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Delvig [45]
3 years ago
10

Una esfera homogénea de radio r y peso W resbala en un piso bajo la acción de una fuerza horizontal constante P aplicada a una c

uerda, como se muestra en la Fig. 1. (a) Demostrar que, si ? es el coeficiente de rozamiento entre la esfera y el piso, la altura h está dada por h = r (1 – ?W/P). (b) Demostrar que la esfera no se encuentra en equilibrio de traslación bajo estas circunstancias. (c) Si se escoge un valor diferente de h, ?se puede lograr que la esfera se encuentre tanto en equilibrio de rotación como de traslación? ?Y si se escoge una dirección diferente para P? Dar una explicación de lo que ocurre.
Physics
1 answer:
bearhunter [10]3 years ago
4 0
Garlic Butter Shrimp Scampi can be enjoyed as an appetizer/light meal OR for dinner with your favourite pasta of choice! You can also keep it low carb and serve over zucchini noodles or with steamed cauliflower! Either way it’s delicious and even better than an Italian restaurant Scampi!
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you are doing an experiment outside on a sunny day you find the temperature of some sand is 28 degress Celsius you also find the
OLga [1]
E the temperature of a substance. Water has a very high specific heat. That means it needs to absorb a lot of energy before its temperature changes. Sand , on the other hand, have lower specific heats. This means that their temperatures change more quickly. When the summer sun shines down on them, they quickly become hot.
3 0
3 years ago
A fighter bomber is making a bombing run flying horizontally at 500 knots (256m/s) at an altitude of 100.0m.
Jobisdone [24]

Answer:

<em>(a) t = 4.52 sec</em>

<em>(b) X = 1,156.49 m</em>

Explanation:

<u>Horizontal Launching </u>

If an object is launched horizontally, its initial speed is zero in the y-coordinate and the horizontal component of the velocity v_o remains the same in time. The distance x is computed as .

\displaystyle x=v_o.t

(a)

The vertical component of the velocity v_y starts from zero and gradually starts to increase due to the acceleration of gravity as follows

v_y=gt

This means the vertical height is computed by

\displaystyle h=h_o-\frac{gt^2}{2}

Where h_o is the initial height. Our fighter bomber is 100 m high, so we can compute the time the bomb needs to reach the ground by solving the above equation for t knowing h=0

\displaystyle t=\sqrt{\frac{2h_o}{g}}

\displaystyle t=\sqrt{\frac{2(100)}{9.8}}=4.52\ sec

(b)

We now compute the horizontal distance knowing v_o=256\ m/s

\displaystyle x=(256).(4.52)

X=1,156.49\ m

4 0
3 years ago
Which of the following would be the magnitude of the vector given a horizontal component of 30 and a
Scorpion4ik [409]

Answer:

50

Explanation:

Use the Pythagorean theorem to find the length of the diagonal, or the hypotenuse of an imaginary triangle. 30^2 + 40^2 = 2500, which is 50^2. So, the magnitude is 50.

Brainliest, please :)

7 0
2 years ago
A series circuit contains a 9-volt battery, a 3-ohm resistor and a 2-ohm resistor. What is the voltage drop across the 2-ohm res
BARSIC [14]
Since everything in the circuit is in series .. .

-- The total resistance is  (3 + 2) = 5 ohms.

-- The voltage across the 3-ohm resistor is 3/5 of the total voltage.

-- The voltage across the 2-ohm resistor is 2/5 of the total voltage.

                  (2/5) of (9 volts)  =  18/5  =  3.6 volts .

7 0
4 years ago
Read 2 more answers
Glycerin is poured into an open U-shaped tube until the height in both sides is 20cm. Ethyl alcohol is then poured into one arm
Liula [17]

Answer:

7.5 cm

Explanation:

In the figure we can see a sketch of the problem. We know that at the bottom of the U-shaped tube the pressure is equal in both branches. Defining \rho_A: Ethyl alcohol density and \rho_G: Glycerin density , we can write:

\rho_A\times g \times h_1 + \rho_G \times g \times h_2 = \rho_G \times g \times h_3

Simplifying:

\rho_A\times h_1 = \rho_G \times (h_3 - h_2) (1)

On the other hand:

h_1 + h_2 = \Delta h + h_3

Rearranging:

h_1 - \Delta h = h_3 - h_2 (2)

Replacing  (2) in (1):

\rho_A\times h_1 = \rho_G \times (h_1 - \Delta h)

Rearranging:

\frac{h_1 \times (\rho_A - \rho_G)}{- \rho_G} = \Delta h

Data:

h_1 = 20 cm; \rho_A = 0.789 \frac{g}{cm^3}; \rho_G = 1.26 \frac{g}{cm^3}

\frac{20 cm \times (0.789 - 1.26) \frac{g}{cm^3}}{- 1.26\frac{g}{cm^3}}  = \Delta h

7.5 cm =  \Delta h

7 0
3 years ago
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