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sleet_krkn [62]
3 years ago
14

When a 0.350-kg package is attached to a vertical spring and lowered slowly, the spring stretches 12.0 cm. The package is now di

splaced from its equilibrium position and undergoes simple harmonic oscillations when released. What is the period of the oscillations?
Physics
1 answer:
AleksandrR [38]3 years ago
4 0

To solve the problem it is necessary to use the concepts related to the calculation of periods by means of a spring constant.

We know that by Hooke's law

F=kx

Where,

k = Spring constant

x = Displacement

Re-arrange to find k,

k= \frac{F}{x}

k= \frac{mg}{x}

k= \frac{(0.35)(9.8)}{12*10^{-2}}

k = 28.58N/m

Perioricity in an elastic body is defined by

T = 2\pi \sqrt{\frac{m}{k}}

Where,

m = Mass

k = Spring constant

T = 2\pi \sqrt{\frac{0.35}{28.58}}

T = 0.685s

Therefore the period of the oscillations is 0.685s

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River basin characteristics:

- tributaries
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- precipitation collects
- water drainage
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Hope I helped. Please vote my answer branliest! Thanks.
7 0
4 years ago
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Suppose you have a pendulum clock which keeps correct time on Earth(acceleration due to gravity = 1.6 m/s2). For ever hour inter
kaheart [24]

The moon clock is A) (9.8/1.6)h compared to 1 hour on Earth

Explanation:

The period of a simple pendulum is given by the equation

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

where

L is the length of the pendulum

g is the acceleration of gravity

In this problem, we want to compare the period of the pendulum on Earth with its period on the Moon. The period of the pendulum on Earth is

T_e=2\pi \sqrt{\frac{L}{g_e}}

where

g_e = 9.8 m/s^2 is the acceleration of gravity on Earth

The period of the pendulum on the Moon is

T_m=2\pi \sqrt{\frac{L}{g_m}}

where

g_m = 1.6 m/s^2 is the acceleration of gravity on the Moon

Calculating the ratio of the period on the Moon to the period on the Earth, we find

\frac{T_m}{T_e}=\frac{g_e}{g_m}=\frac{9.8}{1.6}

Therefore, for every hour interval on Earth, the Moon clock will display a time of

A) (9.8/1.6)h

#LearnwithBrainly

6 0
3 years ago
A 10 g bullet accelerates to 400 m/s after being fired from a gun with a mass of 2.0 kg.
Gala2k [10]

Answer:

See the answers below.

Explanation:

Momentum is defined as the product of mass by velocity, and can be calculated by means of the following expression.

P=m*v

where:

P = Momentum [kg*m/s]

m = mass = 10 [g] = 0.01 [kg]

v = velocity = 400 [m/s]

i)

P=0.01*400\\P=4[kg*m/s]

ii)

The momentum of the gun is equal to zero, because it does not move before being fired, the weapon only moves after having fired the weapon.

P_{gun}=0

iii)

Since the momentum is conserved before and after the shot, the same momentum given to the bullet is equal to the momentum received by the gun.

P=m*v_{recoil}

v_{recoil}=P/m\\v_{recoil}= 4/2\\v_{recoil}=2[m/s]

3 0
3 years ago
Can the kinetic energy of an object be negative? Can the potential energy of an object be negative? Can a potential energy funct
IceJOKER [234]

Answer:

Kinetic energy cannot be negative

potential energy is a reference dependent quantity and it can be positive as well as negative both

Since potential energy is defined only for conservative force so it can not be found for friction force

Explanation:

Kinetic energy of an object is given by the formula

KE = \frac{1}{2}mv^2

here we know that

m = mass of object that can not be negative

v = speed of the object and since its square is given here so it can not be negative

so Kinetic energy is always positive

potential energy is given as the energy due to the virtue of the position of object

so it is

\Delta U = -\int F.dr

so potential energy is a reference dependent quantity and it can be positive as well as negative both

Since potential energy is defined only for conservative force so it can not be found for friction force

5 0
4 years ago
The stemplot below shows the heights (in inches) of students in a class.
masha68 [24]
B.b 70 would be the answer
4 0
4 years ago
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