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3 years ago

Diffrence between UFP&LFP

Physics

1 answer:

BaLLatris [955]3 years ago

8 0

Answer:

In the formula, 'X' is the temperature value to be converted, 'UFP' is the upper fixed point and 'LFP' is the lower fixed point of the temperature scale

Explanation:

LFP means Lower fixed point and UFP means Upper fixed point. Explanation: There are different scales with different lower fixed point and upper fixed point to measure temperature but the distance between these scales is same in any table or scale considered.

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Explanation:

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What is the current when the resistance is 5 ohms and the voltage is 10 volts?

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Ohm's Law states V = IR
So,
I = V/R
The answer is B. 10/5=2 amps

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A vertical scale on a spring balance reads from 0 to 155 N . The scale has a length of 10.0 cm from the 0 to 155 N reading. A fi

Harrizon [31]

Answer:

mass of the fish is 8.11 kg

Explanation:

As we know that the frequency of oscillation of spring block system is given as

$f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}$

here we know that the reading of scale varies from 0 to 155 N from length varies from x = 0 to x = 10 cm

Now we have

$k = \frac{155}{0.10} N/m$

$k = 1550 N/m$

so now we have

$2.20 = \frac{1}{2\pi}\sqrt{\frac{1550}{m}}$

$m = 8.11 kg$

so mass of the fish is 8.11 kg

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3 years ago

In creating his definition of horsepower, James Watt, the inventor of the steam engine, calculated the power output of a horse o

Studentka2010 [4]

Answer:

Part a)

$F = 182.3 Lb$

Part b)

$P = 0.55 HP$

Explanation:

Diameter of the circle = 24 ft

Diameter = 731.52 cm = 7.3152 m

now the horse complete 144 trips in one hour

so time to complete one trip is given as

$t = \frac{3600}{144} s$

$t = 25 s$

now the speed of the horse is given as

$v = \frac{2\pi r}{t}$

$v = \frac{\pi(7.3152)}{25}$

$v = 0.92 m/s$

Part a)

Now we know that the power is defined as rate of work done

it is given as

$P = F v$

$746 = F(0.92)$

$F = 810.9 N$

$F = 182.3 Lb$

Part b)

Work done to climb up to 3 m height is given by

$W = mgh$

now we have

$Power = \frac{Work}{time}$

$P = \frac{mgh}{t}$

$P = \frac{(70kg)(9.81)(3)}{5.0s}$

$P = 412.02 Watt$

now we know that 1 HP = 746 Watt

so we have

$P = \frac{412}{746} = 0.55 HP$

8 0

3 years ago

At what distance from Earth does the force of gravity exerted by Earth on the coasting spacecraft cancel the force of gravity ex

telo118 [61]

Answer:

346 * 10⁶ m

Explanation:

The force of gravity of the earth that will cancel the the force of gravity exerted by the moon will be equal to each other

Let $F_{e}$ be the force of gravity exerted by the earth

and let $F_{m}$ be the force of gravity exerted by the moon

According to Newton's law of universal gravitation, the force of attraction between two different masses, m₁ and m₂ separated by a distance, d, is given by:

$F = \frac{Gm_{1} m_{2} }{d^{2} }$

Mass of the earth, $m_{e} = 5.97 * 10^{24} kg$

Mass of the moon, $m_{m} = 7.348 * 10^{22} kg$

Mass of the satellite, $m_{s} = ?$

$F_{e} = \frac{G*5.97 * 10^{24} M }{d^{2} }$ ...............................(1)

The earth and the moon are separated by a distance, 3.844 * 10⁸ m

$F_{m} = \frac{G*7.348 * 10^{22} M }{(3.844 * 10^{8} - d) ^{2} }$ ............................(2)

Equating equations (1) and (2)

$\frac{5.97 * 10^{24} }{d^{2} } = \frac{7.348 * 10^{24} }{(3.844* 10^{8} -d)^{2} }$

$(5.97 * 10^{24})(14.78 * 10^{16} -7.688*10^{8}d + d^{2}) = 7.348 * 10^{24} d^{2} \\88.24*10^{40} - 45.9 * 10^{32}d + 5.97 * 10^{24}d^{2} = 7.348 * 10^{24} d^{2}\\ 1.378 * 10^{24}d^{2} + 45.9 * 10^{32}d + 88.24*10^{40} = 0\\$

Factorising out $10^{24}$

$1.378d^{2} + 45.9 * 10^{8}d + 88.24*10^{16} = 0$

Solving for d in the quadratic equation above:

d = 346 * 10⁶ m

4 0

3 years ago

Diffrence between UFP&LFP​

Diffrence between UFP&LFP