A 2 Kg Rubber Ball Is Thrown At A Wall Horizontally At 3 M/s, And Bounces Back The Way It Came At An (2024)

Physics College

Answers

Answer 1

Answer:

THE RUBBER BALL

Explanation:

From the question we are told that

The mass of the rubber ball is [tex]m_r = 2 \ kg[/tex]

The initial speed of the rubber ball is [tex]u = 3 \ m/s[/tex]

The final speed at which it bounces bank [tex]v - 3 \ m/s[/tex]

The mass of the clay ball is [tex]m_c = 2 \ kg[/tex]

The initial speed of the clay ball is [tex]u = 3 \ m/s[/tex]

The final speed of the clay ball is [tex]v = 0 \ m/s[/tex]

Generally Impulse is mathematically represented as

[tex]I = \Delta p[/tex]

where [tex]\Delta p[/tex] is the change in the linear momentum so

[tex]I = m(v-u)[/tex]

For the rubber is

[tex]I_r = 2(-3 -3)[/tex]

[tex]I_r = -12\ kg \cdot m/s[/tex]

=> [tex]|I_r| = 12\ kg \cdot m/s[/tex]

For the clay ball

[tex]I_c = 2(0-3)[/tex]

[tex]I_c = -6 \ kg\cdot \ m/s[/tex]

=> [tex]| I_c| = 6 \ kg\cdot \ m/s[/tex]

So from the above calculation the ball with the a higher magnitude of impulse is the rubber ball

Related Questions

An accelerating voltage of 2.25 103 V is applied to an electron gun, producing a beam of electrons originally traveling horizontally north in vacuum toward the center of a viewing screen 36.4 cm away. (a) What is the magnitude of the deflection on the screen caused by the Earth's gravitational field

Answers

Answer:

s= 8.28×10⁻¹⁶m

Explanation:

given

V= 2.25×10³V

from conservation of energy

mv²/2=qΔV

v=√(2qΔV/m)

v= √(2×1.6×10⁻¹⁹×2.25×10³/9.1×10⁻³¹)

=√7.9×10¹⁴m/s

=2.8×10⁷m/s

the deflection of electron beam is

S= gt²/2

recall t= d/v

s=g([tex]\frac{d}{v}[/tex])²/2

s= [tex]\frac{1}{2}[/tex]×9.8×(0.364/2.8×10⁷)²

s= 8.28×10⁻¹⁶m

A rod 16.0 cm long is uniformly charged and has a total charge of -25.0 µC. Determine the magnitude and direction of the electric field along the axis of the rod at a point 42.0 cm from its center.

Answers

Answer:

-1.4x10^6N/C

Explanation:

Pls see attached file

The magnitude of the electric field.

Magnitude is the size of the object in properties that is determines the size of the object. It also displays the result of the order of class of the object. The direction of the electric field tells us about the position of the field in four different directions. As per the question, the answer is 1.4x10^6N/C.

The rod of 16cm of total length is given. Has a charge of a total of -25.0uc. The rod's axis is pointed at 42.0 cm from its center and is given in the question. The rod Length will be then 0.16m and the total change will be 25x10 cm and point where the electricity will be calculated is shown by the axis of the rod at the distance of 42 cms.The magnitude and direction will be calculated based on the measure of the formula of E. This answer to the question will be 1.4x10^6N/C.

Learn more about the uniformly charged.

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A boat develops a leak and, after its passengers are rescued, eventually sinks to the bottom of a lake. When the boat is at the bottom, what is the force of the lake bottom on the boat?

Answers

Answer:

Explanation:

The force of the lake bottom on the boat = force the boat exerts on the bottom

force the boat exerts on the bottom = weight of the boat - buoyant force on the boat

so

The force of the lake bottom on the boat = weight of the boat - buoyant force on the boat

So

The force of the lake bottom on the boat will be less than its weight .

Give some reasons for our knowledge of the solar system has increased considerably in the past few years. Support your response with at least 3 reasons with details regarding concepts from the units learned in this course.

Answers

Answer:

Improvement in observational, and exploratory technology

Rapid increase in knowledge

International collaboration

Explanation:

Our knowledge of the solar system has increased greatly in the past few years due to to some factors which are listed below.

Improvement in observational, and exploratory technology: In recent years, developments in technology has led to the invention of advanced observational instruments and probes, that are used to study the solar system. Also more exploratory units are now developed to go out into the solar system and gather useful data which is then further processed to yield more results about our solar system.

Rapid increase in knowledge: The past few years has seen an increased number of theories proposed to explain phenomena in the solar system. Some of these theories have been seen to be accurate under experimentation, leading to newer and fresher insights into our solar system. Also, new experiments and research are carried out, all these leading to an exponential growth in our knowledge of the solar system.

International Collaboration: The sharing of knowledge by scientists all over has led to a better, quick understanding of the solar system. Also, scientists from different countries, working together on different experiment and data sharing regarding our solar system now allows our knowledge of the solar system to deepen faster.

A wet shirt is put on a clothesline to dry on a sunny day. Do water molecules lose heat and condense, gain heat and condense or gain heat and evaporate

Answers

gain heat energy and evaporate

For a wet shirt is put on a clothesline to dry on a sunny day, water molecules gain heat and evaporate.

When a clothe is placed on a line to dry, the idea is to ensure that the water molecules should evaporate.

For the water molecules to evaporate, they must gain more energy that will enable them to transit from liquid to gaseous state.

Recall that he change from liquid to vapor requires energy, this is why water molecules gain energy when they evaporate.

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The proton from the decay of the neutron is moving at 0.75 c. An observer on the proton measures the radius of the electron as 10 fermi. A Fermi is 10-15 m. What is the radius of the electron measured by someone on the electron?

Answers

Answer:

L₀ = 15.12 fermi

Explanation:

From Einstein's special theory of relativity's consequence of length contraction, we know that:

L₀ = L √(1 - v²/c²)

where,

L₀ = Length of the object (electron) measured by observer at rest = ?

L = Length of the object (electron) measured by observer in relative motion

L = 10 fermi

v = relative speed = 0.75 c

c = speed of light

Therefore,

10 fermi = L₀√(1 - 0.75²c²/c²)

10 fermi = L₀√(1 - 0.5625)

10 fermi = L₀√0.4375

10 fermi/0.66 = L₀

L₀ = 15.12 fermi

A block is released from the top of a frictionless incline plane as pictured above. If the total distance travelled by the block is 1.2 m to get to the bottom, calculate how fast it is moving at the bottom using Conservation of Energy.

Answers

Complete Question

The diagram for this question is showed on the first uploaded image (reference homework solutions )

Answer:

The velocity at the bottom is [tex]v = 11.76 \ m/ s[/tex]

Explanation:

From the question we are told that

The total distance traveled is [tex]d = 1.2 \ m[/tex]

The mass of the block is [tex]m_b = 0.3 \ kg[/tex]

The height of the block from the ground is h = 0.60 m

According the law of energy

[tex]PE = KE[/tex]

Where PE is the potential energy which is mathematically represented as

[tex]PE = m * g * h[/tex]

substituting values

[tex]PE = 3 * 9.8 * 0.60[/tex]

[tex]PE = 17.64 \ J[/tex]

So

KE is the kinetic energy at the bottom which is mathematically represented as

[tex]KE = \frac{1}{2} * m v^2[/tex]

So

[tex]\frac{1}{2} * m* v ^2 = PE[/tex]

substituting values

=> [tex]\frac{1}{2} * 3 * v ^2 = 17.64[/tex]

=> [tex]v = \sqrt{ \frac{ 17.64}{ 0.5 * 3 } }[/tex]

=> [tex]v = 11.76 \ m/ s[/tex]

A 2.25 Ω resistor, a 5.90 Ω resistor, and a 9.91 Ω resistor are connected in series with a 12.0 V battery. What is the equivalent resistance? Submit Answer Tries 0/20 What is the current in each resistor?

Answers

Answer:

(a) The equivalent resistance is 18.06 Ω

(b) The current in each resistor is 0.665 A

Explanation:

Given;

first resistor, R₁ = 2.25 Ω

second resistor, R₂ = 5.90 Ω

third resistor, R₃ = 9.91 Ω

Equivalent resistance;

[tex]R_T = R_1 + R_2+ R_3\\\\[/tex]

[tex]R_T =[/tex] 2.25 Ω + 5.90 Ω + 9.91 Ω

[tex]R_T =[/tex] 18.06 Ω

The current in each resistor;

For series connection, current is the same in every resistor.

V = I(R₁ + R₂ + R₃)

[tex]V = IR_T\\\\I = \frac{V}{R_T} \\\\I = \frac{12}{18.06} \\\\I = 0.665 \ A[/tex]

Two light pulses are emitted simultaneously from a source. The pulses take parallel paths to a detector 7.65 m away, but one moves through air and the other through a block of ice. Determine the difference in the pulses times of arrival at the detector. (Assume the index of refraction of ice is 1.309.) Answer in units of ns.

Answers

Answer:

The difference in the pulses times of arrival at the detector is [tex]\Delta t = 0.79*10^{-8} \ s[/tex]

Explanation:

From the question we are told that

The distance of the detector from the source is [tex]d = 7.65 \ m[/tex]

The index of refraction of ice is [tex]n_i = 1.309[/tex]

Generally the speed of light is a constant with a value [tex]c = 3. *10^{8} \ m/ s[/tex]

So the time taken for the first light source through air is

[tex]t_a = \frac{d}{c}[/tex]

substituting value

[tex]t_a = \frac{7.65}{3.0 *10^{8}}[/tex]

[tex]t_a = 2.55 *10^{8} \ s[/tex]

The time taken to travel through ice is

[tex]t_i = \frac{d}{\frac{c}{n_i} }[/tex]

substituting values

[tex]t_i = \frac{7.65}{\frac{3.0*10^{8}}{1.309} }[/tex]

[tex]t_i = 3.34 *10^{-8}[/tex]

The in pulses time arrival is mathematically evaluated as

[tex]\Delta t = t_2 - t_1[/tex]

substituting values

[tex]\Delta t = (3.34 - 2.55)*10^{-8}[/tex]

[tex]\Delta t = 0.79*10^{-8} \ s[/tex]

A reel of flexible power cable is mounted on the dolly, which is fixed in position. There are 190 ft of cable weighing 0.402 lb per foot of length wound on the reel at a radius of 14 in. The empty spool weighs 65 lb and has a radius of gyration about its axis of 11 in. A tension T of 15 lb is required to overcome frictional resistance to turning. Calculate the angular acceleration α of the reel if a tension of 33 lb is applied to the free end of the cable.

Answers

Answer:

Explanation:

Total mass of cable m = 190 x .402 = 76.38 lb

moment of inertial due to this cable = m r²

= 76.38 x (14/12)²

= 103.96 lb ft²

moment of inertia of empty spoon

= mR² where R is radius of gyration

= 65 x (11 / 12 )²

= 54.61 lb ft²

Total moment of inertia I = 158.57 lb ft²

Net force applied = force applied - frictional force

= 33 - 15 = 18 lb

= 18 x 32 poundal

= 576 poundal

Torque applied = force x radius

= 576 x 14/12

= 672 unit

Angular acceleration = torque / total moment of inertia

= 672 / 158.57

= 4.238 radian / s²

At what minimum angle will you get total internal reflection of light traveling in flint glass and reflected from water?

Answers

Answer:

θ_c = 53.65°

Explanation:

The point after which the light ray had started reflecting internally will be when the reflecting angle is at 90°. The incident angle at this point is called the critical angle and this can be calculated through Snell's law as;

n1 sin θ_c = n2 sin 90

Where;

n1 is the refractive index of the medium through which the incident rays will pass through.

n2 is the Refractive index of the medium through which the refracted rays will pass through.

θ_c is the critical angle at which the incident ray will reflect totally internally.

Now, since sin 90 = 1

Thus;

n1 sin θ_c = n2

θ_c = sin^(-1) (n2/n1)

Now, we are told that the reflection travels in flint glass and reflected from water.

Thus, the first medium is flint glass and the second medium is water.

So, from tables,

Refractive index of flint glass; n1 = 1.655

Refractive index of water; n2 = 1.333

Thus;

θ_c = sin^(-1) (1.333/1.655)

θ_c = 53.65°

how does current change under different polarity?

Answers

Answer:

Due to flipping of polarity

Explanation:

During the changing of polarity, the current on the one side is maximum as the polarity change then the current is gradually reducing toget from another end.

A plastic box has an initial volume of 2.00 m 3 . It is then submerged below the surface of a liquid and its volume decreases to 1.96 m 3. what is the volume strain on the box

Answers

Answer:

Volume strain is 0.02

Explanation:

Volume strain is defined as the change in volume to the original volume.

It is given that,

Initial volume of the plastic box is 2 m³

It is then submerged below the surface of a liquid and its volume decreases to 1.96 m³

We need to find the volume strain on the box. It is defined as the change in volume divided by the original volume. So,

[tex]\delta V=\dfrac{V_f-V_i}{V_i}\\\\\delta V=\dfrac{1.96-2}{2}\\\\\delta V=0.02[/tex]

So, the volume strain on the box is 0.02.

When the pivot point of a balance is not at the center of mass of the balance, how is the net torque on the balance calculated? When a force is applied directly to the pivot point of a balance, what is the torque due to that force?

Answers

Answer:

Explanation:

Net torque is calculated by multiplying the force with distance from the point of application of force to the point of pivot .

If more than 2 forces are present, then we either subtract the product of forces with their respective distances from pivot point or we add them . It depends on whether they both are present on opposite sides of pivot or on same side of pivot .

When a force is applied directly to the pivot point of balance, then the torque on due that force = 0 (zero) .

It is so because the torque is defined as the product of force and perpendicular distance from the pivot point but here the distance is 0 , therefore torque is zero.

At a playground, two young children are on identical swings. One child appears to be about twice as heavy as the other. Part A If you pull them back together the same distance and release them to start them swinging, what will you notice about the oscillations of the two children

Answers

Answer:

The motion of the lighter child would look faster than that of the heavier child, but both have the same period of oscillation.

Explanation:

Oscillation is a type of simple harmonic motion which involves the to and fro movement of an object. The oscillation takes place at a required time called the period of oscillation.

Since the swings are similar, the period of oscillation of the two children are the same and they would complete one oscillation in the same time. Though the oscillation of the lighter child seems faster than that of the heavy child, their masses does not affect the period of oscillation.

When a heavy object oscillates, its mass increases the drag or damping force, but not the period of oscillation. Thus, it oscillate slowly.

Bromine, a liquid at room temperature, has a boiling point

Answers

Yes it does ! The so-called "boiling point" is the temperature at which Bromine liquid can change state and become Bromine vapor, if enough additional thermal energy is provided. The boiling point is higher than room temperature.

In a double-slit interference experiment you are asked to use laser light of different wavelengths and determine the separation between adjacent maxima. You observe that this separation is greatest when you illuminate the double slit with In a double-slit interference experiment you are asked to use laser light of different wavelengths and determine the separation between adjacent maxima. You observe that this separation is greatest when you illuminate the double slit with:_________.
1. yellow light.
2. red light.
3. blue light.
4. green light.
5. The separation is the same for all wavelengths.

Answers

Answer:

Red light

Explanation:

This because All interference or diffraction patterns depend upon the wavelength of the light (or whatever wave) involved. Red light has the longest wavelength (about 700 nm)

A certain corner of a room is selected as the origin of a rectangular coordinate system. If a fly is crawling on an adjacent wall at a point having coordinates (1.2, 1.9), where the units are meters, what is the distance of the fly from the corner of the room?

Answers

Answer:

[tex]D=2.25 m[/tex]

Explanation:

We can use the equation of a distance between to points:

[tex]D=\sqrt{(x_{1}-x_{0})^{2}+(y_{1}-y_{0})^{2}}[/tex]

Now, we know that the origin point (x₀,y₀) = (0,0) and the point (x₁,y₁) = (1.2,1.9)

[tex]D=\sqrt{(1.2-0)^{2}+(1.9-0)^{2}}[/tex]

[tex]D=2.25 m[/tex]

Therefore the distance of the fly will be D = 2.25 m from the corner.

I hope it helps you!

If the person is shaking her hand up-and-down 12 times per second, what is the wave speed?

Answers

Welllll, first of all, it would take incredible muscular development and control to be able to do that, and I don't believe it's actually humanly possible.

But for Math and Physics problems, that's OK. We don't mind suspending our disbelief, accepting a temporary alternate reality, and working with the hand that is dealt.

The speed of a wave doesn't depend on how the wave is created. A puppy wagging its tail, a fly batting its wings, or a person shaking her hand up and down, are moving the air. The wave that travels away from the vibration is a sound wave in air. Its speed depends only on the characteristics of the air it travels through.

For some typical combination of temperature, pressure, and humidity, this speed (of sound) is taken to be 343 meters per second.

Notice that the 'sound' of shaking her hand up and down will not be 'heard' by anyone, no matter how close she stands to them. 12 Hz (12 per second) is not a fast-enough wiggle to be sensed as sound by human ears. If the person senses the wave at all, it will only be as some kind of pulsating breeze.

Imagine two free electrons that collide elastically in an acidic solution where one electron was moving and the other electron was stationary. When the electrons separate the moving electron now has a velocity of 400 m/s and the stationary electron now has a velocity of 200 m/s. What was the initial kinetic energy of the moving electron

Answers

Answer: 9.1 × 10^-26 Joule

Explanation:

Since the collision is elastic. The kinetic energy will be conserved. That is, the sum of kinetic energy before collision will be the same as the sum of the

energy after collision.

Mass of an electron = 9.1 × 10^-31 kg

Given that the velocity of the moving electron = 400 m/s and the stationary electron now has a velocity = 200 m/s.

K.E = 1/2mv^2

Add the two kinetic energies

1/2mV1^2 + 1/2mV2^2

1/2m( V1^2 + V2^2 )

Since they both have common mass

Substitute m and the two velocities

1/2 × 9.1×10^-31( 400^2 + 200^2)

4.55×10^-31 ( 160000 + 40000 )

4.55×10^-31 × 200000

K.E = 9.1 × 10^-26 Joule

Therefore, the initial kinetic energy of the moving electron is 9.1×10^-26 J

Which of the following statements are true?A. The decrease in the amplitude of an oscillation caused by dissipative forces is called damping. B. The increase in amplitude of an oscillation by a driving force is called forced oscillation. C. In a mechanical system, the amplitude of an oscillation diminishes with time unless the lost mechanical energy is replaced. D. An oscillation that is maintained by a driving force is called forced oscillation.

Answers

Answer:

right A, B, C, D

Explanation:

They ask which statements are true

A) Right. The decrease in amplitude is due to the dissipation of energy by friction and is called damping

B) Right. In resonant processes the amplitude of the oscillation increases, being a forced oscillation

C) Right. In a system with energy loss, the amplitude must decrease, therefore energy must be supplied to compensate for the loss.

D) Right. It is a resonant process the driving force keeps the oscillation of the system

Statements that are right as regards oscillation are:

A. The decrease in the amplitude of an oscillation caused by dissipative forces is called damping.

B. The increase in amplitude of an oscillation by a driving force is called forced oscillation.

C. In a mechanical system, the amplitude of an oscillation diminishes with time unless the lost mechanical energy is replaced.

D. An oscillation that is maintained by a driving force is called forced oscillation.

Amplitude can be regarded as magnitude of change that is been experienced by oscillating variable with each oscillation.

When there is a decrease in the amplitude of an oscillation as a result dissipative forces, then it is regarded as damping.

When there is increase in amplitude of an oscillation as a result of driving force then it is termed forced oscillation.

Therefore, the options are correct.

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Two wheels with fixed hubs and radii 0.51 m and 1.9 m, each having a mass of 3 kg, start from rest. Forces 5 N and F2 are applied tangentially to the smaller and larger wheels, respectively. In order to impart the same angular acceleration to each wheel, how large must F2 be? Assume that the hubs and spokes are massless, so that the rotational inertia is given by I = mR2. Answer in units of N.

Answers

Answer:

18.63 N

Explanation:

Assuming that the sum of torques are equal

Στ = Iα

First wheel

Στ = 5 * 0.51 = 3 * (0.51)² * α

On making α subject of formula, we have

α = 2.55 / 0.7803

α = 3.27

If we make the α of each one equal to each other so that

5 / (3 * 0.51) = F2 / (3 * 1.9)

solve for F2 by making F2 the subject of the formula, we have

F2 = (3 * 1.9 * 5) / (3 * 0.51)

F2 = 28.5 / 1.53

F2 = 18.63 N

Therefore, the force F2 has to 18.63 N in order to impart the same angular acceleration to each wheel.

The f2 should be 18.63 N.

Calculation of F2:

here we presume that the sum of torques are equal

So,

Στ = Iα

At First wheel

Στ = 5 * 0.51 = 3 * (0.51)² * α

On making α subject of formula, we have

α = 2.55 / 0.7803

α = 3.27

If we make the α of each one equal to each other

So,

5 / (3 * 0.51) = F2 / (3 * 1.9)

F2 = (3 * 1.9 * 5) / (3 * 0.51)

F2 = 28.5 / 1.53

F2 = 18.63 N

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When a falling meteoroid is at a distance above the Earth's surface of 3.50 times the Earth's radius, what is its acceleration due to the Earth's gravitation?

Answers

Answer:

0.484 m/s²

Explanation:

To solve this question, we would apply the formula for calculating gravitational acceleration at any distance.

g = GM/r², where

g = acceleration due to gravity

G = gravitational constant

r = radius of the meteoroid

Radius of the earth is given as 6371 km

The meteoroid is located at a distance 3.5 times the radius of the earth, so

r = R + 3.5R = 4.5R

r = 28670 km

Mass of the earth is, 5.97*10^24 kg

Now, we proceed to substitute our values into the earlier equation

g = GM/r²

g = (6.67*10^-11 * 5.97*10^24) / 28670²

g = 398.2*10^12 / 822*10^12

g = 0.484 m/s²

Thus, it's acceleration due to earth's gravitation is 0.484 m/s²

A circuit element maintains a constant resistance. If the current through the circuit element is doubled, what is the effect on the power dissipated by the circuit element

Answers

Answer:

This shows that the power dissipated by the circuit element is four times its original power if the current is doubled.

Explanation:

The formula for calculating the power expended in a circuit is P = I²R... 1

i is the current (in amperes)

R is the resistance (in ohms)

If a circuit element maintains a constant resistance and the current through the circuit element is doubled, then new current I₂ = 2I

New power dissipated P₂ = (I₂)²R

P₂ = (2I)²R

P₂ = 4I²R ... 2

Dividing equation 2 by 1 will give;

P₂/P = 4I²R/I²R

P₂/P = 4

P₂ = 4P

This shows that the power dissipated by the circuit element is four times its original power if the current is doubled.

A 170-km-long high-voltage transmission line 2.00 cm in diameter carries a steady current of 1,010 A. If the conductor is copper with a free charge density of 8.50 1028 electrons per cubic meter, how many years does it take one electron to travel the full length of the cable

Answers

Answer:

The number of years is [tex]t_y = 22.8 \ years[/tex]

Explanation:

From the question we are told that

The length of the transmission line is [tex]L = 170 \ km = 170000 \ m[/tex]

The diameter of the transmission line is [tex]d = 2.0 \ cm = 0.02 \ m[/tex]

The current which the transmission line carry is [tex]I = 1,010 \ A[/tex]

The charge density of the transmission line is [tex]j = 8.50 *10^{28 } \ electron/m^3[/tex]

Now the cross-sectional area of the transmission line is mathematically represented as

[tex]A = \pi r^2[/tex]

Here r is the radius which is mathematically evaluated as

[tex]r = \frac{d}{2}[/tex]

substituting values

[tex]r = \frac{0.02}{2}[/tex]

[tex]r = 0.01 \ m[/tex]

Hence

[tex]A = 3.142 * (0.01)^2[/tex]

=> [tex]A = 3.142 *10^{-4} \ m^2[/tex]

Now the drift velocity of electron is mathematically evaluated as

[tex]v = \frac{I}{j* e * A }[/tex]

Where e is the charge on one electron and the values is [tex]e = 1.60 *10^{-19} \ C[/tex]

So

[tex]v = \frac{ 1010}{8.50 *10^{28}* (1.60 *10^{-19}) * 3.142*10^{-4} }[/tex]

[tex]v = 2.363 *10^{-4} \ m/s[/tex]

Now the time taken is mathematically evaluated as

[tex]t = \frac{L}{v }[/tex]

substituting values

[tex]t = \frac{170000}{2.363 *10^{-4} }[/tex]

[tex]t = 7.194*10^{8}\ s[/tex]

Converting to years

[tex]t_y = \frac{t}{365\ days * 24 \ hours * 3600\ seconds}[/tex]

substituting values

[tex]t_y =\frac{7.194 *10^{8}}{365 *24 * 3600}[/tex]

[tex]t_y = 22.8 \ years[/tex]

The predictions of Rutherford's scattering formula failed to correspond with experimental data when the energy of the incoming alpha particles exceeded 32MeV32MeV. This can be explained by the fact that the predictions of the formula apply when the only force involved is the electromagnetic force and will break down if the incoming particles make contact with the nucleus. Use the fact that Rutherford's prediction ceases to be valid for alpha particles with an energy greater than 32MeV32MeV to estimate the radius rrr of the gold nucleus.

Answers

Answer:

r = 7.1 × 10⁻¹⁵

Explanation:

Air is cooled in a process with constant pressure of 150 kPa. Before the process begins, air has a specific volume of 0.062 m^3/kg. The final specific volume is 0.027 m^3/kg. Find the specific work in the process.

Answers

Answer:

The pressure is constant, and it is P = 150kpa.

the specific volumes are:

initial = 0.062 m^3/kg

final = 0.027 m^3/kg.

Then, the specific work can be written as:

[tex]W = \int\limits^{vf}_{vi} {Pdv} \, = P(vf - vi) = 150kPa*(0.0027 - 0.062)m^3/kg = -5.25 kPa*m^3/kg.[/tex]

The fact that the work is negative, means that we need to apply work to the air in order to compress it.

Now, to write it in more common units we have that:

1 kPa*m^3 = 1000J.

-5.25 kPa*m^3/kg = -5250 J/kg.

If an ocean wave takes 2 minutes to generate one full cycle, from crest to crest, then we will call this the frequency of the wave. What would its period be

Answers

Answer:

Explanation:

The period of the wave will be 2 minutes

frequency = 1 / period

= 1 / 2

= .5 per minute

What is the equivalent temperature in kelvin if you have a metal at 50°F?

Answers

Answer:

The required temperature is 283 K.

Explanation:

[tex]T\:=\:\left(50-32\right)\times \frac{5}{9}+273\\\\T=283\:K[/tex]

Best Regards!

If the heating curve is reversed, what would best describe the boiling point?

Answers

Answer:

point of condensation

Explanation:

A 2 Kg Rubber Ball Is Thrown At A Wall Horizontally At 3 M/s, And Bounces Back The Way It Came At An (2024)

FAQs

What happens when a rubber ball thrown with some velocity on a wall? ›

That is because of Newton's third law. The law states that “To every action there is equal and opposite reaction”. This third law gives us the law of conservation of momentum. The ball when it hits the wall exerts a force on the wall and intern the wall exerts force in opposite direction so the ball bounces back.

When a rubber ball bounces off a wall with an initial speed? ›

Option a the momentum of the ball is correct. Explanation: When a rubber ball bounces off of a wall with an initial speed v and then reverses its direction with the same speed after the bounce, momentum is conserved but kinetic energy is not conserved.

When you drop a rubber ball on the floor it bounces back? ›

Answer and Explanation:

When the rubber ball is dropped on the floor, it bounces back to the upward direction. The upward motion was due to the reaction force from the floor and the force due to the elastic nature of the rubber ball.

Why does a ball bounce back when it hits a wall? ›

When the ball hits the wall, the wall's molecule at that projected area will be compressed. And will try to obtain its primary shape. And the conserved momentum inside the wall will push back the ball with the same effect. Thus, the ball bounces back after strikes the wall.

Why does a rubber ball rebound when it hits a wall? ›

Rubber ball has high elasticity, therefore when it hits the wall it gets compressed and due to elasticity, it quickly returns to its original shape. While doing this it pushes back on the wall. Thus, it rebounds back.

When a ball is thrown on the floor or a wall it bounces back? ›

According to Newton's third law, "for every action there is an equal and opposite reaction". When we hit the ball on ground then in this case there is an equal and opposite force on the ball in the upward direction due to ground(Newton's Third Law), which makes it bounce.

What happens to velocity when a ball hits a wall? ›

Clearly, the momentum of the ball is changed by the collision with the wall, since the direction of the ball's velocity is reversed.

What happens when a ball is thrown against a wall? ›

Explanation: When a ball is thrown against a wall, the ball exerts a force on the wall and the wall exerts a force back. This is due to Newton's third law of motion. According to this law, every action have equal and opposite reaction and both action and reaction forces acts on different object.

What is the action and reaction of a ball thrown against a wall? ›

You probably know that when you throw a ball against a wall, the ball exerts a force on the wall. Likewise, the wall puts force on the ball as a result of which the ball bounces off the wall. Similarly, earth pulls you down with gravitational force.

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