A child's toy consists of a spherical object of mass 50 g attached to a spring. One end of the spring is fixed to the side of the baby's crib so that when the baby pulls on the toy and lets go, the object oscillates horizontally with a simple harmonic motion. The amplitude of the oscillation is 6 cm and the maximum velocity achieved by the toy is 3.2 m/s . What is the kinetic energy K of the toy when the spring is compressed 5.1 cm from its equilibrium position?Problem-Solving Strategy: Simple Harmonic Motion II: Energy

The energy equation, E=12mvx2+12kx2=12kA2, is a useful alternative relationship between velocity and position, especially when energy quantities are also required. If the problem involves a relationship among position, velocity, and acceleration without reference to time, it is usually easier to use the equation for simple harmonic motion, ax=d2xdt2=−kmx (from Newton’s second law) or the energy equation above (from energy conservation) than to use the general expressions for x, vx, and ax as functions of time. Because the energy equation involves x2 and vx2, it cannot tell you the sign of x or of vx; you have to infer the sign from the situation. For instance, if the body is moving from the equilibrium position toward the point of greatest positive displacement, then x is positive and vx is positive.

IDENTIFY the relevant concepts

Energy quantities are required in this problem, therefore it is appropriate to use the energy equation for simple harmonic motion.

SET UP the problem using the following steps

Part A

The following is a list of quantities that describe specific properties of the toy. Identify which of these quantities are known in this problem.

Select all that apply.

Select all that apply.

maximum velocity vmax
amplitude A
force constant k
mass m
total energy E
potential energy U at x
kinetic energy K at x
position x from equilibrium

Part B

What is the kinetic energy of the object on the spring when the spring is compressed 5.1 cm from its equilibrium position?

Part C

What is the potential energy U of the toy when the spring is compressed 5.1 cm from its equilibrium position?

Answers

Answer 1
Answer:

Answer:

Part A

Mass = 50g

Vmax = 3.2m/s

Amplitude= 6cm

Position x from the equilibrium= 5.1cm

Part B

Kinetic energy = 0.185J

Part C

Potential energy = 0.185J

Explanation:

Kinetic energy = 1/2mv×2

Vmax = wa

w = angular velocity= 53.33rad/s

Kinetic energy = 1/2mv^2×r^2 = 0.185J

Part c

Total energy = 1/2m×Vmax^2= 0.256J

1/2KA^2= 0.256J

K= 142.22N/m (force constant)

Potential energy = 1/2kx^2

=1/2×142.22×0.051^2

= 0.185J

Answer 2
Answer:

Final answer:

To find the kinetic energy of the toy, we need to use the energy equation for simple harmonic motion and the relationship between velocity and position. We can then substitute the known values to calculate the kinetic energy.

Explanation:

In this problem, we are given the amplitude (A) of the oscillation and the maximum velocity (vmax) achieved by the toy. We need to find the kinetic energy (K) of the toy when the spring is compressed 5.1 cm from its equilibrium position.

To solve for the kinetic energy, we can use the energy equation for simple harmonic motion: K = 1/2mvx2, where m is the mass of the object and vx is the velocity of the object at position x. The mass of the object is given as 50 g, which is equal to 0.05 kg.

Since we know the maximum velocity (vmax = 3.2 m/s), we can use the relationship between velocity and position in simple harmonic motion to find the velocity (vx) at a displacement of 5.1 cm from the equilibrium position. The velocity and position in simple harmonic motion are related by vx = ±ω√(A2 - x2), where ω is the angular frequency of the motion.

Substituting the known values into the equations, we can calculate the kinetic energy of the toy.

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You check the weather and find that the winds are coming are coming from the west at 15 miles per hour. This information describes wind's velocity, direction, acceleration, or speed?

1
Verify the identity. Show your work.

cot θ ∙ sec θ = csc θ

Answers

To verify the identity, we can make use of the basic trigonometric identities:
cot θ = cos θ / sin θ 
sec θ = 1 / cos θ
csc 
θ = 1 / sin θ

Using these identities:
cot θ ∙ sec θ = (cos θ / sin θ ) ( 1 / cos θ)

We can cancel out cos θ, leaving us with
cot θ ∙ sec θ = 1 / sin θ
cot θ ∙ sec θ = = csc θ

At which point in the life cycle of a star does nuclear fusion begin?A. Black hole
B. Main sequence
C. Nebula
D. Protostar

Answers

I would say B : main sequence is the answer . this is the answer i believe because the star will increase in size and than shine brightly and when it's done , it will get smaller turning into  nebula , eventually exploding sometime around the last stage , but not the last stage of b , c, or d

i really hope that this helps you a lot.

Final answer:

The correct answer is D. Protostar.

Nuclear fusion in a star's lifecycle begins during the protostar stage. This is when the temperature and pressure of the protostar ignite nuclear fusion, transitioning into a main sequence star.

Explanation:

In the life cycle of a star, nuclear fusion begins when the star has reached the protostar stage. The initial stage in the development of a star is the nebula, which is a cloud of dust and gas. Over time, the nebula's materials collapse under gravity to form a hot core known as a protostar.

When the protostar's temperature and pressure rises, it triggers nuclear fusion and the star begins turning hydrogen into helium.

This marks the transition from a protostar to a main sequence star, which is the stage of stable nuclear fusion.

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How high does the atmosphere extend from the earth

Answers

The atmosphere extend from the Earth up to  10,000 km.

What is atmosphere?

An atmosphere is the layers of gases surrounding a planet Earth. It's atmosphere is composed of about 78% nitrogen, 21% oxygen, and other gases.

Earth's atmosphere has five major layers. From lowest to highest, the layers are the troposphere, stratosphere, mesosphere, thermosphere and exosphere.

The exosphere is the topmost layer of Earth's atmosphere . It extends from the end of thermosphere, about 700 km above sea level, to about 10,000 km.

Thus, the atmosphere extend from the Earth up to  10,000 km.

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The exosphere is the outermost layer of Earth's atmosphere (i.e. the upper limit of the atmosphere). It extends from the exobase, which is located at the top of the thermosphere at an altitude of about 700 km above sea level, to about 10,000 km (6,200 mi; 33,000,000 ft) where it merges into the solar wind.

A box containing a piece of wood and enough air to burn the wood is measured and found to have a mass of 1.5 kg. If the wood is burned and none of the smoke, ashes, or hot gases are allowed to leave the box, then how will the mass of the box compare to the mass before burning? Will it be more, less, or the same? A. More B. Less C. The same

Answers

Answer: C, the same

Explanation:

This is a practical use of the concept of conservation of mass. Mass cannot be created or destroyed, so if you start with 1.5 kg, as long as you dont lose any (which it says you don't) you will end with 1.5kg.

When object A collides with object B and bounces back, its final momentum is ____ its initial momentum.a. greater than
c. in the same direction as
b. less than
d. in the opposite direction of

Answers

When object A collides with object B and bounces back, its final momentum is in the opposite direction of its initial momentum. The correct option among all the options that are given in the question is the last option or option "d". This type of collision is elastic in nature as the ball bounces back and so the momentum has to be in the opposite direction of the initial momentum. 

Answer : (D) " in the opposite direction of "

Explanation :

It is given that when object A collides with object B and bounces back. This shows that the collision is elastic. There are two conditions for elastic collision :

(1) The momentum remains conserved

(2) Kinetic energy is conserved.

We know that momentum is defined as the product of mass and velocity. After bouncing back, the direction of velocity gets changed. So, the final momentum is in the opposite direction of its initial momentum.

So, the correct option is (D).

Find the slope of the line that passes through the pair of points (6, 7) and (9, 2). Find the slope of the line that passes through the pair of points (17, 9) and (5, 29).

Answers

As you have probably already learned, we calculate slope by dividing the change in vertical position by the change in horizontal position. Δy/Δx
Question 1: (9-6)/(2-7)=-3/5
Question 2: (5-17)/(29-9)= -12/20, which simplifies to -3/5


Answer:

Explanation: