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Alja [10]
2 years ago
10

Meme on newtons law of motion (should me made ur self not from any searh engine)

Physics
1 answer:
k0ka [10]2 years ago
8 0
He will be a pilot and he will fly the plane over bridges fewwww
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A particle executes simple harmonic motion with an amplitude of 2.18 cm.
Bogdan [553]

Answer:

The positive displacement from the midpoint of its motion at the speed equal one half of its maximum speed is 3.56 cm.

Explanation:

Maximum speed is  :

                          v (max) = Aω

Speed v at any displacement y is given by  

v^{2} = w^{2} (A^{2} - y^{2})   ........................................................  i

And,

               v = \frac{1}{2} v (max)  

          or,  2 × v = Aω     ....................................................   ii

Eliminating  ω from equations i and ii,

                       \frac{1}{4} A^{2}  w^{2}  =  w^{2}  ( A^{2}  - y^{2})

                     or, y^{2} =  (\frac{3}{4}) A^{2}  =(\frac{3}{4}) 2.18^{2}

                    or,  y =  3.56 cm.

3 0
3 years ago
What is the SI unit used to measure the temperature of a substance?
ss7ja [257]

Answer:

Kelvin

Explanation:

Kelvin is the universal and scientific unit for temperature as Celsius and Fahrenheit temperatures we use in everyday situations

3 0
3 years ago
A block is resting on a platform that is rotating at an angular speed of 2.4 rad/s. The coefficient of static friction between t
Sloan [31]

Answer:

r = 0m is the Minimum distance from the axis at which the block can remain in place wothout skidding.

Explanation:

From a sum of forces:

Ff = m*a   where Ff = μ * N    and a = \frac{V^2}{r}=\omega^2*r

N - m*g = 0   So, N = m*g.   Replacing everything on the original equation:

\mu*m*g = m*\omega^2*r   (eq2)

Solving for r:

r = \frac{\mu*g}{\omega^2}=1.41m

If we analyze eq2 you can conclude that as r grows, the friction has to grow (assuming that ω is constant), so the smallest distance would be 0 and the greatest 1.41m. Beyond that distance, μ has to be greater than 0.83.

4 0
3 years ago
A grandfather clock has a pendulum that consists of a thin brass disk of radius r = 13.62 cm and mass 1.199 kg that is attached
Feliz [49]

Answer:

Explanation:

Expression for time period of a pendulum is as follows

T = 2\pi\sqrt{\frac{l}{g} }

l is length of pendulum from centre of bob and g is acceleration due to gravity

Given

Time period T = 1.583

g = 9.846

Substituting the values

1.583 = 2\pi\sqrt{\frac{l}{9.846} }

l = \frac{(1.583)^2\times9.846}{4\times(\frac{22}{7})^2 }

l = .6244 m

= 62.44 cm

Length of rod  = length of pendulum - radius of bob

= 62.44 - 13.62

= 48.82 cm

= .488 m

8 0
3 years ago
a u tube contains a liquid of an unknown density an oil of density is poured into the right arm of the tube until the oil column
scoray [572]

Answer:

The answer is "1155\ \frac{kg}{m^3}"

Explanation:

Please find the complete question in the attached file.

p = p_0 + ?gh

pi = pressure only at two liquids' devices

PA = pressure atmosphere.

1 = oil density

2 = uncertain fluid density

h_1 = 11 \ cm\\\\h_2= 3 \ cm

The pressures would be proportional to the quantity 11-3 = 8 cm from below the surface at the interface between both the oil and the liquid.

\to p_A + ?2g(h_1 - h_2) = p_A + ? 1gh_1\\\\\to ?2 = \frac{?1h_1}{(h_1 - h_2)} \\\\

       = \frac{840 \frac{kg}{m^3}}{\frac{11}{8}} \\\\= 1155\ \frac{kg}{m^3}

8 0
3 years ago
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