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2 years ago

7

w strong are you? Could you turn a moving car with just your own strength? No way! If a normal driver inputs about 50 N of force

to turn a car, the output force is 3,750 Newtons! What is the MA of the steering wheel?

1 answer:

malfutka [58]2 years ago

3 0

The mechanical advantage of a simple machine is the measure of its amplified force gain.

The mechanical advantage is defined as the force amplified by a machine to the force required to generate such output.

$Mathematically\ mechanical\ advantage\ MA=\frac{F_{o}} {F_{i}}$

$F_{o} \ and\ F_{i}$ are the amplified force and applied force. We may also consider them as output and input force.

In the given question, the force given to the steering wheel is 50 N.

The output force produced by the steering wheel is 3750 N.

Hence the mechanical advantage will be-

$MA=\frac{F_{o}} {F_{i}}$

$=\frac{3750\ N}{50\ N}$

$=75$ [ans]

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An object of mass m = 5.0 kg hangs from a cord around a light pulley: The length of the cord between the oscillator and the pull

puteri [66]

Answer:

$\mu=0.0049Kg/m$

Explanation:

When a standing wave is formed with six loops means the normal mode of the wave is n=6, the frequency of the normal mode is given by the expression:

$f_n=\frac{nv}{2L}$

Where $L$ is the length of the string and $v$ the velocity of propagation. Use this expression to find the value of $v$ .

$f_6=\frac{6v}{2L}\\(150)=\frac{6v}{2(2)} \\150=\frac{3v}{2} \\3v=150(2)\\ v=\frac{300}{3} \\v=100m/s$

The velocity of propagation is given by the expression:

$v=\sqrt{\frac{T}{\mu }$

Where $\mu$ is the desirable variable of the problem, the linear mass density, and $T$ is the tension of the cord. The tension is equal to the weight of the mass hanging from the cord:

$T=W=mg=(5)(9.81)=49.05N$

With the value of the tension and the velocity you can find the mass density:

$v=\sqrt{\frac{T}{\mu}$

$v^2=\frac{T}{\mu}\\ \mu=\frac{T}{v^2} =\frac{49.05}{(100)^2} =\frac{49.05}{10000} =0.0049Kg/m$

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2 years ago

Jack pushed and pushed, and finally moved the new refrigerator into the kitchen. What is the BEST explanation of what happened?

Anika [276]

Answer:

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Explanation:

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3 years ago

A 5.0 A electric current passes through an aluminum wire of 4.0~\times~10^{-6}~m^2 cross-sectional area. Aluminum has one free e

Serhud [2]

Answer: The electron number density (the number of electrons per unit volume) in the wire is $6.0 \times 10^{28} m^{-3}$ .

Explanation:

Given: Current = 5.0 A

Area = $4.0 \times 10^{-6} m^{2}$

Density = 2.7 $g/cm^{3}$ , Molar mass = 27 g

The electron density is calculated as follows.

$n = \frac{density}{mass per atom}\\= \frac{\rho}{\frac{M}{N_{A}}}\\$

where,

$\rho$ = density

M = molar mass

$N_{A}$ = Avogadro's number

Substitute the values into above formula as follows.

$n = \frac{\rho \times N_{A}}{M}\\= \frac{2.7 g/cm^{3} \times 6.02 \times 10^{23}/mol}{27 g/mol}\\= \frac{16.254 \times 10^{23}}{27} cm^{3}\\= 0.602 \times 10^{23} \times \frac{10^{6} cm^{3}}{1 m^{3}}\\= 6.0 \times 10^{28} m^{-3}$

Thus, we can conclude that the electron number density (the number of electrons per unit volume) in the wire is $6.0 \times 10^{28} m^{-3}$ .

8 0

2 years ago

Consider a series LRC-circuit in which C-120.0 uF. When driven at a frequency w = 200.0 rad s-1 the com ples impedance is given

Lina20 [59]

Answer:

(a). (i). The reactants are $X_{L} =31.66\ \Omega$ .

(II). The inductance of the circuit is 0.1583 Henry.

(b). The resonant angular frequency is 229.4 rad/s.

Explanation:

Given that,

Capacitor = 120.0 μC

Frequency = 200.0 rad/s

Impedance = 100.0 -10j

(I). We need to calculate the $X_{C}$

$X_{C}=\dfrac{1}{C\times\omega}$

Put the value into the formula

$X_{C}=\dfrac{1}{120\times10^{-6}\times200}$

$X_{C}=41.66\ \Omega$

(II). We know that,

Formula of impedance is

$Z=\sqrt{R^2+X_{L}^2+X_{C}^2}$ ...(I)

Given equation of impedance is

$Z=(100-10j)$ ...(II)

On Comparing of equation (I) and (II)

$R = 100$

$X_{L}-X_{C}=-10$

Now, put the value of $X_{C}$

$X_{L=41.66-10$

$X_{L}=31.66\ \Omega$

We need to calculate the inductance

Using formula of inductance

$X_{L}=\omega\times L$

Put the value into the formula

$L=\dfrac{X_{L}}{\omega}$

$L=\dfrac{31.66}{200}$

$L=0.1583\ Henry$

(b). We need to calculate the resonant angular frequency

Using formula of the resonant angular frequency

$angular\ frequency =\dfrac{1}{\sqrt{L\times C}}$

$angular\ frequency =\dfrac{1}{\sqrt{0.1583\times120\times10^{-6}}}$

$angular\ frequency =229.4\ rad/s$

Hence, (a). (i). The reactants are $X_{L} =31.66\ \Omega$ .

(II). The inductance of the circuit is 0.1583 Henry.

(b). The resonant angular frequency is 229.4 rad/s.

4 0

3 years ago

Point charges q1=− 4.40 nC and q2=+ 4.40 nC are separated by distance 3.10 mm , forming an electric dipole. The charges are in a

statuscvo [17]

Answer:

988.39 N/C

Explanation:

First, we start by finding the magnitude of the electric dipole moment, as it's going to be needed

p = 4.40*10^-9 * 3.10*10^-3

p = 1.364*10^-11 Cm

The charges are in a uniform electric field E whose direction makes an angle of 36.4° with the line connecting the charges.

Magnitude of torque exerted on the dipole,

t = 8.0×10−9 N.m

pEsinO = t, making E subject of formula

E = t / psinO

E = 8.0*10^-9/1.364*10^-11 (sin36.4)

E = 8.0*10^-9/1.364*10^-11 * (0.5934)

E = 8.0*10^-9 / 8.094*10^-12

E = 988.39 N/C

the magnitude of this field is E = 988.39 N/C

3 0

2 years ago