A 50-cm-long spring is suspended from the ceiling. A 330g mass is connected to the end and held at rest with the spring unstretched. The mass is released and falls, stretching the spring by 28cm before coming to rest at its lowest point. It then continues to oscillate vertically.

A. What is the spring constant?

B. What is the amplitude of the oscillation?

C. What is the frequency of the oscillation?

this is the sample Answer:    Spring tides occur when the moon is full or new. Earth, the moon, and the Sun are in a line. The moon’s gravity and the Sun’s gravity pull Earth’s crust and ocean water. This causes tides to be higher than normal. At neap tide, the moon and the Sun are at right angles to each other. This happens during the first and third quarters of the lunar cycle. At neap tide, the Sun’s gravity and the moon’s gravity are balanced. High tides are lower; low tides are higher.

Explanation:

just did the assament

Answer:

Spring tides occur when the moon is full or new. Earth, the moon, and the Sun are in a line. The moon’s gravity and the Sun’s gravity pull Earth’s crust and ocean water. This causes tides to be higher than normal.  At neap tide, the moon and the Sun are at right angles to each other. This happens during the first and third quarters of the lunar cycle. At neap tide, the Sun’s gravity and the moon’s gravity are balanced. High tides are lower; low tides are higher.

Explanation:

Just did the quiz.

Answer:

r = 16 Km

Explanation:

given  

m_n= 1.67 x 10^-27 Kg

M_star = 3.88 x 10^30 Kg  

A= M_star/m_n

A= 3.88*10^30/1.67 x 10^-27

A=2.28 *10^57  neutrons                           A = The number of neutrons  

we use the number of neutrons as a mass number because the star has only neutrons. = 1.2 x 10-15 m

r = r_o*A^1/3

r = 1.2*10^-15*2.28 *10^57^1/3

r = 16 Km

Answer:

The tension that must be applied to this string = 477 N

Explanation:

y(x,t)= 0.0321m sin (2.05x-524t + pi/4)

Comparing this to the general wave equation

y(x,t) = A sin (kx - wt + Φ)

where

A = amplitude of the wave = 0.0321 m

k = 2.05 /m

w = 524 /s

Φ = phase angle = pi/4

Velocity of a wave is given by

v = (w/k) = (524/2.05) = 255.61 m/s

Tension in the string is then related to the velocity of wave produced and the linear density through

v = √(T/μ)

v = velocity = 255.61 m/s

T = ?

μ = 0.0073 kg/m

v² = (T/μ)

T = v²μ = (255.61² × 0.0073) = 476.96 N

T = 477 N

Hope this Helps!!!

Answer:

Explanation:

Given that,

A capacitor of capacitance

C = 500pF

Charge on capacitor is

Q = 10μC

Capacitor is then connected to an inductor of inductance 10H

L = 10H

Since we want to calculate the maximum energy stored by the inductor, then, we will assume all the energy from the capacitor is transfer to the inductor

So energy stored in capacitor can be determined by using

U = ½CV²

Then, Q = CV

Therefore V = Q/C

U = ½ C • (Q/C)² = ½ C × Q²/C²

U = ½Q² / C

Then,

U = ½ × (10 × 10^-6)² / (500 × 10^-9)

U = 1 × 10^-4 J

U = 0.1 mJ

So the energy stored in this capacitor is transfers to the inductor.

So, energy stored in the inductor is 0.1mJ

B. Current through the inductor

Energy in the inductor is given as

U = ½Li²

1 × 10^-4 = ½ × 10 × i²

1 × 10^-4 = 5× i²

i² = 1 × 10^-4 / 5

i² = 2 × 10^-5

I = √(2×10^-5)

I = 4.47 × 10^-3 Amps

Then,

I = 4.47 mA

Answer:

Explanation:

Range= U²sin2theta/g

= 102²* sin (2*42)/9.8

= 1054.74m

Yes she will be abable to jump

The distance from the edge will be

(1054.74- 420)m

= 634.74m

Answer:

2.86J

Explanation:

M = 3.0kg

R₁ = 1.0m

R₂ = 0.3m

I₁ = I(mass) + I(student + stool)

I₁ = 2mr₁² + I(student + stool)

I₁ = 2*(3*1²) + 3.0

I₁ = 9.0kgm²

the initial moment of inertia of the system = 9.0kgm²

I₂ = 2mr₂² + I(student + stool)

I₂ = 2*(3 * 0.3²) + 3.0

I₂ = 0.54 + 3.0

I₂ = 3.54kgm²

the final moment of inertia of the system is 3.54kgm²

From conservation of angular momentum

I₁ω₁ = I₂ω₂

ω₂ = (I₁ * ω₁) / I₂

ω₂ = (0.75 * 9) / 3.54

ω₂ = 1.09rad/s

kinetic energy of rotation (k.e) =½ Iω²

K.E = (K.E)₂ - (K.E)₁

k.e = [½ * 3.54 * (1.90)²] - [½ * 9.0 * 0.75²]

K.E = 6.3897 - 2.53125

K.E = 2.85845

K.E = 2.86J

Answer:

Explanation:

Let the mass of the thin block be M = .072 kg .

mass of bullet m = 4.67 x 10⁻³ kg.

speed of bullet v = 556 m/s.

speed of block after billet exits the block = 18 m/s.

We shall apply law of conservation of momentum

momentum of bullet block system

.072 x 0 + 4.67 x 10⁻³  x 556 = .072 x 18 +  4.67 x 10⁻³  x V here V is velocity of bullet after exit

= 2.59652 = 1.296 + 4.67 x 10⁻³  x V

4.67 x 10⁻³  x V = 1.30052

V = .278 x 10³ m /s

= 278 m /s

Answer:

[tex]E=\frac{\lambda}{2\pi r\epsilon_0}[/tex]

Explanation:

We are given that

Linear charge density of wire=[tex]\lambda[/tex]

Radius of hollow cylinder=R

Net linear charge density of cylinder=[tex]2\lambda[/tex]

We have to find the expression for the magnitude of the electric field strength inside the cylinder r<R

By Gauss theorem

[tex]\oint E.dS=\frac{q}{\epsilon_0}[/tex]

[tex]q=\lambda L[/tex]

[tex]E(2\pi rL)=\frac{L\lambda}{\epsilon_0}[/tex]

Where surface area of cylinder=[tex]2\pi rL[/tex]

[tex]E=\frac{\lambda}{2\pi r\epsilon_0}[/tex]

Answer:

A. 2.57 K

Explanation:

From specific heat capacity,

Q = cmΔT........................ Equation 1

Where Q = Amount of heat that entered into silver, m = mass of silver, c = specific heat capacity of silver, ΔT = change in temperature of the silver.

make ΔT the subject of the equation

ΔT = Q/cm................... Equation 2

Given: Q = 300 J, m = 500 g = 0.5 kg

Constant: c = 233 J/kg.K

Substitute into equation 2

ΔT = 300/(0.5×233)

ΔT  = 300/116.5

ΔT = 2.57 K

Hence the right option is A. 2.57 K

The change in the temperature of the silver should be considered as the 2.57 K.

From specific heat capacity,

Q = cmΔT........................ Equation 1

Here

Q = Amount of heat that entered into silver,

m = mass of silver,

c = specific heat capacity of silver,

ΔT = change in temperature of the silver.

Now

ΔT = Q/cm................... Equation 2

Since

Q = 300 J, m = 500 g = 0.5 kg

Constant: c = 233 J/kg.K

So,

ΔT = 300/(0.5×233)

ΔT  = 300/116.5

ΔT = 2.57 K

hence, The change in the temperature of the silver should be considered as the 2.57 K.

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Answer:

Option C. The distance between interference fringes increases.

Explanation:

The distance between interference fringes for small angles can be given by the formula:

[tex]y = \frac{\lambda D}{d}[/tex]..........(1)

Where D = Distance between the slits and the screen

[tex]\lambda =[/tex] the wavelength of light

d = separation between the two slits

from the formula given in equation (1)

[tex]y \alpha 1/d[/tex]

It is obvious from the relationship above that the distance between interference fringes is inversely proportional to the separation between the slits.

Therefore, if the separation between slits is increased, the distance between interference fringes is increased.

The pressure difference (∆P) would be the atmospheric pressure in the room (1.01 x 10⁵ Pa) minus the pressure inside the container (0.41 x 10⁵ Pa).

To calculate the minimum upward applied force required to lift the lid of a partially evacuated vertical cylindrical container, you would need to know two additional quantities: the radius of the lid, and consequently the surface area of the lid.

The force required to lift the lid can be calculated using the equation F = P x A, where F is the force, P is the difference in pressure (external minus internal), and A is the surface area of the lid. Recall that the area of the circular lid would be calculated using the formula A = πr², where r is the radius of the circular lid.

In this case, the pressure difference (∆P) would be the atmospheric pressure in the room (1.01 x 10⁵ Pa) minus the pressure inside the container (0.41 x 10⁵ Pa).

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Complete question is:

Two uniform solid spheres have the same mass of 1.65 kg, but one has a radius of 0.206 m and the other has a radius of 0.804 m. Each can rotate about an axis through its center. (a) What is the magnitude τ of the torque required to bring the smaller sphere from rest to an angular speed of 367 rad/s in 14.5 s? (b) What is the magnitude F of the force that must be applied tangentially at the sphere’s equator to give that torque? What are the corresponding values of (c) τ and (d) F for the larger sphere?

Answer:

A) τ = 0.709 N.m

B) F = 3.44 N

C) τ = 10.8 N.m

D) F = 13.43N

Explanation:

We are given;

Mass if each sphere = 1.65kg

Radius of the first sphere; r1 = 0.206m

Radius of second sphere; r2 = 0.804m

A) initial angular speed of smaller sphere; ω_i = 0 rad/s

Final angular speed of smaller sphere; ω_f = 367 rad/s

Time;t = 14.5 s

The constant angular acceleration is calculated from;

ω_f = ω_i + αt

367 = 0 + α(14.5)

Thus,

α = 367/14.5 = 25.31 rad/s²

The torque is given by the formula;

τ = Iα

Where τ is torque ; I is moment of inertia given as (2/5)Mr²

α is angular acceleration

Thus;

τ = (2/5)(1.65)(0.206)² x 25.31

τ = 0.709 N.m

B) The magnitude of the force that must be applied to give the torque τ is gotten from the formula;

τ = F•r•sin90°

0.709 = F x 0.206 x 1

F = 0.709/0.206

F = 3.44 N

C) Now for the larger sphere, we'll repeat the same procedure in a above. Thus;

The torque is given by the formula;

τ = Iα

Where τ is torque ; I is moment of inertia given as (2/5)Mr²

α is angular acceleration

Thus;

τ = (2/5)(1.65)(0.804)² x 25.31

τ = 10.8 N.m

D) Now for the larger sphere, we'll repeat the same procedure in b above. Thus, τ is gotten from the formula;

τ = F•r•sin90°

10.8 = F x 0.804 x 1

F = 10.8/0.804

F = 13.43N

Answer:

E. All of the answers are correct

Explanation:

Overload principle in fitness training is associated with a gradual development of an athlete's abilities by progressively increasing the athlete's load and training.

In order to do this, the athlete's limit must be surpassed albeit gradually at first then picked up later over time.

Final answer:

Overload refers to using a demand (load) above the normal demand faced by a muscle, fundamental for muscle growth and strength increase. It involves progressively increasing the workload to stimulate muscle adaptation and improvements. Overload, progression, and specificity are key principles in effective training programs.

Explanation:

Overload refers to B. Using a demand (load) above the normal demand faced by a muscle. This is a fundamental concept in strength training and physical conditioning. Overload is the principle that for muscles to grow stronger or for endurance to increase, they must be challenged with a demand that is greater than what they are accustomed to. This concept involves progressively increasing the workload on the muscle to stimulate adaptation and improvements in muscle strength, endurance, and size.

The principle of overload is based on the body's ability to adapt to increased demands. When you perform exercises that are more challenging than your muscles are used to, your body adapts by making those muscles stronger, which in turn makes them capable of handling the increased load. This can be achieved through various methods such as increasing the weight lifted, the number of repetitions performed, or the intensity of the exercise.

It is also important to note that overload, along with progression and specificity, form the three foundational principles of training that guide the development of effective strength and conditioning programs. These principles ensure that exercises not only challenge the body but do so in a way that promotes optimal growth, strength, and performance enhancements over time.

Answer:

I didn't see any statement. But here is the answer:--

Explanation:

But objects are pulled towards earth's center by gravity. classical mechanics can not define the gravity but general relativity describe it as a curve of space-time by a concentrated energy or mass. As the mass of an object increases the more space time will be curved in the direction of positive energy so the center of the earth is the minimum point of the curved space-time by earth.That's why smaller objects than earth will fall to the center of the    earth.

The statement best explains why objects are pulled toward Earth’s center is : Earth has a much greater mass than objects on its surface .Gravity is the attractive force acting between masses. This force is why objects fall to the ground when dropped.

This phenomenon is explained by gravity. As Newton's Law of Universal Gravitation says, every object with mass attracts every other object with mass. The Earth, having a much greater mass compared to objects on its surface, exerts a significant gravitational force pulling objects towards its center. This is why when you drop an object, it falls to the ground. The gravitational force of Earth keeps us grounded and is responsible for attracting objects towards its center.

Complete Question

Which statement best explains why objects are pulled toward Earth’s center?

Earth has a magnetic force that is strongest at its core.

Earth has a much greater mass than objects on its surface.

The weight of the atmosphere pushes objects toward Earth's surface.

The strength of the Sun’s gravity pushes down on objects at Earth's surface.

Answer:

b. 0.25cm

Explanation:

You can solve this question by using the formula for the position of the fringes:

[tex]y=\frac{m\lambda D}{d}[/tex]

m: order of the fringes

lambda: wavelength 500nm

D: distance to the screen 5 m

d: separation of the slits 1mm=1*10^{-3}m

With the formula you can calculate the separation of two adjacent slits:

[tex]\Delta y=\frac{(m+1)(\lambda D)}{d}-\frac{m\lambda D }{d}=\frac{\lambda D}{d}\\\\\Delta y=\frac{(500*10^{-9}nm)(5m)}{1*10^{-3}m}=2.5*10^{-3}m=0.25cm[/tex]

hence, the aswer is 0.25cm

Answer:

C. The maximum speed of emitted electrons increase

D. The stopping potential increases

Explanation:

Albert Einstein provided a successful explanation of the photoelectric effect on the basis of quantum theory. He proposed that,

“An electron either absorbs one whole photon or it absorbs none. After absorbing a photon, an electron either leaves the surface of metal or dissipate its energy within the metal in such a short time interval that it has almost no chance to absorb a second photon. An increase in intensity of light source simply increase the number of photons and thus, the number of electrons, but the energy of electron remains same. However, increase in frequency of light increases the energy of photons and hence, the energy of electrons too.”

Therefore, due to this increase in energy of photons, the kinetic energy of emitted electrons also increase. And the increase in kinetic energy follows with the increase in velocity.

Now, the stopping potential is directly proportional to Kinetic Energy of the electrons. Thus, the increase in Kinetic Energy, results in an increase in stopping potential.

Therefore, two options apply here:

C. The maximum speed of emitted electrons increases

D. The stopping potential increases

If the frequency of the incident light in the photoelectric effect is increased, C. The maximum speed of the emitted electrons increases and D. The stopping potential increases; however, the work function does not change and the number of electrons emitted per second remains the same at a constant light intensity.

For the question about what happens if the frequency of the incident light in the photoelectric effect is increased, the correct options are:

The work function of the metal (Option A) is a material property and does not change with the frequency of the incident light. Therefore, this option is incorrect. The number of electrons emitted from the metal per second (Option B) depends on the intensity of the light, not its frequency, so increasing the frequency at constant intensity does not increase the emission rate. Thus, this option is also incorrect. When the frequency of the incident light is increased, the kinetic energy of the emitted electrons increases (as per Einstein's equation for the photoelectric effect), which correlates with Option C. The stopping potential, which is the electric potential needed to stop the fastest photoelectrons, also increases with the electron's kinetic energy, confirming Option D as correct.

Answer:

The spaceship is placed at a distance of [tex]1.28\times 10^6\[/tex] meters from the Earth.

Explanation:

Distance between two astronauts, d = 1.4 m  

The time it takes for sound waves to travel at 328 m/s through the air between the astronauts equals the time it takes for the electromagnetic waves to travel to the earth.

For sound, time taken is given by :

[tex]t=\dfrac{1.4}{328}\ ........(1)[/tex]

For electromagnetic wave, time taken to travel the Earth is :

[tex]t=\dfrac{d}{3\times 10^8}\ ........(2)[/tex]

d is the distance from the Earth to the spaceship.

As time are same :

[tex]\dfrac{d}{3\times 10^8}=\dfrac{1.4}{328}[/tex]

[tex]d=\dfrac{1.4}{328}\times 3\times 10^8\\\\d=1.28\times 10^6\ m[/tex]

So, the spaceship is placed at a distance of [tex]1.28\times 10^6[/tex] meters from the Earth.

Answer:

2. 17.7cm

3. -7 that is magnification

Explanation:

See attached handwritten document

Answer:

Explanation:

The weight of the wheelbarrow will act downwards . Its component parallel to the plank will act downwards along the plank .

The value of component = mgsinθ

21.6 x 9.8 x sin20

= 72.4 N

The person shall have to apply force equal to this component so that the barrow moves with uniform velocity.

Force required = 72.4 N .

Answer:

Explanation:

DetaM=4 x 1.02875 - 4.002603

DetaM= 0.028697u

Using E= mc²

= 0.028697 x 1.49x*10^-10

= 4.2x10^-12J

At the highest point in the ball's trajectory, the magnitude of its velocity vector is at its minimum nonzero value, while the magnitude of its acceleration vector is a constant.

When the ball reaches the highest point in its trajectory, the magnitude of its velocity vector is at its minimum nonzero value, while the magnitude of its acceleration vector is a constant. Since the ball is at the highest point, it momentarily comes to rest in the vertical direction, resulting in its velocity magnitude being zero in that direction. However, the ball still has a horizontal velocity component, which is constant throughout its motion.

Answer:

three halves of a wavelength

Explanation:because a

path difference between waves at minimum is (2n+1)*wavelength /2 therefore at second minimum it is for n=1=1.5 times a wavelength. or three halves a wavelength

The path difference in the distance traveled by the wave is three halves of a wavelength. Therefore, option (B) is correct.

Interference can be defined as a phenomenon that takes place in every place. It is a phenomenon in which two waves superpose to create the resultant wave of the higher, lower, or same amplitude.

There are two kinds of Interference which are constructive interference and destructive interference. In constructive interference, the crest of one wave falls on the crest of another wave, and the amplitude increases. In destructive interference, the crest of one wave falls on the trough of another wave, and the amplitude decreases.

For the minima, the path difference is given by (2n -1)/λ.

Here the value of n is given, n = 2

Path difference, = (2×2 - 1)/λ = (3/2)λ

Therefore, the path difference in the distance is (3/2)λ.

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After collision, truck remains at 14.0 m/s, car stops (0 m/s) due to elastic collision.

To find the velocities of both vehicles after the collision, we can use the principle of conservation of momentum, which states that the total momentum before the collision is equal to the total momentum after the collision.

The momentum [tex]\( p \) of an object is given by its mass \( m \) multiplied by its velocity \( v \)[/tex]:

[tex]\[ p = mv \][/tex]

Before the collision, the total momentum is the sum of the momentum of the truck and the momentum of the car. After the collision, the total momentum remains the same.

Let [tex]\( v_t \) be the velocity of the truck after the collision, and \( v_c \)[/tex] be the velocity of the car after the collision.

Before the collision:

[tex]\[ \text{Total momentum} = \text{momentum of truck} + \text{momentum of car} \][/tex]

[tex]\[ (1400 \, \text{kg}) \times (14.0 \, \text{m/s}) + (641 \, \text{kg}) \times (0 \, \text{m/s}) = (1400 \, \text{kg}) \times (v_t) + (641 \, \text{kg}) \times (v_c) \][/tex]

[tex]\[ 1400 \times 14.0 = 1400v_t + 641v_c \][/tex]

[tex]\[ 19600 = 1400v_t + 641v_c \][/tex]

Since the collision is approximately elastic, we also have the condition that the relative velocity of approach equals the relative velocity of separation.

[tex]\[ v_t - v_c = 14.0 \, \text{m/s} \][/tex]

We now have a system of two equations:

[tex]\[ 19600 = 1400v_t + 641v_c \][/tex]

[tex]\[ v_t - v_c = 14.0 \][/tex]

We can solve this system of equations to find the values of [tex]\( v_t \)[/tex] and [tex]\( v_c \)[/tex].

[tex]\[ v_t = \frac{19600 - 641v_c}{1400} \][/tex]

Substitute this expression for [tex]\( v_t \)[/tex] into the second equation:

[tex]\[ \frac{19600 - 641v_c}{1400} - v_c = 14.0 \][/tex]

Multiply both sides by 1400 to eliminate the denominator:

[tex]\[ 19600 - 641v_c - 1400v_c = 14.0 \times 1400 \][/tex]

[tex]\[ 19600 - 2041v_c = 19600 \][/tex]

Now, isolate [tex]\( v_c \)[/tex] :

[tex]\[ -2041v_c = 0 \][/tex]

[tex]\[ v_c = 0 \][/tex]

Substitute [tex]\( v_c = 0 \)[/tex] into [tex]\( v_t = \frac{19600 - 641v_c}{1400} \)[/tex] :

[tex]\[ v_t = \frac{19600 - 641 \times 0}{1400} \][/tex]

[tex]\[ v_t = \frac{19600}{1400} \][/tex]

[tex]\[ v_t = 14.0 \, \text{m/s} \][/tex]

So, after the collision, the truck's speed remains 14.0 m/s, and the car's speed becomes 0 m/s.

The speed of the truck and car after the collision is approximately 9.47 m/s

Use the principle of conservation of momentum to solve this problem. The total momentum before the collision is equal to the total momentum after the collision. We can express this as:

[tex]$$ m_{\text{truck}} \cdot v_{\text{truck, initial}} + m_{\text{car}} \cdot v_{\text{car, initial}} = (m_{\text{truck}} + m_{\text{car}}) \cdot v_{\text{final}} $$[/tex]

Given the following information:

- Mass of the truck, [tex]\(m_{\text{truck}}\)[/tex]: 1400 kg

- Initial velocity of the truck, [tex]\(v_{\text{truck, initial}}\)[/tex]: 14.0 m/s

- Mass of the car, [tex]\(m_{\text{car}}\):[/tex] 641 kg

- Initial velocity of the car, [tex]\(v_{\text{car, initial}}\)[/tex]: 0 m/s (since the car is stopped)

We can solve for the final velocity, [tex]\(v_{\text{final}}\):[/tex]

[tex]$$ 1400 \cdot 14.0 + 641 \cdot 0 = (1400 + 641) \cdot v_{\text{final}} $$[/tex]

Solving for [tex]\(v_{\text{final}}\):[/tex]

[tex]$$ v_{\text{final}} = \frac{1400 \cdot 14.0}{1400 + 641} $$[/tex]

Calculating the value:

[tex]$$ v_{\text{final}} \approx 9.47 \, \text{m/s} $$[/tex]

Therefore, the speed of the truck and car after the collision is approximately 9.47 m/s.

Answer:

Explanation:

Amount of heat required can be found from the following relation

Q = mcΔT

m is mass of the body , c is specific heat and ΔTis rise in temperature .

Here m = 300 kg

c = 3350 J /kg k

ΔT = 30 - 25

= 5 °C

Putting the values in the expression above

Q = 300 x 3350 x 5

= 5025000 J

Rate at which energy is absorbed = 1200 J /s

Time required

= 5025000 / 1200

= 4187.5 S

= 69.8 minute

= 1 hour 9.8 mimutes.

Answer: 2000 J.

Explanation: Since work is force*displacement, we just have to multiply the force by the distance: w = f*d = 400 N*5.0 m = 2000 J.

Answer:

This is because The energies of atoms are quantized.

Electrons are allowed "in between" quantized energy levels, and, thus, only specific lines are observed

Answer

Given,

Rotational inertia = 2.0 x 10-3 kg·m²

Torque = 11 N.m

time, t = 19 ms

a) Angular momentum

  [tex]\tau = \dfrac{\Delta L}{\Delta t}[/tex]

 L is angular momentum

  [tex]\Delta L = \tau \Delta t[/tex]

  [tex]\Delta L = 11\times 19 \times 10^{-3}[/tex]

  [tex]\Delta L = 0.209\ Kg m^2/s[/tex]

b) Angular velocity

  We know.

     [tex]L = I \omega[/tex]

     [tex]\omega = \dfrac{L}{I}[/tex]

     [tex]\omega = \dfrac{0.209}{2\times 10^{-3}}[/tex]

     [tex]\omega = 104.5\ rad/s[/tex]

Answer:

If necessary, readjust the f-stop of the CCD camera until only the LED’s from the pucks

are visible. You may find that a vertical stripe appears associated with the LED. This

is called streaking or ‘blooming’ and is a limitation of CCD technology in the presence of

localized bright spots. Some amount of streaking is acceptable, and can be compensated for

later in the computer analysis.

Once the exposure level of the camera has been set you should capture a ‘dark frame’.

For this, first remove all pucks, hands, etc., from the air table and click on the box to the

right of Background Frame. The dark frame can later be subtracted from your images of

collisions to suppress any constant background such as the edge of the table.

Answer: The average kinetic energy is 2.3 × 10^-20 J

Explanation: Please see the attachments below

Answer:

The chunk went as high as

2.32m above the valley floor

Explanation:

This type of collision between both ice is an example of inelastic collision, kinetic energy is conserved after the ice stuck together.

Applying the principle of energy conservation for the two ice we have based on the scenery

Momentum before impact = momentum after impact

M1U1+M2U2=(M1+M2)V

Given data

Mass of ice 1 M1= 5.20kg

Mass of ice 2 M2= 5.20kg

velocity of ice 1 before impact U1= 13.5 m/s

velocity of ice 2 before impact U2= 0m/s

Velocity of both ice after impact V=?

Inputting our data into the energy conservation formula to solve for V

5.2*13.5+5.2*0=(10.4)V

70.2+0=10.4V

V=70.2/10.4

V=6.75m/s

Therefore the common velocity of both ice is 6.75m/s

Now after impact the chunk slide up a hill to solve for the height it climbs

Let us use the equation of motion

v²=u²-2gh

The negative sign indicates that the chunk moved against gravity

And assuming g=9.81m/s

Initial velocity of the chunk u=0m/s

Substituting we have

6.75²= 0²-2*9.81*h

45.56=19.62h

h=45.56/19.62

h=2.32m