D. Calculate the electric force (F.) between an electron and
proton that are 5.29 x 10-11 meters apart.

Answer:

it can be used for communication, industrial application and also for instrumentation application

Answer:

The circular turning of roads

Explanation:

please mark brainliest

Answer:

Explanation:

The circular turning of roads

Driving on Curves

Answer:

True

Explanation:

Answer:

1) asking a question about something you observe, 2) doing background research to learn what is already known about the topic, 3) constructing a hypothesis, 4) experimenting to test the hypothesis

Answer:

1) make an observation that describes a problem - see something you can fix or improve and try to describe the problem like which type of vegetables does rabbit like

2) create a hypothesis - What do you think will happen - If the rabbit eat the lettuce than .........

3) test the hypothesis - Do your experiment with the variables and follow the procedures

4) draw conclusions and refine the hypothesis - see if your hypothesis was correct - In conclusion my hypothesis was not correct because.......

Question :-

Answer :-

Explanation :-

As per the provided information in the given question, The Force is given as 10 Newton . The Distance is given as 30 cm [ 0.3 m ] . And, we have been asked to calculate the Amount of Force .

For calculating the Force , we will use the Formula :-

[tex] \bigstar \: \: \: \boxed{ \: \sf {Moment \: of \: Force \: = \: Force \: \times \: Distance} \: } [/tex]

Therefore , by Substituting the given values in the above Formula :-

[tex] \dag \: \: \: \sf {Moment \: of \: Force \: = \: Force \: \times \: Distance} [/tex]

[tex] \longmapsto \: \: \: \sf {Moment \: of \: Force \: = \: 10 \: \times \: 0.3} [/tex]

[tex] \longmapsto \: \: \: \textbf {\textsf {Moment \: of \: Force \: = \: 3}} [/tex]

Hence :-

[tex] \underline {\rule {185pt}{4pt}} [/tex]

build a fence.....................

If steam in a cylinder expands against a piston, exerting 10 atm of external pressure, the change in energy of the steam is mathematically given as

dE= -3010 kJ

Generally, the equation for the workdone  is mathematically given as

Work = - (pressure * Volume)

Therefore

w= - (10 x 10)

w= -100 L atm

In conclusion

dEnergy = q + work

dE = -3000 + -10.13

dE= -3010 kJ

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Answer:

(B)

Explanation:

Time = change of velocity ÷ acceleration

= (6-0) ÷ 5

= 1.2

1.2 s it take for an object accelerating at a constant rate of 5 m/s to change its velocity from 0 to 6 m/s.

Hence, the correct option is D.

To calculate the time it takes for an object to change its velocity from 0 to 6 m/s, we can use the formula:

time = change in velocity / acceleration

Given that the change in velocity (Δv) is 6 m/s and the acceleration (a) is 5 m/s², we can plug these values into the formula:

time = 6 m/s / 5 m/s²

time = 1.2 seconds

Therefore, 1.2 s it take for an object accelerating at a constant rate of 5 m/s to change its velocity from 0 to 6 m/s.

Hence, the correct option is D.

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Answer:

B: The object will start moving

Explanation:

If energy is transferred the object will definitely change so it can't be a. If you add energy the object will have more force so it cant be c. The mass of an object can't increase just by giving an object energy so it cant be d

Answer:B

Explanation:sorry they removed my answer for some reason.

Answer: B

Explanation: Sorry these brainly trolls deleted my first one

Answer:

.002273 seconds

Explanation:

1/f = period

1/ 440 c/s = .002273  seconds

The speed of the block after it has moved the given distance away from the initial position is 1.1 m/s.

The angular speed of the pulley after the block m1 fall through a distance, d, is obatined from conservation of energy and it is given as;

K.E = P.E

[tex]\frac{1}{2} mv^2 + \frac{1}{2} I\omega^2 = mgh\\\\\frac{1}{2} m_2v_0^2 + \frac{1}{2} \omega^2(m_1R^2_2 + m_2R_2^2) + \frac{1}{2} \omega^2( \frac{1}{2} MR_1^2 + \frac{1}{2} MR_2^2) = m_1gd- \mu_km_2gd\\\\\frac{1}{2} m_2v_0^2 + \frac{1}{2} \omega^2[R_2^2(m_1 + m_2)+ \frac{1}{2} M(R_1^2 + R_2^2)] = gd(m_1 - \mu_k m_2)\\\\[/tex]

[tex]\frac{1}{2} m_2v_0 + \frac{1}{4} \omega^2[2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] = gd(m_1 - \mu_k m_2)\\\\2m_2v_0 + \omega^2 [2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] = 4gd(m_1 - \mu_k m_2)\\\\\omega^2 [2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] = 4gd(m_1 - \mu_k m_2) - 2m_2v_0^2\\\\\omega^2 = \frac{ 4gd(m_1 - \mu_k m_2) - 2m_2v_0^2}{2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)} \\\\\omega = \sqrt{\frac{ 4gd(m_1 - \mu_k m_2) - 2m_2v_0^2}{2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)}} \\\\[/tex]

Substitute the given parameters and solve for the angular speed;

[tex]\omega = \sqrt{\frac{ 4(9.8)(0.7)(0.365 \ - \ 0.25\times 0.825) - 2(0.825)(0.82)^2}{2(0.03)^2(0.365 \ + \ 0.825)\ \ +\ \ 0.35(0.02^2\ + \ 0.03^2)}} \\\\\omega = \sqrt{\frac{3.25}{0.00214\ + \ 0.000455 } } \\\\\omega = 35.39 \ rad/s[/tex]

The linear speed of the block after travelling 0.7 m;

v = ωR₂

v = 35.39 x 0.03

v = 1.1 m/s

Thus, the speed of the block after it has moved the given distance away from the initial position is 1.1 m/s.

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Based on law of conservation of angular momentum, the rotational speed of the star is equal to 6,000 rev/s.

Given the following data:

First of all, we would determine the initial angular speed of the neutron star as follows:

[tex]\omega_i = \frac{1 \;Rev}{10 \;days} \\\\\omega_i = \frac{1 \;Rev}{10 \times 24 \times 60 \times 60}\\\\\omega_i = 1.157 \times 10^{-6}\;rev/s[/tex]

Mathematically, the moment of inertia of a uniform solid sphere is given by this formula:

[tex]I=\frac{2}{5} mr^2[/tex]

Where:

In order to determine the rotational speed of this neutron star, we would apply the law of conservation of angular momentum:

[tex]L_1 = L_2\\\\I_1\omega_1 = I_2\omega_2\\\\\frac{2}{5} m_1r_1^2 \omega_1 = \frac{2}{5} m_2r_2^2\omega_2\\\\m_1r_1^2 \omega_1 = m_2r_2^2\omega_2\\\\\omega_2 =\frac{r_1^2 \omega_1}{r_2^2}[/tex]

Substituting the given parameters into the formula, we have;

[tex]\omega_2 = \frac{ (7 \times 10^5 )^2 \times 1.157 \times 10^{-6}}{ (10 )^2}\\\\\\\omega_2 = \frac{ (7 \times 10^5 )^2 \times 1.157 \times 10^{-6}}{ 10 0}\\\\\omega_2 =\frac{566,930}{100}[/tex]

Final angular speed = 5,669 ≈ 6,000 rev/s.

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it has no energy because its being held still

it has motion energy because it will fall if let go

Answer:

PE (relative to water) = M g h

PE = 1200 kg * 9.8 m/s^2 * 24 m = 2.82E5 Joules

KE = PE when vehicle strikes water

1/2 M V^2 = 2.82E5

V = (2.82E5 * 2 / 1200)^1/2 = 21.7 m/s

Check:

M g h = 1/2 M V^2

V = (2 g h)^1/2 = (2 * 9.8 * 24)^1/2 = 21.7 m/s

(1.a) The surface area being vibrated by the time the sound reaches the listener is 5,026.55 m².

(1.b) The intensity of the sound wave as it reaches the person listening is 0.02 W/m².

(1.c) The relative intensity of the sound as heard by the listener is 103 dB.

(2.a) The speed of sound if the air temperature is 15⁰C is 340.3 m/s.

(2.b) The frequency of the sound heard by the suspect is 614.3 Hz.

The surface area being vibrated by the time the sound reaches the listener is calculated as follows;

A = 4πr²

A = 4π x (20)²

A = 5,026.55 m²

The intensity of the sound is calculated as follows;

I = P/A

I = (100) / (5,026.55)

I = 0.02 W/m²

[tex]B = 10log(\frac{I}{I_0} )\\\\B = 10 \times log(\frac{0.02}{10^{-12}} )\\\\B = 103 \ dB[/tex]

[tex]v= 331.3\sqrt{1 + \frac{T}{273} } \\\\v = 331.3\sqrt{1 + \frac{15}{273} } \\\\v = 340.3 \ m/s[/tex]

The frequency of the sound heard is determined by applying Doppler effect.

[tex]f_o = f_s(\frac{v \pm v_0}{v \pm v_s} )[/tex]

where;

[tex]f_0 = f_s(\frac{v-v_0}{v- v_s} )[/tex]

[tex]f_0 = 512(\frac{340.3 - 10}{340.3 - 65} )\\\\f_0 = 614.3 \ Hz[/tex]

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Answer:

Chapter 1

1. Show that the Lorentz transformation is such that the velocity of a light ray

travelling in the x direction is the same for the observer in the frame S and for

the observer in the frame S

.

Solution: Consider a light ray travelling in the x direction. If the light ray

connects two space–time points {t1, x1} and {t2, x2}, we have

c = x2 − x1

t2 − t1

The speed of light observed in the frame S will be

c = x

2 − x

1

t

2 − t

1

= c

γ ((x2 − x1) − βc(t2 − t1))

γ (c(t2 − t1) − β(x2 − x1))

= c

x2 − x1

t2 − t2

− βc

c − β x2 − x1

t2 − t2

= c

2. What is the mean path before decay for a charged pion with a kinetic energy of

1 GeV?

Solution: The pion has a lifetime 2.6 × 10–8 s and a mass of 139.6 MeV. If the

energy is 1 GeV, the velocity of the pion is 99% of the velocity of light (Eq. 1.4).

The mean path before decay is

= 0.99 c γ τ

= 0.99 c

1000 + 139.6

139.6

2.6 10−8 = 63 m

S. Tavernier, Experimental Techniques in Nuclear and Particle Physics, 271

DOI 10.1007/978-3-642-00829-0, C Springer-Verlag Berlin Heidelberg 2010

272 Solutions to Exercises

3. Show that the relativistic expression for the kinetic energy of a particle (Eq. 1.2)

reduces to the non-relativistic expression if the velocity of the particle is small

compared to the velocity of light.

Solution:

E = Ekinetic + m0c2 = m0c2

1 − (v/c)2

≈ m0c2

(1 − 1/2(v/c)2) ≈ m0c2(1 + 1/2(v/c)

2)

= m0c2 +

1

2

m0v2

4. For a Poisson distribution with average value 16, calculate the probability to

observe 12, 16 and 20 as measured value. Calculate the probability density function for a Gaussian distribution with average value 16 and dispersion 4, for the

values x = 12, 16 and 20. Compare the results.

Solution: For a Poisson distribution P(12) = 0.0829, P(16) = 0.1024, P(20) =

0.0418

For a Gaussian distribution, f(12) = 0.0605, f(16) = 0.0997, f(20) = 0.0605

5. Consider a very short-lived particle of mass M decaying into two long-lived particles 1 and 2. Assume you can measure accurately the energies and momenta of

the two long-lived particles. How will you calculate the mass of the short-lived

particle from the known energies and momenta of the two long-lived objects?

Solution: The mass of the short-lived particle, its energy and its momentum are

related by Eq. (1.1). The energy and momentum of the particle are equal to the

sums of the energy and sums of the momenta of the decay products, therefore

M2c4 = (E1 + E2)

2 − c2(P1 + P2)

2

6. Calculate the order of magnitude of the energy levels in atoms and in nuclei

using the ‘particle in a box’ approximation, Eq. (1.9). Use for the dimension of

the atom 10–10 m and for the dimension of the nucleus 10−15 m.

Solution: Atomic energy levels: ≈40 eV; nuclear energy levels: ≈400 MeV.

7 . Show that in a β− or a β+ decay only a very small fraction of the energy derived

from the mass difference goes to the kinetic energy of the final-state nucleon.

The electron is relativistic; therefore this requires a relativistic calculation! Hint:

the 3-body problem can be reduced to a 2-body problem by considering the

electron–neutrino system as one object with a mass of a few MeV.

Solution. Consider the 2-body decay of some heavy object with mass M into

two objects with masses m1 and m2. The kinetic energy of each of the final-state

particles in the overall centre of mass system is found as follows.

Solutions to Exercises 273

Consider two particles with energy and momentum four vectors p1 and p2.

The symbol pi stands for the four-vector {Ei,cpi}. The energy E appearing in this

expression is the total energy E, i.e. the rest energy mc2 plus the kinetic energy.

The four-vector product (p1.p2) is defined as

(p1.p2) =

(

E1E2 − c2 p1 p2

)

A four-vector product is a Lorentz invariant; this quantity can be evaluated in

any reference frame, and the result is the same. Consider now the quantity

(p1.p2)

m1c2

This is a Lorentz invariant. Evaluating this expression in the rest frame of

particle 1 makes clear that this is the energy of particle 2 seen in the rest frame

of particle 1. This remains true also if one of the particles is in fact a system

of particles, for example the system of the two particles 1 and 2. The energy of

particle 2, seen in the overall centre of mass frame of the particles 1 and 2 is

therefore

E∗

2 = (p1 + p2).p2

(p1 + p2)2

We have the following relations:

(p1 + p2)

2 = M2c4

(p1.p2) = 1

2

(

(p1 + p2)

2 − (p1)

2 − (p2)

2

)

= M2c4 − m2

1c4 − m2

2c4

And therefore finally

E∗

2 = M2c4 + m2

2c4 − m2

1c4

2Mc2

Let us now apply the above relation to the decay

N∗ → N + e− + ¯νe + Q

The symbol Q represents the energy liberated in the reaction. Let us denote

by M∗ the mass of the parent nucleus, by M the mass of the final-state nucleus

and by m the mass of the electron–neutrino system. The kinetic energy of the

nucleus in the final state is given by

274 Solutions to Exercises

Ekin = M∗2c4 + M2c4 − m2c4

2M∗c2 − Mc2

= M∗2c4 + M2c4 − m2c4 − 2M∗c2Mc2

2M∗c2

= (M∗ − M)

2 c4 − m2c4

2M∗c2

=

!

mc2 + Q

The energy of the cosmic ray photon is zero. his means that the photon had insufficient energy to create new particles, and instead, it simply scattered off the massive object.

Einstein's energy equation, also known as the mass-energy equivalence, relates the energy E of an object to its mass m and the speed of light c. The equation is:

E = mc^2

where:

E is the energy of the object in joules (J)

m is the mass of the object in kilograms (kg)

c is the speed of light in meters per second (m/s)

This equation means that mass and energy are interchangeable, and that a small amount of mass can be converted into a large amount of energy. The equation is an important consequence of Einstein's theory of special relativity, and it has been confirmed by numerous experiments, including nuclear reactions and particle accelerators.

Here in the Question,

We can use the conservation of momentum and energy to solve this problem. Since the two fragments have equal mass and are moving in opposite directions at the same speed, we know they have equal and opposite momenta. Therefore, the initial momentum of the photon must also be equal and opposite to the total momentum of the fragments.

Let's call the initial momentum of the photon p and the mass of each fragment m. The total momentum of the fragments is:

p' = 2mv

where v is the speed of each fragment, which we know is 0.6c. Therefore, we can write:

p' = 2m(0.6c) = 1.2mc

By conservation of momentum, we have:

p = -p'

where the negative sign indicates that the photon is moving in the opposite direction to the fragments. Therefore:

p = -1.2mc

Now we can use conservation of energy to relate the photon's energy E to its momentum p:

E^2 = p^2c^2 + m^2c^4

Substituting the expression we found for p, we get:

E^2 = (1.2mc)^2c^2 + m^2c^4

E^2 = 1.44m^2c^4 + m^2c^4

E^2 = 1.45m^2c^4

Solving for E, we get:

E = mc^2 * sqrt(1.45)

Plugging in the values for m and c, we get:

E = (0 * 9.0 × 10^16 kg) * sqrt(1.45) = 0

Therefore, The photon from a cosmic ray has no energy. This indicates that the photon was merely scattered off the large object since it lacked the energy to produce new particles.

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​

The linear mass density of the rope in the given motion of the wave is 0.15 kg/m.

The linear mass density of the rope in the given motion of the wave is determined by dividing the mass of the rope with the length of the entire rope.

The linear mass density of the rope is calculated as follows;

μ = m/L

μ = 0.6 kg / 4 m

μ = 0.15 kg/m

Thus, the linear mass density of the rope in the given motion of the wave is 0.15 kg/m.

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The force the horse and the rider exerts on the wall is equal to the weight combined acting in the opposite direction:

Given Data

When a body of mass rests on a surface, it exerts a force equal to the weight of the mass but opposite in direct on the mass/object

hence the force is computed as

Force = mass * acceleration

Force = 575 * 9.81

Force = 5640.75N

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The power delivered to the truck of mass 1000 kg that was lifted by a garage hoist, 2 m high above the ground in 15 seconds is 1308 W.

Power can be deifned as the rate at which work is done.

To calculate the power delivered to the truck, we use the formula below.

Formula:

Where:

From the question,

Given:

Subsitute these values into equation 1

Hence, The power delivered to the truck of mass 1000 kg that was lifted by a garage hoist, 2 m high above the ground in 15 seconds is 1308 W.

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Answer:

Explanation:

The efficiency can be calculated by:

[tex]\eta = \frac{Q_{E}-Q_{W}}{Q_{E}}=\frac{6000-2000}{6000}=\frac{2}{3}[/tex]

or we can say it becomes approximately 66.7%

For a class of 10 students taking an exam has a power output per student of about 200W, the temperature is mathematically given as

dT=25C

Generally, the equation for  the temperature is mathematically given as

dT=Q/mc

Where

Q=Pt

Q=2000*3600

Q=7200000J

And m

m=pv

m=1.3*270

m=351kg

Therefore

dT=7200000J/351*837

dT=25C

In conclusion, The temperature of the room at the end of 1.0 h

dT=25C

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Answer:

A   generates electricity, the others do not

Explanation:

The relative velocity of the ball with respect to the truck is same as that of the truck with respect to the ball at equal time.

The relative velocity of an object is the velocity of the object observed with respect to rest frame of another object.

The distance traveled by each object is determined from the prouduct of velocity and time of motion.

d = vt

where;

Thus, if the ball moved the same distance when it was on the flat bed truck, then the relative velocity of the ball with respect to the truck is same as that of the truck with respect to the ball at equal time.

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Explanation:

they generate electricity by contact

Answer:

This is because the force of gravity is much less on the moon than on the earth, therefore the person wont be pulled down much and will jump higher

Answer: This is because the force of gravity is much less on the moon than on the earth, therefore the person won't be pulled down much and will jump higher.