Earth has a mass of 5.98 1024 kg and a radius of 6.38 106 m. Assume it is a uniform solid sphere. The distance of Earth from the Sun is 1.50 1011 m. (Assume Earth completes a single rotation in 24.0 hours and orbits the Sun once every 365 Earth days.) (a) What is the rotational kinetic energy of Earth on its axis

Answer:

[tex]\triangle U=-e (V_2-V_1)[/tex]

[tex]\triangle U=130eV[/tex]

[tex]V_2=\sqrt{ \frac{2}{me}(\frac{1}{2}meV_1^2+e(V_2-V_1)}[/tex]

Explanation:

From the question we are told that

The potential at point 1, [tex]V_1 = 24V[/tex]

The potential at point 2, [tex]V_2 = 154V[/tex]

a)Generally work done by proton is given as

 [tex]w=-\triangle U[/tex]

 [tex]e\triangle V=-\triangle U[/tex]

 [tex]\triangle U=-e (V_2-V_1)[/tex]  

Generally the Equation for the change of electric potential energy AU of the proton, in terms of the symbols given and the charge of the proton e is mathematically given as

 [tex]\triangle U=-e (V_2-V_1)[/tex]

b)Generally the electric potential energy in electron volts (eV). is mathematically given as

 [tex]\triangle U=-e (154-24)V[/tex]

 [tex]|\triangle U| =|-e (130)V|[/tex]

 [tex]\triangle U=130eV[/tex]

c) Generally according to the law of conservation of energy

[tex](K.E+P.E)_1=(K.E+P.E)_2[/tex]

[tex]\frac{1}{2}meV_1^2+eV_1 =\frac{1}{2}mev_2^2+eV_2[/tex]

[tex]V_2^2=\frac{2}{me}(\frac{1}{2}meV_1^2+e(V_2-V_1)[/tex]

[tex]V_2=\sqrt{ \frac{2}{me}(\frac{1}{2}meV_1^2+e(V_2-V_1)}[/tex]

Answer:

LAW 1 :  For a given metal and frequency, the number of photoelectrons emitted is directly proportional to the intensity of the incident radiation.  

---------------------------------------------

LAW 2: For a given metal, there exists a certain frequency below which the photoelectric emission does not take place. This frequency is called threshold frequency.

-----------------------------------------------

LAW 3: For a frequency greater than the threshold frequency, the kinetic energy of photoelectrons is dependent upon frequency or wavelength but not on the intensity of light.

-----------------------------------------------

LAW 4: Photoelectric emission is an instantaneous process. The time lag between incidence of radiations and emission of electron is 10^-9 seconds.

Explanation:

Answer:

LAW 1 : For a given metal and frequency, the number of photoelectrons emitted is directly proportional to the intensity of the incident radiation. ... LAW 4: Photoelectric emission is an instantaneous process.

Answer:

2.16 inch

Explanation:

area under water = 66 km²

= 66 x ( 3280.84 x 12 )² inch²

= 1.023 x 10¹¹ sq inch

volume of rain = 9.57 x 10⁸  gallon = 9.57 x 10⁸ x 231 inch³

= 2.21 x 10¹¹ inch³

If depth of rainfall be t

volume of rain = surface area x depth

= 1.023 x 10¹¹ x t

So ,

1.023 x 10¹¹ x t  = 2.21 x 10¹¹

t = 2.16 inch

Answer:

[tex]v=0.8m/s[/tex]

Explanation:

From the question we are told that

Distance [tex]d=1.5m/sm/s[/tex]

Time  [tex]t_1=60s[/tex]  

Time  [tex]t_2=90s[/tex]  

Generally the  the equation for the distance traveled is mathematically given as

[tex]d=vt[/tex]

[tex]d=1.5*90[/tex]

[tex]d=138m[/tex]

Generally equation for speed of side walk is mathematically given as

[tex]d=(v+u)t[/tex]

[tex]v=\frac{d}{t}-u[/tex]

[tex]v=\frac{138}{60}-1.5[/tex]

[tex]v=0.8m/s[/tex]

Answer:

Macroscopic energy is energy at a level of system while microscopic energy is energy at the level of atoms and molecules

Explanation:

1. Macroscopic energy is possessed by a system as whole while microscopic energy is possessed by its constituents’ atoms or molecules.  

2. The common form of macroscopic energy is Kinetic and potential energy while the microscopic form of energy are atomic forces due its random, disordered motion and due to intermolecular forces

3. At microscopic level we consider behaviour of every molecule and in macroscopic approach we consider gross or average effects of various molecular infractions

Answer:

1.4999

Explanation:

Efficiency can be calculated using below expresion

Efficiency = W/Q.............eqn(1)

Where W= work = 50 J

Q= thermal energy= 150 J

But

W/Q= (Th-Tc)/Th ...........Eqn(2)

Th= temperature of the hot

Tc= temperature of the cold

Where Th/ Tc= ratio of the temperature hot and cold reservoirs?

If we simplify eqn(2) we have

W/Q = 1-Tc/Th.........eqn(3)

If we make the ratio subject of the formula we have

Tc/Th = 1-(W/Q)

Th/Tc = 1/(1-W/Q )

Then substitute the values

= 1/(1-50/150) = 1.4999

Hence, the smallest possible ratio of the temperatures (in kelvin) of the hot and cold reservoirs is 1.4999

Answer:

The magnitude of the average acceleration of the ball during this time interval is 1.238 x 10⁴ m/s².

Explanation:

Given;

mass of the super ball, m = 50 g = 0.05 kg

initial velocity of the ball, u = 29.5 m/s

final velocity of the ball, v = -20.0 m/s (negative because it rebounds)

time of contact of the ball and the wall, t = 4 ms = 4 x 10⁻³ s

The force exerted on the brick wall by the ball is given as;

[tex]F = ma\\\\ma = \frac{m(v-u)}{t} \\\\a = \frac{v-u}{t} \\\\a = \frac{(-20) - 29.5}{4.0 \ \times \ 10^{-3}} \\\\a = \frac{-49.5}{4.0 \ \times \ 10^{-3}} \\\\a = -1.238 \times 10^4 \ m/s^2\\\\|a| = 1.238 \times 10^4 \ m/s^2[/tex]

Therefore, the magnitude of the average acceleration of the ball during this time interval is 1.238 x 10⁴ m/s².

Answer:

x = 1.04866

Explanation:

Force can be defined from power energy by the expressions

          F = [tex]- \frac{ dU}{ dx}[/tex]

in this case we are the expression of the potential energy

          U = [tex]\frac{2.6}{x^{8} } - \frac{4.3}{ x^{4} }[/tex]

let's find the derivative

         dU / dx = 2.6 ( [tex]\frac{-8}{x^{9} }[/tex]) - 4.3 ([tex]\frac{-4}{ x^{5} }[/tex])

         dU / dx = [tex]- \frac{20.8}{ x^{9} } + \frac{17.2 }{ x^{5} }[/tex]

we substitute

          F = + \frac{20.8}{ x^{9} }  - \frac{17.2 }{ x^{5} }

at the equilibrium point the force is zero, so

           [tex]\frac{20.8}{ x^{9} } = \frac{17.2}{ x^{5} }[/tex]

           20.8 / 17.2 = x⁴

            x⁴ = 1.2093

             x = [tex]\sqrt[4]{ 1.2093}[/tex]

             x = 1.04866

Answer:

the inertia of the crate is (49.67 kg)r²

Explanation:

Given the data in the question;

First; we will use the law of conservation of momentum to determine the mass of the crate;

m₁v₁ - m₂v₂ = 0

given that; m₁ = 0.60 kg and v₂ = 0.058 m/s

we substitute

0.60 × v₁ = m₂ × 0.058 = 0

m₂ = 0.60v₁ / 0.058 ----------- EQU 1

Next, we use the energy conservation relation to find the velocity

According to conservation of energy;

1/2m₁v₁² + 1/2m₂v₂² = 7 J

we substitute

1/2×0.60×v₁² + 1/2×m₂×(0.058)² = 7 J

0.3v₁² + 0.001682m₂ = 7 J ----- EQU 2

substitute value of m₂ form equ 1 into equ 2

0.3v₁² + 0.001682(0.60v₁ / 0.058) = 7 J

0.3v₁² + 0.0174v₁ = 7 J

0.3v₁² + 0.0174v₁ - 7 J = 0

we solve the quadratic equation;

{  x =  [-b±√( b² - 4ac)] / 2a   }

v₁  =  [-0.0174 ±√( 0.0174² - 4×0.3×-7)] / 2×0.3

=  [-0.0174 ±√(8.4003)] / 0.6

= [-0.0174 ± 2.8983 ] / 0.6  

= -4.8595 or 4.8015     but{ v₁ ≠ - }

so v₁ = 4.8015 m/s ≈ 4.802 m/s

next we input value of  v₁ into equation 1

m₂ = (0.60×4.8015) / 0.058

m₂ =  2.8809 / 0.058

m₂ =  49.67 kg

So, the moment of inertia of the crate will be;

I₂ = m₂r²

we substitute value of m₂

I₂ = (49.67 kg)r²

Therefore, the inertia of the crate is (49.67 kg)r²

Answer:

difficulty concentrating

Answer:

a. 10.5 s b. 6.6 s

Explanation:

a. The driver's perception/reaction time before drinking.

To find the driver's perception time before drinking, we first find his deceleration from

v² = u² + 2as where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 0 m/s (since he stops), a = deceleration of driver and s = distance moved by driver = 385 ft = 385 × 0.3048 m = 117.35 m

So, a = v² - u²/2s

substituting the values of the variables into the equation, we have

a = v² - u²/2s

a = (0 m/s)² - (22.35 m/s)²/2(117.35 m)

a =  - 499.52 m²/s²/234.7 m

a = -2.13 m/s²

Using a = (v - u)/t where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 0 m/s (since he stops), a = deceleration of driver = -2.13 m/s² and t = reaction time

So, t = (v - u)/a

Substituting the values of the variables into the equation, we have

t = (0 m/s - 22.35 m/s)/-2.13 m/s²

t = - 22.35 m/s/-2.13 m/s²

t = 10.5 s

b. The driver's perception/reaction time after drinking.

To find the driver's perception time after drinking, we first find his deceleration from

v² = u² + 2as where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 30 mi/h = 30 × 1609 m/3600 s = 13.41 m/s, a = deceleration of driver and s = distance moved by driver = 385 ft = 385 × 0.3048 m = 117.35 m

So, a = v² - u²/2s

substituting the values of the variables into the equation, we have

a = v² - u²/2s

a = (13.41 m/s)² - (22.35 m/s)²/2(117.35 m)

a = 179.83 m²/s² - 499.52 m²/s²/234.7 m

a = -319.69 m²/s² ÷ 234.7 m

a = -1.36 m/s²

Using a = (v - u)/t where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 30 mi/h = 30 × 1609 m/3600 s = 13.41 m/s, a = deceleration of driver = -1.36 m/s² and t = reaction time

So, t = (v - u)/a

Substituting the values of the variables into the equation, we have

t = (13.41 m/s - 22.35 m/s)/-1.36 m/s²

t = - 8.94 m/s/-1.36 m/s²

t = 6.6 s

I believe the answer would be protentional because they have the potential energy in them to lift the hammer.

Answer:

Explanation:

75 N and 90 N are acting in opposite direction so net force = 90 - 75 = 15 N .

Friction force will act in the direction opposite to the direction of net force .

So friction force will act in the direction in which 75 N is acting .

Total force acting in the direction of 75 =  75 + 12 = 87 N

Net force acing on the third child = 90 - 87 = 3 N  

Its direction will be that in the direction of 90 N .

Answer:

yes

Explanation:

yes

According to question this is a riddle and the doctor was the boy's mother so she could not operate on him.

A statement, question, or phrase that is presented as a problem to be solved and has a dual or disguised meaning is called a riddle. Enigmas, which are difficulties typically presented in metaphoric or allegorical language that call for inventiveness and careful thought to solve, and conundra, which are problems that rely on puns either in the question or the answer, are two different forms of riddles.

Across many nations and even entire continents, many riddles take on a similar format. Riddles might be borrowed close to home as well as long distances. A man and his son were rock climbing on a particularly dangerous mountain when they slipped and fell. the man was killed, but the son lived and was rushed to a hospital.

Therefore, According to question this is a riddle and the doctor was the boy's mother so she could not operate on him.

Learn more about riddle here:

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Assuming the particle is in free fall once it is shot up, its vertical velocity v at time t is

v = 30 m/s - g t

where g = 9.8 m/s² is the magnitude of the acceleration due to gravity, and its height y is given by

y = (30 m/s) t - 1/2 g t ²

(a) At its maximum height, the particle has 0 velocity, which occurs for

0 = 30 m/s - g t

t = (30 m/s) / g ≈ 3.06 s

at which point the particle's maximum height would be

y = (30 m/s) (3.06 s) - 1/2 g (3.06 s)² ≈ 45.9184 m ≈ 46 m

(b) It takes twice the time found in part (a) to return to 0 height, t ≈ 6.1 s.

(c) The particle falls 35 m below its starting point when

-35 m = (30 m/s) t - 1/2 g t ²

Solve for t to get a time of about t ≈ 7.1 s

This question is incomplete, the complete question is;

The vector sum of the forces acting on the beam is zero, and the sum of the moments about the left end of the beam is zero.

(a) Determine the forces and and the couple

(b) Determine the sum of the moments about the right end of the beam.

(c) If you represent the 600-N force, the 200-N force, and the 30 N-m couple by a force F acting at the left end of the beam and a couple M, what is F and M?

Answer:

a)

the x-component of the force at A is [tex]A_{x}[/tex] = 0

the y-component of the force at A is [tex]A_{y}[/tex]  = 400 N

the couple acting at A is; [tex]M_{A}[/tex] = 146 N-m

b)

the sum of the momentum about the right end of the beam is;  ∑[tex]M_{R}[/tex]  = 0

c)

the equivalent force acting at the left end is; F = -400J ( N)

the couple acting at the left end is; M = - 146 N-m

Explanation:

Given that;

The sum of the forces acting on the beam is zero ∑f = 0

Sum of the moments about the left end of the beam is also zero ∑[tex]M_{L}[/tex] = 0

Vector force acting at A, [tex]F_{A}[/tex] = [tex]A_{x}i[/tex] + [tex]A_{y}j[/tex]

Now, From the image, we have;

a)

∑f = 0

[tex]F_{A}[/tex] - 600j + 200j = 0i + 0j

[tex]A_{x}i[/tex] + [tex]A_{y}j[/tex] - 600j + 200j = 0i + 0j

[tex]A_{x}i[/tex] + ([tex]A_{y}[/tex] - 400)j = 0i + 0j

now by equating i- coefficients'

[tex]A_{x}[/tex] = 0

so, the x-component of the force at A is [tex]A_{x}[/tex] = 0

also by equating j-coefficient

[tex]A_{y}[/tex] - 400 = 0

[tex]A_{y}[/tex]  = 400 N

hence, the y-component of the force at A is [tex]A_{y}[/tex]  = 400 N

we also have;

∑[tex]M_{L}[/tex] = 0

[tex]M_{A}[/tex]  - ( 30 N-m ) - ( 0.380 m )( 600 N ) + ( 0.560 m )( 200 N ) = 0

[tex]M_{A}[/tex] - 30 N-m - 228 N-m + 112 Nm = 0

[tex]M_{A}[/tex] - 146 N-m = 0

[tex]M_{A}[/tex] = 146 N-m

Therefore, the couple acting at A is; [tex]M_{A}[/tex] = 146 N-m

b)

The sum of the moments about right end of the beam is;

∑[tex]M_{R}[/tex] = (0.180 m)(600N) - (30 N-m) - ( 0.56 m)([tex]A_{y}[/tex] ) + [tex]M_{A}[/tex]

∑[tex]M_{R}[/tex] = (108  N-m) - (30 N-m) - ( 0.56 m)(400 N ) + 146 N-m

∑[tex]M_{R}[/tex] = (108 N-m) - (30 N-m) - ( 224 N-m ) + 146 N-m

∑[tex]M_{R}[/tex]  = 0

Therefore, the sum of the momentum about the right end of the beam is;  ∑[tex]M_{R}[/tex]  = 0

c)

The 600-N force, the 200-N force and the 30 N-m couple by a force F which is acting at the left end of the beam and a couple M.

The equivalent force at the left end will be;

F = -600j + 200j (N)

F = -400J ( N)

Therefore, the equivalent force acting at the left end is; F = -400J ( N)

Also couple acting at the left end

M = -(30 N-m) + (0.560 m)( 200N) - ( 0.380 m)( 600 N)

M = -(30 N-m) + (112 N-m) - ( 228 N-m))

M = 112 N-m - 258 N-m

M = - 146 N-m

Therefore, the couple acting at the left end is; M = - 146 N-m

Answer:

exerts force

Explanation:

The accumulation of excess electric charge on an object is called static electricity. ... An electric field surrounds every electric charge and exerts the force that causes other electric charges to attract or repel. Electric fields are represented by arrows showing the electric field would make a positive charge move.

Answer:

[tex]0.05\ \text{kJ/kg}[/tex]

[tex]3141.6\ \text{kW}[/tex]

Explanation:

v = Velocity of wind = 10 m/s

A = Swept area of blade = [tex]\dfrac{\pi}{4}d^2[/tex]

d = Diameter of turbine = 80 m

[tex]\rho[/tex] = Density of air = [tex]1.25\ \text{kg/m}^3[/tex]

Wind energy per unit mass of air is given by

[tex]E=\dfrac{v^2}{2}\\\Rightarrow E=\dfrac{10^2}{2}\\\Rightarrow E=50\ \text{J/kg}[/tex]

The mechanical energy of air per unit mass is [tex]0.05\ \text{kJ/kg}[/tex]

Power is given by

[tex]P=\rho AvE\\\Rightarrow P=1.25\times \dfrac{\pi}{4}\times 80^2\times 10\times 50\\\Rightarrow P=3141592.65=3141.6\ \text{kW}[/tex]

The power generation potential of the wind turbine is [tex]3141.6\ \text{kW}[/tex].

Answer:

the work done by the net external force acting on the skier is 3046.12 J.

Explanation:

Given;

initial speed of the water skier, u = 6.3 m/s

final speed of the water skier, v = 10.9 m/s

mass of the water skier, m = 77 kg

The work done by the net external force is calculated as;

W = ΔK.E

[tex]W = \frac{1}{2} m(v^2 - u^2)\\\\W = \frac{1}{2} \times \ 77.0(10.9^2 - 6.3^2)\\\\ W= 3046.12 \ J[/tex]

Therefore, the work done by the net external force acting on the skier is 3046.12 J.

Answer:

the speed of electron is 5.021 x 10⁶ m/s

the speed of proton is 2733.91 m/s

Explanation:

Given;

magnitude of electric field, E = 554 N/C

charge of the particles, Q = 1.6 x 10⁻¹⁹ C

time of motion, t = 51.6 ns = 51.6 x 10⁻⁹ s

The force experienced by each particle is calculated as;

F = EQ

F = (554)(1.6 x 10⁻¹⁹)

F = 8.864 x 10⁻¹⁷ N

The speed of the particles are calculated as;

[tex]F = ma\\\\F = \frac{mv}{t} \\\\v = \frac{Ft}{m} \\\\v_e = \frac{Ft}{m_e}\\\\v_e = \frac{(8.864 \times 10^{-17})(51.6\times 10^{-9})}{9.11 \times \ 10^{-31}}\\\\v_e = 5.021 \times 10^{6} \ m/s[/tex]

[tex]v_p = \frac{Ft}{m_p}\\\\v_p = \frac{(8.864 \times 10^{-17})(51.6\times 10^{-9})}{1.673 \times \ 10^{-27}}\\\\v_p = 2733.91 \ m/s[/tex]

Answer: C

Explanation: 60 divided by 4 =15

Velocity can be defined as the rate of change of distance with time

Given data

Acceleration = 4/ms^2

Distance = 60m

Initial Velocity U= 0

Final Velocity V= ?

The expression for velocity is given by

V^2= U^2+2as

Let us substitute our given data into the expression

V^2 = 0^2 + 2*4*60

V^2 = 480

Square both sides

V= √480

V= 21.9 meters/second

V= 22 meters/seconds Approx.

The correct answer is option B

For more information on velocity kindly visit

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Answer:

D = -4/7 = - 0.57

C = 17/7 = 2.43

Explanation:

We have the following two equations:

[tex]3C + 4D = 5\ --------------- eqn (1)\\2C + 5D = 2\ --------------- eqn (2)[/tex]

First, we isolate C from equation (2):

[tex]2C + 5D = 2\\2C = 2 - 5D\\C = \frac{2 - 5D}{2}\ -------------- eqn(3)[/tex]

using this value of C from equation (3) in equation (1):

[tex]3(\frac{2-5D}{2}) + 4D = 5\\\\\frac{6-15D}{2} + 4D = 5\\\\\frac{6-15D+8D}{2} = 5\\\\6-7D = (5)(2)\\7D = 6-10\\\\D = -\frac{4}{7}[/tex]

D = - 0.57

Put this value in equation (3), we get:

[tex]C = \frac{2-(5)(\frac{-4}{7} )}{2}\\\\C = \frac{\frac{14+20}{7}}{2}\\\\C = \frac{34}{(7)(2)}\\\\C = \frac{17}{7}\\[/tex]

C = 2.43

Answer:

To measure work, you must multiply the force by the distance through which it acts.

Answer:

I think the answers March 21

Answer:

Once a year, mean solar time and sidereal time will be the same.

Answer:

see below :)

Explanation:

Radiometric dating.

Answer:

Follows are the solution to this question:

Explanation:

Please find the correct question in the attachment file.

Let:

[tex]\overrightarrow{R}= R_i\hat{i}+R_j\hat{j}+R_k\hat{k}\\\\\overrightarrow{S}= S_i\hat{i}+S_j\hat{j}+S_k\hat{k}\\\\[/tex]

Calculating the value of  [tex]\overrightarrow{R} \times \overrightarrow{S}:[/tex]

[tex]\to \left | \begin{array}{ccc}\hat{i}&\hat{j}&\hat{K}\\R_i&R_j&R_k\\S_i&S_j&S_k\end{array}\right | = \hat{i}[R_j S_k-S_jR_k]-\hat{j}[R_i S_k-S_iR_k]+\hat{k}[R_i S_j-S_iR_j][/tex]

Calculating the value of [tex]\overrightarrow{R} \cdot (\overrightarrow{R} \times \overrightarrow{S}):[/tex]

[tex]\to (R_i\hat{i}+R_j\hat{j}+R_k\hat{k}) \cdot ( \hat{i}[R_j S_k-S_jR_k]-\hat{j}[R_i S_k-S_iR_k]+\hat{k}[R_i S_j-S_iR_j])[/tex]

by solving this value it is equal to 0.

Why words are more important than numbers: ... Words on the other hand are harder to manipulate, they also tell you why someone voted a particular way and to improve your delivery and thus your customer satisfaction you need to understand the why's...

#pglubestiehere