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umka21 [38]
3 years ago
8

An object moves in one dimensional motion with constant acceleration a = 4.2 m/s^2. At time t = 0 s, the object is at x0 = 3.9 m

and has an initial velocity of v0 = 2.2 m/s. How far will the object move before it achieves a velocity of v = 7.2 m/s? Your answer should be accurate to the nearest 0.1 m.
Physics
1 answer:
solong [7]3 years ago
8 0

Answer:

The object will move to Xfinal = 7.5m

Explanation:

By relating the final velocity of the object and its acceleration, I can obtain the time required to reach this velocity point:

Vf= a × t ⇒ t= (7.2 m/s) / (4.2( m/s^2)) = 1,7143 s

With the equation of the total space traveled and the previously determined time I can obtain the end point of the object on the x-axis:

Xfinal= X0 + /1/2) × a × (t^2) = 3.9m + (1/2) × 4.2( m/s^2) × ((1,7143 s) ^2) =

= 3.9m + 3.6m = 7.5m

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Answer:

Mutual inductance, M=2.28\times 10^{-5}\ H

Explanation:

(a) A toroidal solenoid with mean radius r and cross-sectional area A is wound uniformly with N₁ turns. A second thyroidal solenoid with N₂ turns is wound uniformly on top of the first, so that the two solenoids have the same cross-sectional area and mean radius.

Mutual inductance is given by :

M=\dfrac{\mu_oN_1N_2A}{2\pi r}

(b) It is given that,

N_1=550

N_2=290

Radius, r = 10.6 cm = 0.106 m

Area of toroid, A=0.76\ cm^2=7.6\times 10^{-5}\ m^2

Mutual inductance, M=\dfrac{4\pi \times 10^{-7}\times 550\times 290\times 7.6\times 10^{-5}}{2\pi \times 0.106}

M=0.0000228\ H

or

M=2.28\times 10^{-5}\ H

So, the value of mutual inductance of the toroidal solenoid is 2.28\times 10^{-5}\ H. Hence, this is the required solution.

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Can we use position as another word for phase?
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You serve a volleyball with a mass of 2.1 kg. The ball leaves your hand with a speed of 35 m/s. The ball has __________________
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The basic barometer can be used to measure the height of a building. If the barometric readings at the top and at the bottom of
LekaFEV [45]

Answer:

   (h₁-h₂) = 2.30 10² m

Explanation:

The pressure depends on the height with the formula

          P = P_atm + rho g h

Let's apply this expression for the building

         P₁ = P_atm + rho_air g h₁

        P₂ = P_atm + rho_air g h₂

Subtract

        P₁ - P₂ = roh_air g (h₁ –h₂)

         

         

The measured pressure is in mm Hg to take this unit to units of pressure must be multiplied by the density of mercury and the acceleration of gravity

           P₁- P₂ = rho_Hg g (h₁-h₂) _Hg

         rho_Hg g (h₁-h₂) _Hg = roh_air g (h₁ –h₂)

          (h₁ –h₂) = rho_Hg / rho_air (h₁-h₂) _ Hg

Let's calculate

         (h₁-h₂) = 13600 / 1.18 (695-675)

         (h₁-h₂) = 2.30 10⁵ mm

Let's reduce to meter

         (h₁-h₂) = 2.30 10⁵ mm (1 m / 10³ mm)

         (h₁-h₂) = 2.30 10² m

4 0
3 years ago
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