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3 years ago

Mr.pointer has as mass of 50kg. his is at the top of a 10m hill. what is his PE

Physics

1 answer:

iragen [17]3 years ago

6 0

Answer:

PE=4900J

Hope it helps please mark me as brainliest

Explanation:

P.E=m×g×h

Given m = 50 Kg , h = 10 m.

P.E=50×10×9.8

P.E=50×98=4900J

Hence increase in potential energy is 4900J.

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3 years ago

A ball A of mass 0.5 kg moving with a Velacity of 10 m/s a head on Collision with a ball B of mass 2kg moving with a Velocity of

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2 years ago

In an experiment the mass of a calorimeter is 36.35 g . Express in micrometer ,millimetre and kg.

Andru [333]

Answer:

1. 36.35 g = 36.35E15 micrometer.

II. 36.35 g = 363.5 millimetre.

III. 36.35 g = 0.03635 kilogram.

Explanation:

Given the following data;

Mass of calorimeter = 36.35 grams

To convert the mass in grams (g) to;

I. Micrometer

Conversion:

1 g = 1 exp 15 um

36.35 g = X um

Cross-multiplying, we have;

X = 36.35 * 1 exp 15 = 36.35 exp 15 um

36.35 g = 36.35E15 micrometer

II. Millimetre

Conversion:

1 g = 1 milliliter

36.35 g = X milliliter

Cross-multiplying, we have;

X = 36.35 * 1 = 36.35 milliliter

Next, we would convert milliliter to millimetre;

1 milliliter = 10 millimetre

36.35 milliliter = X millimetre

Cross-multiplying, we have;

X = 36.35 * 10 = 363.5 millimetre

36.35 g = 363.5 millimetre

III. Kilogram

Conversion:

1000 grams = 1 kilogram

36.35 g = X kilogram

Cross-multiplying, we have;

X * 1000 = 36.35 * 1

Dividing both sides by 1000, we have;

X = 36.35/1000 = 0.03635 kilogram

36.35 g = 0.03635 kilogram

Note:

g is the symbol for grams.
Exp (E) means exponential = 10
um is the symbol for micrometer.

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3 years ago

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3 years ago

Read 2 more answers

Measurements made on an operating centrifugal pump indicate that for a flow rate of 240 gpm, 6 HP are being consumed. The operat

Ludmilka [50]

Answer:

62.1 ft

Explanation:

We know that $1 gpm= 0.0022 ft^{3}/s$ hence 240 gpm will be

$240*0.0022=0.528 ft^{3}/s$

From the formula of obtaining efficiency

$\eta=\frac {\gamma Q h_a}{500W_{in}}$ and making $h_a$ the subject we have

$h_a=\frac {\eta*550*W_{in}}{\gamma Q}$ where Q is flow rate, $\gamma$ is specific weight of water, $h_a$ is actual head rise and $W_{in}$ is the power input to the compressor

Substituting 0.62 for $\eta$ , 6hp for $W_{in}$ , $62.4 lb/ft^{3}$ for $\gamma$ and $0.528 ft^{3}/s$ for Q then

$h_a=\frac {0.62*550*6}{62.4*0.528}=\frac {2046}{32.9472}=62.09936\approx 62.1 ft$

Therefore, actual head rise of the water being pumped is 62.1 ft

6 0

3 years ago