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kykrilka [37]
2 years ago
5

A 600kg car is at rest, and then it accelerates to 5 m/s.

Physics
1 answer:
uranmaximum [27]2 years ago
6 0

Answer:

1. 0 J

2. 7500 J

3. 7500 J

Explanation:

From the question given above, the following data were obtained:

Mass (m) of car = 600 Kg

Initial velocity (v₁) of car = 0 m/s

Final velocity (v₂) of car = 5 m/s

Original kinetic energy (KE₁) =?

Final kinetic energy (KE₂) =?

Work used =?

1. Determination of the original kinetic energy.

Mass (m) of car = 600 Kg

Initial velocity (v₁) of car = 0 m/s

Original kinetic energy (KE₁) =?

KE₁ = ½mv₁²

KE₁ = ½ × 600 × 0²

KE₁ = 0 J

Thus, the original kinetic energy of the car is 0 J.

2. Determination of the final kinetic energy.

Mass (m) of car = 600 Kg

Final velocity (v₂) of car = 5 m/s

Final kinetic energy (KE₂) =?

KE₂ = ½mv₂²

KE₂ = ½ × 600 × 5²

KE₂ = 300 × 25

KE₂ = 7500 J

Thus, the final kinetic energy of the car is 7500 J

3. Determination of the work used.

Original kinetic energy (KE₁) = 0

Final kinetic energy (KE₂) = 7500 J

Work used =?

Work used = KE₂ – KE₁

Work used = 7500 – 0

Work used = 7500 J

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Complete Question

An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field in the wire changes with time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is measured in seconds.

I = 1.2 A at time 5 secs.

Find the charge Q passing through a cross-section of the conductor between time 0 seconds and time 5 seconds.

Answer:

The charge is  Q =2.094 C

Explanation:

From the question we are told that

    The diameter of the wire is  d =  0.205cm = 0.00205 \ m

     The radius of  the wire is  r =  \frac{0.00205}{2} = 0.001025  \ m

     The resistivity of aluminum is 2.75*10^{-8} \ ohm-meters.

       The electric field change is mathematically defied as

         E (t) =  0.0004t^2 - 0.0001 +0.0004

     

Generally the charge is  mathematically represented as

       Q = \int\limits^{t}_{0} {\frac{A}{\rho} E(t) } \, dt

Where A is the area which is mathematically represented as

       A =  \pi r^2 =  (3.142 * (0.001025^2)) = 3.30*10^{-6} \ m^2

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       \frac{A}{\rho} =  \frac{3.3 *10^{-6}}{2.75 *10^{-8}} =  120.03 \ m / \Omega

Therefore

      Q = 120 \int\limits^{t}_{0} { E(t) } \, dt

substituting values

      Q = 120 \int\limits^{t}_{0} { [ 0.0004t^2 - 0.0001t +0.0004] } \, dt

     Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] }  \left | t} \atop {0}} \right.

From the question we are told that t =  5 sec

           Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] }  \left | 5} \atop {0}} \right.

          Q = 120 [ \frac{0.0004(5)^3 }{3} - \frac{0.0001 (5)^2}{2} +0.0004(5)] }

         Q =2.094 C

     

5 0
3 years ago
A ride-sharing car moving along a straight section of road starts from rest, accelerating at 2.00 m/s2 until it reaches a speed
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Answer:

Explanation:

Time taken to accelerate to 28 m /s

= 28 / 2 = 14 s

a ) Total length of time in motion

= 14 + 41 + 5

= 60 s .

b )

Distance covered while accelerating

s = ut + 1/2 at²

= 0 + .5 x 2 x 14²

= 196 m .

Distance covered while moving in uniform motion

= 28 x 41

= 1148 m

distance covered while decelerating

v = u - at

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a = 5.6 m / s²

v² = u² - 2 a s

0 = 28² - 2 x 5.6 x s

s = 28² / 2 x 5.6

= 70 m .

Total distance covered

= 196 + 1148 + 70

= 1414 m

total time taken = 60 s

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Answer:

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Explanation:

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\frac{T_{1}^{2}}{T_{2}^{2}} = \frac{r_{1}^{3}}{r_{2}^{3}}

<em>where T₁: is the orbital period of the geosynchronous Earth satellite = 1 d, T₂: is the orbital period of the moon = 0.07481 y, r₁: is the radius of such an orbit and r₂: is the orbital radius of the moon = 3.84x10⁵ km.                           </em>                              

From equation (1), r₁ is:

r_{1} = r_{2} \sqrt[3] {(\frac{T_{1}}{T_{2}})^{2}}                            

r_{1} = 3.84\cdot 10^{5} km \sqrt[3] {(\frac{1 d}{0.07481 y \cdot \frac{365 d}{1 y}})^{2}}      

r_{1} = 4.24 \cdot 10^{4} km      

Therefore, the radius of such an orbit is 4.24x10⁴ km.

I hope it helps you!

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Answer:

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