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ehidna [41]
3 years ago
13

**100 POINTS**

Physics
1 answer:
lawyer [7]3 years ago
6 0

Answer:

B

Explanation:

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Two loudspeakers are placed next to each other and driven by the same source at 500 Hz. A listener is positioned in front of the
stellarik [79]

To solve this problem we will apply the concepts related to wavelength as the rate of change of the speed of the wave over the frequency. Mathematically this is

\lambda = \frac{v}{f}

Here,

v = Wave velocity

f = Frequency,

Replacing with our values we have that,

\lambda = \frac{340}{500}

\lambda = 0.68m

The distance to move one speaker is half this

\lambda/2 = 0.34m

Therefore the minimum distance will be 0.34m

7 0
3 years ago
What must be true about a surface in order for diffuse reflection to occur?
balu736 [363]

Answer:

carpet

Explanation:

Diffuse reflection is the reflection of light from a surface such that an incident ray is reflected at many angles rather than at just one angle as in the case of specular reflection.

The structure of carpet's surface is as shown. Thus it shows large amount of diffuse reflection.

4 0
3 years ago
An object accelerates in a direction that is always perpendicular to its motion.
anzhelika [568]
The object will not be able to accelerate perpendicular to direction of motion
7 0
3 years ago
A mechanic needs to replace the motor for a merry-go-round. The merry-go-round should accelerate from rest to 1.5 rad/s in 6.0s
pashok25 [27]

Answer:

109656.25 Nm

Explanation:

\omega_f = Final angular velocity = 1.5 rad/s

\omega_i = Initial angular velocity = 0

\alpha = Angular acceleration

t = Time taken = 6 s

m = Mass of disk = 29000 kg

r = Radius = 5.5 m

\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\dfrac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\dfrac{1.5-0}{6}\\\Rightarrow \alpha=0.25\ rad/s^2

Torque is given by

\tau=I\alpha\\\Rightarrow \tau=\dfrac{1}{2}mr^2\alpha\\\Rightarrow \tau=\dfrac{1}{2}29000\times 5.5^2\times 0.25\\\Rightarrow \tau=109656.25\ Nm

The torque specifications must be 109656.25 Nm

5 0
3 years ago
The initial kinetic energy imparted to a 0.25 kg bullet is 1066 J. The acceleration of gravity is 9.81 m/s 2 . Neglecting air re
lubasha [3.4K]

Answer:

The range of the bullet is 0.435 kilometers.

Explanation:

According to the problem, maximum height is equal to the range of the bullet. That is:

\Delta x = \Delta y

Where:

\Delta x - Range of the bullet, measured in meters.

\Delta y - Maximum height of the bullet, measured in meters.

By the Principle of Energy Conservation, gravitational potential energy reaches its maximum at the expense of the initial kinetic energy. That is to say:

K_{1} = U_{2}

Where:

K_{1} - Kinetic energy at point 1, measured in joules.

U_{1} - Gravitational potential energy at point 2, measured in joules, and:

U_{2} = m\cdot g \cdot \Delta y

Where:

m - Mass of the bullet, measured in kilograms.

g - Gravitational constant, measured in meters per square second.

The maximum height is now cleared:

K_{1} = m\cdot g \cdot \Delta y

\Delta y = \frac{K_{1}}{m\cdot g}

If K_{1} = 1066\,J, m = 0.25\,kg and g = 9.81\,\frac{m}{s^{2}}, the maximum height is now computed:

\Delta y = \frac{1066\,J}{(0.25\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}

\Delta y = 434.791\,m

\Delta y = 0.435\,km

Lastly, the range of the bullet is 0.435 kilometers.

3 0
3 years ago
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