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3 years ago

I need help with the graphs

Physics

1 answer:

Viefleur [7K]3 years ago

8 0

For heating Solid, Liquid, Gas and for cooling the opposite Gas, Liquid, Solid

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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

3 years ago

When looking at a food label, to be considered a good source of a nutrient, the nutrient must be? A.More than 20% of your daily

aleksklad [387]

Answer:

A. More than 20% of your daily recommended amount.

Explanation:

Reading food labels can be tricky. The percent daily value listed on the right of all food labels lets you know what percent out of the recommended daily intake of each nutrient you are consuming in that specific food.

To check if the food you're consuming is a good source of that nutrient you need in higher amount, the nutrient must be labeled 20% or higher.

The rule used here is called the 5/20 rule. According to this rule, A nutrient that is 5% or below is considered less and a nutrient which is labeled 20% or higher is considered good enough in that food source.

8 0

3 years ago

A factory worker pushes a crate of mass 31.0 kg a distance of 4.35 m along a level floor at constant velocity by pushing horizon

Debora [2.8K]

Answer:

a. 79.1 N

b. 344 J

c. 344 J

d. 0 J

e. 0 J

Explanation:

a. Since the crate has a constant velocity, its net force must be 0 according to Newton's 1st law. The push force $F_p$ by the worker must be equal to the friction force $F_f$ on the crate, which is the product of friction coefficient μ and normal force N:

Let g = 9.81 m/s2

$F_p = F_f = \mu N = \mu mg = 0.26 * 31 * 9.81 = 79.1 N$

b. The work is done on the crate by this force is the product of its force $F_p$ and the distance traveled s = 4.35

$W_p = F_ps = 79.1*4.35 = 344 J$

c. The work is done on the crate by friction force is also the product of friction force and the distance traveled s = 4.35

$W_f = F_fs = -79.1*4.35 = -344 J$

This work is negative because the friction vector is in the opposite direction with the distance vector

d. As both the normal force and gravity are perpendicular to the distance vector, the work done by those forces is 0. In other words, these forces do not make any work.

e. The total work done on the crate would be sum of the work done by the pushing force and the work done by friction

$W_p + W_f = 344 - 344 = 0 J$

8 0

3 years ago

Read 2 more answers

A seagull flying horizontally over the ocean at a constant speed of 2.60 m/s carries a small fish in its mouth. It accidentally

Ivenika [448]

(a) +2.60 m/s

The motion of the fish dropped by the seagul is a projectile motion, which consists of two independent motions:

- a horizontal uniform motion, at constant speed

- a vertical motion, at constant acceleration (acceleration of gravity, $g=-9.8 m/s^2$ , downward)

In this part we are only interested in the horizontal motion. As we said the horizontal component of the fish's velocity does not change, therefore its value when the fish reaches the ocean is equal to its initial value, which is the speed at which the seagull was flying (because it was flying horizontally):

$v_x = +2.60 m/s$

(b) -17.2 m/s

The vertical component of the fish's velocity instead follows the equation:

$v_y = u_y +gt$

where

$u_y = 0$ is the initial vertical velocity, which is zero

$g=-9.8 m/s^2$ is the acceleration of gravity

t is the time

Since the fish reaches the ocean at t = 1.75 s, we can substitute this time into the formula to find the final vertical velocity:

$v_y = 0+(-9.8)(1.75)=-17.2 m/s$

where the negative sign indicates the direction (downward).

(c)

The horizontal component of the fish's velocity would increase

The vertical component of the fish's velocity would stay the same.

As we said from part (a) and (b):

- The horizontal component of the fish's velocity is constant during the motion and it is equal to the initial velocity of the seagull -> so if the seagull's initial speed increases, the horizontal velocity of the fish will increase too

- The vertical component of the fish's velocity does not depend on the original speed of the seagull, therefore it is not affected.

4 0

3 years ago

A child looks at his reflection in a spherical Christmas tree ornament 8.0 cm in diameter in season that the image of his face i

malfutka [58]

From the information given,

diameter of ornament = 8

radius = diameter/2 = 8/2

radius of curvature, r = 4

Recall,

focal length, f = radius of curvature/2 = 4/2

f = 2

Recall,

magnification = image d

8 0

2 years ago