A weightless spring is stretched 10 cm by a suspended 1-kg block. If two such springs are used to suspend the block, one spring above the other, to effectively provide one double-length spring, then the total stretch of the double-length spring will be

"When The motional emf generated between the ends of the wire is e = 17745.1 V".

What is Motional emf generated?

When Motional EMFs in the Earth’s weak magnetic occupation are not ordinarily very large, we would notice voltage along metal rods, Than such as a screwdriver, during ordinary motions. For illustration, a simple computation of the motional emf of a 1 m rod moving at 3. 0 m/s perpendiculars to the Earth’s field gives emf is = Bℓv

Then Length, l is = 4.32 x 10^4 m

Then speed, v is = 7.07 x 10^3 m/s

Now the magnetic field, B is = 5.81 x 10^-5 T

Now, The formula for the motion emf is given by

e is = B x v x l

e is = 5.81 x 10^-5 x 7.07 x 10^3 x 4.32 x 10^4

Therefore, e = 17745.1 V

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Answer:

Explanation:

Length, l = 4.32 x 10^4 m

speed, v = 7.07 x 10^3 m/s

magnetic field, B = 5.81 x 10^-5 T

The formula for the motion emf is given by

e = B x v x l

e = 5.81 x 10^-5 x 7.07 x 10^3 x 4.32 x 10^4

e = 17745.1 V

Answer:

The average angular acceleration of the Earth, α  = 6.152 X 10⁻²⁰ rad/s²

Explanation:

Given data,

The period of 365 rotation of Earth in 2006, T₁ = 365 days, 0.840 sec

                                                                                  = 3.1536 x 10⁷ +0.840

                                                                                 = 31536000.84 s

The period of 365 rotation of Earth in 2006, T₀ = 365 days

                                                                               = 31536000 s

Therefore, time period of one rotation on 2006, Tₐ = 31536000.84/365

                                                                                   = 86400.0023 s

The time period of rotation is given by the formula,

                                Tₐ = 2π /ωₐ

                                 ωₐ = 2π / Tₐ

Substituting the values,

                                  ωₐ = 2π /  365.046306        

                                      = 7.272205023 x 10⁻⁵ rad/s

Therefore, the time period of one rotation on 1906, Tₓ = 31536000/365

                                                                                    = 86400 s

Time period of rotation,

                                   Tₓ = 2π /ωₓ

                                    ωₓ = 2π / T

                                           =  2π /86400

                                          = 7.272205217  x 10⁻⁵ rad/s

The average angular acceleration

                                   α  = (ωₓ -   ωₐ) /  T₁

             = (7.272205217  x 10⁻⁵ - 7.272205023 x 10⁻⁵) / 31536000.84

                                    α  = 6.152 X 10⁻²⁰ rad/s²

Hence the average angular acceleration of the Earth, α  = 6.152 X 10⁻²⁰ rad/s²

Final answer:

The average angular acceleration of the Earth from the year 1906 to 2006 would be -5.73 x 10^-20 rad/s^2. This value was obtained by finding the change in angular velocity and then dividing it by the elapsed time.

Explanation:

The question is asking for the average angular acceleration of the Earth from the year 1906 to 2006, during which the Earth's rotation rate decreased, causing the day to increase in duration by about 0.840 seconds.

To find the average angular acceleration, you first need to calculate the change in angular velocity, which can be found from the change in rotation time. One revolution (one day) is 2π radians, so the change in angular velocity is Δω = 2π/86400 s - 2π/(86400+0.840) s = -1.81 x 10^-10 rad/s.

The time interval from 1906 to 2006 is 100 years or about 3.16 x 10^9 seconds. Therefore, the average angular acceleration, α, which is the change in angular velocity divided by time, would be α = Δω/Δt = -1.81 x 10^-10 rad/s / 3.16 x 10^9 s = -5.73 x 10^-20 rad/s^2.

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By definition, we have that the speed of an object is given by:

Where,

d: distance [in units of length: meters, feet, miles]

t: time [in units of time: minutes, seconds, hours]

Therefore, knowing the distance traveled, and the time to travel this distance, we can know the speed of an object.

Then, since velocity is a vector, then we need the direction of the vector.

Therefore, the velocity vector can be written as:

Answer:

two things you need to know to describe the velocity of an object are:

1) Magnitude (speed)

2) direction

Answer:

all the options listed are renewable sources of energy

Explanation:

renewable sources of energy are energy sources that are easily replenished naturally such as solar energy, wind energy , wave energy even energy from rain. These sources are usually environmental friendly that non renewable energy sources

non renewable energy however, are sources of energy that will eventually run out. examples are coal, crude oil, natural gas, nuclear energy. these sources mostly pose a high risk to the environment when being used.

hope this is quite helpful?

Explanation:

They will be alerted of the traffic conditions. They can choose the route according to the weather condition. They will be aware if there is any road construction going on or accident happened. These awareness will save their time and avoid delays. their journey can be more comfortable and less trouble some.

Explanation:

Researching these options time, routes and weather conditions and even more gives the driver ample information about the journey, and places the driver in better position for decision making, with this information and data afore hand he can decides the best route to take should there be any gridlock, he even decides where or not to embark on the journey based on the prevailing weather conditions,

Regarding time of the journey he is well disposed for proper planning and scheduling.