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Kitty [74]
2 years ago
6

To compare the hearing capacities of 5 of his friends, Ravi designs a simple experiment. He places a CD player at the end of a l

ong hall containing a CD with 5 well-known songs.
One by one each person hears the same 5 songs he plays softly but at a pre-determined volume.

Each person starts at a distance of 1 metre from the player and moves away from it. The distance at which a person says that he cannot hear the song any more, is noted. Others are not in the room when 1 person is hearing the songs.
Which of these steps may increase the reliability of the results obtained?

A)having each person listen to just 1 song instead of 5
B)allowing all people to stay in the room for the entire experiment
C)having each person start from far and walk towards the C
D )player Dvarying the volume for the different songs being played
Physics
1 answer:
aliya0001 [1]2 years ago
8 0

Answer:

A) having each person listen to just 1 song instead of 5.

Explanation:

Ravi could accurately use the outcome to compare the hearing capacities of his friends by this experiment if appropriate precautions are observed.

Playing 5 well known songs simultaneously could result to the interference of sounds when each participant moves away from the player. Which could naturally cause the variation in volume of the songs played with respect to the increasing distance.

Therefore, the reliability of the result from this experiment can be increased if each person listen to just 1 song instead of 5 at a predetermined volume. Each participant would be able to focus on hearing a song during the experiment.

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A vertical cylinder is divided into two parts by a movable piston of mass m. The piston and cylinder system is well insulated (t
Mekhanik [1.2K]

Answer:

Final temperature will be 438.076 K

Explanation:

We have given temperature T_1=323K

Volume V_1=V\ and\ V_2=\frac{V}{2}

As there is no heat transfer so this is an adiabatic process

For and adiabatic process TV^{\gamma -1}=constant

Here \gamma =1.4

So T_1V_1^{\gamma -1}=T_2V_2^{\gamma -1}

T_2=\left ( \frac{V_1}{V_2} \right )^{\gamma -1}\times T_1

T_2=\left ( \frac{V}{\frac{V}{2}} \right )^{1.4 -1}\times 332=2^{0.4}\times 332=438.076K

4 0
2 years ago
An inventive child named Nick wants to reach an apple in a tree without climbing the tree. Sitting in a chair connected to a rop
Oduvanchick [21]

UHHH WHAT? I DONT GET THAT AT ALLOW

5 0
2 years ago
Un niño de 25 kg corre por un jardín con una velocidad de 2.5 m/s de forma que su trayectoria es tangente al borde de un carruse
Schach [20]

Answer:

La velocidad angular del niño y del carrusel cuando se mueven juntos es 0.208 radianes por segundo.

Explanation:

Asumamos que tanto el niño como el carrusel no tienen carga externa aplicada sobre aquellos, de modo que se puede aplicar el Principio de Conservación de la Cantidad de Movimiento Angular:

m\cdot v \cdot R = (m\cdot R^{2}+I)\cdot \omega (1)

Donde:

m - Masa del niño, medida en kilogramos.

v - Velocidad lineal inicial del niño, medida en metros por segundo.

R - Radio máximo del carrusel, medida en metros.

I - Momento de inercia del carrusel, medida en kilogramo-metros cuadrados.

\omega - Velocidad angular final del sistema niño-carrusel, medida en radianes por segundo.

Si sabemos que m = 25\,kg, v = 2.5\,\frac{m}{s}, R = 2\,m y I = 500\,kg\cdot m^{2}, tenemos que la velocidad angular final es:

\omega = \frac{m\cdot v\cdot R}{m\cdot R^{2}+I}

\omega = \frac{(25\,kg)\cdot \left(2.5\,\frac{m}{s} \right)\cdot (2\,m)}{(25\,kg)\cdot (2\,m)^{2}+500\,kg\cdot m^{2}}

\omega = 0.208\,\frac{rad}{s}

La velocidad angular del niño y del carrusel cuando se mueven juntos es 0.208 radianes por segundo.

4 0
2 years ago
A skateboarder travels on a horizontal surface with an initial velocity of 3.2 m/s toward the south and a constant acceleration
Naddika [18.5K]

Answer:

A. 0.432

B. -1.92

C. 1.44 units/second

D. -3.2 units/second

Explanation:

A. To calculate her x position, we just use the following equation of motion to find the distance traveled:

    s=u*t+\frac{1}{2} (a*t^2)

here s = displacement

t = time (in seconds)

a = acceleration

Solving for the distance, we get:

s = 0 * 0.6 + \frac{1}{2}(2.4 * 0.6^2)

s = 0.432 m

Since 0.432 meters east is equals to 0.432 meter in the positive x-direction, the x position is also 0.432.

B. Since the skater has a constant v - velocity of -3.2 m/s, (south means negative y axis), the total distance traveled is:

Distance = speed * time = -3.2 * 0.6 = -1.92 m

The answer is -1.92 units in the y-axis.

C. The x velocity component is the final speed in the east direction, which is going to be:

v^2 - u^2=2*a*s

v^2 = 2*2.4*0.432

v = 1.44 units/second (in positive x direction)

D. Her y velocity component does not change, since the velocity towards the south is a constant 3.2 m/s

Thus the answer is -3.2 units/second in the y-axis.

8 0
2 years ago
A homeowner is trying to move a stubborn rock from his yard. By using a a metal rod as a lever arm and a fulcrum (or pivot point
finlep [7]

Answer:

1.17894 m

Explanation:

The rock is at one end of the rod which is 0.211 m from the fulcrum

F = Force

d = Distance

L = Length of rod

M = Mass of rock = 325 kg

g = Acceleration due to gravity = 9.81 m/s²

Torque

\tau=F\times d

Torque of man

\tau_m=F(L-d)\\\Rightarrow \tau_m=695(L-0.211)

Torque of rock

\tau_r=Mg\times d\\\Rightarrow \tau=325\times 9.81\times 0.211\\\Rightarrow \tau=672.72075\ Nm

The torques acting on the system is conserved

\tau_m=\tau_r\\\Rightarrow 695(L-0.211)=672.72075\\\Rightarrow L-0.211=\frac{672.72075}{695}\\\Rightarrow L-0.211=0.96794\\\Rightarrow L=0.96794+0.211\\\Rightarrow L=1.17894\ m

The length of the rod is 1.17894 m

5 0
2 years ago
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