how do a proton and neutron compare?

Answer:

I = 27kg.mi/h

Explanation:

In order to calculate the impulse of the ball, you use the following formula:

[tex]I=m\Delta v[/tex]  [tex]=m(v-v_o)[/tex]      (1)

m: mass of the ball = 0.3kg

v: speed of the ball after the bat hit it = 60mi/h

vo: speed of the ball before the bat hit it = 30mi/h

You replace the values of all parameters in the equation (1):

[tex]I=(0.3kg)(60mi/h-(-30mi/h))=27kg\frac{mi}{h}[/tex]

where the minus sign of the initial velocity means that the motion of the ball is opposite to the final direction of such a motion.

The imulpse of the ball is 27 kg.miles/hour

Answer:

8 ft/s

Explanation:

This is a straight forward question without much ado.

It is given from the question that she walks with a speed of 8 ft/s

Answer:

5000N

Explanation:

According to Newton's second law of motion, the net force (∑F) acting on a body is the product of the mass (m) of the body and the acceleration (a) of the body caused by the force. i.e

∑F = m x a             -------------(i)

From the question, the net force is the combined effect of the thrust (F) and the friction force (Fₓ). i.e

∑F = F + Fₓ             -------------(ii)

Where;

Fₓ = -200N       [negative sign because the friction force opposes motion]

Combine equations(i) and (ii) together to get;

F + Fₓ = m x a

F = ma - Fₓ         -------------(iii)

Where;

m = mass of car = 1200kg

a = acceleration of the car = 4m/s²

Now substitute the values of m, a and Fₓ into equation (iii) as follows;

F = (1200 x 4) - (-200)

F = 4800 + 200

F = 5000N

Therefore, the force the thrust is providing is 5000N

Answer:

I = 81.5 W/m^2

Explanation:

In order to calculate the intensity of the sound at the speaker, you use the following formula:

[tex]I=\frac{P}{A}[/tex]          (1)

P: power of the speaker's sound = 1.63W

A: surface area of the stereo set = 0.07m^2

You assume that the intensity of the sound at the speaker depends only of the surface area of the stereo set. Furthermore, you consider that the wave front of the sound is approximately plane.

You replace the values of the parameters in the equation (1):

[tex]I=\frac{1.63W}{0.02m^2}=81.5\frac{W}{m^2}[/tex]

The intensity of the speaker's sound at the speaker is 81.5 W/m^2

Answer:

h = 1.94 m

Explanation:

When the bull dog and skate board reach the bottom of the well, all of its potential energy is converted to the kinetic energy:

Kinetic Energy Gained by Bull Dog and Skate Board = Potential Energy Lost by Bull Dog and Skate Board

K.E = P.E

K.E = mgh

h = K.E/mg

where,

h = depth of well = ?

K.E = Kinetic Energy at bottom = 380 J

m = mass of bull dog and skate board = 20 kg

g = 9.8 m/s²

Therefore,

h = 380 J/(20 kg)(9.8 m/s²)

h = 1.94 m

Question is incomplete and image is not attached ti the question. The required image is attached below, so the complete question is:

The diagram shows a device that uses radio waves.

What is the role of the part in the diagram labeled Y?

Answer:

2. capture, amplify, and demodulate waves

Explanation:

The part Y labeled in the diagram refers to radio receiver which capture, amplify and demodulate the radio waves.

The radio receiver seperates required radio frequency signals through antenna and consist of an amplifier that amplify or increase the power of receiving signal. At the end, demodulators present in receivers recover the information from the modulated wave.

Hence, the correct option is 2.

Answer:

B

Explanation:

edge 2020

Answer:

The index of refraction of this material is 1.7241

Explanation:

Recall that the index of refraction (n) of a medium is defined as the quotient between the speed of light in vacuum divided the speed of light in the medium whose index of refraction is being calculated. In mathematical terms:

[tex]n=\frac{c}{v}[/tex]

Therefore, in our case, since we know the speed of light in the medium ([tex]v=1.74\,\,10^8\,\,m/s[/tex]) and the speed of light in vacuum ([tex]c=3\,\,10^8\,\,m/s[/tex]), we can estimate the index of refraction of the medium:

[tex]n=\frac{c}{v} \\n=\frac{3\,\,10^8}{1.74\,\,10^8} \\n=1.7241[/tex]

Answer:

Explanation:

Mass ( m ) = 2 kg

Speed of the object (v) = 6 metre per second

Kinetic energy =?

Now,

We have,

Kinetic Energy = [tex] \frac{1}{2} \times m \times {v}^{2} [/tex]

Plugging the values,

[tex] = \frac{1}{2} \times 2 \times {(6)}^{2} [/tex]

Reduce the numbers with Greatest Common Factor 2

[tex] = {(6)}^{2} [/tex]

Calculate

[tex] = 36 \: joule[/tex]

Hope this helps...

Good luck on your assignment...

The Kinetic energy of the object will be "36 joules".

The excess energy of moving can be observed as that of the movement of an object, component, as well as the group of components. There would never be a negative (-) amount of kinetic energy.

According to the question,

Mass of object, m = 2 kg

Speed of object, v = 6 m/s

As we know the formula,

→ Kinetic energy (K.E),

= [tex]\frac{1}{2}[/tex] × m × v²

By substituting the values, we get

= [tex]\frac{1}{2}[/tex] × 2 × (6)²

=  [tex]\frac{1}{2}[/tex] × 2 × 36

= 36 joule

Thus the above answer is appropriate.

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Answer:

1 yr = 24 * 3600 * 365 = 3.2 * 10E7 sec

C = 6 R = 1.5 * 10E8 * 6 = 9 * 10E8 km     circumference of orbit

v = C / t = 9 * 10E8 km / 3 * 10E7  sec = 30 km / sec = 18 mi/sec

The correct answer is B. thrill and adventure seeking, experience seeking, disinhibition, and susceptibility to boredom

Explanation:

Marvin Zuckerman was an important American Psychologists mainly known for his research about personality and the creation of a model to study this aspect of human psychology. This model purposes five factors define personality, these are the thrill and adventure-seeking that involves seeking for adventures and danger; experience seeking that implies a strong interest in participating in new activities; disinhibition that implies being open and extrovert; and susceptibility to boredom that implies avoiding boredom or repetition. Thus, option B correctly describes the characteristics used in Zuckerman's test.

Answer:

a. 4N and 7N

Explanation:

Draw a free body diagram.

Sum of the forces in the x direction:

∑F = ma

T₂ cos 30° − T₁ cos 60° = 0

T₂ cos 30° = T₁ cos 60°

T₂ (½√3) = T₁ (½)

T₁ = T₂ √3

Sum of the forces in the y direction:

∑F = ma

T₂ sin 30° + T₁ sin 60° − mg = 0

T₂ sin 30° + (T₂ √3) sin 60° − mg = 0

½ T₂ + 1.5 T₂ − mg = 0

2 T₂ = mg

T₂ = 4 N

T₁ = 4√3 N

T₁ ≈ 7 N

Answer:

It was safe to handle the circuit with dry hands because dry skin body resistance is very high, measuring up to 500,000 ohms.

Explanation:

Given;

Current of 0.001 A to be felt

Current of 0.005 A can produce a pain response

Current of 0.015 A can cause a loss of muscle control

Total current that traveled in the three-battery circuit = 0.03 A

Thus, we can conclude that, it was safe to handle the above mentioned circuit with dry hands because dry skin body resistance is very high, measuring up to 500,000 ohms.

Answer:

The final speed of the lion-gazelle system immediately after the attack is 69.862 kilometers per hour.

Explanation:

Let suppose that lion and Thomson's gazelle are running at constant speed before and after collision and that collision is entirely inelastic. Given the absence of external force, the Principle of Momentum Conservation is applied such that:

[tex]\vec p_{L} + \vec p_{G} = \vec p_{F}[/tex]

Where:

[tex]\vec p_{L}[/tex] - Linear momentum of the lion, measured in kilograms-meters per second.

[tex]\vec p_{G}[/tex] - Linear momentum of the Thomson's gazelle, measured in kilograms-meters per second.

[tex]\vec p_{F}[/tex] - Linear momentum of the lion-Thomson's gazelle, measured in kilograms-meters per second.

After using the definition of momentum, the system is expanded:

[tex]m_{L}\cdot \vec v_{L} + m_{G}\cdot \vec v_{G} = (m_{L} + m_{G})\cdot \vec v_{F}[/tex]

Vectorially speaking, the final velocity of the lion-gazelle system is:

[tex]\vec v_{F} = \frac{m_{L}}{m_{L}+m_{G}}\cdot \vec v_{L} + \frac{m_{G}}{m_{L}+m_{G}}\cdot \vec v_{G}[/tex]

Where:

[tex]m_{L}[/tex], [tex]m_{G}[/tex] - Masses of the lion and the Thomson's gazelle, respectively. Measured in kilograms.

[tex]\vec v_{L}[/tex], [tex]\vec v_{G}[/tex], [tex]\vec v_{F}[/tex] - Velocities of the lion, Thomson's gazelle and the lion-gazelle system. respectively. Measured in meters per second.

If [tex]m_{L} = 177\,kg[/tex], [tex]m_{G} = 32\,kg[/tex], [tex]\vec v_{L} = 81.8\cdot j\,\left[\frac{km}{h} \right][/tex] and [tex]\vec v_{G} = 59.0\cdot i\,\left[\frac{km}{h} \right][/tex], the final velocity of the lion-gazelle system is:

[tex]\vec v_{F} = \frac{177\,kg}{177\,kg+32\,kg}\cdot \left(81.8\cdot j\right)\,\left[\frac{km}{h} \right] + \frac{32\,kg}{177\,kg+32\,kg}\cdot \left(59.0\cdot i\right)\,\left[\frac{km}{h} \right][/tex]

[tex]\vec v_{F} = 9.033\cdot i + 69.276\cdot j\,\left[\frac{km}{h} \right][/tex]

The speed of the system is the magnitude of the velocity vector, which can be found by means of the Pythagorean theorem:

[tex]\|\vec v_{F}\| = \sqrt{\left(9.033\frac{km}{h} \right)^{2}+\left(69.276\frac{km}{h} \right)^{2}}[/tex]

[tex]\|\vec v_{F}\| \approx 69.862\,\frac{km}{h}[/tex]

The final speed of the lion-gazelle system immediately after the attack is 69.862 kilometers per hour.

Answer:

17.69 m

Explanation:

The time it takes the brick to strike the ground is 1.90 seconds.

We can apply one of Newton's equation of linear motion to find the height of the building:

[tex]s = ut + 0.5gt^2[/tex]

where s = distance (in this case height)

u = initial velocity = 0 m/s

t = time = 1.90 s

g = acceleration due to gravity = 9.8 m/s^2

Therefore:

s = (0 * 1.9) + (0.5 * 9.8 * 1.9 * 1.9)

s = 0 + 17.68

s = 17.69 m

The height of the building is 17.69 m.

Answer:

D is determined by distance between the telescopes.

Explanation:

Answer:

2000 N

Explanation:

20 km/h = 5.56 m/s

100 km/h = 27.78 m/s

F = ma

F = m Δv/Δt

F = (200 kg + 80 kg) (27.78 m/s − 5.56 m/s) / (3 s)

F = 2074 N

Rounded to one significant figure, the force is 2000 N.

Answer:

The heat rate produced from the motor is 84.216 watts.

Explanation:

The electric motor receives power from electric current and releases power in the form of mechanical energy (torque) and waste heat and can be considered an stable-state system. The model based on the First Law of Thermodynamics for the electric motor is:

[tex]\dot W_{e} - \dot W_{T} -\dot Q = 0[/tex]

Where:

[tex]\dot Q[/tex] - Heat transfer from the electric motor, measured in watts.

[tex]\dot W_{e}[/tex] - Electric power, measured in watts.

[tex]\dot W_{T}[/tex] - Mechanical power, measured in watts.

The heat transfer rate can be calculated in terms of electric and mechanic powers, that is:

[tex]\dot Q = \dot W_{e} - \dot W_{T}[/tex]

The electric and mechanic powers are represented by the following expressions:

[tex]\dot W_{e} = i \cdot V[/tex]

[tex]\dot W_{T} = T \cdot \omega[/tex]

Where:

[tex]i[/tex] - Current, measured in amperes.

[tex]V[/tex] - Steady-state voltage, measured in volts.

[tex]T[/tex] - Torque, measured in newton-meters.

[tex]\omega[/tex] - Angular speed, measured in radians per second.

Now, the previous expression for heat transfer rate is expanded:

[tex]\dot Q = i \cdot V - T \cdot \omega[/tex]

The angular speed, measured in radians per second, can be obtained by using the following expression:

[tex]\omega = \frac{\pi}{30}\cdot \dot n[/tex]

Where:

[tex]\dot n[/tex] - Rotational rate of change, measured in revolutions per minute.

If [tex]\dot n = 1000\,rpm[/tex], then:

[tex]\omega = \left(\frac{\pi}{30} \right)\cdot (1000\,rpm)[/tex]

[tex]\omega \approx 104.720\,\frac{rad}{s}[/tex]

Given that [tex]i = 10\,A[/tex], [tex]V = 110\,V[/tex], [tex]T = 9.7\,N\cdot m[/tex] and [tex]\omega \approx 104.720\,\frac{rad}{s}[/tex], the heat transfer rate from the electric motor is:

[tex]\dot Q = (10\,A)\cdot (110\,V) -(9.7\,N\cdot m)\cdot \left(104.720\,\frac{rad}{s} \right)[/tex]

[tex]\dot Q = 84.216\,W[/tex]

The heat rate produced from the motor is 84.216 watts.

Answer:

The next resonance will be 150 Hz.

Explanation:

The frequency of the sound produced by a tube, both open and closed, is directly proportional to the speed of propagation. Hence, to produce the different harmonics of a tube, the wave propagation speed must be increased.

The frequency of the sound produced by a tube, both open and closed, is inversely proportional to the length of the tube. The greater the length of the tube, the frequency is lower.

Frecuency of the standing sound wave modes in a open-closed tube is:

fₙ=n*f₁ where m is an integer and f₁ is the first frecuency (30 Hz)

The next resonance is at 90 Hz. This means that it occurs when n = 3:

f₃=3*30 Hz= 90 Hz

This means that the next resonance occurs when n = 5:

f₅=5*30 Hz= 150 Hz

The next resonance will be 150 Hz.

Answer:

  E = 1/9  E₀

Explanation:

In this exercise we are told that the electric field is Eo when the diameter of the balloon is D, the expression

we are asked to shorten the electric field when the diameter is 3D with the same eclectic charge

For this we can use the gauss law to find the field in the new diameter, for this we create a Gaussian surface in the form of a sphere

                Ф = ∫ E. dA = [tex]q_{int}[/tex] /ε₀

In this case the lines of the electric field and the radii of the sphere are parallel, therefore the scalar product is reduced to the algebraic product and the charge inside the sphere is the initial charge Q

               A = 4π r²

               E 4π r² = Q /ε₀

               E = 1 /4πε₀     Q / r²

the value of the indicated distance is 3 times the initial diamete

                 r  = 3 D / 2

we substitute

              E = 1/4 πε₀ Q (2/ 3D)²

for the initial conditions

              E₀ = 1 / 4πε₀ Q  (2/D)²

subtitled in the equation above

         E = 1/9  E₀

Answer:

Horizontal component: [tex]F_x = 58\ N[/tex]

Vertical component: [tex]F_y = 33.5\ N[/tex]

Explanation:

To find the horizontal and vertical components of the force, we just need to multiply the magnitude of the force by the cosine and sine of the angle with the horizontal, respectively.

Therefore, for the horizontal component, we have:

[tex]F_x = F * cos(angle)[/tex]

[tex]F_x = 67 * cos(30)[/tex]

[tex]F_x = 58\ N[/tex]

For the vertical component, we have:

[tex]F_y = F * sin(angle)[/tex]

[tex]F_y = 67 * sin(30)[/tex]

[tex]F_y = 33.5\ N[/tex]

So the horizontal component of the tension force is 58 N and the vertical component is 33.5 N.

Answer:

20,000 kg m/s

Explanation:

Given:

v₀ = 50 m/s

a = -5 m/s²

t = 2 s

Find: v

v = at + v₀

v = (-5 m/s²) (2 s) + (50 m/s)

v = 40 m/s

p = mv

p = (500 kg) (40 m/s)

p = 20,000 kg m/s

Answer:

I hope it is correct ✌️

Complete Question

The complete question is  shown on the first uploaded image (reference for Photobucket )

Answer:

The  electric field is  [tex]E = -1.3 *10^{-4} \ N/C[/tex]

Explanation:

 From the question we are told that

    The linear charge density on the inner conductor is  [tex]\lambda _i = -26.8 nC/m = -26.8 *10^{-9} C/m[/tex]

    The linear charge density on the outer conductor is

  [tex]\lambda_o = -60.0 nC/m = -60.0 *10^{-9} \ C/m[/tex]

     The position of interest is r =  37.3 mm =0.0373 m

Now this position we are considering is within the outer conductor so the electric field  at this point is due to the inner conductor (This is because the charges on the conductor a taken to be on the surface of the conductor according to Gauss Law )

Generally according to Gauss Law

         [tex]E (2 \pi r l) = \frac{ \lambda_i }{\epsilon_o}[/tex]

=>    [tex]E = \frac{\lambda _i }{2 \pi * \epsilon_o * r}[/tex]

substituting values  

       [tex]E = \frac{ -26 *10^{-9} }{2 * 3.142 * 8.85 *10^{-12} * 0.0373}[/tex]

       [tex]E = -1.3 *10^{-4} \ N/C[/tex]

The  negative  sign tell us that the direction of the electric field is radially inwards

    =>   [tex]|E| = 1.3 *10^{-4} \ N/C[/tex]

Answer:

the result is the quantization of __Energy__ of the particle

Explanation:

Answer:

The  total number of maxima produced is   [tex]m_T = 7[/tex] maxima

Explanation:

From the question we are told that

    The number of lines per cm is  [tex]n = 5000 \ lines/cm[/tex]

      The wavelength of the light is  [tex]\lambda = 633 nm = 633 *10^{-9} \ m[/tex]

Now the distance between the lines is mathematically evaluated as

           [tex]d = \frac{1}{n}[/tex]

substituting values  

          [tex]d = \frac{1}{5000}[/tex]

          [tex]d = \frac{1 *10^{-2}}{5000}[/tex]     N/B - this  statement convert it from  cm to m

         [tex]d = 2 *10^{ -6} \ m[/tex]

Generally the condition for diffraction i mathematically represented as  

          [tex]dsin(\theta ) = m \lambda[/tex]

at maximum  [tex]\theta = 90 ^o[/tex]

             [tex]d sin (90) = \lambda m[/tex]

here m is the  number of  maxima  

      Thus  making  m the subject we have

          [tex]m = \frac{d sin (90)}{ \lambda }[/tex]

So     [tex]m = \frac{2*10^{-6} sin (90)}{ 633 *10^{-9}}[/tex]

          [tex]m = 3.2[/tex]

=>          m  =3  

  Now the total number of maxima would include the bright fringe(3) and  dark fringe (3) plus the central maxima (1)

Thus  

      [tex]m_T = 3 + 3 +1[/tex]

       [tex]m_T = 7[/tex] maxima

Answer:

W = 12.96 J

Explanation:

The force acting in the direction of motion of the sand paper is the frictional force. So, we first calculate the frictional force:

F = μR

where,

F = Friction Force = ?

μ = 0.92

R = Normal Force = 2.6 N

Therefore,

F = (0.92)(2.6 N)

F = 2.4 N

Now, the displacement is given as:

d = (0.12 m)(45)

d = 5.4 m

So, the work done will be:

W = F d

W = (2.4 N)(5.4 m)

W = 12.96 J

Answer:

156KC

Explanation:

Pls see attached file

156 C is the total charge on the Earth is implied by measurement.

It states that the electric flux Φ across any closed surface is proportional to the net electric charge q enclosed by the surface;

i.e.  Φ = q/ε0

According to the question,

Electric field of Earth is 139 N/C

ε[tex]_{o}[/tex][tex]= 8.85 *10^{-12} C^{2} /Nm^{2}[/tex]

[tex]R_{e} = 6.37 * 10^{6}[/tex]

Using the formula by Gauss's Law

Φ = q/ε[tex]_{o}[/tex] = EA

q = ε[tex]_{o}[/tex]EA = ε[tex]_{o}[/tex]E*r²

= [tex]= 8.85*10^{12} *139*(6.37*10^{6} )^2\\\\= 156,905.9*10^0 = 156C[/tex]

Therefore ,

The total charge on the Earth is implied by this measurement is 156 C.

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Answer:

e = -0.00031 ( the -ve sign is due to the increase in length)

The error depends on the distance measured

Explanation:

Cross Sectional Area of the tape, A = 0.0625 in²

Length of the steel tape, L = 3600 in

Normal room temperature, T₁ = 68°F

Temperature of the hot day, T₂ = 130°F

ΔT = T₂  - T₁ = 130 - 68

ΔT = 62°F

Coefficient of Linear expansion [tex]\alpha = 5 * 10^{-6} F^{-1}[/tex]

The coefficient of linear expansion is given by the formula:

[tex]\alpha = \frac{\triangle L}{L* \triangle T} \\\triangle L = \alpha L \triangle T\\\triangle L = 5* 10^{-6} * 3.6* 10^3 * 62\\\triangle L = 1.11 6 in[/tex]

Since the length is increased, the error will be given by the formula:

[tex]e = \frac{-\triangle L}{L} \\\\e = \frac{-1.116}{3600}[/tex]

e = -0.00031 ( the -ve sign is due to the increase in length)

Since the error is a function of length and change in length, it depends on the distance measured

Answer:

The two small charged spheres are now 4.382 cm apart

Explanation:

Given;

distance between the two small charged sphere, r = 7.59 cm

The force on each of the charged sphere can be calculated by applying Coulomb's law;

[tex]F = \frac{kq_1q_2}{r^2}[/tex]

where;

F is the force on each sphere

q₁ and q₂ are the charges of the spheres

r is the distance between the spheres

[tex]F = \frac{kq_1q_2}{r^2} \\\\kq_1q_2 = Fr^2 \ \ (keep \ kq_1q_2 \ constant)\\\\F_1r_1^2 = F_2r_2^2\\\\r_2^2 = \frac{F_1r_1^2}{F_2} \\\\r_2 = \sqrt{\frac{F_1r_1^2}{F_2}} \\\\r_2 = r_1\sqrt{\frac{F_1}{F_2}}\\\\(r_1 = 7.59 \ cm, \ F_2 = 3F_1)\\\\r_2 = 7.59cm\sqrt{\frac{F_1}{3F_1}}\\\\r_2 = 7.59cm\sqrt{\frac{1}{3}}\\\\r_2 = 7.59cm *0.5773\\\\r_2 = 4.382 \ cm[/tex]

Therefore, the two small charged spheres are now 4.382 cm apart.

A simple random sample is a sample drawn in such a way that each member of the population has equal chance for being included in the sample.

The mass percent of hydrogen in CH₄O is 12.5%.

Mass percent is the mass of the element divided by the mass of the compound or solute.

Step 1: Calculate the mass of the compound.

mCH₄O = 1 mC + 4 mH + 1 mO = 1 (12.01 amu) + 4 (1.00 amu) + 1 (16.00 amu) = 32.01 amu

Step 2: Calculate the mass of hydrogen in the compound.

mH in mCH₄O = 4 mH = 4 (1.00 amu) = 4.00 amu

Step 3: Calculate the mass percent of hydrogen in the compound.

%H = (mH in mCH₄O / mCH₄O) × 100%

%H = 4.00 amu / 32.01 amu × 100% = 12.5%

The mass percent of hydrogen in CH₄O is 12.5%.

CO2 = 1.580 grams H2O = 0.592 grams Lookup the molar mass of each element in the compound Carbon = 12.0107 Hydrogen = 1.00794 Oxygen = 15.999 Calculate the molar mass of CH4O by adding the total masses of each element used. 12.0107 + 4 * 1.00794 + 15.999 = 32.04146 Now calculate how many moles of CH4O you have by dividing by the molar mass. m = 1.15 g / 32.04146 g/mole = 0.035891 mole Now figure out how many moles of carbon and hydrogen you have. Carbon = 0.035891 moles Hydrogen = 0.035891 moles *

Therefore, The mass percent of hydrogen in CH₄O is 12.5%.

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