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BabaBlast [244]
3 years ago
8

What is the formula for pulley

Physics
1 answer:
o-na [289]3 years ago
8 0

Answer:

Insert the tension and gravitational force you just calculated into the original equation: -F = T + G = 18N + 88.2N = 106.2N. The force is negative because the object in the pulley system is accelerating upwards. The negative from the force is moved over to the solution so F= -106.2N.

Explanation:

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Which of the following are true for E = mc2?
lozanna [386]

The question is very poor.

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6 0
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Express 9.39 x 109 seconds in terms of days
Allisa [31]

                     (9.39 x 10⁹ sec) x (1 day / 8.64 x 10⁴ sec)

                 =    (9.39 / 8.64) x 10⁵ days

                 =        108,680.56 days
4 0
3 years ago
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Which sequence shows all the spectral colors of visible light arranged in an increasing order of their frequency? A. red, yellow
Ad libitum [116K]

Answer:

The answer is B. red, orange, yellow, green, blue, indigo, violet

Explanation:

Most textbooks have the acronym ROYGBV to express the order in which colors appear on the spectrum of light, indigo is included in your list, and that's not a problem, although it's not typical. This spectrum of light is the same order in which colors appear in rainbows.

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An open 1-m-diameter tank contains water at a depth of 0.7 m when at rest. As the tank is rotated about its vertical axis the ce
Mamont248 [21]

Answer:

Explanation:

To find the angular velocity of the tank at which the bottom of the tank is exposed

From the information given:

At rest, the initial volume of the tank is:

V_i = \pi R^2 h_i --- (1)

where;

height h which is the height for the free surface in a rotating tank is expressed as:

h = \dfrac{\omega^2 r^2}{2g} + C

at the bottom surface of the tank;

r = 0, h = 0

∴

h = \dfrac{\omega^2 r^2}{2g} + C

0 = 0 + C

C = 0

Thus; the free surface height in a rotating tank is:

h=\dfrac{\omega^2 r^2}{2g} --- (2)

Now; the volume of the water when the tank is rotating is:

dV = 2π × r × h × dr

Taking the integral on both sides;

\int \limits ^{V_f}_{0} \ dV = \int \limits ^R_0 \times 2 \pi \times r \times h \ dr

replacing the value of h in equation (2); we have:

V_f} = \int \limits ^R_0 \times 2 \pi \times r \times ( \dfrac{\omega ^2 r^2}{2g} ) \ dr

V_f = \dfrac{ \pi \omega ^2}{g} \int \limits ^R_0 \ r^3 \ dr

V_f = \dfrac{ \pi \omega ^2}{g} \Big [  \dfrac{r^4}{4} \Big]^R_0

V_f = \dfrac{ \pi \omega ^2}{g} \Big [  \dfrac{R^4}{4} \Big] --- (3)

Since the volume of the water when it is at rest and when the angular speed rotates at an angular speed is equal.

Then V_f  =  V_i

Replacing equation (1) and (3)

\dfrac{\pi \omega^2}{g}( \dfrac{R^4}{4}) = \pi R^2 h_i

\omega^2 = \dfrac{4g \times h_i }{R^2}

\omega =\sqrt{ \dfrac{4g \times h_i }{R^2}}

\omega = \sqrt{\dfrac{4 \times 9.81 \ m/s^2 \times 0.7 \ m}{(0.5)^2} }

\omega = \sqrt{109.87 }

\mathbf{\omega = 10.48 \ rad/s}

Finally, the angular velocity of the tank at which the bottom of the tank is exposed  = 10.48 rad/s

6 0
3 years ago
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