What is a reasonable measurement for the distance to Neptune?
30 light years
30 kilometers
30 parsecs
30 Astronomical Units

Answer:

1.5 m is ur answer

Explanation:

Answer:

a) the distance between her and the wall is 13 m

b) the period of her up-and-down motion is 6.5 s

Explanation:

Given the data in the question;

wavelength λ = 26 m

velocity v = 4.0 m/s

a) How far from the wall is she?

Now, The first antinode is formed at a distance λ/2 from the wall, since the separation distance between the person and wall is;

x = λ/2

we substitute

x = 26 m / 2

x = 13 m

Therefore, the distance between her and the wall is 13 m

b) What is the period of her up-and-down motion?

we know that the relationship between frequency, wavelength and wave speed is;

v = fλ

hence, f = v/λ

we also know that frequency is expressed as the reciprocal of the time period;

f = 1/T

Hence

1/T = v/λ

solve for T

Tv = λ

T = λ/v

we substitute

T = 26 m / 4 m/s

T = 6.5 s

Therefore, the period of her up-and-down motion is 6.5 s

Answer:

It will go to the right way 500N.

Explanation:

Because, up and down is 300 and 300 so it minus each other and get zero, that made the resultant force will never go up or down. And now we are focusing on the left and right. 900 is more than 400, so it will go to the right. But we also pull to the left too!! That make the resultant force is 500 to the right.

Complete Question

A system consists of a disk of mass 2.0 kg and radius 50 cm upon which is mounted an annular cylinder of mass 1.0 kg with inner radius 20 cm and outer radius 30 cm (see below). The system rotates about an axis through the center of the disk and annular cylinder at 10 rev/s.

(a) What is the moment of inertia of the system

(b) What is its rotational kinetic energy? axis 50 cm 30 cm 20 cm

Answer:

a) [tex]M.I=0.32kg.m^2[/tex]

b) [tex]K.E=621.8J[/tex]

Explanation:

From the question we are told that:

Mass [tex]m=2.0kg[/tex]

Disk Radius [tex]r_d=50cm=0.5m[/tex]

Mass of annular cylinder [tex]M_c=1.0kg[/tex]

Inner Radius of cylinder [tex]R_i=20cm=0.2m[/tex]

Outer Radius of cylinder [tex]R_o=0.3m[/tex]

Angular  Velocity [tex]\omega=10rev/s=2 \pi*10=62.83[/tex]

Generally the equation for moment of inertia is mathematically given by

[tex]M.I=0.5M r_d^2+0.5M_c(R_i^2+R_o^2)[/tex]

[tex]M.I=0.5(2*0.50)^2)+0.5*1*(0.2^2+0.30^)[/tex]

[tex]M.I=0.32kg.m^2[/tex]

Generally the equation for Rotational Kinetic Energy  is mathematically given by

[tex]K.E=0.5M.I \omega^2[/tex]

[tex]K.E=0.5 *0.32*62.83[/tex]

[tex]K.E=621.8J[/tex]

Answer:

An Object is positively charged if it has more Positive Electrons in that object

Answer:

Economists, I guess.

Explanation:

The speed is in feet per seconds so change the length of the field from yards to feet.

1 yard = 3 feet

100 yards x 3 = 300 feet

The field is 300 feet long

Time = distance / speed

Time = 300 feet / 5 feet per second = 60 seconds = 1 minute

It will take 1 minute

Answer:

A person will take 60 Seconds to walk the distance of 100 yards.

Explanation:

Data Given ;

Distance ( d ) = 100 yards = 300 Ft

Speed ( v ) = 5 Ft/s

Time ( t ) = ?

The distance travelled in unit time is called speed.

formula ; [tex]v = \frac{d}{t} \\[/tex]

On putting values,

[tex]5 = \frac{300}{t}[/tex]

[tex]t =\frac{300}{5}[/tex]

[tex]t = 60 sec[/tex]

Hence the time taken by the person is 60 sec.

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Answer:

α = 13.7 rad / s²

Explanation:

Let's use Newton's second law for rotational motion

         ∑ τ = I α

we will assume that the counterclockwise turns are positive

         F₁  0 + F₂ R₂ - F₃ R₃ = I α

give us the cylinder moment of inertia

        I = ½ M R₂²

        α = (F₂ R₂ - F₃ R₃)  [tex]\frac{2}{M R_2^2}[/tex]

let's calculate

        α = (24  0.22 - 13  0.10) [tex]\frac{2}{12 \ 0.22^2}[/tex]2/12 0.22²

        α = 13.7 rad / s²

The boy pushed the sled for 16 seconds.

We have a boy who pushes his little brother on a sled.

We have to determine for how long time does boy push the sled.

The Work - Energy theorem states that the work done by the sum of all forces acting on a particle equals the change in the kinetic energy of the particle.

According to the question -

The sled is initially at rest → initial velocity (u) = 0.

Final velocity (v) = 4 m/s

Mass of boy and sled (M) = 40 kg

Power developed (P) = 20 W = 20 Joules/sec

According to work - energy theorem -

Work done (W) = Δ E(K) = E(f) - E(i)

Therefore -

W = ([tex]\frac{1}{2} \times 40 \times 4 \times 4 - \frac{1}{2}[/tex] x 40 x 0) = 320 Joule

Now, Power is defined as the rate of doing work -

P = [tex]\frac{dW}{dt}[/tex] = [tex]\frac{W}{t}[/tex]

20 = [tex]\frac{320}{t}[/tex]

t = 16 seconds

Hence, the boy pushed the sled for 16 seconds.

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Answer:

the law of conservation of energy.

Explanation:

An electric circuit can be defined as an interconnection of electrical components which creates a path for the flow of electric charge (electrons) due to a driving voltage.

Generally, an electric circuit consists of electrical components such as resistors, capacitors, battery, transistors, switches, inductors, etc.

In Physics, a point where at least three circuit paths (wires) meet is referred to as a junction.

The Kirchhoff’s circuit laws are two(2) equations first published by Gustav Kirchhoff in 1845.

Fundamentally, they address the conservation of energy and charge in the context of electrical circuits.

One of the laws known as Kirchoff's Current Law (KCL) deals with the principle of application of conserved energy in electrical circuits.

Kirchoff's Current Law (KCL) states that the sum of all currents entering a junction must equal the sum of all currents leaving the junction.

This simply means that the algebraic sum of currents in a network of conductors(wires) meeting at a point is equal to zero.

The Law of Conservation of Energy states that energy cannot be destroyed but can only be transformed or converted from one form to another.

This ultimately implies that, Kirchhoff's junction rule is a statement of the law of conservation of energy.

Answer:

[tex]V_2=3.3m/s[/tex]

Explanation:

From the question we are told that:

Distance [tex]d_1=1.4m[/tex]

Tangential speed [tex]V=2.2m/s[/tex]

Distance 2 [tex]d_2=2.1m[/tex]

Generally the equation for Angular velocity is mathematically given by

 [tex]w=\frac{v}{r}[/tex]

Therefore

 [tex]\frac{v_1}{r_1}=\frac{v_2}{r_2}[/tex]

 [tex]V_2=\frac{2.2*2.1}{1.4}[/tex]

 [tex]V_2=3.3m/s[/tex]

Answer:

mmmlbdhdjdkekkdnxnfjkckckcklclglglvkglglvkgkvkvkvkvkvkvkvkvkvkvkbkkbkbkkbbkkbkbkbkkbkbkblbkkbkb

because they work on the same object

Explanation:

A push and pull factor

Answer:

We know that the change in electric potential energy is defined as:

q*ΔV = ΔP

So, the change in the electric potential energy is the charge times the change in the electric potential.

For the case of an electron gas, we have:

q = -e

where -e is the charge of an electron (remember that is negative).

So, if the electron gains electric potential then:

ΔP > 0

this means that the final potential energy is larger than the initial one, then we have:

-e*ΔV > 0

This means that ΔV must be negative.

V₂ = electric potential at point 2, so it is the final electric potential

V₁ = electric potential at point 1, so it is the final electric potential

Then we should get:

ΔV = V₂ - V₁ < 0.

This means that:

V₂ < V₁

The correct option is b.

Answer:

Angle of reflection

Explanation:

The angle between the reflected ray and the normal line is called the angle of reflection

Answer:

Activity 2 Directions: Solve the following problems using Charles' Law,

1. A 3.8 mL of a gas in a container has a temperature of 2.0K. What would the volume be if the temperature becomes 8.0K?

2. If at 27°C the volume of a dry gas is 3.0 ml. What is the new temperature if the volume increased to 7.5 ml?

3. The temperature of the gas is 460,0 mL at 3.5°C. Calculate the new volume at 1.5°C

Answer:45,000

Explanation:

:TIL the average pencil holds enough graphite to draw a line about 35 miles long or to write roughly 45,000 words

Answer:

4 ohm

Explanation:

The equivalent resistance (Re) of three resistors in parallel is given by;

1/Re = 1/R1 + 1/R2 + 1/R3

Where; R1 = 16 ohm, R2 = 16 ohm, R3 = 8 ohm

1/Re= 1/16 + 1/16 + 1/8

1/Re= (0.0625) + (0.0625) + (0.125)

Re= 4 ohm

Answer:

venus - 2

earth - 3

mars - 4

mercury - 1

Answer:

Explanation:

From the given information:

total mass = 25 kg

distance d = 100 m

height = 4.0 m

Force F = 9.0 N

The speed at (bottom) u = 5.0 m/s

Using the concept of energy conservation;

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

divide both sides by m

[tex]\dfrac{1}{2}u^2 + \dfrac{W }{m}= \dfrac{1}{2}v^2 + gh[/tex]

multiply both sides by 2

[tex]\dfrac{1}{2}u^2\times 2 + \dfrac{W }{m}\times 2= \dfrac{1}{2}v^2\times 2 + gh\times 2[/tex]

[tex]u^2 +2 \dfrac{W }{m}=v^2 + 2gh[/tex]

[tex]v^2 =u^2 - 2gh+ 2 \dfrac{W }{m}[/tex]

Recall that:

W = Fd

∴

[tex]v^2 =u^2 - 2gh+ 2 \dfrac{Fd }{m}[/tex]

[tex]v^2 =(5.0 \ m/s)^2 - 2(9.81 \ m/s)(4.0 \ m)+ 2 \dfrac{( 9.0 \ N \times 100 \ m) }{25 \ kg}[/tex]

[tex]v^2 =(25.0 ) - 78.48 +72[/tex]

[tex]v^2 = 18.52 \ m^2/s^2[/tex]

[tex]v = \sqrt{18.52 \ m^2/s^2}[/tex]

v = 4.30 m/s

Answer:

The speed of the train before and after slowing down is 22.12 m/s and 11.06 m/s, respectively.

Explanation:

We can calculate the speed of the train using the Doppler equation:

[tex] f = f_{0}\frac{v + v_{o}}{v - v_{s}} [/tex]        

Where:

f₀: is the emitted frequency

f: is the frequency heard by the observer  

v: is the speed of the sound = 343 m/s

[tex] v_{o}[/tex]: is the speed of the observer = 0 (it is heard in the town)

[tex] v_{s}[/tex]: is the speed of the source =?

The frequency of the train before slowing down is given by:

[tex] f_{b} = f_{0}\frac{v}{v - v_{s_{b}}} [/tex]  (1)                  

Now, the frequency of the train after slowing down is:

[tex] f_{a} = f_{0}\frac{v}{v - v_{s_{a}}} [/tex]   (2)  

Dividing equation (1) by (2) we have:

[tex] \frac{f_{b}}{f_{a}} = \frac{f_{0}\frac{v}{v - v_{s_{b}}}}{f_{0}\frac{v}{v - v_{s_{a}}}} [/tex]

[tex] \frac{f_{b}}{f_{a}} = \frac{v - v_{s_{a}}}{v - v_{s_{b}}} [/tex]   (3)  

Also, we know that the speed of the train when it is slowing down is half the initial speed so:

[tex] v_{s_{b}} = 2v_{s_{a}} [/tex]     (4)

Now, by entering equation (4) into (3) we have:

[tex] \frac{f_{b}}{f_{a}} = \frac{v - v_{s_{a}}}{v - 2v_{s_{a}}} [/tex]  

[tex] \frac{300 Hz}{290 Hz} = \frac{343 m/s - v_{s_{a}}}{343 m/s - 2v_{s_{a}}} [/tex]

By solving the above equation for [tex]v_{s_{a}}[/tex] we can find the speed of the train after slowing down:

[tex] v_{s_{a}} = 11.06 m/s [/tex]

Finally, the speed of the train before slowing down is:

[tex] v_{s_{b}} = 11.06 m/s*2 = 22.12 m/s [/tex]

Therefore, the speed of the train before and after slowing down is 22.12 m/s and 11.06 m/s, respectively.                        

I hope it helps you!                                                        

Solution :

The distance between the starting point and the end point, [tex]L_0[/tex] = 10 light years

But due to the relativistic motion of Bob and Charlie, the distance will be reduced following the Lorentz contraction. The contracted length will be different since they are moving with different speeds.

For Bob,

Speed of Bob's rocket with respect to Alice, [tex]L_b = 0.7 \ c[/tex]

So the distance appeared to Bob due to the length contraction,

[tex]$L_b=L_0\sqrt{1-\frac{V_b^2}{c^2}$[/tex]

[tex]$L_b=10\times \sqrt{1-0.49} \ Ly$[/tex]

    [tex]$=7.1 \ Ly$[/tex]

Therefore, the time required to finish the race by Bob is

[tex]$t_b = \frac{L_b}{V_b}$[/tex]

  [tex]$=\frac{7.1 \ c}{0.7 \ c}$[/tex]

  = 10.143 year

For Charlie,

Speed of Charlie's rocket with respect to Alice, [tex]L_c = 0.866 \ c[/tex]

So the distance appeared to Charlie due to the length contraction,

[tex]$L_b=L_0\sqrt{1-\frac{V_c^2}{c^2}$[/tex]

[tex]$L_b=10\times \sqrt{1-0.75} \ Ly$[/tex]

    [tex]$=5 \ Ly$[/tex]

The time required to finish the race by Charlie is

[tex]$t_b = \frac{L_c}{V_c}$[/tex]

  [tex]$=\frac{5 \ c}{0.866 \ c}$[/tex]

  = 5.77 year

Answer:

40479.6 J

Explanation:

Applying,

q = cm(t₂-t₁).................... Equation 1

Where q = change in heat content of the system, c = specific heat capacity of the system, m = mass of the system, t₁ = initial temperature, t₂ = final temperature.

From the question,

Given: m = 79 g = 0.079 kg, t₁ = 21°C, t₂ = 143°C

Constant: c = 4200 J/kg.°C

Substitute these values into equation 1

q = 4200(0.079)(143-21)

q = 331.8(122)

q = 40479.6 J

Answer:

[tex]I=0.902kg*m^2[/tex]

Explanation:

From the question we are told that:

Mass [tex]m=92.0kg[/tex]

Radius [tex]r=0.140m[/tex]

Generally the equation for moment of Inertia is mathematically given by

 [tex]I = 0.5*m*r^2[/tex]

 [tex]I=0.5*(92)*0.14^2[/tex]

 [tex]I=0.902kg*m^2[/tex]

Answer:

559.5 N

Explanation:

Applying,

v² = u²+2gs............. Equation 1

Where v = final velocity,

From the question,

Given: s = 5.10 m, u = 0 m/s ( from rest)

Constant: 9.8 m/s²

Therefore,

v² = 0²+2×9.8×5.1

v² = 99.96

v = √(99.96)

v = 9.99 m/s

As the diver eneters the water,

u = 9.99 m/s, v = 0 m/s

Given: t = 1.34 s

Apply

a = (v-u)/t

a = 9.99/1.34

a = -7.46 m/s²

F = ma.............. Equation 2

Where F = force, m = mass

Given: m = 75 kg, a = -7.46 m/s²,

F = 75(-7.46)

F = -559.5 N

Hence the average force exerted on the diver is 559.5 N

Answer:

Distance

Explanation:

Work can be defined as the energy transferred to a physical object by exertion of a force on the object to cause a displacement of the object. Thus, work is typically done when a person or simple machine move an object over a distance through the application of a force.

Mathematically, work done is given by the formula;

[tex] W = F * d[/tex]

Where,

W is the work done

F represents the force acting on a body.

d represents the distance covered by the body.

A dependent variable is the event expected to change when an independent variable is manipulated.

Hence, distance is the dependent variable because its value changes with respect to the amount of force exerted on an object.

Answer:

the magnitude of the average force on the bumper is 3189.8 N

Explanation:

Given the data in the question;

In terms of force and displacement, work done is;

W =[tex]F^>[/tex] × [tex]x^>[/tex]

W = [tex]Fxcos\theta[/tex]    ------- let this be equation 1

where F is force applied, x is displacement and θ is angle between force and displacement.

Now, since the displacement of the bumper and force acting on it is in the same direction,

hence, θ = 0°

we substitute into equation 1

W = [tex]Fxcos([/tex] 0° [tex])[/tex]

W = [tex]Fx[/tex] ------- let this be equation 2

Now, using work energy theorem,

total work done on the system is equal to the change in kinetic energy of the system.

[tex]W_{net[/tex] = ΔKE

= [tex]\frac{1}{2}[/tex]mv² -  [tex]\frac{1}{2}[/tex]mu² --------- let this be equation 3

where m is mass of object, v is final velocity, u is initial velocity.

from equation 2 and 3

[tex]Fx[/tex] = [tex]\frac{1}{2}[/tex]mv² -  [tex]\frac{1}{2}[/tex]mu²

we make F, the subject of formula

F = [tex]\frac{m}{2x}[/tex]( v² - u² )

given that mass of car m = 830 kg, x = 0.255 m, v = 0 m/s, and u = 1.4 m/s

so we substitute

F = [tex]\frac{830}{(2)(0.255)}[/tex]( (0)² - (1.4)² )

F = 1627.45098 ( 0 - 1.96 )

F = 1627.45098 ( - 1.96 )

F = -3189.8 N

The negative sign indicates that the direction of the force was in opposite compare to the direction of the velocity of the car.

Therefore, the magnitude of the average force on the bumper is 3189.8 N

Explanation:

bosu

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