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3 months ago

Compare the time period of two simple pendulums of length 4m and 16m at a place.

Physics

1 answer:

Ostrovityanka [3.2K]3 months ago

5 0

Answer:

The period of the pendulum measuring 16 m is double that of the 4 m pendulum.

Explanation:

Recall that the period (T) of a pendulum with length (L) is defined by:

$T=2\,\pi\,\sqrt{ \frac{L}{g} }$

where "g" denotes the local gravitational acceleration.

Since both pendulums are positioned at the same location, the value of "g" will be consistent for both, and when we compare the periods, we find:

$T_1=2\,\pi\,\sqrt{\frac{4}{g} } \\T_2=2\,\pi\,\sqrt{\frac{16}{g} } \\ \\\frac{T_2}{T_1} =\sqrt{\frac{16}{4} } =2$

Thus, the duration of the 16 m pendulum is two times that of the 4 m one.

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An air hockey game has a puck of mass 30 grams and a diameter of 100 mm. The air film under the puck is 0.1 mm thick. Calculate

inna [3103]

Answer:

the time it takes after impact for the puck is 2.18 seconds

Explanation:

initially given information

mass = 30 g = 0.03 kg

diameter = 100 mm = 0.1 m

thickness = 0.1 mm = 1 × $10^{-4}$ m

dynamic viscosity = 1.75 × $10^{-5}$ Ns/m²

temperature of air = 15°C

to determine

time needed for the puck to reduce its speed by 10%

solution

we note that velocity changes from 0 to v

assuming initial velocity = v

therefore final velocity = 0.9v

implying a change in velocity is du = v

and clearance dy = h

shear stress acting on the surface is expressed as

= µ $\frac{du}{dy}$

therefore

= µ $\frac{v}{h}$ ............1

substituting the values

= 1.75 × $10^{-5}$ × $\frac{v}{10^{-4}}$

= 0.175 v

the area between the air and puck is given by

Area = $\frac{\pi }{4} d^{2}$

area = $\frac{\pi }{4} 0.1^{2}$

area = 7.85 × $\frac{v}{10^{-3}}$ m²

thus, the force on the puck can be represented as

Force = × area

force = 0.175 v × 7.85 × $10^{-3}$

force = 1.374 × $10^{-3}$ v

now applying Newton's second law

force = mass × acceleration

- force = $mass \frac{dv}{dt}$

- 1.374 × $10^{-3}$ v = $0.03 \frac{0.9v - v }{t}$

solving for t = $\frac{0.1 v * 0.03}{1.37*10^{-3} v}$

the time needed after impact for the puck is 2.18 seconds

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3 months ago

The flight of a kicked football follows the quadratic function f(x)=−0.02x2+2.2x+2, where f(x) is the vertical distance in feet

Keith_Richards [3271]

The ball covers a horizontal distance of 0.902 meters. The trajectory of a kicked football adheres to a quadratic equation expressed as: f(x), where f(x) indicates the vertical distance in feet, and x signifies how far the ball travels horizontally. To compute the distance the ball will advance before striking the ground, we set the condition f(x) = 0. Upon solving this quadratic equation, we find that the horizontal distance traveled by the ball is: x = -0.902 meters, leading us to conclude that it travels 0.902 meters across the field.

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2 months ago