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2 years ago

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Las ovejas son animales sociales que seguirán a la oveja guía a cualquier lugar que decida ir. Pensando en esto, describa un mom

ento en el que puede ser ventajoso para nosotros no ser el líder sino ser un seguidor

1 answer:

saw5 [17]2 years ago

8 0

Un momento podría ser como, por ejemplo, la oveja podría llevarlo a algo como dar de beber a otra oveja entera o lo mismo con otros animales o incluso digamos que no sabe qué hacer para una tarea que se asignó pero su amigo lo hace, eso podría ser un problema. siguiente momento en el que tu amigo te puede enseñar qué hacer, con suerte, eso ayudó a que mi español no sea tan bueno, lo siento :/

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An object of mass 6 kg. is resting on a horizontal surface. A horizontal force

son4ous [18]

Answer:

a) The work done by the applied force is 1500 joules.

b) The kinetic energy of the block after 10 seconds is 1200 joules.

c) The magnitude of the force of friction is 3 newtons and its direction is against motion.

d) 300 joules of energy are lost during motion.

Explanation:

a) Since the object has a constant mass, on which a constant horizontal force is exerted. The work done by the force ( $W$ ), measured in joules, is defined by the following expression:

$W = F\cdot \Delta x$ (1)

Where:

$F$ - Force, measured in newtons.

$\Delta x$ - Distance, measured in meters.

If we know that $F = 15\,N$ and $\Delta x = 100\,m$ , then the work done by the force exerted on the object is:

$W = (15\,N)\cdot (100\,m)$

$W = 1500\,J$

The work done by the applied force is 1500 joules.

b) At first we need to calculate the net acceleration of the object ( $a$ ), measured in meters per square second. By assuming a constant acceleration, we use the following kinematic formula:

$\Delta x = v_{o}\cdot t +\frac{1}{2}\cdot a\cdot t^{2}$ (2)

Where $v_{o}$ is the initial velocity of the object, measured in meters per second.

We clear the acceleration within the equation above:

$\frac{1}{2}\cdot a \cdot t^{2} = \Delta x-v_{o}\cdot t$

$a = \frac{2\cdot (\Delta x - v_{o}\cdot t)}{t^{2}}$

If we know that $\Delta x = 100\,m$ , $v_{o} = 0\,\frac{m}{s}$ and $t = 10\,s$ , then the net acceleration experimented by the object is:

$a = \frac{2\cdot \left[100\,m-\left(0\,\frac{m}{s} \right)\cdot (10\,s)\right]}{(10\,s)^{2}}$

$a = 2\,\frac{m}{s^{2}}$

By the 2nd Newton's Law, we construct the following equation of equilibrium under the consideration of a friction force acting against the motion of the object:

$\Sigma F = F - f = m\cdot a$ (3)

Where:

$F$ - External force exerted on the object, measured in newtons.

$f$ - Kinetic friction force, measured in newtons.

If we know that $F = 15\,N$ , $m = 6\,kg$ and $a = 2\,\frac{m}{s^{2}}$ , the kinetic friction force is:

$f = F-m\cdot a$

$f = 15\,N-(6\,kg)\cdot \left(2\,\frac{m}{s^{2}} \right)$

$f = 3\,N$

The work done by friction ( $W'$ ), measured in joules, is:

$W' = f\cdot \Delta x$ (4)

$W' = (3\,N) \cdot (100\,m)$

$W' = 300\,J$

And the net work experimented by the object is:

$\Delta W = 1500\,J - 300\,J$

$\Delta W = 1200\,J$

By the Work-Energy Theorem we understand that change in translational kinetic energy ( $\Delta K$ ), measured in joules, is equal to the change in net work. That is:

$\Delta K = \Delta W$ (5)

If we know that $\Delta W = 1200\,J$ , then the change in translational kinetic energy is:

$\Delta K = 1200\,J$

The kinetic energy of the block after 10 seconds is 1200 joules.

c) The magnitude of the force of friction is 3 newtons and its direction is against motion.

d) The energy lost by the object is equal to the work done by the force of friction. Therefore, 300 joules of energy are lost during motion.

7 0

3 years ago

Room temperature water is placed in an Erlenmyer flask and heated to the boiling point. After the flask is removed from the heat

Vaselesa [24]

Explanation:

When water is boiled in the flask . Some portion of it is evaporated out . Now when cork is placed on it and is placed in the ice box . It cools down , by which the pressure inside decreases .

Due to decrease of pressure , the boiling point of water also decreases . Now it can boil at lower temperature . Thus it starts boiling at lower temperature even , when placed in the ice box .

7 0

3 years ago

A spring is 20cm long is stretched to 25cm by a load of 50N. What will be its length when stretched by 100N. assuming that the e

IgorLugansk [536]

Answer:

Final Length = 30 cm

Explanation:

The relationship between the force applied on a string and its stretching length, within the elastic limit, is given by Hooke's Law:

F = kΔx

where,

F = Force applied

k = spring constant

Δx = change in length of spring

First, we find the spring constant of the spring. For this purpose, we have the following data:

F = 50 N

Δx = change in length = 25 cm - 20 cm = 5 cm = 0.05 m

Therefore,

50 N = k(0.05 m)

k = 50 N/0.05 m

k = 1000 N/m

Now, we find the change in its length for F = 100 N:

100 N = (1000 N/m)Δx

Δx = (100 N)/(1000 N/m)

Δx = 0.1 m = 10 cm

but,

Δx = Final Length - Initial Length

10 cm = Final Length - 20 cm

Final Length = 10 cm + 20 cm

<u>Final Length = 30 cm</u>

6 0

3 years ago

An airplane with an airspeed of 120 km/h has a heading of 30 degree east of North in a wind that is blowing toward the east at 6

lesya692 [45]

Answer:

Explanation:

Velocity of plane relative to ground V_pg = ?

Given the velocity in vector form ,

velocity of plane relative to air V_pw = 120 cos30 i + 120sin30j

V_wg = 60 i

V_pg = V_pw +V_wg

= 120 cos30 i + 120sin30j + 60i

= 164 i + 60 j

magnitude

=251 km / h

=

8 0

3 years ago

What is the relationship between temperature and altitude in the stratosphere? (2 points).

stealth61 [152]

As altitude increases, temperature increases. The stratosphere is the part of the atmosphere that starts in the tropopause and ends in the estratopause. In the troposphere, the air is close to the Earth surface. The air surface can absorb more sunlight energy than the air, so the Earth surface heats the air. As you go higher, the distance to the Earth surface is higher, so the temperature is lower. The troposphere ends in the tropopause, where this trend changes. In the estratopause, there is a lot of ozone, which absorbs the dangerous UV radiation and converts into heat. That heat warms the air. So the air which is close to the estratopause is warm because of the heat released by the ozone reactions. The tropopause is far from the Earth surface and far from the ozone layer, that’s why it is cold. So the tropopause is cold and the estratopause is warm, which means: the air becomes warmer <span>as you rise above the tropopause until you get to the estratopause.</span>

8 0

3 years ago