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3 years ago

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Below are birds-eye views of six identical toy cars moving to the right at 2 m/s. Various forces act on the cars with magnitudes

and directions indicated below. All forces act in the horizontal plane and are either parallel or at 45 or 90 degrees to the car's motion. Rank these cars on the basis of their speed a short time (ie. before any car's speed can reach zero) after the forces are applied. Rank from largest to smallest. To rank items as equivalent, overlap them.

1 answer:

Firdavs [7]3 years ago

3 0

All cars have the same speed before the forces are applied ( v = 2 m/s ).
After that rank is:
1. Largest: 6th car ; Fr = 10 N,
2. 1st car ; Fr = 2.95 N.
3. After that: 2nd, 3rd and 5th car ; Fr = 0 N
4. Smallest : 4th car ; Fr = - 2.04 N.

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A transformer has a primary coil with 106 turns and a secondary coil of 340 turns. The AC voltage across the primary coil has a

UkoKoshka [18]

To solve this problem it is necessary to apply the concepts related to transformers, that is to say passive electrical device that transfers electrical energy from one electrical circuit to one or more circuits.

From the mathematical definition we have that the relationship between the voltage of the first coil and the second coil is proportional to the number of loops of the first and second loop, that is:

$\frac{V_s}{V_p} = \frac{N_s}{N_p}$

Where

$V_p =$ input voltage on the primary coil.

$V_s=$ input voltage on the secondary coil.

$N_p=$ number of turns of wire on the primary coil.

$N_s =$ number of turns of wire on the secondary coil.

Replacing our values we have:

$V_p = 128V$

$N_p = 106$

$N_s = 340$

Replacing,

$\frac{V_s}{128} = \frac{340}{106}$

$V_s = 410.56V$

From the same relations of number of turns and the voltage of the first and second coil we also have the relation of electricity and voltage whereby:

$V_s I_s = V_p I_p$

Where

$I_p$ = Current Primary Coil

$I_s$ = Current secundary Coil

Therefore:

$I_s = \frac{V_p I_p}{V_s}$

$I_s = \frac{(128)(6)}{410.56}$

$I_s = 1.87 A$

Therefore the maximum values for the secondary coil of the voltage is 410.56V and Current is 1.87A

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Cocaine, which is derived from the coca leaf, is grown mostly in Colombia and its neighboring countries while heroin, which is d

nevsk [136]

Answer:

Because of the location, humidity and temperatures.

Explanation:

Coca is grown in humid and very humid subtropical forests, called yungas and

they form the lower floor of the upper Jungle, in the Central Andes, mostly in Peru and Bolivia. The yungas are in contact with the rainforests of the lowlands in Amazonia, where it has been started to expand coca cultivation recently (Dourojeanni, 1988). The optimum altitude is 1000 a 2000 meters (where cocaine content is higher), with optimal annual average precipitation, is 2000 meters mm, but it is grown between 700 and 2000 msnm and with an average annual rainfall of 1000 to 4200 mm.

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vladimir2022 [97]

Answer:

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2 years ago

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Ultraviolet rays are used to _____.

NeTakaya

Answer:

Grow plants where little light is available

Explanation:

The plants need the ultraviolet rays in order to be able to survive and develop. The need mainly comes from the dependence of these rays for production of food, in a process known as photosynthesis. The plants are producers, thus they create their own food. In order to be able to do that they are using the ultraviolet rays, as well as water, and carbon dioxide. By combining them, the plants manage to create glucose for them, and that is their food source. The plants that are kept at places where there's not enough light are often exposed to ultraviolet rays so that they are able to perform the process of photosynthesis and grow properly.

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2 years ago

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A 4.87-kg ball of clay is thrown downward from a height of 3.21 m with a speed of 5.21 m/s onto a spring with k = 1570 N/m. The

Yuki888 [10]

Answer:

Approximately $0.560\; {\rm m}$ , assuming that:

the height of $3.21\; {\rm m}$ refers to the distance between the clay and the top of the uncompressed spring.
air resistance on the clay sphere is negligible,
the gravitational field strength is $g = 9.81\; {\rm m\cdot s^{-2}}$ , and
the clay sphere did not deform.

Explanation:

Notations:

Let $k$ denote the spring constant of the spring.
Let $m$ denote the mass of the clay sphere.
Let $v$ denote the initial speed of the spring.
Let $g$ denote the gravitational field strength.
Let $h$ denote the initial vertical distance between the clay and the top of the uncompressed spring.

Let $x$ denote the maximum compression of the spring- the only unknown quantity in this question.

After being compressed by a displacement of $x$ , the elastic potential energy $\text{PE}_{\text{spring}}$ in this spring would be:

$\displaystyle \text{PE}_{\text{spring}} = \frac{1}{2}\, k\, x^{2}$ .

The initial kinetic energy $\text{KE}$ of the clay sphere was:

$\displaystyle \text{KE} = \frac{1}{2}\, m \, v^{2}$ .

When the spring is at the maximum compression:

The clay sphere would be right on top of the spring.
The top of the spring would be below the original position (when the spring was uncompressed) by $x$ .
The initial position of the clay sphere, however, is above the original position of the top of the spring by $h = 3.21\; {\rm m}$ .

Thus, the initial position of the clay sphere ( $h = 3.21\; {\rm m}$ above the top of the uncompressed spring) would be above the max-compression position of the clay sphere by $(h + x)$ .

The gravitational potential energy involved would be:

$\text{GPE} = m\, g\, (h + x)$ .

No mechanical energy would be lost under the assumptions listed above. Thus:

$\text{PE}_\text{spring} = \text{KE} + \text{GPE}$ .

$\displaystyle \frac{1}{2}\, k\, x^{2} = \frac{1}{2}\, m\, v^{2} + m\, g\, (h + x)$ .

Rearrange this equation to obtain a quadratic equation about the only unknown, $x$ :

$\displaystyle \frac{1}{2}\, k\, x^{2} - m\, g\, x - \left[\left(\frac{1}{2}\, m\, v^{2}\right)+ (m\, g\, h)\right] = 0$ .

Substitute in $k = 1570\; {\rm N \cdot m^{-1}}$ , $m = 4.87\; {\rm kg}$ , $v = 5.21\; {\rm m\cdot s^{-1}}$ , $g = 9.81\; {\rm m \cdot s^{-2}}$ , and $h = 3.21\; {\rm m}$ . Let the unit of $x$ be meters.

$785\, x^{2} - 47.775\, x - 219.453 \approx 0$ (Rounded. The unit of both sides of this equation is joules.)

Solve using the quadratic formula given that $x \ge 0$ :

$\begin{aligned}x &\approx \frac{-(-47.775) + \sqrt{(-47.775)^{2} - 4 \times 785 \times (-219.453)}}{2 \times 785} \\ &\approx 0.560\; {\rm m}\end{aligned}$ .

(The other root is negative and is thus invalid.)

Hence, the maximum compression of this spring would be approximately $0.560\; {\rm m}$ .

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2 years ago