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Natalija [7]
3 years ago
14

In a supermarket, you place a 22.3-N (around 5 lb) bag of oranges on a scale, and the scale starts to oscillate at 2.7 Hz. What

is the force constant (spring constant) of the spring of the scale?
Physics
1 answer:
allsm [11]3 years ago
7 0

Answer:

Force constant, k = 653.3 N/m

Explanation:

It is given that,

Weight of the bag of oranges on a scale, W = 22.3 N

Let m is the mass of the bag of oranges,

m=\dfrac{W}{g}

m=\dfrac{22.3}{9.8}

m = 2.27 kg

Frequency of the oscillation of the scale, f = 2.7 Hz

We need to find the force constant (spring constant) of the spring of the scale. We know that the formula of the frequency of oscillation of the spring is given by :

f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}

k=4\pi^2 f^2m

k=4\pi^2 \times (2.7)^2\times 2.27

k = 653.3 N/m

So, the force constant of the spring of the scale is 653.3 N/m. Hence, this is the required solution.

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The howler monkey is the loudest land animal and, under some circumstances, can be heard up to a distance of 8.9 km. Assume the
exis [7]

Answer:

113.7

Explanation:

maximum distance (s) = 8.9 km

reference intensity (I0) = 1 x 10^{-12} W/m^{2}

power of a juvenile howler monkey (p) = 63 x 10^{-6} W

distance (r) = 210 m

intensity (I) = power/area

where we assume the area of a sphere due to the uniformity of the output in all directions

area = 4πr^{2} =  4π x 210^{2} = 554,176.9 m^{2}

intensity (I) = \frac{63 x 10^{-6} }{554,176.9} = 113.7 x 10^{-12}

therefore the desired ratio I/I0 = \frac{113.7 x 10^{-12}}{1 x 10^{-12}} = 113.7

7 0
2 years ago
How would you solve for x I cant remember right now 4x+6x=9x-10
-Dominant- [34]
Combine all of the x's on one side of the equation and then finish the problem!
8 0
2 years ago
Read 2 more answers
The dial of a scale looks like this: 00.0kg. A physicist placed a spring on it. The dial read 00.6kg. He then placed a metal cha
saveliy_v [14]

Answer:

d. The scale's resolution is too low to read the change in mass

Explanation:

If we want to find the change in energy of the spring, we will have to use the Hooke's Law. Hooke's Law states that:

F = kx

since,

w = Fd

dw = Fdx

integrating and using value of F, we get:

ΔE = (0.5)kx²

where,

ΔE = Energy added to spring

k = spring constant

x = displacement

The spring constant is typically in range of 4900 to 29400 N/m.

So if we take the extreme case of 29400 N/m and lets say we assume an unusually, extreme case of 1 m compression, we get the value of energy added to be:

ΔE = (0.5)(29400 N/m)(1 m)²

ΔE = 1.47 x 10⁴ J

Now, if we convert this energy to mass from Einstein's equation, we get:

ΔE = Δmc²

Δm = ΔE/c²

Δm = (1.47 x 10⁴ J)/(3 x 10⁸ m/s)²

<u>Δm =  4.9 x 10⁻¹³ kg</u>

As, you can see from the answer that even for the most extreme cases the value of mass associated with the additional energy is of very low magnitude.

Since, the scale only gives the mass value upto 1 decimal place.

Thus, it can not determine such a small change. So, the correct option is:

<u>d. The scale's resolution is too low to read the change in mass</u>

8 0
2 years ago
refrigerant 134a enters a compressor operating at steady state as saturated vapor at 0.12 MPa and exits at 1.2 MPa and 70 C at a
Afina-wow [57]

Answer:

the power input to the compressor is 7.19Kw

Explanation:

Hello!

To solve this problem follow the steps below.

1. We will call 1 the refrigerant state at the compressor inlet and 2 at the outlet.

2. We use thermodynamic tables to determine enthalpies in states 1 and 2.

(note: Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)  

through prior knowledge of two other properties such as pressure and temperature.  )

h1[quality=1, P=0.12Mpa)=237KJ/Kg

h2(P=1.2Mpa, t=70C)=300.6KJ/kg

3. uses the first law of thermodynamics in the compressor that states that the energy that enters a system is the same that must come out

Q=heat=0.32kJ/s

W=power input to the compressor

m=mass flow=0.108kg/S

m(h1)+W=Q+m(h2)

solving for W

W=Q+m(h2-h1)

W=0.32+0.108(300.6-237)=7.19Kw

the power input to the compressor is 7.19Kw

7 0
2 years ago
Juan is standing on the street. An ambulance moves toward him and then passes by. What best describes the pitch that Juan hears?
ozzi

Answer: The correct answer is "the pitch drops to a lower pitch once the ambulance passes by Juan".

Explanation:

Doppler effect is the phenomenon in which there is an increase or decrease in the frequency when there is relative motion between the source and listener.

Pitch depends on the frequency. if the frequency increases then pitch increases.

When the source and the listener are moving towards each other then there is increase in the frequency. When the source and the listener are moving away each other then there is decrease in the frequency.

In the given problem, Juan is standing on the street. An ambulance moves toward him and then passes by the pitch drops to a lower pitch once the ambulance passes by Juan.

5 0
2 years ago
Read 2 more answers
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