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Simora [160]
1 year ago
15

The sound level at 1.0 m from a certain talking person talking is 60 dB. You are surrounded by five such people, all 1.0 m from

you and all talking equally loud at the same time. The threshold of hearing is 1.0 × 10-12 W/m2. What sound level are you being exposed to? You can neglect any absorption, reflection, or interference of the sound. The threshold of hearing is 1.0 × 10-12 W/m2.
Physics
1 answer:
Hunter-Best [27]1 year ago
7 0

Answer:

66.98 db

Explanation:

We know that

L_T=L_S+10log(n)

L_T= Total signal level in db

n= number of sources

L_S= signal level from signal source.

L_T=60+10 log(5)

= 66.98 db

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A student practicing for a cross country meet runs 250 m in 30 s. What is the average speed
Kaylis [27]

Answer:8.3m/sec 30 sec,

Explanation:

A student practicing for a track meet, ran 250 m in 30 sec. a. What was her average speed? 250 m = 8.3 m/sec 30 sec.

5 0
1 year ago
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How long does it take for the velocity of the rain drop to reach 99% of its terminal velocity? (assume the conditions from part
vodomira [7]
If you think about it its part a and b 
3 0
2 years ago
A densly wound cylindrical coil has 210 turns per meter, a 5 cm radius, and carries 38 mA. What is the magnitude of the uniform
kaheart [24]

Answer:

(a). The magnitude of the uniform magnetic field is 10.02 μT

(b). The the magnitude of the total magnetic field is 10.78 μT.

Explanation:

Given that,

Number of turns = 210

Radius = 5 cm

Current = 38 mA

Current in the wire = 500 mA

We need to calculate the magnetic field inside a coil

Using formula of magnetic field

B=\mu_{0} ni

Put the value into the formula

B_{c}=4\pi\times10^{-7}\times210\times38\times10^{-3}

B_{c}=0.0000100\ T

B_{c}=10.02\times10^{-6}\ T

Now a straight wire is inserted into the coil that carries 500 mA. It travels down the central axis of the coil

Distance d=\dfrac{5}{2}

d=2.5\ cm

We need to calculate the magnetic field from the straight wire

Using formula of magnetic field

B_{w}=\dfrac{\mu_{0}I}{2\pi d}

Put the value into the formula

B_{w}=\dfrac{4\pi\times10^{-7}\times500\times10^{-3}}{2\times\pi\times2.5\times10^{-2}}

B_{w}=4.0\times10^{-6}\ T

This field is perpendicular to the wire.

The magnitude of magnetic field is

B=\sqrt{B_{c}^2+B_{w}^2}

B=\sqrt{(10.02\times10^{-6})^2+(4.0\times10^{-6})^2}

B=0.00001078\ T

B=10.78\times10^{-6}\ T

B=10.78\ \mu\ T

Hence, (a). The magnitude of the uniform magnetic field is 10.02 μT

(b). The the magnitude of the total magnetic field is 10.78 μT.

6 0
2 years ago
A 10. g cube of copper at a temperature T1 is placed in an insulated cup containing 10. g of water at a temperature T2. If T1 &g
Anna35 [415]

Answer:

a. The temperature of the copper changed more than the temperature of the water.

Explanation:

Because we're only considering the isolated system cube-water, the heat of the system should be constant, that implies the heat the cube loses is equal the heat the water gains (because by zero law of thermodynamics heat (Q) flows from hot body to cold body until reach thermal equilibrium and T1>T2). So:

Q_{cube}=Q_{water} (1)

But Q is related with mass (m), specific heat (c) and changes in temperature (\varDelta T)in the next way:

Q=cm\varDelta T(2)

Using (2) on (1):

c_{cooper}*m_{cooper}*\varDelta T_{cooper}=c_{water}*m_{waterer}*\varDelta T_{water}

(10g)(0.385 \frac{J}{g\,C})(\varDelta T_{cooper})=(10g)(4.186 \frac{J}{g\,C})(\varDelta T_{water})

(0.385 \frac{J}{g\,C})(\varDelta T_{cooper})=(4.186 \frac{J}{g\,C})(\varDelta T_{water})

Because we have an equality and 0.385 < 4.186 then \varDelta T_{cooper}>\varDelta T_{waterer} to conserve the equality

4 0
2 years ago
The U.S. Department of Energy had plans for a 1500-kg automobile to be powered completely by the rotational kinetic energy of a
navik [9.2K]

Answer:

230

Explanation:

\omega = Rotational speed = 3600 rad/s

I = Moment of inertia = 6 kgm²

m = Mass of flywheel = 1500 kg

v = Velocity = 15 m/s

The kinetic energy of flywheel is given by

K=\dfrac{1}{2}I\omega^2\\\Rightarrow K=\dfrac{1}{2}6\times 3600^2\\\Rightarrow K=38880000\ J

Energy used in one acceleration

K=\dfrac{1}{2}mv^2\\\Rightarrow K=\dfrac{1}{2}1500\times 15^2\\\Rightarrow K=168750\ J

Number of accelerations would be given by

n=\dfrac{38880000}{168750}\\\Rightarrow n=230.4

So the number of complete accelerations is 230

8 0
2 years ago
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