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2 years ago

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A 2.0 kg pendulum has an initial total energy of 20 J. Calculate the energy lost as heat if the pendulum is 0.10 m high and is t

raveling at 4.0 m/s.
A) 2.0J
B) 18J
C) 14J
D)16J

1 answer:

Maurinko [17]2 years ago

4 0

The correct answer is (A) 2.0 J

Total energy of the pendulum is the sum of its kinetic and potential energy. At the instant of time, when the pendulum is at a height <em>h</em> and has a speed <em>v, </em>Its energy is given by,

$E=mgh+\frac{1}{2} mv^2$

Substitute 2.0 kg for <em>m</em>, the mass of the pendulum, 9.81 m/s² for <em>g</em>, the acceleration due to gravity, 0.10 m for <em>h and 4.0 m/s for </em>v<em>.</em>

$E=mgh+\frac{1}{2} mv^2\\ =(2.0kg)(9.81m/s^2)(0.10 m)+\frac{1}{2}(2.0kg)(4.0m/s)^2\\ =17.962J$

The pendulum has an initial energy of 20 J. the energy lost is given by,

$\Delta E=(20J)-(17.962J)\\ =2.038J=2.0J$

Thus, the energy lost by the pendulum is (A) 2.0 J

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a solution made by dissolving 116 g of CaCl2 in 64 g of water has a density of 1.180 g/ml at 20 degrees celsius what is the mola

Makovka662 [10]

Answer:

The molarity of CaCl2 in the solution is 4.94 M

Explanation:

First of all you need to calculate CaCl2 mass.

You have one atom of Ca = 40.07 g/mol and two atoms of Cl = 35.45 g/mol so the molecule has a mass of 110.97 g/mol.

Now, knowing that your solution will have a mass of 64 grams of water + 116 grams of CaCl2 = 180 grams, you can calculate its volume, knowing that density = mass/volume

density x mass = volume --> 1,180 g/ml x 180 g = 212.4 ml

In 212.4 ml, you have 116 grams of CaCl2. You can calculate how many moles of CaCl2 you have:

110.97 g ------ 1 mol

116 g -------- x = (116 g x 1 mol) / 110.97 g = 1.05 moles

The molarity in a solution equals how many moles of a certain solute you have in 1000 ml of solution. In this solution, you have 1.05 moles in 212.4 ml, so in 1000 ml you will have:

212.4 ml ------- 1.05 moles

1000 ml -------- x = (1000 ml x 1.05 moles) / 212.4 ml = 4.94 moles.

This means the molarity of CaCl2 in the solution is 4.94 M.

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3 years ago

A railroad car moving at a speed of 3.46 m/s overtakes, collides and couples with two coupled railroad cars moving in the same d

I am Lyosha [343]

Answer:

$2.09\ \text{m/s}$

$22298.4\ \text{J}$

Explanation:

m = Mass of each the cars = $1.6\times 10^4\ \text{kg}$

$u_1$ = Initial velocity of first car = 3.46 m/s

$u_2$ = Initial velocity of the other two cars = 1.4 m/s

v = Velocity of combined mass

As the momentum is conserved in the system we have

$mu_1+2mu_2=3mv\\\Rightarrow v=\dfrac{u_1+2u_2}{3}\\\Rightarrow v=\dfrac{3.46+2\times 1.4}{3}\\\Rightarrow v=2.09\ \text{m/s}$

Speed of the three coupled cars after the collision is $2.09\ \text{m/s}$ .

As energy in the system is conserved we have

$K=\dfrac{1}{2}mu_1^2+\dfrac{1}{2}2mu_2^2-\dfrac{1}{2}3mv^2\\\Rightarrow K=\dfrac{1}{2}\times 1.6\times 10^4\times 3.46^2+\dfrac{1}{2}\times 2\times 1.6\times 10^4\times 1.4^2-\dfrac{1}{2}\times 3\times 1.6\times 10^4\times 2.09^2\\\Rightarrow K=22298.4\ \text{J}$

The kinetic energy lost during the collision is $22298.4\ \text{J}$ .

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3 years ago

Why do nations should establish a set of rules and principles for responsible lunar/moon explortions? ASAP pleaseeee :(

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So that we do not contaminate it with microorganisms or garbage or other human stuff.

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2 years ago

You are listening to the radio when one of your favorite songs comes on, so you turn up the volume. If you managed to increase t

andrew-mc [135]

To solve this problem we need to apply the corresponding sound intensity measured from the logarithmic scale. Since in the range of intensities that the human ear can detect without pain there are large differences in the number of figures used on a linear scale, it is usual to use a logarithmic scale. The unit most used in the logarithmic scale is the decibel yes described as

$\beta_{dB} = 10log_{10} \frac{I}{I_0}$

Where,

I = Acoustic intensity in linear scale

$I_0$ = Hearing threshold

The value in decibels is 17dB, then

$17dB = 10log_{10} \frac{I}{I_0}$

Using properties of logarithms we have,

$\frac{17}{10} = log_{10} \frac{I}{I_0}$

$log_{10} \frac{I}{I_0} = 1.7$

$\frac{I}{I_0} = 10^{1.7}$

$\frac{I}{I_0} = 50.12 W/m^2$

Therefore the factor that the intensity of the sound was $50.12W/m^2$

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3 years ago

A rocket is launched at an angle of 53.0° above the horizontal with an initial speed of 103 m/s. The rocket moves for 3.00 s alo

Serggg [28]

Before the engines fail $(0\le t\le3.00\,\rm s)$ , the rocket's horizontal and vertical position in the air are

$x=\left(103\,\frac{\rm m}{\rm s}\right)\cos53.0^\circ\,t+\dfrac12\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\cos53.0^\circ t^2$

$y=\left(103\,\frac{\rm m}{\rm s}\right)\sin53.0^\circ\,t+\dfrac12\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\sin53.0^\circ t^2$

and its velocity vector has components

$v_x=\left(103\,\frac{\rm m}{\rm s}\right)\cos53.0^\circ+\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\cos53.0^\circ t$

$v_y=\left(103\,\frac{\rm m}{\rm s}\right)\sin53.0^\circ+\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\sin53.0^\circ t$

After $t=3.00\,\rm s$ , its position is

$x=273\,\rm m$

$y=362\,\rm m$

and the rocket's velocity vector has horizontal and vertical components

$v_x=120\,\frac{\rm m}{\rm s}$

$v_y=159\,\frac{\rm m}{\rm s}$

After the engine failure $(t>3.00\,\rm s)$ , the rocket is in freefall and its position is given by

$x=273\,\mathrm m+\left(120\,\frac{\rm m}{\rm s}\right)t$

$y=362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)t-\dfrac g2t^2$

and its velocity vector's components are

$v_x=120\,\frac{\rm m}{\rm s}$

$v_y=159\,\frac{\rm m}{\rm s}-gt$

where we take $g=9.80\,\frac{\rm m}{\mathrm s^2}$ .

a. The maximum altitude occurs at the point during which $v_y=0$ :

$159\,\frac{\rm m}{\rm s}-gt=0\implies t=16.2\,\rm s$

At this point, the rocket has an altitude of

$362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)(16.2\,\rm s)-\dfrac g2(16.2\,\rm s)^2=1650\,\rm m$

b. The rocket will eventually fall to the ground at some point after its engines fail. We solve $y=0$ for $t$ , then add 3 seconds to this time:

$362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies t=34.6\,\rm s$

So the rocket stays in the air for a total of $37.6\,\rm s$ .

c. After the engine failure, the rocket traveled for about 34.6 seconds, so we evalute $x$ for this time $t$ :

$273\,\mathrm m+\left(120\,\frac{\rm m}{\rm s}\right)(34.6\,\rm s)=4410\,\rm m$

5 0

3 years ago