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3 years ago

8

The specific heat of water is 4.19 x 103 J/kg°C. How much energy will it take to raise the temperature of 0.5 kg of water by 2 d

egrees?

1 answer:

fredd [130]3 years ago

5 0

Energy required = mass x specific heat x temperature difference
= 0.5 x 4.19x10^3 x 2
= 4190J

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Hi, Please help me with the 2 questions, will mark and 5 stars and THANKS

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A ski gondola is connected to the top of a hill by a steel cable of length 620 m and diameter 1.5 cm. As the gondola comes to th

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Answer:

(a) 89 m/s

(b) 11000 N

Explanation:

Note that answers are given to 2 significant figures which is what we have in the values in the question.

(a) Speed is given by the ratio of distance to time. In the question, the time given was the time it took the pulse to travel the length of the cable twice. Thus, the distance travelled is twice the length of the cable.

$v=\dfrac{2\times 620 \text{ m}}{14\text{ s}} = \dfrac{1240\text{ m}}{14\text{ s}}=88.571428\ldots \text{ m/s}= 89\text{ m/s}$

(b) The tension, $T$ , is given by

$v =\sqrt{\dfrac{T}{\mu}}$

where $v$ is the speed, $T$ is the tension and $\mu$ is the mass per unit length.

Hence,

$T = \mu\cdot v^{2}$

To determine $\mu$ , we need to know the mass of the cable. We use the density formula:

$\rho = \dfrac{m}{V}$

where $m$ is the mass and $V$ is the volume.

$m=\rho\cdot V$

If the length is denoted by $l$ , then

$\mu = \dfrac{m}{l} = \dfrac{\rho\cdot V}{l}$

$T = \dfrac{\rho\cdot V}{l} v^{2}$

The density of steel = 8050 kg/m3

The cable is approximately a cylinder with diameter 1.5 cm and length or height of 620 m. Its volume is

$V = \pi \dfrac{d^{2}}{4} l$

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4 years ago

300N effort is applied to lift the load of 900N by using a first class lever. If the distance between the load and fulcrum is 20

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Answer:

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Ed=(900*20)/300

Ed=1800/300

Ed=6m

Explanation:

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With no friction, you can use the relationship between potential and kinetic energy to predict the speed of the car at the botto

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The speed of the car at the bottom of the hill is obtained as, $v = \sqrt{2gh}$

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where;

<em>v </em><em>is the speed of the car at the bottom of the hill</em>
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Thus, the speed of the car at the bottom of the hill is obtained as, $v = \sqrt{2gh}$

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