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3 years ago

It takes 185 kj of work to accelerate a car from 23.0 m/s to 28.0 m/s. what is the car's mass?

1 answer:

6 0

<span>This question is based on conservation of energy as the work done would lead to change in kinetice energy of car change in KE = 1/2 mv(f)^2 - 1/2mv(i)^2 = 1/2m(v(f)^2-v(i)^2) where v(f) and v(i) are the final and initial speeds change in KE = 185kJ = 185,000J = 1/2 m((28m/s)^2-(23m/s)^2) 185,000=1/2 m(255m^2/s^2) solving for m m=1451kg</span>

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Arlecino [84]

Answer:

Entonces seria 127 para vencer.

Explanation:

espero averte ayudado:-)

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3 years ago

1. Bone has a Young’s modulus of about

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As we know that

$Y = \frac{stress}{strain}$

now plug in all data into this

$1.8\times 10^{10} = \frac{1.68 \times 10^8}{strain}$

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now from the formula of strain

$strain = \frac{\Delta L}{L}$

$9.33 \times 10^{-3} = \frac{\Delta L}{0.54}$

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As we know that

pressure * area = Force

here we know that

$Area = 3.53 \times 11.6 = 40.95 m^2$

$P = 0.2 atm = 0.2 \times 1.01 \times 10^5 = 2.02\times 10^4 Pa$

now force is given as

$F = 40.95 \times (2.02\times 10^4) = 8.27 \times 10^5 N$

As we know that density of water will vary with the height as given below

$\rho = \frac{\rho_0}{1 - \frac{\Delta P}{B}}$

here we know that

$\Delta P = 2600 atm = 2600 \times 1.01 \times 10^5 = 2.63\times 10^8 Pa$

$B = 2.3 \times 10^9 N/m^2$

now density is given as

$\rho = \frac{1050}{1 - \frac{2.63\times 10^8}{2.3 \times 10^9}}$

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as we know that pressure changes with depth as per following equation

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here we know that

$P = 3 P_0$

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here we will have

$h = 20.1 m$

so it is 20.1 m below the surface

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so here we will have

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$A(0.0476)979 = 922(A)(0.0476 - x) + 1000(A)(x)$

$46.6 = 43.89 - 922x + 1000x$

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3 years ago

If Juan upgrades from a sports car to a large truck, what will happen to the

vovikov84 [41]

Answer:

The force will have to increase

Explanation:

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Force = mass x acceleration

A truck has a larger mass compared to a sports car.

By virtue of this, to make sure both automobiles attain the same speed, the force powering them to accelerate must be the same.

Therefore, the force from the engine must increase.

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2 years ago

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I believe the answer is C - 78.4 m/s.

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2 years ago