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2 years ago

8

Water is pumped from a lower reservoir to a higher reservoir by a pump that provides 20 kW of shaft power. The free surface of t

he upper reservoir is 45 m higher than that of the lower reservoir. If the flow rate of water is measured to be 0.03 m3/s, determine mechanical power that is converted to thermal energy during this process due to frictional effects.

1 answer:

kifflom [539]2 years ago

7 0

Answer:

6.5 kW

Explanation:

Input power = 20 kW = 20000 W

h = 45 m

Volume flow per second = 0.03 m^3 /s

mass flow per second = volume flow per second x density of water

= 0.03 x 1000 = 30 kg/s

Output power = m g h / t = 30 x 10 x 45 = 13500 W

Power converted in form of heat = Input power - Output power

= 20000 - 13500 = 6500 W = 6.5 kW

Thus, mechanical power converted into heat is 6.5 kW.

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Ken received a 66 on his first math exam, which counted for 20% of his final grade; he now believes that he won't be able to pas

pogonyaev

Answer:

His conclusion best illustrates a pessimistic outlook.

Explanation:

As seen in the question above, Ken got 20% of his final grade in the first test he did for this class, that is, there will be other tests that can provide him to reach the grade needed to pass the class. However, even if there are possibilities, he believes that he will not pass the class, he does not have a positive and optimistic view of his future in this class and is sure that he will fail. This negative view of the future is an example of a pessimistic outlook.

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2 years ago

A motorcyclist heading east through a small Iowa town accelerates after he passes a signpost at x=0 marking the city limits. His

adoni [48]

Answer:

1) $v_2=23\ m.s^{-1}$ & $x_2=43\ m$ east of sign post

2) $x'=55\ m$ east of sign post

3) $x_n=205\ m$ east of the signpost.

4) $v_z=35\ m.s^{-1}$

Explanation:

Given:

position of motorcyclist on entering the city at the signpost, $x_0=0\ m$

time of observation after being at x=5m east of the signpost, $t_m=0\ s$

constant acceleration of the on entering the city, $a=4\ m.s^{-2}$

distance of the motorcyclist moments later after entering, $s_m=5\ m$

velocity of the motorcyclist moments later after entering, $u_m=15\ m.s^{-1}$

<u>Now the initial velocity on at the sign board:</u>

$u_m^2=u^2+2.a.x_m$

where:

$u=$ initial velocity of entering the city at the signpost

Putting respective values:

$15^2=u^2+2\times 4\times 5$

$u=13.6015\ m.s^{-1}$

1)

Position at time $t_2=2\ s$ sec.:

Using equation of motion,

$x_2=u_m.t_2+\frac{1}{2} a.(t_2)^2+5$ because it has already covered 5m before that point

$x_2=15\times 2+0.5\times 4\times 2^2+5$

$x_2=43\ m$ east of sign post

Velocity at time $t_2=2\ s$ sec.:

$v_2=u_m+a.t_2$

$v_2=15+4\times 2$

$v_2=23\ m.s^{-1}$

2)

Position when the velocity is $v'=25\ m.s^{-1}$ :

using equation of motion,

$v'^2=u_m^2+2.a.x'+5$

$25^2=15^2+2\times 4\times x'+5$

$x'=55\ m$ east of sign post

3)

Given that:

acceleration be, $a_n=2\ m.s^{-2}$

time, $t_n=5\ s$

Position after the new acceleration and the new given time:

using equation of motion,

$x_n=u_m.t_n+\frac{1}{2} a_n.(t_n)^2+5$

$x_n=15\times 5+0.5\times 2\times 5^2+5$

$x_n=205\ m$ east of the signpost.

4)

now time of observation, $t_z=5\ s$

$v_z=u_m+a.t_z$

$v_z=15+4\times 5$

$v_z=35\ m.s^{-1}$

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2 years ago

A 4 kg box is on a frictionless 35° slope and is connected via a massless string over a massless, frictionless pulley to a hangi

Anarel [89]

Answer:

(a) 19.62 N

(b) Box moves down the slope

(c) 24.43 N

Explanation:

(a)

2 Kg box causes tension

$T=mgwhere m is mass, g is gravitational force taken as 9.81T=2*9.81 =19.62 N (b) Block mass of 4 Kg [tex]T'-mg sin \theta=0$ hence $T'=mg sin \theta$ where m is mass and g is gravitational force

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Since T' is greater than $mg sin\theta$ , then the box moves down the slope

(c)

Acceleration a= $\frac {forward force-backward force}{Total mass}= \frac {mg sin \theta -mg}{m1 + m2}$

$a= \frac {22.51-19.62}{2+4}=0.48$

When moving, the box will exert force T"= $mgsin \theta + ma$

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2 years ago

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Answer:

Kathmandu

Explanation:

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2 years ago

A car is pulled with a force of 10,000 N. The car's mass is 1267 kg. But, the car covers 394.6 m in 15 seconds

solong [7]

A. Formula: F=ma or F/m=a

10,000N/1,267kg≈7.9m/ $s^{2}$

B. Formula: a= $\frac{V-V_{0} }{t}$ and s=d/t

speed= 394.6/15

s=26.3m/s

a= $\frac{26.3-0}{15}$

a=1.75m/ $s^{2}$

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D. The force that most likely caused this difference is friction forces

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1 year ago

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