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Tcecarenko [31]

3 years ago

15

A rectangular tank that is 4 meters long, 3 meters wide and 6 meters deep is filled with a rubbing alcohol that has density 786

kilograms per cubic meter. In each part below, assume that the tank is initially full, and that gravity is 9.8 meters per second squared. Your answers must include the correct units.
(a) How much work is done pumping all of the liquid out over the top of the tank?
(b) How much work is done pumping all of the liquid out of a spout 2 meters above the top of the tank?
(c) How much work is done pumping two-thirds of the liquid out over the top of the tank?
(d) How much work is done pumping two-thirds of the liquid out of a spout 2 meters above the top of the tank?

1 answer:

ki77a [65]3 years ago

4 0

Answer:

Explanation:

The explanation is given in the attached document.

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A locomotive accelerates a 25-car train along a level track. Every car has a mass of 7.7 ✕ 104 kg and is subject to a friction f

MaRussiya [10]

To solve the problem it is necessary to apply the concepts related to Force of Friction and Tension between the two bodies.

In this way,

The total mass of the cars would be,

$m_T = 25(7.7*10^4)Kg$

$m_T = 1.925*10^6Kg$

Therefore the friction force at 29Km / h would be,

$f=250v$

$f= 250*29Km/h$

$f = 250*29*(\frac{1000m}{1km})(\frac{1h}{3600s})$

$f = 2013.889N$

In this way the tension exerts between first car and locomotive is,

$T=m_Ta+f$

$T=(1.925*10^6)(0.2)+2013.889$

$T= 3.8701*10^5N$

Therefore the tension in the coupling between the car and the locomotive is $3.87*10^5N$

6 0

3 years ago

One cycle of the power dissipated by a resistor ( R = 800 Ω R=800 Ω) is given by P ( t ) = 60 W , 0 ≤ t < 5.0 s P(t)=60 W, 0≤

OLga [1]

Answer:

42.5W

Explanation:

To solve this problem we must go back to the calculations of a weighted average based on the time elapsed thus,

$Power_{avg} = \frac{P_1(t_1)+P_2(t_2)}{t_1+t_2}$

We need to calculate the average power dissipated by the 800\Omega resistor.

Our values are given by:

$P(t)=60 W, 0\leq t$

$P(t)=25 W, 5.0\leq t$

Aplying the values to the equation we have:

$Power_{avg} = \frac{P_1(t_1)+P_2(t_2)}{t_1+t_2}$

$Power_{avg} = \frac{60(5-0)+25(10-5)}{(5-0)+(10-5)}$

$Power_{avg} = 42.5W$

5 0

3 years ago

Unpolarizedlight of intensity I_0 is incident on three polarizingfilters. The axis of the first is vertical, that of the secondi

Marina86 [1]

To solve this problem it is necessary to apply the concepts related to the law of Malus which describe the intensity of light passing through a polarizer. Mathematically this law can be described as:

$I = I_0 cos^2\theta$

Where,

$I_0 =$ Indicates the intensity of the light before passing through the polarizer

I = Resulting intensity

$\theta$ = Indicates the angle between the axis of the analyzer and the polarization axis of the incident light

From the law of Malus when the light passes at a vertical angle through the first polarizer its intensity is reduced by half therefore

$I_1= \frac{I_0}{2}$

In the case of the second polarizer the angle is directly 60 degrees therefore

$I_2 = I_1 cos^2\theta$

$I_2 = (\frac{I_0}{2} ) cos^2(60)$

$I_2 = 0.125I_0$

In the case of the third polarizer, the angle is reflected on the perpendicular, therefore, its angle of index would be

$\theta_3 = 90-60 = 30$

Then,

$I_3 = I_2 cos^2\theta_3$

$I_3 = 0.125I_0 cos^2 (30)$

$I_3 = 0.09375I_0$

Then the intensity at the end of the polarized lenses will be equivalent to 0.09375 of the initial intensity.

5 0

3 years ago

What units is the Metric System of Measurement based on?

Alex787 [66]

Answer:

metre for length and the kilogram for Mass

3 0

3 years ago

Frim the castle wall 20 m high shot an arrow. The initial speed of the bow is 45 m/s directed 40 degrees above horizontal. Find

True [87]

Answer:

Range of arrow = 225.09 meter

Final horizontal velocity = 34.47 m/s

Explanation:

We have equation of motion $s=ut+\frac{1}{2} at^2$ , where u is the initial velocity, t is the time taken, a is the acceleration and s is the displacement.

Considering the vertical motion of arrow ( up direction as positive)

We have u = 45 sin40 = 28.93 m/s, s = -20 m, a = acceleration due to gravity = $-9.8m/s^2$ .

$-20=28.93*t-\frac{1}{2} *9.8*t^2\\ \\ 4.9t^2-28.93t-20=0$

t = 6.53 seconds or t = -0.63 seconds

So time = 6.53 seconds.

Considering the horizontal motion of arrow

u = 45 cos 40 = 34.47 m/s, t = 6.53 s, a = $0m/s^2$

$s=34.47*6.53+\frac{1}{2} *0*6.53^2\\ \\ s=225.09m$

So range of arrow = 225.09 meter

Horizontal velocity will not change , final horizontal velocity = 34.47 m/s.

7 0

3 years ago