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artcher [175]
2 years ago
7

Un pendul este suspendat de un ax cu o tijă subțire foarte ușoară.

Physics
1 answer:
FromTheMoon [43]2 years ago
4 0

Answer:

A)

B)

C)

Explanation:

Given that a pendulum is suspended by a shaft with a very light thin rod.

Followed by the given information: m = 100 g, I = 0.5 m, g = 9.8 m / s²

We can determine the answer to these questions using angular kinematics.

Angular kinematics is just derived from linear kinematics but in different symbols, and expressions.

Here are the formulas for angular kinematics:

  • θ = ωt
  • ∆w =
  • L [Angular momentum] = mvr [mass × velocity × radius]

A) What is the minimum speed required for the pendulum to traverse the complete circle?

We can use the formula v = √gL derived from

B) The same question if the pendulum is suspended with a wire?

C) What is the ratio of the two calculated speeds?

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A pitcher exerts a force on a baseball that is 30 times the balls weight. How fast is the pitcher accelerating the ball?
iVinArrow [24]

Answer:

 Pitcher is accelerating the ball at 30 times of acceleration due to gravity = 294 m/s²

Explanation:

  Force applied on baseball = 30 times weight of the ball.

   Weight of ball = mg, where m is the mass of ball and g is acceleration due to gravity value.

  We have force applied is also equal to product of mass and acceleration.

                            F = ma = 30 x mg

                                 a = 30g

   So, pitcher is accelerating the ball at 30 times of acceleration due to gravity = 294 m/s²

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3 years ago
Calculate the potential difference across a 10 ohm register carrying a current of 2.5 ampere​
Bas_tet [7]

Answer:

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V=IR

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V = 2 Volt.

Explanation:

4 0
2 years ago
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3 years ago
A single slit 1.4 mmmm wide is illuminated by 460-nmnm light. Part A What is the width of the central maximum (in cmcm ) in the
mamaluj [8]

Answer:

<h2> 1.643*10⁻⁴cm</h2>

Explanation:

In a single slit experiment, the distance on a screen from the centre point is expressed as y = \frac{\delta m \lambda d}{a} where;

\delta m is the first two diffraction minima = 1

\lambda is light wavelength

d is the distance of diffraction pattern from the screen

a is the width of the slit

Given \lambda = 460-nm = 460*10⁻⁹m

d = 5.0mm = 5*10⁻³m

a = 1.4mm = 1.4*10⁻³m

Substituting this values into the formula above to get width of the central maximum y;

y = 1*460*10⁻⁹ * 5*10⁻³/1.4*10⁻³

y = 2300*10⁻¹²/1.4*10⁻³

y = 1642.86*10⁻⁹

y = 1.643*10⁻⁶m

Converting the final value to cm,

since 100cm = 1m

x = 1.643*10⁻⁶m

x = 1.643*10⁻⁶ * 100

x = 1.643*10⁻⁴cm

Hence, the width of the central maximum in the diffraction pattern on a screen 5.0 mm away is  1.643*10⁻⁴cm

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3 years ago
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