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3 years ago

At highway speeds a car can accelerate at 1.7 m/s2.

Physics

1 answer:

Vaselesa [24]3 years ago

6 0

Answer: 2.5 seconds

Explanation:

We know that the acceleration is:

a(t) = 1.7 m/s^2

To get the velocity function, we must integrate over time, and we will get:

v(t) = (1.7m/s^2)*t + v0

Where v0 is the initial velocity, in this case, we assume that we start at 23.6m/s, then the initial velocity is:

v0 = 23.6 m/s

Then the velocity equation is:

v(t) = (1.7m/s^2)*t + 23.6 m/s

Now we want to find the value of t such v(t) = 27.8 m/s

Then:

v(t) = 27.8 m/s = (1.7m/s^2)*t + 23.6 m/s

27.8 m/s - 23.6 m/s = (1.7m/s^2)*t

4.2 m/s = (1.7m/s^2)*t

4.2m/s/(1.7m/s^2) = t = 2.5 s

Then at that acceleration, you need 2.5 seconds.

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Explanation:

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Chapter 14, Problem 042 A flotation device is in the shape of a right cylinder, with a height of 0.588 m and a face area of 4.19

Alisiya [41]

Answer:

The workdone is $W = 9.28 * 10^{3} J$

Explanation:

From the question we are told that

The height of the cylinder is $h = 0.588\ m$

The face Area is $A = 4.19 \ m^2$

The density of the cylinder is $\rho = 0.346 * \rho_w$

Where $\rho_w$ is the density of freshwater which has a constant value

$\rho_w = 1000 kg/m^3$

Now

Let the final height of the device under the water be $= h_f$

Let the initial volume underwater be $= V_n$

Let the initial height under water be $= h_i$

Let the final volume under water be $= V_f$

According to the rule of floatation

The weight of the cylinder = Upward thrust

This is mathematically represented as

$\rho_c g V_n = \rho_w gV_f$

$\rho_c A h = \rho A h_f$

So $\frac{0.346 \rho_w}{\rho_w} = \frac{h_f}{h}$

=> $\frac{h_f}{h_c} = 0.346$

Now the work done is mathematically represented as

$W = \int\limits^{h_f}_{h} {\rho_w g A (-h)} \, dh$

$= \rho_w g A [\frac{h^2}{2} ] \left | h_f} \atop {h}} \right.$

$= \frac{g A \rho}{2} [h^2 - h_f^2]$

$= \frac{g A \rho}{2} (h^2) [1 - \frac{h_f^2}{h^2} ]$

Substituting values

$W = \frac{(9.8 ) (4.19) (10^3)}{2} (0.588)^2 (1 - 0.346)$

$W = 9.28 * 10^{3} J$

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3 years ago

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Answer:

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. Common household appliances are rated at 110 V, but power companies deliver voltage in the kilovolt range and then step the vo

natulia [17]

Answer: High voltage transmission minimizes energy losses during electricity transmission.

Explanation:

When electricity is to be transmitted over a long distance, high voltage transmission is preferred to minimize energy losses due to heat.

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High voltage transmission in kilowatts enables light weight cables to be used for long distance electricity transmission.

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