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xxTIMURxx [149]

3 years ago

13

An electric pump rated 1.5 KW lifts 200kg of water through a vertical height of 6m in 10 secs: way is the efficiency of the pump

?

1 answer:

ruslelena [56]3 years ago

8 0

Answer:

80%

Explanation:

Efficiency = Power output / Power input × 100 %

To calculate efficiency we need to find power output of electric pump.

We can use,

Work done = Energy change

Work done per second = Energy change per second

Work done per second = Power

Therefore, Power = Energy change per second

= Change in potential energy of water per second

=mgh / t

= 200× 10×6 / 10

= 1200 W = 1.2 kW

Now use the first equation to find efficiency,

Efficiency = $\frac{1.2}{1.5}$ × 100%

= 80 %

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At what minimum speed must a roller coaster be traveling when upside down at the top of a circle so that the passengers do not f

Basile [38]

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3 years ago

A rock is suspended by a light string. When the rock is in air, the tension in the string is 51.9 N . When the rock is totally i

Luden [163]

Answer:

$\rho _{liquid}=1995.07kg/m^{3}$

Explanation:

When the rock is immersed in unknown liquid the forces that act on it are shown as under

1) Tension T by the string

2) Weight W of the rock

3) Force of buoyancy due to displaced liquid B

For equilibrium we have $T_{3}+B = W_{rock}$

$T_{3}+\rho _{Liquid}V_{rock}g$ = $W_{rock}.....(\alpha)$

When the rock is suspended in air for equilibrium we have

$T_{1}=W_{rock}....(\beta)$

When the rock is suspended in water for equilibrium we have

$T_{2}$ + $\rho _{water}V_{rock}g$ = $W_{rock}.....(\gamma)$

Using the given values of tension and solving α,β,γ simultaneously for $\rho _{Liquid}$ we get

$W_{rock}=51.9N\\31.6+1000\times V_{rock}\times g=51.9N\\\\11.4+\rho _{liquid}V_{rock}g=51.9N\\\\$

Solving for density of liquid we get

$\rho _{liquid}=\frac{51.9-11.4}{51.9-31.6}\times 1000$

$\rho _{liquid}=1995.07kg/m^{3}$

5 0

2 years ago

Co (5. 00 g) and co2 (5. 00 g) were placed in a 750. 0 ml container at 50. 0 °c. The partial pressure of co in the container was

konstantin123 [22]

Use PV = nRT and solve for P.
n = grams/molar mass = 5.00/molar mass CO2
Ignore the SO2.

But I think the answer it 6.74 atm

But I’m not saying but is but I’m just thinking this is the answer.

Goodluck!

6 0

2 years ago

Light travels 300 000 000 m/s and one year has approximately 32 000 000 second a light year is the distance light travels in one

Lelu [443]

Explanation:

It is given that,

Speed of light, $v=300 000 000\ m/s=3\times 10^8\ m/s$

Seconds in 1 year, $t=32 000 000=32\times 10^6\ s$

We need to find the distance traveled by light in one year. Speed of an object is given by :

$v=\dfrac{d}{t}$

So,

$d=v\times t\\\\d=3\times 10^8\times 32\times 10^6\\\\d=9.6\times 10^{15}\ m$

Since,

$1\ \text{light year}=9.46\times 10^{15}\ m\\\\1\ m=\dfrac{1}{9.46\times 10^{15}}\ \text{ly}\\\\9.6\times 10^{15}\ m=\dfrac{9.6\times 10^{15}}{9.46\times 10^{15}}\\\\d=1.01\ \text{ly}$

So, the distance covered by light is 1.01 light years.

8 0

2 years ago

A particle of mass 2.37 kg is subject to a force that is always pointed towards East or West but whose magnitude changes sinusoi

Pavlova-9 [17]

Answer:

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Explanation:

<u>Dynamics</u>

When a particle of mass m is subject to a net force F, it moves at an acceleration given by

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$\displaystyle a=\frac{F}{m}=\frac{2cos1.1t}{2.37}$

$a=0.8439cos1.1t$

The instant velocity is the integral of the acceleration:

$\displaystyle v(t)=\int_{t_o}^{t_1}a.dt$

$\displaystyle v(t)=\int_{0}^{1.5}0.8439cos1.1t.dt$

Integrating

$\displaystyle v(1.5)=0.7672sin1.1t \left |_0^{1.5}$

$\displaystyle v(1.5)=0.7672(sin1.1\cdot 1.5-sin1.1\cdot 0)$

$\boxed{v(1.5)=0.7648\ m/s}$

3 0

3 years ago