Ask question

Ask question

All categories

3 years ago

5

In 1898, the world land speed record was set by Gaston Chasseloup-Laubat driving a car named Jeantaud. His speed was 39.24 mph (

63.15 km/h), much lower than the limit on our interstate highways today. Repeat the calculations of Example 2.7 (assume the car accelerates for 6 miles to get up to speed, is then timed for a one-mile distance, and accelerates for another 6 miles to come to a stop) for the Jeantaud car. (Assume the car moves in the +x direction.)
Find the acceleration for the first 6 miles.

1 answer:

FromTheMoon [43]3 years ago

7 0

Answer:

the acceleration $a^{\to} = (0.0159 \ \ m/s^2 )i$

Explanation:

Given that:

the initial speed v₁ = 0 m/s i.e starting from rest ; since the car accelerates at a distance Δx = 6 miles in order to teach that final speed v₂ of 63.15 km/h.

So; the acceleration for the first 6 miles can be calculated by using the formula:

v₂² = v₁² + 2a (Δx)

Making acceleration a the subject of the formula in the above expression ; we have:

v₂² - v₁² = 2a (Δx)

$a = \dfrac{v_2^2 - v_1^2 }{2 \Delta x}$

$a = \dfrac{(63.15 \ km/s)^2 - (0 \ m/s)^2 }{2 (6 \ miles)}$

$a = \dfrac{(17.54 \ m/s)^2 - (0 \ m/s)^2 }{2 (9.65*10^3 \ m)}$

$a =0.0159 \ m/s^2$

Thus;

Assume the car moves in the +x direction;

the acceleration $a^{\to} = (0.0159 \ \ m/s^2 )i$

You might be interested in

Aliya deposited half as much money in a savings account earning 1.9% simple interest as she invested in a money market account t

yanalaym [24]

Let <em>A</em> be the amount of money that Aliya deposited in the savings account. Since <em>A</em> is half as much as money as she invested in a money market account, then the amount that she invested in the market account is <em>2A.</em>

<em />

Express the interest that Aliya earned in terms of A. Set it equal to the amount of $297.60 and then solve for <em>A</em>.

Since the savings account gives 1.9% simple interest, the total amount of interest that she will earn from the savings account is 1.9% of A, which is equal to:

$\frac{1.9}{100}\times A$

Since the money market account gives 3.7% simple interest, the total amount of interest that she will earn from the money market account, is 3.7% of <em>2A</em>, which is equal to:

$\frac{3.7}{100}\times2A$

Add both interests in terms of A and simplify the expression:

$\begin{gathered} \frac{1.9}{100}\times A+\frac{3.7}{100}\times2A \\ =\frac{1.9}{100}\times A+\frac{7.4}{100}\times A \\ =(\frac{1.9}{100}+\frac{7.4}{100})\times A \\ =\frac{1.9+7.4}{100}\times A \\ =\frac{9.3}{100}\times A \end{gathered}$

The expression (9.3/100)*A represents the total interest after one year. Then:

$\begin{gathered} \frac{9.3}{100}\times A=297.60 \\ \Rightarrow A=\frac{100}{9.3}\times297.60 \\ \Rightarrow A=\frac{100\times297.60}{9.3} \\ \Rightarrow A=\frac{29760}{9.3} \\ \Rightarrow A=3200 \end{gathered}$

Use the value of <em>A</em> to find the amount that was invested in the money market account:

$2A=2\times3200=6400$

Therefore, Aliya deposited 3200 in a savings account and 6400 in a money market account.

3 0

1 year ago

What is the name for all the electromagnetic waves that exist

zheka24 [161]

The Electromagnetic spectrum.

8 0

3 years ago

A 40 kg girl and an 8.4 kg sled are on the surface of a frozen lake, 15 m apart. By means of a rope, the girl exerts a 5.2 N for

stealth61 [152]

Answer:

(a) $a_s=0.62\frac{m}{s^2}$

(b) $a_s=0.13\frac{m}{s^2}$

(c) $x_f=2.6m$

Explanation:

(a) According to Newton's second law, the acceleration of a body is directly proportional to the force exerted on it and inversely proportional to it's mass.

$a_s=\frac{F}{m_s}\\a_s=\frac{5.2N}{8.4kg}\\a_s=0.62\frac{m}{s^2}$

(b) According to Newton's third law, the force that the sled exerts on the girl is equal in magnitude but opposite in the direction of the force that the girl exerts on the sled:

$a_g=\frac{F}{m_g}\\a_g=\frac{5.2N}{40kg}\\a_g=0.13\frac{m}{s^2}$

(c) Using the kinematics equation:

$x_f=x_0+v_0t \pm \frac{at^2}{2}$

For the girl, we have $x_0=0$ and $v_0=0$ . So:

$x_f_g=\frac{a_gt^2}{2}(1)$

For the sled, we have $v_0=0$ . So:

$x_f_s=x_0_s-\frac{a_st^2}{2}(2)$

When they meet, the final positions are the same. So, equaling (1) and (2) and solving for t:

$x_0_s-\frac{a_st^2}{2}=\frac{a_st^2}{2}\\t^2(a_g+a_s)=2x_0_s\\t=\sqrt{\frac{2x_s_0}{a_g+a_s}}\\t=\sqrt{\frac{2(15m)}{0.13\frac{m}{s^2}+0.62\frac{m}{s^2}}}\\t=6.32s$

Now, we solve (1) for $x_f_g$

$x_f_g=\frac{0.13\frac{m}{s^2}(6.32s)^2}{2}\\x_f_g=2.6m\\x_f=2.6m$

5 0

3 years ago

What are gamma rays and what are its uses??<br>

sukhopar [10]

"Gamma rays" is the name that we call the shortest of all electromagnetic waves. They're shorter than radio waves, microwaves, infrared waves, heat waves, visible light waves, ultraviolet waves, and X-rays. They extend all the way down to waves that are as short as the distance across an atom.

Being so short, they carry lots of energy. They can penetrate many materials, and they can damage living cells and DNA. They're dangerous.

The sun puts out a lot of gamma radiation. The atmosphere (air) filters out a lot of it, otherwise there couldn't even be any life on Earth.

As soon as astronauts fly out of the atmosphere, they need a lot of shielding from gamma rays.

You know the precautions we take when we're around X-rays. The same precautions apply around gamma rays, only a lot more so.

It's only in the past several years that we've learned how to MAKE gamma rays without blowing things up. Also, how to control them, and how to use them for medical and industrial applications.

3 0

2 years ago

HELP - What are some good topics to add to a Pipeline presentation? So yes, a presentation about pipelines, like for example in

In-s [12.5K]

Talk about the jobs that pipeline construction provides to citizens and wether or not it’s effects are ultimately eco-friendly

6 0

3 years ago