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3 years ago

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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?

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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A curtain hangs straight down in front of an open window. A sudden gust of wind blows past the window; and the curtain is pulled

VikaD [51]

Answer:

option B.

Explanation:

The correct answer is option B.

The phenomenon of the curtains to pull out of the window can be explained using Bernoulli's equation.

According to Bernoulli's Principle when the speed of the moving fluid increases the pressure within the fluid decrease.

When wind flows in the outside window the pressure outside window decreases and pressure inside the room is more so, the curtain moves outside because of low pressure.

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2 years ago

Fill in the blanks to complete each statement about weathering. Weathering is the breakdown of rocks into smaller particles call

nevsk [136]

Answer:

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Explanation:

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sediments

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chemical weathering

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Hope this Helps! c:

8 1

2 years ago

Read 3 more answers

Two insulated copper wires of similar overall diameter have very different interiors. One wire possesses a solid core of copper,

balandron [24]

Answer with Explanation:

We are given that

Radius of solid core wire=r=2.28 mm= $2.28\times 10^{-3} m$

$1mm=10^{-3} m$

Radius of each strand of thin wire=r'=0.456 mm= $0.456\times 10^{-3} m$

Current density of each wire= $J=3750 A/m^2$

a.Area = $\pi r^2$

Where $\pi=3.14$

Using the formula

Cross section area of copper wire has solid core = $3.14\times (2.28\times 10^{-3})^2=16.3\times 10^{-6} m^2$

Current density = $J=\frac{I}{A}$

Using the formula

$3750=\frac{I}{16.3\times 10^{-6}}$

$I=3750\times 16.3\times 10^{-6}=0.061 A$

Total number of strands=19

Area of strand wire= $A'=19\times 3.14\times (0.456\times 10^{-3})^2=12.4\times 10^{-6} m^2$

$J'=\frac{I'}{A'}$

$3750=\frac{I'}{19\times 3.14(0.456\times 10^{-3})^2}$

$I'=3750\times 19\times 3.14(0.456\times 10^{-3})^2$

$I'=0.047 A$

b.Resistivity of copper wire= $\rho=1.69\times 10^{-8}\Omega-m$

Length of each wire =6.25 m

Resistance, R= $\frac{\rho l}{A}$

Using the formula

Resistance of solid core wire= $R=\frac{1.69\times 10^{-8}\times 6.25}{16.3\times 10^{-6}}=6.5\times 10^{-3}\Omega$

Resistance of strand wire= $R'=\frac{1.69\times 10^{-8}\times 6.25}{12.4\times 10^{-6}}=8.5\times 10^{-3}\Omega$

7 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

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2 years ago

Three identical train cars, coupled together, are rolling east at speed v0. A fourth car traveling east at 2v0 catches up with t

aliina [53]

Answer: $v_o$

Explanation:

It is given that three cars has same mass m with speed $v_o$

suppose rest two cars also has same mass m

As there is no external force therefore momentum is conserved

Initial Momentum $P_i$

$P_i=3mv_0+m(2v_0)+m\times 0$

Final momentum $P_f$

$P_f=5m\times v$

where v=final velocity

$P_i=P_f$

$5mv_o=5mv$

$v=v_o$

thus final velocity is $v_o$

8 0

2 years ago