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2 years ago

4

When you read a poem, you might pick up on a central message that readers can apply to their own lives. What is this central mes

sage called?

2 answers:

Finger [1]2 years ago

8 0

I, at my school, was taught that something that the reader reads and can connect to is called the central theme or the universal theme this is because many people can relate to it and it's not specifically directed towards the text but more of a general theme

Ulleksa [173]2 years ago

8 0

Answer:

the theme

Explanation:

You might be interested in

A hiker caught in a rainstorm absorbs 1.00 L of water in her clothing. If it is windy so that the water evaporates quickly at 20

masya89 [10]

Answer:

(A) Q = 2.26×10⁶J

(B) ΔT = 9°C

(C)

Explanation:

We have been given the mass of the hiker, the volume of water from which we can calculate the mass knowing that the density if water is 1000kg/m³.

Evaporation is a phase change and occurs at a constant temperature. We would use the latent heat of vaporization to calculate the amount of heat evaporated.

We would then equate this to the heat change it brings about in the hiker's body and then calculate the temperature drop.

See the attachment below for full solution.

6 0

2 years ago

A bar 4.0m long weights 400 N. Its center of gravity is 1.5m from on end. A weight of 300N is attached at the light end. What ar

Debora [2.8K]

Answer:

The resulting, needed force for equilibrium is a reaction from a support, located at 2.57 meters from the heavy end. It is vertical, possitive (upwards) and 700 N.

Explanation:

This is a horizontal bar.

For transitional equilibrium, we just need a force opposed to its weight, thus vertical and possitive (ascendent). Its magnitude is the sum of the two weights, 400+300 = 700 N, since weight, as gravity is vertical and negative.

Now, the tricky part is the point of application, which involves rotational equilibrium. But this is quite simple if we write down an equation for dynamic momentum with respect to the heavy end (not the light end where the additional weight is placed). The condition is that the sum of momenta with respect to this (any) point of the solid bar is zero:

$0=\Sigma_{i}M=400\cdot1.5+300\cdot4-d\cdot700$

Where momenta from weights are possitive and the opposed force creates an oppossed momentum, then a negative term. Solving our unknown d:

$d=\frac{400\cdot1.5+300\cdot4}{700} =2.57 m$

So, the resulting force is a reaction from a support, located at 2.57 meters from the heavy end (the one opposed to the added weight end).

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2 years ago

Kate is working on a project in her tech education class. She plans to assemble a fan motor. Which form of energy does the motor

grin007 [14]

The job that the fan is designed and built to do is to convert the electrical energy it uses into the kinetic (motion) energy of moving air.

I can't really guarantee that it accomplishes that with MOST of the electrical energy it uses, because I don't know how efficient your fan is. For example, if it's a really old fan, and one blade has the end broken off, and a lot of dust and mosquitoes have gotten into the motor, and it shakes and vibrates and makes a lot of noise when it's running, then it's converting a lot of the electrical energy into thermal energy (it gets hot when it runs) and some into sound energy too.

If you can live without the word "most" in the question, then we can assume that the fan is well designed and running like a top, and the answer is definitely choice-B .

4 0

2 years ago

Anna applies a force of 19.5 newtons to push a book placed on a table. If the normal force of the book is 51.7 newtons, what is

GarryVolchara [31]

that would be given by

[email protected]

@ representing coefficient of kinetic friction.

thus 19.5/51.7 = 0.377

6 0

2 years ago

A robotic rover on Mars finds a spherical rock with a diameter of 10 centimeters [cm]. The rover picks up the rock and lifts it

Makovka662 [10]

Answer: 5166.347

Explanation:

The specific gravity of a solid $SG$ (also called relative density) is the ratio of the density of that solid $\rho_{rock}$ to the density of water $\rho_{water}=1 kg/m^{3}$ (normally at $4\°C$ ):

$SG=\frac{\rho_{rock}}{\rho_{water}}$ (1)

On the other hand, the density of the rock is calculated by:

$\rho_{rock}=\frac{m_{rock}}{V_{rock}}$ (2)

Where:

$m_{rock}$ is the mass of the rock

$V_{rock}=\frac{4}{3} \pi r^{3}$ is the volume of the rock, since is spherical

Well, we already know the value of $\rho_{water}$ , but we need to find $\rho_{rock}$ in order to find the rock's specific gravity; and in order to do this, we firsly have to find $m_{rock}$ and then calculate $V_{rock}$ :

In the case of the mass of the rock, we can calclate it by the following equation:

$W_{rock}=m_{rock}g_{mars}$ (3)

Where:

$W_{rock}$ is the weight if the rock in mars

$g_{mars}=3.7 m/s^{2}$ is the acceleration due gravity in Mars

Isolating $m_{rock}$ :

$m_{rock}=\frac{W_{rock}}{g_{mars}}$ (4)

$m_{rock}=\frac{W_{rock}}{3.7 m/s^{2}}$ (5)

To find $W_{rock}$ we can use the following equation of the potential gravitational energy $U$ :

$U=W_{rock}H$ (6)

Where:

$U=2 J=2 Nm$ is the potential energy

$H=20 cm \frac{1m}{100 cm}=0.2 m$ is the height at which the rock has the mentioned potential energy

Isolating $W_{rock}$ :

$W_{rock}=\frac{U}{H}$ (7)

$W_{rock}=\frac{2 Nm}{0.2 m}$ (8)

$W_{rock}=10 N$ (9)

Substituting (9) in (5):

$m_{rock}=\frac{10 N}{3.7 m/s^{2}}$ (10)

$m_{rock}=2.702 kg$ (11)

Substituting (11) in (2):

$\rho_{rock}=\frac{2.702 kg}{V_{rock}}$ (12) At this point we only need to find the volume of the rock, knowing its diameter is $d=10 cm$ , hence its radius is $r=\frac{d}{2}=5 cm$

$V_{rock}=\frac{4}{3} \pi (5 cm)^{3}$ (13)

$V_{rock}=523.59 cm^{3} \frac{1 m^{3}}{(100 cm)^{3}}=0.000523 m^{3}$ (14)

Substituting (14) in (12):

$\rho_{rock}=\frac{2.702 kg}{0.000523 m^{3}}$ (15)

$\rho_{rock}=5166.34 kg/m^{3}$ (16)

Substituting (16) in (1):

$SG=\frac{5166.34 kg/m^{3}}{1 kg/m^{3}}$ (17)

Finally we obtain the specific gravity of the rock:

$SG=5166.347$

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2 years ago