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mina [271]
2 years ago
8

An object is dropped from a bridge. A second object is thrown downwards 1.0 s later. They both reach the water 20 m below at the

same instant. What was the initial speed of the second object? Neglect air resistance.
Physics
1 answer:
Lerok [7]2 years ago
7 0
Answer:

Explanation:
Kinematics equation for first Object:

but:
The initial velocity is zero

it reach the water at in instant, t1, y(t)=0:


Kinematics equation for the second Object:
The initial velocity is zero

but:

it reach the water at in instant, t2, y(t)=0. If the second object is thrown 1s later, t2=t1-1=1.02s


The velocity is negative, because the object is thrown downwards
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Given Information:

Pendulum 1 mass = m₁ = 0.2 kg

Pendulum 2 mass = m₂ = 0.6 kg

Pendulum 1 length = L₁ = 5 m

Pendulum 2 length = L₂ = 1 m

Required Information:

Affect of mass on the frequency of the pendulum = ?

Answer:

The mass of the ball will not affect the frequency of the pendulum.

Explanation:

The relation between period and frequency of pendulum is given by

f = 1/T

The period of pendulum is given by

T = 2π√(L/g)

Where g is the acceleration due to gravity and L is the length of the string

As you can see the period (and frequency too) of pendulum is independent of the mass of the pendulum. Therefore, the mass of the ball will not affect the frequency of the pendulum.

Bonus:

Pendulum 1:

T₁ = 2π√(L₁/g)

T₁ = 2π√(5/9.8)

T₁ = 4.49 s

f₁ = 1/T₁

f₁ = 1/4.49

f₁ = 0.22 Hz

Pendulum 2:

T₂ = 2π√(L₂/g)

T₂ = 2π√(1/9.8)

T₂ = 2.0 s

f₂ = 1/T₂

f₂ = 1/2.0

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So we can conclude that the higher length of the string increases the period of the pendulum and decreases the frequency of the pendulum.

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Two fixed charges, q1 = +1.07µC and q2 = -3.28µC, are 61.8cm apart. Where may a third charge be located so that no net force act
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Answer:

1.4E-3J

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Given that

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But we know that

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So E= 5.02E-8 x 28.8E3

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Answer:

(a) t = 22.9 s

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Explanation:

The uniformly accelerated circular movement, is a circular path movement in which the angular acceleration is constant.

We apply the equations of circular motion uniformly accelerated :

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Where:

θ : angle that the body has rotated in a given time interval (rad)

α : angular acceleration (rad/s²)

t : time interval (s)

ω₀ : initial angular speed ( rad/s)

ωf : final angular speed ( rad/s)

Data

θ =  19.5 revolutions  : angular displacement of each wheel or angle that the  wheel has rotated in a given time interval

ω₀= 10.7 rad/s :  initial angular speed of the Wheel ( rad/s)

ωf = 0 : final angular speed  of the Whee( rad/s)

Calculating of the angular acceleration (α )

We replace data in the fómula (1),considering that 1 revolution is equal to 2π radians :

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α= (-114.49) /(245.04)

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α= -0.467 rad/s²

Time does it take for the bike to come to rest

We replace data in the formula (2)

ωf = ω₀ + α*t

0 =  10.7 + -0.467*t

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10.7  = 0.467*t  

t = 10.7 / 0.467

t = 22.9 s

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