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Gnom [1K]
3 years ago
12

Two cylinders with the same mass density rhoC = 713 kg / m3 are floating in a container of water (with mass density rhoW = 1025

kg / m3). Cylinder #1 has a length of L1 = 20 cm and radius r1 = 5 cm. Cylinder #2 has a length of L2 = 10 cm and radius r2 = 10 cm. If h1 and h2 are the heights that these cylinders stick out above the water, what is the ratio of the height of Cylinder #2 above the water to the height of Cylinder #1 above the water (h2 / h1)? h2 / h1 =
Physics
1 answer:
Tanzania [10]3 years ago
5 0

Answer:

Explanation:

Given

density of cylinder is \rho _c=713 kg/m^3

Length of first cylinder is L_1=20 cm

radius r_1=5 cm

For cylinder 2 L_2=10 cm

r_2=10 cm

h_1 and h_2 are the height above water

E

as object is floating so its weight must be balanced with buoyant force

\rho _c\frac{\pi }{4}d_1^2L_1g=\rho _w\frac{\pi }{4}d_1^2(L_1-h_1)g----1

For 2nd cylinder

\rho _c\frac{\pi }{4}d_2^2L_2g=\rho _w\frac{\pi }{4}d_2^2(L_2-h_2)g----2

Dividing 1 and 2 we get

\frac{L_1}{L_2}=\frac{L_1-h_1}{L_2-h_2}

\frac{20}{10}=\frac{20-h_1}{10-h_2}

2h_2=h_1

\\\Rightarrow\frac{h_2}{h_1}=\frac{1}{2}                            

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If an object accelerates from rest, what will its velocity be after 1.3 s if it has a constant acceleration of 9.1 m/s^2?
HACTEHA [7]

\text{Given that,}\\\\\text{Initial velocity,} ~v_0 = 0~ \text{m~s}^{-1}\\\\\text{Time,  t = 1.3~sec}\\\\\text{Acceleration, a = 9.1 m s}^{-2}\\\\\\\\\text{Velocity,}\\\\v = v_0  +at\\\\\implies v = 0 + 9.1 \times 1.3 = 11.83~~ \text{m~s}^{-1}

5 0
2 years ago
Se golpea una pelota de golf de manera que su velocidad inicial forma un ángulo de 45° con la horizontal. La pelota alcanza el s
nordsb [41]

Answer:

42m/s

6.06s

Explanation:

To find the initial velocity and time in which the ball is fling over the ground you use the following formulas:

x_{max}=\frac{v_o^2sin(2\theta)}{g}\\\\x_{max}=vt_{max}

θ: angle = 45°

vo: initial velocity

g: gravitational constant = 9.8m/s^2

x_max: max distance = 180 m

t_max: max time

by replacing the values of the parameters and do vo the subject of the first formula you obtain:

v_o=\sqrt{\frac{gx_{max}}{sin(2\theta)}}\\\\v_o=\sqrt{\frac{(9.8m/s^2)(180m)}{sin(2(45\°))}}=42\frac{m}{s}

with this value of vo you calculate the max time:

t_{max}=\frac{x_{max}}{v}=\frac{x_{max}}{v_ocos(45\°)}\\\\t_{max}=\frac{180m}{(42m/s)cos(45\°)}=6.06s

hence, the initial velocity of the ball is 42m/s and the time in which the ball is in the air is 6.06s

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TRANSLATION:

Para encontrar la velocidad inicial y el tiempo en el que la pelota está volando sobre el suelo, use las siguientes fórmulas:

θ: ángulo = 45 °

vo: velocidad inicial

g: constante gravitacional = 9.8m / s ^ 2

x_max: distancia máxima = 180 m

t_max: tiempo máximo

reemplazando los valores de los parámetros y haciendo el tema de la primera fórmula que obtiene:

con este valor de vo usted calcula el tiempo máximo:

por lo tanto, la velocidad inicial de la pelota es de 42 m / sy el tiempo en que la pelota está en el aire es de 6.06 s

4 0
2 years ago
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Ivanshal [37]

Answer:

B. 0.552

Explanation:

To find the resistance in the circuit above, u simply divide the current in the circuit by the voltage to get the resistance.

7 0
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Earth is about 93 million miles from the Sun but only 239,000 miles from the Moon. Likewise, the Sun has a mass that is over 20
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Answer:

Gravity is dependent on the mass of two bodies and the distance between them. There is a strong gravitational attraction between Earth and the Moon because they’re relatively close to one another. There is a strong gravitational attraction between Earth and the Sun because the Sun is so massive

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Liquified matter maybe.
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