.67 * 10^{-8 } * 5350 ^4$

=> $Q = 2.332 *10^{26} \ J$

So

$s_r = \frac{2.332 *10^{26}}{2.73}$

=> $s_r = 8.5415 *10^{25}\ J/K$

Here $s_x$ is the entropy of the rest of the universe which is mathematically represented as

$s_x = \frac{Q}{T_x}$

=> $s_x = \frac{2.332 *10^{26} }{5350}$

=> $s_x = 4.359 *10^{22} \ J/K$

So

$\Delta s = 8.5415 *10^{25} - 4.359 *10^{22}$

=> $\Delta s = 8.537 *10^{25 } \ J/K$

7 0

1 year ago