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3 years ago

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A crowbar is being used with an actual mechanical advantage of 8. If the input force is 400 Newton’s on the 0.3 meter long end o

f the crowbar, what is the output force on the short end?
a) 960 N
b) 50N
C) 3,200 n
D) 120 n

1 answer:

finlep [7]3 years ago

7 0

Answer:

C. 3,200

Explanation:

MA= F out/ F in.

multiply F in on both sides to cancel it out so you can solve for F out.

Fin*MA=F out

400 * 8 = F out

3,200= F out

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Alternative sources of energy include : geothermal

baherus [9]

Answer:

All of the above

Explanation:

The correct answer is option E (All of the above)

All the options are alternative source of energy.

The option given are not the traditional way (energy production from coal) of extracting energy as the loss of natural resources does not take place in these source of energy.

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3 years ago

A 0.40 kg mass hangs on a spring with a spring constant of 12 N/m. The system oscillated with a constant amplitude of 12 cm. Wha

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Answer:

The maximum acceleration of the system is 359.970 centimeters per square second.

Explanation:

The motion of the mass-spring system is represented by the following formula:

$x(t) = A\cdot \cos (\omega \cdot t + \phi)$

Where:

$x(t)$ - Position of the mass with respect to the equilibrium position, measured in centimeters.

$A$ - Amplitude of the mass-spring system, measured in centimeters.

$\omega$ - Angular frequency, measured in radians per second.

$t$ - Time, measured in seconds.

$\phi$ - Phase, measured in radians.

The acceleration experimented by the mass is obtained by deriving the position equation twice:

$a (t) = -\omega^{2}\cdot A \cdot \cos (\omega\cdot t + \phi)$

Where the maximum acceleration of the system is represented by $\omega^{2}\cdot A$ .

The natural frequency of the mass-spring system is:

$\omega = \sqrt{\frac{k}{m} }$

Where:

$k$ - Spring constant, measured in newtons per meter.

$m$ - Mass, measured in kilograms.

If $k = 12\,\frac{N}{m}$ and $m = 0.40\,kg$ , the natural frequency is:

$\omega = \sqrt{\frac{12\,\frac{N}{m} }{0.40\,kg} }$

$\omega \approx 5.477\,\frac{rad}{s}$

Lastly, the maximum acceleration of the system is:

$a_{max} = \left(5.477\,\frac{rad}{s})^{2}\cdot (12\,cm)$

$a_{max} = 359.970\,\frac{cm}{s^{2}}$

The maximum acceleration of the system is 359.970 centimeters per square second.

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3 years ago

Science Net Forces. Could somebody help me?

tensa zangetsu [6.8K]

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3) Balanced. Both boys are pulling with the same force. Neither is winning.

4) Unbalanced. The rope will move with 10 N to the west. The teachers are winning.

5) Unbalanced. The kids are pulling 220 N to the east. The kids are winning.

6) Balanced. You and the dog are pulling with the same force.

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3 years ago

A stationary police car emits a sound of frequency 1240 HzHz that bounces off of a car on the highway and returns with a frequen

Tju [1.3M]

Answer

given,

frequency from Police car= 1240 Hz

frequency of sound after return = 1275 Hz

Calculating the speed of the car = ?

Using Doppler's effect formula

Frequency received by the other car

$f_1 = \dfrac{f_0(u + v)}{u}$ ..........(1)

u is the speed of sound = 340 m/s

v is the speed of the car

Frequency of the police car received

$f_2= \dfrac{f_1(u)}{u-v}$

now, inserting the value of equation (1)

$f_2= f_0\dfrac{u+v}{u-v}$

$1275=1240\times \dfrac{340+v}{340-v}$

1.02822(340 - v) = 340 + v

2.02822 v = 340 x 0.028822

2.02822 v = 9.799

v = 4.83 m/s

hence, the speed of the car is equal to v = 4.83 m/s

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3 years ago

A uranium and iron atom reside a distance R = 44.10 nm apart. The uranium atom is singly ionized; the iron atom is doubly ionize

Ad libitum [116K]

Answer:

distance r from the uranium atom is 18.27 nm

Explanation:

given data

uranium and iron atom distance R = 44.10 nm

uranium atom = singly ionized

iron atom = doubly ionized

to find out

distance r from the uranium atom

solution

we consider here that uranium electron at distance = r

and electron between uranium and iron so here

so we can say electron and iron distance = ( 44.10 - r ) nm

and we know single ionized uranium charge q2= 1.602 × $10^{-19}$ C

and charge on iron will be q3 = 2 × 1.602 × $10^{-19}$ C

so charge on electron is q1 = - 1.602 × $10^{-19}$ C

and we know F = $k\frac{q*q}{r^{2} }$

so now by equilibrium

Fu = Fi

$k\frac{q*q}{r^{2} }$ = $k\frac{q*q}{r^{2} }$

put here k = $9*10^{9}$ and find r

$9*10^{9}\frac{1.602 *10^{-19}*1.602 *10^{-19}}{r^{2} }$ = $9*10^{9}\frac{1.602 *10^{-19}*1.602 *10^{-19}}{(44.10-r)^{2} }$

$\frac{1}{r^{2} } = \frac{2}{(44.10 -r)^2}$

r = 18.27 nm

distance r from the uranium atom is 18.27 nm

6 0

3 years ago