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saw5 [17]
2 years ago
8

According to US government regulations, the maximum sound intensity level in the workplace is 90.0 dB. Within one factory, 32 id

entical machines produce a sound intensity level of 92.0 dB. How many machines must be removed to bring the factory into compliance with the regulation?
Physics
1 answer:
sergejj [24]2 years ago
3 0

Answer:

12 machines

Explanation:

I_0 = Threshold of sound = 10^{-12}\ W/m^2

90=10log\dfrac{I}{I_0}\\\Rightarrow 9=log\dfrac{I}{10^{-12}}\\\Rightarrow 10^9=\dfrac{I}{10^{-12}}\\\Rightarrow I=10^9\times 10^{-12}\\\Rightarrow I=10^{-3}\ W/m^2

92=10log\dfrac{I}{I_0}\\\Rightarrow 9.2=log\dfrac{I}{10^{-12}}\\\Rightarrow 10^{9.2}=\dfrac{I}{10^{-12}}\\\Rightarrow I=10^{9.2}\times 10^{-12}\\\Rightarrow I=10^{-2.8}\ W/m^2

One machine makes

I_m=\dfrac{10^{-2.8}}{32}=0.0000495279122644\ W/m^2

Number of machines

n=\dfrac{10^{-3}}{\dfrac{10^{-2.8}}{32}}=20.19\approx 20\ machines

Number of machines that are needed to be removed = 32-20 = 12 machines

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eimsori [14]

The force of friction is 19.1 N

Explanation:

According to Newton's second law, the net force acting on the bag is equal to the product between its mass and its acceleration:

\sum F = ma

where

\sum F is the net force

m is the mass

a is the acceleration

The bag is moving at constant speed, so its acceleration is zero:

a=0

Therefore the net force is zero as well:

\sum F = 0

Here we are interested only in the forces acting along the horizontal direction, therefore the net force is given by:

\sum F = F cos \theta - F_f = 0

where

F cos \theta is the horizontal component of the applied force, with

F = 22.5 N

\theta=32.0^{\circ}

F_f is the force of friction

And solving for F_f, we find

F_f =Fcos \theta=(22.5)(cos 32.0^{\circ})=19.1 N

Learn more about friction:

brainly.com/question/6217246

brainly.com/question/5884009

brainly.com/question/3017271

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7 0
2 years ago
Jade and her roommate Jari commute to work each morning, traveling west on I-10. One morning Jade left for work at 6:45 A.M., bu
Veronika [31]

Answer:

Jari

Explanation:

The question requires to know who is traveling faster. This is done by comparing the gradients. The steeper the slope (high gradient), the faster the speed and vice versa.

From Jari's line, the starting point is (0, 0) and another point is (6, 7)

The gradient being change in y to change in x

Change in y=7-0=7

Change in x=6-0=6

Slope is 7/6

For Jade, first point is (0, 10) then another point is (6, 16)

Change in y=16-10=6

Change in x=6-0=6

Slope is 6/6=1

Clearly, 7/6 is greater than 6/6 or 1 hence Jari is faster than Jade

3 0
2 years ago
Two long straight wires enter a room through a window. One carries a current of 2.9A into the room, while the other carries a cu
Degger [83]

Answer and Explanation:

curents i = 2.9 A

           i ' = 4.4 A

the magnitude (in T.m) of the path integral of B.dl around the window frame = μo * current enclosed

          = μo* ( i '- i )

Since from Ampere's law

where μ o = permeability of free space = 4π * 10 ^-7 H / m

plug the values we get the magnitude (in T.m) of the path integral of B.dl = ( 4π*10^-7 ) (2.9+4.4)

                                 = 1.884 * 10^-6 Tm

4 0
2 years ago
A farmer wants to determine which of two brands of cow feed is best for the cows on a farm. Before using one of the feeds on all
Effectus [21]

Answer:

the answer is A

Explanation:

6 0
2 years ago
Read 2 more answers
Each shot of the laser gun most favored by Rosa the Closer, the intrepid vigilante of the lawless 22nd century, is powered by th
mote1985 [20]

Answer:

U = 1794.005 × 10⁶ J

Explanation:

Data provided;

Capacitance of the original capacitor, C = 1.27 F

Potential difference applied to the original capacitor, V = 59.9 kV

= 59.9 × 10³ V

Now,

The Potential energy (U) for the capacitor is calculated as:

Potential energy of the original capacitor, U = \frac{\textup{1}}{\textup{2}}  × C × V²

on substituting the respective values, we get

U = \frac{\textup{1}}{\textup{2}}  × 1.27 × ( 59.9 × 10³ )²

or

U = 1794.005 × 10⁶ J

7 0
2 years ago
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