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Lana71 [14]
3 years ago
15

How much work must be done to stop a 1075-kg car traveling at 115 km/h ? Express your answer to two significant figures and incl

ude the appropriate units
Physics
1 answer:
jok3333 [9.3K]3 years ago
6 0

Answer:

Work done on the car will be 548440.9463 J

Explanation:

We have given mass of the car m = 1075 kg

As the car finally stops so final velocity of the car v_f=0m/sec

Initial velocity of car u = 115 km/hr =115\times \frac{5}{18}=31.943m/sec

Work done is equal to change in kinetic energy

So work done =\frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2

==\frac{1}{2}m(v_f^2-v_i^2)=\frac{1}{2}\times 1075\times (0^2-31.943^2)=-548440.946J

As the work done is negative so work is done on the car to stop it .

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Which of the following choices defines energy in scientific terms?
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1) B. Energy is the ability to do work

2) C. Energy is conserved, it just goes from one form to another.

3) Work = Force x displacement
= 300 x 100 = 30,000 Joules

4) leaning a brick because no displacement is taking place.

5) They change the amount/strength or direction of the force needed.

6) Less force is needed and applied over a longer distance.

7) Heat is the flow of thermal energy from one object to another.
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Calculate the speed for a car that went a distance of 125 kilometers in 2 hours time.
Murljashka [212]
S=125km
t=2h
v=s/t=125/2=62,5km/h
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3 years ago
Three charges lie along the x axis. The positive charge q1 = 15 μC is at x= 2.0 m and the positive charge q2 = 6.0μC is at the o
Delicious77 [7]

Answer:

x = 0.775m

Explanation:

Conceptual analysis

In the attached figure we see the locations of the charges. We place the charge q₃ at a distance x from the origin. The forces F₂₃ and F₁₃ are attractive forces because the charges have an opposite sign, and these forces must be equal so that the net force on the charge q₃ is zero.

We apply Coulomb's law to calculate the electrical forces on q₃:

F_{23} = \frac{k*q_{2}*q_{3}}{x^2} (Electric force of q₂ over q₃)

F_{13} = \frac{k*q_{1}*q_{3}}{(2-x)^2} (Electric force of q₁ over q₃)

Known data

q₁ = 15 μC = 15*10⁻⁶ C

q₂ = 6 μC = 6*10⁻⁶ C

Problem development

F₂₃ = F₁₃

\frac{k*q_{2}*q_{3}}{x^2} = \frac{k*q_{1}*q_{3}}{(2-x)^2} (We cancel k and q₃)

\frac{q_2}{x^2}=\frac{q_1}{(2-x)^2}

q₂(2-x)² = q₁x²

6×10⁻⁶(2-x)² = 15×10⁻⁶(x)² (We cancel 10⁻⁶)

6(2-x)² = 15(x)²

6(4-4x+x²) = 15x²

24 - 24x + 6x² = 15x²

9x² + 24x - 24 = 0

The solution of the quadratic equation is:

x₁ = 0.775m

x₂ = -3.44m

x₁ meets the conditions for the forces to cancel in q₃

x₂ does not meet the conditions because the forces would remain in the same direction and would not cancel

The negative charge q₃ must be placed on x = 0.775 so that the net force is equal to zero.

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At a rock concert, the sound intensity 1.0 m in front of the bank of loudspeakers is 0.10 W/m². A fan is 30 m from the loudspeak
Klio2033 [76]

To solve this problem we will apply the concepts related to the Area, the power and the proportionality relationships between intensity and distance.

The expression for sound power is,

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Now the intensity is inversely proportional to the square of the distance from the source, then

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The expression for the intensity at different distance is

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Here,

I_1 = Intensity at distance 1

I_2 = Intensity at distance 2

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If we rearrange the expression to find the intensity at second position we have,

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