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2 years ago

HELP PLEASE 100 POINTS!

Physics

2 answers:

erica [24]2 years ago

4 0

Yes the relationship between them is direct linear ..

Evgen [1.6K]2 years ago

4 0

Yes, the relationship between them is direct linear because linearity is the property of a mathematical relationship or function which means that it can be graphically represented as a straight line.

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Find the quantity of heat needed

krok68 [10]

Answer:

Approximately $3.99\times 10^{4}\; \rm J$ (assuming that the melting point of ice is $0\; \rm ^\circ C$ .)

Explanation:

Convert the unit of mass to kilograms, so as to match the unit of the specific heat capacity of ice and of water.

$\begin{aligned}m&= 100\; \rm g \times \frac{1\; \rm kg}{1000\; \rm g} \\ &= 0.100\; \rm kg\end{aligned}$

The energy required comes in three parts:

Energy required to raise the temperature of that $0.100\; \rm kg$ of ice from $(-10\; \rm ^\circ C)$ to $0\; \rm ^\circ C$ (the melting point of ice.)
Energy required to turn $0.100\; \rm kg$ of ice into water while temperature stayed constant.
Energy required to raise the temperature of that newly-formed $0.100\; \rm kg$ of water from $0\; \rm ^\circ C$ to $10\;\ rm ^\circ C$ .

The following equation gives the amount of energy $Q$ required to raise the temperature of a sample of mass $m$ and specific heat capacity $c$ by $\Delta T$ :

$Q = c \cdot m \cdot \Delta T$ ,

where

$c$ is the specific heat capacity of the material,
$m$ is the mass of the sample, and
$\Delta T$ is the change in the temperature of this sample.

For the first part of energy input, $c(\text{ice}) = 2100\; \rm J \cdot kg \cdot K^{-1}$ whereas $m = 0.100\; \rm kg$ . Calculate the change in the temperature:

$\begin{aligned}\Delta T &= T(\text{final}) - T(\text{initial}) \\ &= (0\; \rm ^\circ C) - (-10\; \rm ^\circ C) \\ &= 10\; \rm K\end{aligned}$ .

Calculate the energy required to achieve that temperature change:

$\begin{aligned}Q_1 &= c(\text{ice}) \cdot m(\text{ice}) \cdot \Delta T\\ &= 2100\; \rm J \cdot kg \cdot K^{-1} \\ &\quad\quad \times 0.100\; \rm kg \times 10\; \rm K\\ &= 2.10\times 10^{3}\; \rm J\end{aligned}$ .

Similarly, for the third part of energy input, $c(\text{water}) = 4200\; \rm J \cdot kg \cdot K^{-1}$ whereas $m = 0.100\; \rm kg$ . Calculate the change in the temperature:

$\begin{aligned}\Delta T &= T(\text{final}) - T(\text{initial}) \\ &= (10\; \rm ^\circ C) - (0\; \rm ^\circ C) \\ &= 10\; \rm K\end{aligned}$ .

Calculate the energy required to achieve that temperature change:

$\begin{aligned}Q_3&= c(\text{water}) \cdot m(\text{water}) \cdot \Delta T\\ &= 4200\; \rm J \cdot kg \cdot K^{-1} \\ &\quad\quad \times 0.100\; \rm kg \times 10\; \rm K\\ &= 4.20\times 10^{3}\; \rm J\end{aligned}$ .

The second part of energy input requires a different equation. The energy $Q$ required to melt a sample of mass $m$ and latent heat of fusion $L_\text{f}$ is:

$Q = m \cdot L_\text{f}$ .

Apply this equation to find the size of the second part of energy input:

$\begin{aligned}Q_2&= m \cdot L_\text{f}\\&= 0.100\; \rm kg \times 3.36\times 10^{5}\; \rm J\cdot kg^{-1} \\ &= 3.36\times 10^{4}\; \rm J\end{aligned}$ .

Find the sum of these three parts of energy:

$\begin{aligned}Q &= Q_1 + Q_2 + Q_3 = 3.99\times 10^{4}\; \rm J\end{aligned}$ .

3 0

2 years ago

A 0.311 kg tennis racket moving 30.3 m/s east makes an elastic collision with a 0.0570 kg ball moving 19.2 m/s west. Find the ve

Troyanec [42]

The velocity of tennis racket after collision is 14.96m/s

<u>Explanation:</u>

Given-

Mass, m = 0.311kg

u1 = 30.3m/s

m2 = 0.057kg

u2 = 19.2m/s

Since m2 is moving in opposite direction, u2 = -19.2m/s

Velocity of m1 after collision = ?

Let the velocity of m1 after collision be v

After collision the momentum is conserved.

Therefore,

m1u1 - m2u2 = m1v1 + m2v2

$v1 = (\frac{m1-m2}{m1+m2})u1 + (\frac{2m2}{m1+m2})u2$

$v1 = (\frac{0.311-0.057}{0.311+0.057})30.3 + (\frac{2 X 0.057}{0.311 + 0.057}) X-19.2\\\\v1 = (\frac{0.254}{0.368} )30.3 + (\frac{0.114}{0.368}) X -19.2\\ \\v1 = 20.91 - 5.95\\\\v1 = 14.96$

Therefore, the velocity of tennis racket after collision is 14.96m/s

7 0

2 years ago

A motorcycle is moving at a constant velocity of 15 meters/second. Then it starts to accelerate and reaches a velocity of 24 met

galben [10]

3 meters per second

5 0

3 years ago

A vacuum tube can be used to__. A. change alternating current into direct current B. increase the strength of a signal.. C. turn

vesna_86 [32]

The correct answer of this question is : A) Change alternating current into direct current.

EXPLANATION :

As per the question, we are given vacuum tube. Vacuum tube can be of various types. Normally it contains two electrodes called cathode and anode which are enclosed in an evacuated glass chamber . There are also other types of vacuum tubes which contain extra electrodes like control grid .

The vacuum tube can be used as a rectifier. It means that it can be used as an electronic device which will convert alternating current into direct current. It may be a half wave rectifier or a full wave rectifier. Actually the direct current obtained during the rectification of alternating current is pulsating in nature.

Hence, the correct answer is that a vacuum tube can be used to change alternating current into direct current.

4 0

3 years ago

Read 2 more answers

A rope is being used to pull a mass of 10 kg vertically upward. Determine the tension on the rope, if starting from rest, the ma

Sergio [31]

Answer:

mgh= 10 x 8 x 10

= 800

but you can try 10 x 8 x 4^-1 x 10

6 0

2 years ago