A very elastic rubber ball is dropped from a certain height and hits the floor with a (2pts) downward speed v. Since it is so elastic, the ball bounces back with the same speed v going upward. Which of the following statements about the bounce are correct? (There could be more than one correct choice.) 1. The balrs mpmentum was conserved during the bounce. 2. The ball ha the same momentum just before and just after the bounce. 3. The magnitude of the ball's momentum was the same just before and just after the bounce. 4. All of the above 5. None of the above

Answer:

speed of sound in unknown gas will be 360m/sec

Explanation:

We have given length of the pipe l = 0.770 m

It is given that pipe is open at one end and closed at one end

Frequency of third harmonic f = 750

Third harmonic frequency of pipe which one end is open and one end is closed is given by

[tex]f=\frac{5v}{4L}[/tex], here v is the speed of the sound, L is the length of the pipe

So  [tex]750=\frac{5v}{4\times 0.770}[/tex]

v = 360 m /sec

So speed of sound in unknown gas will be 360m/sec

Answer:

Constant pressure heat addition in a boiler,

Isentropic compression in a pump,

constant pressure heat rejection in a condenser and Isentropic expansion in a turbine

B

Explanation:

The Rankine cycle is a model used to predict the performance of steam turbine systems. It involves the use of these four components

1. Pump

2. Boiler

3. Turbine

4. Condenser

It includes the following processes;

1. Isentropic compression in pump

2. Constant pressure heat addition in boiler

3. Isentropic expansion in turbine

4. Constant pressure heat rejection in condenser

All this processes make up a Rankine cycle.

Answer:

So amount of work produced will be [tex]10{-4}J[/tex]

Explanation:

We have given diameter of ammonia bubble is changes from 1 cm to 3 cm

So radius changes from 0.5 cm to 1.5 cm

Surface area of bubble[tex]=4\pi r^2[/tex]

So change in area of bubble [tex]=4\pi (0.015^2-0.005^2)=8\times 3.14\times (0.015^2-0.005^2)=0.00251m^2[/tex]

Surface tension of ammonia = 0.04 N/m

So work done will be [tex]Work\ done=surface\ tension\times change\ in\ area=0.04\times 0.00251= 10^{-4}J[/tex]

Answer:

B_m = 3.186 x 10⁻³ T

Explanation:

given,

diameter of the beam = 0.863 mm

Amplitude = 0.955 MV/m

speed of light = 2.99792 × 10⁸ m/s

he permeability of free space= 4π × 10⁻⁷ T· N/A

Amplitude of the magnetic field is given by

[tex]B_m = \dfrac{E_m}{c}[/tex]

E_m is amplitude of the electric field.

[tex]B_m = \dfrac{0.955\times 10^{6}}{2.99792\times 10^8}[/tex]

     B_m = 0.31855 x 10⁻² T

     B_m = 3.186 x 10⁻³ T

The amplitude of magnetic field is equal to 3.186 x 10⁻³ T

Answer:

I= 12.09×10^8 W/m^2

Explanation:

Em = amplitude of electric field = 0.955 MV/m = 0.955 x 10^6 V/m

B_m = amplitude of magnetic field = ?

c = speed of light = 2.99792 x 10^8 m/s

amplitude of magnetic field is given as

B_m = E_m /c

B_m = (0.955 x 10^6)/(2.99792 x 10^8 )

B_m = 0.00318 T

b)

intensity is given as

[tex] I=(0.5)\epsilon\times E_m^2 c [/tex]

[tex]I=0.5(8.85\times10^{-12})(0.955\times10^6)^2(2.99792\times10^8)[/tex]

I= 1209873759.69

I= 12.09×10^8 W/m^2

Answer:

32.4289 N

Explanation:

A = Amplitude = 20 cm

[tex]v_m[/tex] = Maximum velocity = 44 cm/s

k = Spring constant = 16 N/m

g = Acceleration due to gravity = 9.81 m/s²

m = Mass of object

Maximum velocity is given by

[tex]v_m=A\omega[/tex]

Angular velocity is given by

[tex]\omega=\sqrt{\dfrac{k}{m}}[/tex]

[tex]v_m=A\sqrt{\dfrac{k}{m}}\\\Rightarrow m=\dfrac{A^2k}{v_m^2}\\\Rightarrow m=\dfrac{0.2^2\times 16}{0.44^2}\\\Rightarrow m=3.3057\ kg[/tex]

Weight is given by

[tex]W=mg\\\Rightarrow W=3.3057\times 9.81\\\Rightarrow W=32.4289\ N[/tex]

The weight of the bananas is 32.4289 N

Answer:

speed of eight ball speed after the collision is 3.27 m/s

Explanation:

given data

initially moving v1i = 3.4 m/s

final speed is v1f = 0.94 m/s

angle = θ w.r.t. original line of motion

solution

we assume elastic collision

so here using conservation of energy

initial kinetic energy = final kinetic energy .............1

before collision kinetic energy = 0.5 × m× (v1i)²

and

after collision kinetic energy =  0.5 × m× (v1f)²  + 0.5 × m× (v2f)²

put in equation 1

0.5 × m× (v1i)² =  0.5 × m× (v1f)²  + 0.5 × m× (v2f)²

(v2f)² = (v1i)² - (v1f)²

(v2f)² = 3.4² - 0.94²

(v2f)² = 10.68

taking the square root both

v2f = 3.27 m/s

speed of eight ball speed after the collision is 3.27 m/s

Final answer:

To find the eight ball's speed after the collision, we can use the principle of conservation of momentum and the fact that the collision is elastic. By solving equations for momentum and kinetic energy, we can determine the final velocity of the eight balls. Given the initial velocity and final velocity of the cue ball, and assuming equal masses for both balls, we can substitute these values and solve for the final velocity of the eight balls.

Explanation:

To find the eight ball's speed after the collision, we can use the principle of conservation of momentum. In an elastic collision, the total momentum before the collision is equal to the total momentum after the collision. Since the cue ball initially strikes the stationary eight ball, the total initial momentum is the momentum of the cue ball, and the total final momentum is the momentum of both the cue ball and the eight ball after the collision.

Let's denote the mass of both balls as 'm'. The initial momentum of the cue ball is given by: p_initial = m * v_cue (where v_cue is the initial velocity of the cue ball).

The final momentum of both balls is given by: p_final = m * v_cue_final + m * v_eight_final (where v_cue_final is the final velocity of the cue ball and v_eight_final is the final velocity of the eighth ball).

Since the collision is elastic, there is no loss of kinetic energy, so the total kinetic energy before the collision is equal to the total kinetic energy after the collision. The initial kinetic energy of the cue ball is given by: KE_initial = (1/2) * m * v_cue^2.

The final kinetic energy of both balls is given by: KE_final = (1/2) * m * v_cue_final^2 + (1/2) * m * v_eight_final^2.

We can solve these equations to find the final velocity of the eight balls. Given that the cue ball's initial velocity (v_cue) is 3.4 m/s, its final velocity (v_cue_final) is 0.94 m/s at an angle θ (concerning its original line of motion), and the masses of both balls are the same, we can substitute these values into the equations and solve for the final velocity of the eight ball.

Answer:

Timmy is 1 meters away from the center.

Explanation:

Given:

Distance of Suzie from center = 2 m

Weight of Suzie = 400 N

Weight of Timmy = 800 N

The see saw is balanced

To find the distance of Timmy from the center.

Solution:

Since the see saw is balanced,

So, the sum of clockwise moment = Sum of anticlockwise moments

Moment is given by:

⇒ [tex]Force\times Perpendicular\ distance\ from\ the\ center[/tex]

Let Suzies have moment in clockwise direction.

Moment of Suzie would be = [tex]400\ N \times 2\ m= 800\ Nm[/tex]

Timmy would have moment in anticlockwise direction.

Let Timmy be [tex]l[/tex] meters away from center

Moment of Timmy would be = [tex]800\ N\times l\ m=800l\ Nm[/tex]

Since, the see saw is balanced, the system is in equilibrium.

Thus, we have,

Anticlockwise Moment = Clockwise moment

[tex]800l=800[/tex]

Dividing both sides by 800.

[tex]\frac{800l}{800}=\frac{800}{800}[/tex]

[tex]l=1\ m[/tex]

Thus, Timmy is 1 meters away from the center.

Complete question

A student measures the mass of a 1.0 kg standard bar. He obtains measurements of 0.77 kg, 0.78 kg, and 0.79 kg. Which describes his measurements

a)precise but not accurate

b)accurate but not precise

c)neither precise nor accurate

d)both precise and accurate

Answer:

The measurement is precise but not accurate

Explanation:

A measurement can either be precise or accurate.

In this question, the measured values (0.77 kg, 0.78 kg and 0.79 kg) are far from the true value (1.0 kg), therefore the measurement is not accurate.

However, the measured values (0.77 kg, 0.78 kg and 0.79 kg) are close to each other, therefore the measurement is precise.

Therefore the correct option is 'a' the measurement is precise but not accurate

Answer:

precise but not accurate

Explanation:

precise but not accurate

Answer:

A) The ball's gravitational potential energy is greatest at the instant when the ball is at its highest point.

Explanation:

Suppose you kick a soccer ball straight up to a height of 10 meters. Which of the following is true about the gravitational potential energy of the ball during its flight?

A) The ball's gravitational potential energy is greatest at the instant when the ball is at its highest point. (true)

B) The ball's gravitational potential energy is always the same. (false)

    because the gravitational potential energy is changed as the height changed.

C) The ball's gravitational potential energy is greatest at the instant the ball leaves your foot. (false)

    Because The ball's gravitational potential energy is greatest at the instant when the ball is at its highest point.

D) The ball's gravitational potential energy is greatest at the instant it returns to hit the ground. (false)

     The ball's gravitational potential energy is greatest at the instant when the ball is at its highest point. not when return to the ground

Final answer:

The gravitational potential energy of a soccer ball kicked straight up is A) The ball's gravitational potential energy is greatest at the instant when the ball is at its highest point.

Explanation:

If you kick a soccer ball straight up to a height of 10 meters, the gravitational potential energy of the ball is greatest at the instant when the ball is at its highest point. At the highest point, all the energy in the system is gravitational potential energy.

This is because gravitational potential energy is calculated by the formula PE = mgh, where m is the mass of the ball, g is the acceleration due to gravity, and h is the height above the reference point, which in this case is the ground. At the ball's highest point, the h value is at its maximum of 10 meters. Therefore, the correct answer to the question is A) The ball's gravitational potential energy is greatest at the instant when the ball is at its highest point.

Answer with Explanation:

We are given that

Dipole moment=[tex]0.5 e nm=0.5\times 1.6\times 10^{-19}\times 10^{-9}=0.8\times 10^{-28}[/tex]C-m

Because [tex]1 e=1.6\times 10^{-19} C[/tex]

[tex]1 nm=10^{-9} m[/tex]

[tex]a^x\cdot a^y=a^{x+y}[/tex]

Magnitude of electric field,E=[tex]8\times 10^4[/tex] N/C

a.We have to find the magnitude of torque on the dipole when

dipole is parallel to the electric field

i.e[tex]\theta=0[/tex]

We know that

[tex]\tau=PEsin\theta[/tex]

Substitute the values then we get

[tex]\tau=0.8\times 10^{-28}\times 8\times 10^{4} sin0=0[/tex]

Because sin 0=0

[tex]\tau=0[/tex]

b.[tex]\theta=90^{\circ}[/tex]

[tex]\tau=0.8\times 10^{-28}\times 8\times 10^4sin90[/tex]

By using [tex]sin90^{\circ}=1[/tex]

[tex]\tau=6.4\times 10^{-28+4}=6.4\times 10^{-24}Nm[/tex]

[tex]\tau=6.4\times 10^{-24}[/tex]Nm

c.[tex]\theta=30^{\circ}[/tex]

[tex]\tau=0.8\times 10^{-28}\times 8\times 10^4sin 30[/tex]

[tex]\tau=6.4\times 10^{-24}\times \frac{1}{2}[/tex] Nm

By using [tex] sin30^{\circ}=\frac{1}{2}[/tex]

[tex]tau=3.2\times 10^{-24} Nm[/tex]

d.Potential energy, U=[tex]-PEcos\theta[/tex]

[tex]\theta=0[/tex]

[tex]U=-0.8\times 10^{-28}\times 8\times 10^4cos 0=-6.4\times 10^{-28}J[/tex]

Because cos 0degree=1

By having the p, E, and anglёs, we can calculate the τ value and the U values. a) 0 Nm, b) 6.4x10⁻²⁴ Nm, c) 3.2x10⁻²⁴ Nm. d) U = 6.4x10⁻²⁴, U = 0, U = 5.57x10⁻²⁴

First, we need to change the units of the given dip0lё momёnt, p.

The result is

p = 0.8 x 10 ⁻²⁸ Cm

Now we can calculate the t0rquё values.

T0rque, t = p x E x sёn θ

The results are

p0tёntial energy, U = p x E x c0s θ

You will find the complete explanation in the attached files

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The package takes approximately 0.117 seconds to reach the ground. The velocity just before it lands is 25.364 m/s (upwards).

To find the time it takes for the package to reach the ground, we can use the equation for uniform acceleration: acceleration (a) = g - (-g) = 2g, where g is the acceleration due to gravity (9.8 m/s^2) and the negative sign represents the direction of motion. Using v = u + at, where v is the final velocity (0 m/s), u is the initial velocity (2.3 m/s), and a is the acceleration, we can find the time (t) it takes for the package to reach the ground. Rearranging the equation, we have t = (v - u)/a. Plugging in the values, we have t = (0 - 2.3)/(2g) = -2.3/19.6 = 0.117 seconds. The package takes approximately 0.117 seconds to reach the ground.

To find the velocity just before it lands, we can use the equation v = u + at, where u is the initial velocity, a is the acceleration, and t is the time. Rearranging the equation, we have v = u + at. Plugging in the values, we have v = 2.3 + (2g)(0.117) = 2.3 + 23.064 =  25.364 m/s (upwards).

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Answer:To ensure significant dialogue which will guarantee that the needed Products meets specifications

It will help to develop a better Operational model for the users.

Explanation: The design phase is one of the most critical phase in product development,it is in this phase that the product structure, contents and specifications are put into action. It is necessary that the end users have a significant time to dialogue with the project or product development team to guarantee that the end product meets specifications.

It is also necessary to spent adequate time in order to make the product user friendly (easy to operate and maintain).

The sympathetic and parasympathetic divisions of the autonomic nervous system release different neurotransmitters.

In the autonomic nervous system, there are two main divisions - the sympathetic and parasympathetic. These divisions have different effects on the body due to the neurotransmitters they release. Sympathetic preganglionic fibers release acetylcholine (ACh), whereas parasympathetic preganglionic fibers also release ACh. However, parasympathetic postganglionic fibers release norepinephrine (NE), and sympathetic postganglionic fibers also release NE.

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In the autonomic nervous system, preganglionic fibers release ACh and target ganglionic neurons through nicotinic receptors. Postganglionic fibers release NE, except for certain fibers that release ACh.

In the autonomic nervous system, both sympathetic and parasympathetic preganglionic fibers release ACh. The ganglionic neurons, which are the targets of these preganglionic fibers, have nicotinic receptors. These receptors are ligand-gated cation channels that cause depolarization of the postsynaptic membrane. On the other hand, postganglionic sympathetic fibers release norepinephrine (NE), except for those that project to sweat glands and blood vessels associated with skeletal muscles, which release ACh.

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Final answer:

To find the speed of the eight ball after an elastic collision with the cue ball in motion, apply conservation laws of momentum and kinetic energy.

Explanation:

A cue ball initially moving at 2.5 m/s strikes a stationary eight ball of the same size and mass. After the collision, the cue ball’s final speed is 1.2 m/s at an angle of ? With respect to its original line of motion. To find the eight ball’s speed after the collision, we can apply the principles of conservation of momentum and conservation of kinetic energy.

Let's denote the velocities: V1 (initial cue ball velocity), V2 (initial eight ball velocity), V1' (final cue ball velocity), and V2' (final eight ball velocity). Considering an elastic collision, where momentum and kinetic energy are conserved, we can set up equations to solve for V2'.

By using the conservation laws, we can determine that the eight ball's speed after the collision is approximately 2.3 m/s.

Answer:

a=19.8977 m/s²

Explanation:

Given data

ω=3.00 rad/s

r=1.40 m

α=11.0 rad/s²

To find

Acceleration

Solution

As the object moves in a circle so it has tangential acceleration also due to circular motion  is  has centripetal acceleration

The total acceleration can be found by

[tex]a=\sqrt{(a_{c})^{2}+(a_{T})^{2}}[/tex]

where

at is tangential acceleration

ac is centripetal acceleration

First we need to find centripetal acceleration

so

[tex]a_{c}=rw^{2}[/tex]

put the values or r and ω

[tex]a_{c}=(1.40m)*(3.00rad/s)^{2}\\a_{c}=12.6 m/s^{2}[/tex]

Now for tangential acceleration

[tex]a_{t}=ra\\a_{t}=(1.40m)*(11.0rad/s^{2} )\\a_{t}=15.4 m/s^{2}[/tex]

Put values of ac and at to find total acceleration

So

[tex]a=\sqrt{(a_{t})^{2} +(a_{c})^{2} }\\ a=\sqrt{(15.4m/s^{2} )^{2}+(12.6m/s^{2} )^{2}  }\\ a=19.8977m/s^{2}[/tex]

The magnitude of the total acceleration at a point on the rotating platform is approximately 19.79 m/s².

To find the magnitude of the total acceleration at a point on the rotating platform, we need to consider both the tangential acceleration and the centripetal acceleration. The tangential acceleration is given by the product of the angular acceleration and the radius, while the centripetal acceleration is given by the square of the angular speed times the radius. Adding these two accelerations together will give us the magnitude of the total acceleration at that point.

First, calculate the tangential acceleration:

Tangential acceleration = angular acceleration × radius

Tangential acceleration = 11.0 rad/s² × 1.40 m = 15.4 m/s²

Then, calculate the centripetal acceleration:

Centripetal acceleration = (angular speed)² × radius

Centripetal acceleration = (3.00 rad/s)² × 1.40 m = 12.60 m/s²

Finally, find the vector sum of the tangential acceleration and the centripetal acceleration:

Total acceleration = √((tangential acceleration)² + (centripetal acceleration)²)

Total acceleration =√((15.4 m/s²)² + (12.60 m/s²)²)

Total acceleration ≈ 19.79 m/s²

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A baseball bat hits a ball is an example of contact force.

Option A.

Contact force is defined as those forces which are acting between two bodies by actually touch between two bodies.

In case of the book falling to ground, the force acting is the gravitational force. Gravitational force acts even when two bodies aren't in contact. So its a non contact force.

Similar is the condition of the leaf and ground where the gravitational force acts.

The magnetic force between the paper clip and the magnet is also non contact force, which acts even when two bodies aren't in contact.

But the bat hitting the ball is having direct contact between the two bodies, and their contact is what makes the ball fly off. So its a contact force.

Contact force is the type of force that occurs when the external force act on the system act physically. A baseball bat hitting a ball is the right example of contact forces.

Any force that requires touch is referred to as a contact force. The majority of apparent interactions between macroscopic groupings of matter are caused by contact forces.

Contact forces are used in everyday circumstances. For example, a baseball bat hits a ball.

Hence a baseball bat hitting a ball is the right example of contact forces.

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Answer:

-3.769 m/min

Explanation:

[tex]\dfrac{dA}{dt}[/tex] = Rate of change of area = -49 m²/min (negative due to shrinking)

s = Side length = 13 m

[tex]\dfrac{ds}{dt}[/tex] = Rate of change of side

Area of a square is given by

[tex]A=s^2[/tex]

Differentiating with respect to time

[tex]\dfrac{dA}{dt}=\dfrac{ds^2}{dt}\\\Rightarrow \dfrac{dA}{dt}=2s\dfrac{ds}{dt}\\\Rightarrow \dfrac{ds}{dt}=\dfrac{dA}{dt}\times \dfrac{1}{2s}\\\Rightarrow \dfrac{ds}{dt}=-49\times \dfrac{1}{13}\\\Rightarrow \dfrac{ds}{dt}=-3.769\ m/min[/tex]

The rate of change of the sides of the square is -3.769 m/min

Complete question

The complete question is shown on the first and second uploaded image

Answer:∈

Answer to first question is shown on the second uploaded image.

Part B the Answer is:

The ratio  [tex]\frac{R}{a}[/tex] is evaluated to be 49.99

Explanation:

The explanation is shown on the third ,fourth and fifth image.

The ratio  [tex]\frac{R}{a}[/tex] is "49.9975".

For the last part, we should have

[tex]\to E = 0.99 \frac{\eta }{\varepsilon_0 }[/tex]

Therefore we should have

[tex]\to \frac{\eta}{\varepsilon_0}(1-\frac{1}{\sqrt{ (\frac{2R}{a})^2+1}}) = 0.99 \frac{\eta}{\varepsilon_0}[/tex]

[tex]\to (1-\frac{1}{\sqrt{ (\frac{2R}{a})^2+1}}) = 0.99 \\\\\to 1-0.99= \frac{1}{\sqrt{ (\frac{2R}{a})^2+1}}\\\\ \to 0.01= \frac{1}{\sqrt{ (\frac{2R}{a})^2+1}} \\\\\to \sqrt{ (\frac{2R}{a})^2+1}= 100\\\\\to (\sqrt{ (\frac{2R}{a})^2+1})^2= 100^2\\\\\to (\frac{2R}{a})^2+1= 10000\\\\\to (\frac{2R}{a})^2= 10000-1\\\\\to (\frac{2R}{a})^2= 9999\\\\\to \frac{2R}{a}= 99.9995\\\\\to \frac{R}{a}= 49.9975\\\\[/tex]

Note:

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Answer:

By suspending the needles on in the air, the needle which orients itself in the north-south direction is magnetized

Explanation:

Answer:

I = 1.525 kg.m/s

Explanation:

given,

mass of the ball = 0.165 Kg

height of drop, h = 1.25 m

ball rebound and reach to height, h' = 0.940 m

impulse = ?

using conservation of energy

Potential energy is converted into kinetic energy

[tex]mgh = \dfrac{1}{2}mv^2[/tex]

[tex]v=\sqrt{2gh}[/tex]

[tex]v=\sqrt{2\times 9.8 \times 1.25}[/tex]

  v = 4.95 m/s

velocity of the ball after rebound

again using conservation of energy

[tex]mgh = \dfrac{1}{2}mv'^2[/tex]

[tex]v'=\sqrt{2gh}[/tex]

[tex]v'=\sqrt{2\times 9.8 \times 0.94}[/tex]

  v' = 4.29 m/s

impulse is equal to change in momentum

I = m ( v' - v )

I = 0.165 x ( 4.29 - (-4.95))

I = 1.525 kg.m/s

There is one mistake in the question as unit of height of building is not given.So I assume it as meter.The complete question is here

You are on the roof of the physics building, 46.0 m above the ground. Your physics professor, who is 1.80 m tall, is walking alongside the building at a constant speed of 1.20 m/s. If you wish to drop an egg on your professor’s head, where should the professor be when you release the egg? Assume that the egg is in free fall.  

Answer:

d=3.67 m

Explanation:

Height of building=46.0 m

First we need to find time taken by egg to reach 1.80 m above the surface

So to find time use below equation

[tex]S=vt+\frac{1}{2} gt^{2}\\ (46.0-1.80)m=(om/s)t+\frac{1}{2}(9.8m/s^{2} )t^{2}\\t=\sqrt{\frac{(46.0-1.80)m}{4.9} }\\ t=3.06s[/tex]

As velocity 1.20m/s is given and we have find time.So we can easily find the distance

So

[tex]distance=velocity*time\\d=v*t\\d=(1.20m/s)*(3.06s)\\d=3.67m[/tex]

Answer:

The impulse needed to stop the child = -200 Ns

Explanation:

Impulse: This can be defined as the product of force and time of a body. The S.I unit of impulse is kgm/s or Ns

From newton's second law,

I = mΔv ................... Equation 1.

Where I = Impulse needed to stop the child, m = mass of the child, Δ = change in velocity of the child.

Given: m = 20.0 kg,

Δv = v-u where v =0 m/s, u = 10.0 m/s

Δv = 0-10 = -10 m/s.

Substituting into equation 1

I = 20(-10)

I = -200 Ns.

Thus the impulse needed to stop the child = -200 Ns

Note: The negative sign shows that the impulse act against the motion of the child

The correct impulse needed to stop the child is 200 N·s.

The impulse needed to stop the child is 16 N·s.To find the impulse needed to stop the child, we can use the formula for impulse, which is the change in momentum of an object. The momentum of an object is given by the product of its mass and velocity. Impulse is the product of the average force applied to the object and the time interval during which the force is applied. The impulse is equal to the change in momentum.

The initial momentum of the child (p_initial) is given by the mass of the child (m) multiplied by the initial velocity of the car (v_initial), since the child's velocity is assumed to be the same as that of the car. The final momentum of the child (p_final) is zero because the child comes to a stop.

Given:

- Mass of the child, m = 20.0 kg

- Initial velocity of the car (and the child), v_initial = 10.0 m/s

- Time taken to stop, Aat = 0.050 s

- The change in momentum (Aap) is:

Aap = [tex]p_final - p_initial[/tex]

Aap = [tex]0 - (m * v_initial)[/tex]

Aap = - (20.0 kg * 10.0 m/s)

Aap= - 200 kgA·m/s

The negative sign indicates that the momentum is decreasing (the child is stopping).

 The impulse (J) is equal to the change in momentum:

J = Aap

J = - 200 kgA·m/s

Since impulse is a vector quantity and we are interested in the magnitude of the impulse needed to stop the child, we take the absolute value:

J = | - 200 kgA·m/s |

J = 200 kgA·m/s

Now, we can express the impulse in Newtons-seconds (NA·s) by noting that 1 kgA·m/s is equivalent to 1 NA·s:

J = 200 NA·s

However, the answer provided initially (16 NA·s) seems to be incorrect. Let's re-evaluate the calculation to ensure accuracy.

The impulse J is the change in momentum Aap over the time interval Aat. The average force F_avg applied to the child is then the impulse divided by the time interval:

F_avg = J / Aat

 Rearranging for J gives:

J = F_avg / Aat

Since we know the change in momentum Aap is equal to the impulse J, we can write:

J = Aap = m * Aav

The change in velocity Aav of the car (and thus the child) is the final velocity [tex]v_final[/tex]minus the initial velocity [tex]v_initial[/tex]. Since the car stops, [tex]v_final[/tex]= 0, and Aav is simply [tex]-v_initial[/tex].

Aav = v[tex]_final - v_initial[/tex]

Aav = 0 - 10.0 m/s

Aav= - 10.0 m/s

Now we can calculate the impulse needed to stop the child:

J = m * Aav

J = 20.0 kg * (- 10.0 m/s)

J = - 200 kgA·m/s

Taking the magnitude:

J = | - 200 kgA·m/s |

J = 200 kgA·m/s

Since 1 kgA·m/s is equivalent to 1 NA·s, the impulse in Newtons-seconds is:

J = 200 NA·s

This confirms that the initial answer provided (16 N·s) is indeed incorrect, and the correct impulse needed to stop the child is 200 N·s.

Explanation:

Force is given by rate of change of momentum.

Mass of object = 200 kg

Initial velocity = 5 m/s

Final velocity = 25 m/s

a) Change in momentum = 200 x 25 - 200 x 5

Change in momentum = 4000 kg m/s in the direction of motion

b) Time taken = 50 s

Rate of change of momentum = Change in momentum ÷ Time

Rate of change of momentum = 4000 ÷ 50

Rate of change of momentum = 80 N

Force = 80 N

Magnitude of the force is 80 N

In 1932, James Chadwick proved the existence of neutrons by bombarding Beryllium-9 with alpha particles. The reaction resulted in the production of neutrons and likely also Carbon-12.

In the experiment conducted by James Chadwick in 1932, he bombarded a sample of Beryllium-9 with alpha particles. The reaction resulted in two products: a neutron, which was significant as it directly demonstrated the existence of neutrons, and another product. This missing product has not been mentioned in the question, but based on the options provided, one can infer that the most likely answer is Carbon-12. This was identified because the carbon-12 isotope is a stable one usually formed in nuclear reactions like Chadwick's.

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The orbital speed of the satellite can be found using the equations for gravitational force and centripetal acceleration. By equating these two equations, we can solve for the orbital speed. Plugging in the given values, we find that the orbital speed of the satellite is approximately 6.43 x 10^3 m/s.

To find the orbital speed of the satellite, we can use the equation for gravitational force: F = ma, where F is the gravitational force, m is the mass of the satellite, and a is the centripetal acceleration. We can also use the equation for centripetal acceleration: a = v^2 / r, where v is the orbital speed and r is the distance between the satellite and the center of Jupiter. By equating the two equations, we can solve for v.

The gravitational force acting on the satellite is given by the formula F = G * (m1 * m2) / r^2, where G is the gravitational constant, m1 is the mass of Jupiter, and r is the distance between the satellite and the center of Jupiter.

Plugging in the values given in the question, we can solve for v:

v^2 = (G * m1) / r

v = sqrt((G * m1) / r)

Using the given values of m1 = 1.90 x 10^27 kg, r = 8.52 x 10^5 m, and the gravitational constant G = 6.67 x 10^-11 Nm^2/kg^2, we can calculate the orbital speed of the satellite:

v = sqrt((6.67 x 10^-11 Nm^2/kg^2 * 1.90 x 10^27 kg) / (8.52 x 10^5 m))

v ≈ 6.43 x 10^3 m/s

Answer:

[tex]g_n=\dfrac{2}{9}g[/tex]

Explanation:

M = Mass of Earth

G = Gravitational constant

R = Radius of Earth

The acceleration due to gravity on Earth is

[tex]g=\dfrac{GM}{R^2}[/tex]

On new planet

[tex]g_n=\dfrac{G2M}{(3R)^2}\\\Rightarrow g_n=\dfrac{2GM}{9R^2}[/tex]

Dividing the two equations we get

[tex]\dfrac{g_n}{g}=\dfrac{\dfrac{2GM}{9R^2}}{\dfrac{GM}{R^2}}\\\Rightarrow \dfrac{g_n}{g}=\dfrac{2}{9}\\\Rightarrow g_n=\dfrac{2}{9}g[/tex]

The acceleration due to gravity on the other planet is [tex]g_n=\dfrac{2}{9}g[/tex]

Answer:

2 g/9

Explanation:

mass of planet, Mp = 2 x Me

radius of planet, Rp = 3 x Re

Where, Me is the mass of earth and Re is the radius of earth.

The formula for acceleration due to gravity on earth is given by

[tex]g = \frac{GM_{e}}{R_{e}^{2}}[/tex]     .... (1)

The acceleration due to gravity on the planet is given by

[tex]g' = \frac{GM_{p}}{R_{p}^{2}}[/tex]

By substituting the values, we get

[tex]g' = \frac{2GM_{e}}{9R_{e}^{2}}[/tex]    ..... (2)

Divide equation (2) by equation (1), we get

g'/g = 2/9

g' = 2 g/9

Thus, the acceleration due to gravity on th enew planet is 2 g/9.  

Answer:

2.23 × 10^6 g of F- must be added to the cylindrical reservoir in order to obtain a drinking water with a concentration of 0.8ppm of F-

Explanation:

Here are the steps of how to arrive at the answer:

The volume of a cylinder = ((pi)D²/4) × H

Where D = diameter of the cylindrical reservoir = 2.02 × 10^2m

H = Height of the reservoir = 87.32m

Therefore volume of cylindrical reservoir = (3.142×202²/4)m² × 87.32m = 2798740.647m³

1ppm = 1g/m³

0.8ppm = 0.8 × 1g/m³

= 0.8g/m³

Therefore to obtain drinking water of concentration 0.8g/m³ in a reservoir of volume 2798740.647m³, F- of mass = 0.8g/m³ × 2798740.647m³ = 2.23 × 10^6 g must be added to the tank.

Thank you for reading.

Answer:

0.09852 seconds

Explanation:

[tex]\mu[/tex] = Linear density = 0.0292 kg/m

[tex]T_A[/tex] = Tesnsion in string A = [tex]33\times 10^2\ N[/tex]

[tex]T_B[/tex] = Tesnsion in string B = [tex]3.76\times 10^2\ N[/tex]

t = Time taken till the pulses pass each other

Velocity of wave in a string is given by

[tex]v_A=\sqrt{\dfrac{T_A}{\mu}}\\\Rightarrow v_A=\sqrt{\dfrac{33\times 10^2}{0.0292}}[/tex]

[tex]v_B=\sqrt{\dfrac{T_2}{\mu}}\\\Rightarrow v_B=\sqrt{\dfrac{3.76\times 10^2}{0.0292}}[/tex]

[tex]Distance=Speed\times Time[/tex]

[tex]L=(v_A+v_B)t\\\Rightarrow t=\dfrac{L}{v_A+v_B}\\\Rightarrow t=\dfrac{44.3}{\sqrt{\dfrac{33\times 10^2}{0.0292}}+\sqrt{\dfrac{3.76\times 10^2}{0.0292}}}\\\Rightarrow t=0.09852\ s[/tex]

The time taken is 0.09852 seconds

Answer:

[tex]t=0.0985\ s[/tex]

Explanation:

Given:

Now, the velocity of the wave pulse in the stretched spring is given as:

FOR A:

[tex]v_a=\sqrt{\frac{T_a}{\mu_a} }[/tex]

[tex]v_a=\sqrt{\frac{3300}{0.0292} }[/tex]

[tex]v_a=336.175\ m.s^{-1}[/tex]

FOR B:

[tex]v_b=\sqrt{\frac{T_b}{\mu_b} }[/tex]

[tex]v_b=\sqrt{\frac{376}{0.0292} }[/tex]

[tex]v_b=113.476\ m.s^{-1}[/tex]

Let the distance at which the wave of A meets the wave of B be x from left then the distance from B is (44.3-x) meters.

Now the time taken is constant:

[tex]t=\frac{x}{v_a} =\frac{x-44.3}{v_b}[/tex]

[tex]\frac{x}{336.175} =\frac{x-44.3}{113.476}[/tex]

[tex]x=33.12\ m[/tex] is the distance travelled by pulse in wire A in when the pulse in the wire B meets

Now the time taken to travel this distance by pulse in wire A:

[tex]t=\frac{x}{v_a}[/tex]

[tex]t=\frac{33.12}{336.175}[/tex]

[tex]t=0.0985\ s[/tex] is the time taken by the two waves to pass each other.

Answer:

a. 0.0 m/s

Explanation:

According to the law of conservation of energy, the maximum potential energy of the body is equal to the minimum kinetic energy:

P+K=const; Pmax=mgH Kmin=mv^2/2=0 (because v=0)

At the maximum height, the speed of the ball is equal to zero.  Therefore, option (A) is correct.

Speed can be explained as a scalar measurement of the motion of an object with time. The speed of an object can be described as the change in position w.r.t. time.

Speed is a scalar parameter as it exhibits only magnitude and no direction. A formula that can be used to calculate the speed of a moving body.

S = d / t

where S is the speed, d is the distance the object moved, and t is the time.

Although the SI unit for speed is m/s and can also represent in miles per hour (mph), and kilometers per hour (kph).

The speed of a ball thrown upwards decreases with time because the ball is moving against gravity, and the eventually becomes zero when the ball reaches maximum height.

Thus, the speed of the ball thrown above the surface of the Earth is zero when the ball reaches maximum height.

Learn more about Speed, here:

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a) The jogger has run approximately 7.978 miles after 1 hour. b) The jogger's acceleration at t = 0 is approximately 0.41 ft/s^2. c) The equation does not provide a direct relationship between distance and time, so we cannot determine the time required to run 6 miles.

(a) To find the distance the jogger has run when t = 1 h, we need to substitute t = 1 into the given equation. Plugging in t = 1, we have: v = 7.5(1 + 2(0.04(1)))^0.3. Simplifying this expression, we get v ≈ 7.5(1.08)^0.3. Evaluating this expression, we find v ≈ 7.978 mi/h. To find the distance, we multiply the speed by the time: d = v × t = 7.978 × 1 = 7.978 mi.

(b) The acceleration can be found by taking the derivative of the velocity equation with respect to time. Differentiating the given equation, we have dv/dt = 7.5(1 + 2(0.04x))^(-0.7) × 2(0.04). Substituting t = 0 into this expression, we get: a = 7.5(1 + 2(0.04(0)))^(-0.7) × 2(0.04). Simplifying this expression, we find a ≈ 0.28 mi/h^2. To convert this to ft/s^2, we multiply by 5280 ft/mi and divide by 3600 s/h: a = 0.28 × 5280 / 3600 ≈ 0.41 ft/s^2.

(c) To find the time required for the jogger to run 6 mi, we need to solve the equation v = 7.5(1 + 2(0.04x))^0.3 for t. Plugging in d = 6 and v = 7.5(1 + 2(0.04x))^0.3, we can solve for t. However, the given equation does not provide a direct relationship between distance and time, so we cannot solve it.