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artcher [175]
3 years ago
7

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

Physics
1 answer:
FromTheMoon [43]3 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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natima [27]

Answer:

by Multiply Force,change direction and speed multiply

5 0
3 years ago
The free-fall acceleration on Mars is 3.7 m/s2.
hichkok12 [17]

Answer:

(a) The length of the pendulum on Earth is 36.8cm

(b) The length of the pendulum on Mars is 13.5cm

(c) Mass suspended from the spring on Earth is 0.37kg

(d) Mass suspended from the spring on Mars is 0.36kg

Explanation:

Period = 1.2s, free fall acceleration on Earth = 9.8m/s^2, free fall acceleration on Mars = 3.7m/s^2

( a) Length of pendulum on Earth = [( period ÷ 2π)^2] × acceleration = (1.2 ÷ 2×3.142)^2 × 9.8 = 0.0365×9.8 = 0.358m = 35.8cm

(b) Length of the pendulum on Mars = (1.2÷2×3.142)^2 × 3.7 = 0.0365×3.7 = 0.135cm = 13.5m

(c) Mass suspended from the spring on Earth = (force constant×length in meter) ÷ acceleration = (10×0.358) ÷ 9.8 = 0.37kg

(d) Mass suspended from the spring on Mars = (10×0.135)÷3.7 = 0.36kg

6 0
3 years ago
Define the term energy density of a body under strain​
vodka [1.7K]

Answer:

Please mark as Brainliest!!

Explanation:

Strain energy is defined as the energy stored in a body due to deformation. The strain energy per unit volume is known as strain energy density and the area under the stress-strain curve towards the point of deformation. When the applied force is released, the whole system returns to its original shape.

8 0
3 years ago
A student conduct an experiment to determine habitation of salt to water affect the density of the water the student feels three
Nikolay [14]
Any like answer choices would be helpful '-'
8 0
3 years ago
A girl (mass M) standing on the edge of a frictionless merry-go-round (radius R, rotational inertia I) that is not moving. She t
vladimir1956 [14]

a) \omega=\frac{-mvR}{I+MR^2}

b) v=\frac{-mvR^2}{I+MR^2}

Explanation:

a)

Since there are no external torques acting on the system, the total angular momentum must remain constant.

At the beginning, the merry-go-round and the girl are at rest, so the initial angular momentum is zero:

L_1=0

Later, after the girl throws the rock, the angular momentum will be:

L_2=(I_M+I_g)\omega +L_r

where:

I is the moment of inertia of the merry-go-round

I_g=MR^2 is the moment of inertia of the girl, where

M is the mass of the girl

R is the distance of the girl from the axis of rotation

\omega is the angular speed of the merry-go-round and the girl

L_r=mvR is the angular momentum of the rock, where

m is the mass of the rock

v is its velocity

Since the total angular momentum is conserved,

L_1=L_2

So we find:

0=(I+I_g)\omega +mvR\\\omega=\frac{-mvR}{I+MR^2}

And the negative sign indicates that the disk rotates in the direction opposite to the motion of the rock.

b)

The linear speed of a body in rotational motion is given by

v=\omega r

where

\omega is the angular speed

r is the distance of the body from the axis of rotation

In this problem, for the girl, we have:

\omega=\frac{-mvR}{I+MR^2} is the angular speed

r=R is the distance of the girl from the axis of rotation

Therefore, her linear speed is:

v=\omega R=\frac{-mvR^2}{I+MR^2}

5 0
3 years ago
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