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3 years ago

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Cylinder A has a mass of 2kg and cylinder B has a mass of 10kg. Determinethe velocity of A after it has displaced 2m from its or

iginal starting position. Neglect the mass of the cable and pulleys and assume that both cylinders start at res

1 answer:

Marta_Voda [28]3 years ago

5 0

The velocity of A is 5.16m/s²

<u>Explanation:</u>

Given-

mass of cylinder A, mₐ = 2kg

mass of cylinder B, mb = 10kg

Distance, s = 2m

Velocity of A, v = ?

Let acceleration due to gravity, g = 10m/s²

We know,

$a = \frac{mb * g - ma * g}{ma + mb} \\\\a = \frac{10 * 10 - 2 * 10}{ 2 + 10} \\\\a = \frac{80}{12} \\\\a = 6.67m/s^2$

We know,

$v = \sqrt{2as}$

$v = \sqrt{2 X 6.67 X 2} \\\\v = \sqrt{26.68} \\\\v = 5.16m/s^2$

Therefore, the velocity of A is 5.16m/s²

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Calcular la longitud del faldón de una Rampa de Acceso , que en planta tiene una longitud de 20 m y la pendiente es 27%.

seraphim [82]

La longitud del faldón de la rampa es de 5.4 m.

La pendiente expresada en porcentaje sigue la siguiente ecuación:

$m=\frac{y}{x}*100$ (1)

Donde:

y es la elevacion de la rampa (faldón)
x es la longitud de la ramapa (20 m)

Sabemos que la pendiente es de 27%. Por lo tanto, usando la ecuación 1, despejamos y.

$27=\frac{y}{20}*100$

$y=\frac{27*20}{100}$

$y=5.4\: m$

La longitud del faldón es 5.4 m

Pudes ver más sobre el tema aquí:

brainly.com/question/8906330

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3 years ago

What objects do balanced forces act on?

leva [86]

Answer: Stationary or constant velocity

Explanation:

Objects with balanced forces acting on them experience no change in motion, or no acceleration. So these objects could either be stationary at rest or have a constant velocity. These include a hanging object, a floating object, an object on a table that doesn't move, and a car moving at a constant 10 mph

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3 years ago

A wave has a wavelength of 10mm and a frequency of 5 hz what is the speed?

geniusboy [140]

V=fλ
v=5*0.01
Therefore v=0.05

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4 years ago

Read 2 more answers

A 0.150 kg stone rests on a frictionless, horizontal surface. A bullet of mass 9.50 g, traveling horizontally at 380 m/s, strike

Anvisha [2.4K]

Answer:

(a)Magnitude=28.81 m/s

Direction=33.3 degree below the horizontal

(b) No, it is not perfectly elastic collision

Explanation:

We are given that

Mass of stone, M=0.150 kg

Mass of bullet, m=9.50 g= $9.50\times 10^{3} kg$

Initial speed of bullet, u=380 m/s

Initial speed of stone, U=0

Final speed of bullet, v=250m/s

a. We have to find the magnitude and direction of the velocity of the stone after it is struck.

Using conservation of momentum

$mu+ MU=mv+ MV$

Substitute the values

$9.5\times 10^{-3}\times 380 i+0.150(0)=9.5\times 10^{-3} (250)j+0.150V$

$3.61i=2.375j+0.150V$

$3.61 i-2.375j=0.150V$

$V=\frac{1}{0.150}(3.61 i-2.375j)$

$V=24.07i-15.83j$

Magnitude of velocity of stone

= $\sqrt{(24.07)^2+(-15.83)^2}$

|V|=28.81 m/s

Hence, the magnitude and direction of the velocity of the stone after it is struck, |V|=28.81 m/s

Direction

$\theta=tan^{-1}(\frac{y}{x})$

= $tan^{-1}(\frac{-15.83}{24.07})$

$\theta=tan^{-1}(-0.657)$

=33.3 degree below the horizontal

(b)

Initial kinetic energy

$K_i=\frac{1}{2}mu^2+0=\frac{1}{2}(9.5\times 10^{-3})(380)^2$

$K_i=685.9 J$

Final kinetic energy

$K_f=\frac{1}{2}mv^2+\frac{1}{2}MV^2$

= $\frac{1}{2}(9.5\times 10^{-3})(250)^2+\frac{1}{2}(0.150)(28.81)^2$

$K_f=359.12 J$

Initial kinetic energy is not equal to final kinetic energy. Hence, the collision is not perfectly elastic collision.

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3 years ago

How much heat is absorbed by a 450g gold block as energy from the sun causes its temperature to change from 27°C to 32°C? (Speci

kotegsom [21]

Answer:

Q = 1057.5 [cal]

Explanation:

In order to solve this problem, we must use the following equation of thermal energy.

$Q=m*C_{p}*(T_{final}-T_{initial})$

where:

Q = heat energy [cal]

Cp = specific heat = 0.47 [cal/g*°C]

T_final = final temperature = 32 [°C]

T_initial = initial temperature = 27 [°C]

m = mass of the substance = 450 [g]

Now replacing:

$Q=450*0.47*(32-27)\\Q=1057.5[cal]$

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3 years ago