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3 years ago

15

A massless spring with force constant ????=200N/m hangs from the ceiling. A 2.0-kg block is attached to the free end of the spri

ng and released. If the block falls 17 cm before starting back upwards, how much work is done by friction during its descent?

1 answer:

Makovka662 [10]3 years ago

8 0

Answer:

-0.4454 Joules

Explanation:

m = Mass of block = 2 kg

h = Height of extension = 17 cm = x

g = Acceleration due to gravity = 9.81 m/s²

Potential energy of the spring

$P=mgh\\\Rightarrow P=2\times 9.81\times 0.17\\\Rightarrow P=3.3354\ J$

The kinetic energy of the spring

$K=\frac{1}{2}mx^2\\\Rightarrow K=\frac{1}{2}\times 200\times 0.17^2\\\Rightarrow K=2.89\ J$

In this system as the potential and kinetic energy is conserved from work energy equivalence we get

$W=P-K\\\Rightarrow W=2.89-3.3354\\\Rightarrow W=-0.4454\ J$

The work done by friction is -0.4454 Joules

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Water flows at 0.850 m/s from a hot water heater, through a 450-kPa pressure regulator. The pressure in the pipe supplying an up

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Answer:

The velocity is $v_2= 0.45 \ m/s$

Explanation:

From the question we are told that

The initial speed of the hot water is $v_1 = 0.85 \ m/s$

The pressure from the heater $P_1 = 450 \ KPa = 450 *10^{3} \ Pa$

The height of the hot water before flowing is $h_1 = 0 \ m$

The height of bathtub above the heater is $h_2 = 3.70 \ m$

The pressure in the pipe is $P_2 = 414 KPa = 414 *10^{3} \ Pa$

The density of water is $\rho = 1000 \ kg/m^3$

Apply Bernoulli equation

$P_1 + \rho gh_1 +\frac{1}{2} \rho v_1^2 = \rho g h_2 + \frac{1}{2}\rho v_2 ^2$

Substituting values

$(450 *10^{3}) + (1000 * 9.8 * 0) + (0.5 * 1000 * 0.85^2) = (1000 * *9.8*3.70) + (0.5*1000*v_2^2 )$

=> $v_2^2 = \frac{ (450 *10^{3}) + (1000 * 9.8 *0 ) + (0.5 * 1000 * 0.85^2) -[ (1000 * *9.8*3.70) ]}{0.5*1000}$

=> $v_2= \sqrt{ \frac{ (450 *10^{3}) + (1000 * 9.8 * 0) + (0.5 * 1000 * 0.85^2) -[ (1000 * *9.8*3.70) ]}{0.5*1000}}$

=> $v_2= 0.45 \ m/s$

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2 years ago

A student using a tuning fork of frequency 512 Hz observes that the speed of sound is 340.0 m/s. What is the wavelength of this

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Answer:

0.066m

Explanation:

Step one:

given

frequency =512 Hz

The speed of sound is 340.0 m/s.

Required

The wavelength

Step two:

the formula for wavelength is

$v=f \lambda$

$\lambda= v/f$

substitute the given data

$\lambda= 340/514\\\\\lambda= 0.66m$

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3 years ago

Write the following number in scientific notation: 658000

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6.58 x 105 is the answer

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3 years ago

An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The el

Tresset [83]

Complete Question

An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field in the wire changes with time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is measured in seconds.

I = 1.2 A at time 5 secs.

Find the charge Q passing through a cross-section of the conductor between time 0 seconds and time 5 seconds.

Answer:

The charge is $Q =2.094 C$

Explanation:

From the question we are told that

The diameter of the wire is $d = 0.205cm = 0.00205 \ m$

The radius of the wire is $r = \frac{0.00205}{2} = 0.001025 \ m$

The resistivity of aluminum is $2.75*10^{-8} \ ohm-meters.$

The electric field change is mathematically defied as

$E (t) = 0.0004t^2 - 0.0001 +0.0004$

Generally the charge is mathematically represented as

$Q = \int\limits^{t}_{0} {\frac{A}{\rho} E(t) } \, dt$

Where A is the area which is mathematically represented as

$A = \pi r^2 = (3.142 * (0.001025^2)) = 3.30*10^{-6} \ m^2$

So

$\frac{A}{\rho} = \frac{3.3 *10^{-6}}{2.75 *10^{-8}} = 120.03 \ m / \Omega$

Therefore

$Q = 120 \int\limits^{t}_{0} { E(t) } \, dt$

substituting values

$Q = 120 \int\limits^{t}_{0} { [ 0.0004t^2 - 0.0001t +0.0004] } \, dt$

$Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | t} \atop {0}} \right.$

From the question we are told that t = 5 sec

$Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | 5} \atop {0}} \right.$

$Q = 120 [ \frac{0.0004(5)^3 }{3} - \frac{0.0001 (5)^2}{2} +0.0004(5)] }$

$Q =2.094 C$

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3 years ago

Vince weighs 160 pounds, and his friend Nadir weighs 140 pounds. Nadir calculated that his weight on another planet would be abo

prisoha [69]

Answer:

C) 64lb

Explanation:

use the linearity method to find the weight of nadir on another planet, it is applied as follows.

Nadir Weight in earth ⇒ Nadir weight in another planet

Vince Weigh in eart ⇒ X

our goal is to find the weight of vince in another planet (X), for this we multiply the diagonal that continents the data and divide among the remaining

140pounds ⇒ 56lb

160pounds ⇒ X

X= $\frac{(160)(56)}{140} =64lb$

Vince weigh on the other planet is C) 64lb

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3 years ago

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