Ask question

Ask question

All categories

3 years ago

11

Two drag cars race. They line up at the starting line at rest. The winning car accelerates at a constant rate a and reaches the

finish line with a final velocity v. The losing car accelerates at a constant rate a/4. How far had the losing car traveled, d, when the winning car crossed the finish line?

1 answer:

Mila [183]3 years ago

5 0

Answer: $d=\frac{v^2}{8a}$

Explanation:

Given

winning car accelerates with a and its final velocity is v

considering they both start from rest

time taken by winning car is

v=u+at

where u=initial velocity

a=acceleration

t=time

$v=at$

$t=\frac{v}{a}$

Now loosing car is accelerating with $\frac{a}{4}$

Distance traveled by loosing car in time t

$s_1=ut+\frac{at^2}{2\cdot 4}$

$s_1=0+\frac{a}{8}\times (\frac{v}{a})^2$

$s_1=\frac{v^2}{8a}$

Thus distance d traveled by loosing car is given by $d=\frac{v^2}{8a}$

You might be interested in

A normal mode of a closed system is an oscillation of the system in which all parts oscillate at a single frequency. In general

valentina_108 [34]

Answer:

Explanation:

(A)

The string has set of normal modes and the string is oscillating in one of its modes.

The resonant frequencies of a physical object depend on its material, structure and boundary conditions.

The free motion described by the normal modes take place at the fixed frequencies and these frequencies is called resonant frequencies.

Given below are the incorrect options about the wave in the string.

• The wave is travelling in the +x direction

• The wave is travelling in the -x direction

• The wave will satisfy the given boundary conditions for any arbitrary wavelength $\lambda_i$

• The wave does not satisfy the boundary conditions $y_i(0;t)=0
$

Here, the string of length L held fixed at both ends, located at x=0 and x=L

The key constraint with normal modes is that there are two spatial boundary conditions, $y(0,1)=0
$

and $y(L,t)=0$

.The spring is fixed at its two ends.

The correct options about the wave in the string is

• The wavelength $\lambda_i$ can have only certain specific values if the boundary conditions are to be satisfied.

(B)

The key factors producing the normal mode is that there are two spatial boundary conditions, $y_i(0;t)=0$ and $y_i(L;t)=0$ , that are satisfied only for particular value of $\lambda_i$ .

Given below are the incorrect options about the wave in the string.

• $A_i$ must be chosen so that the wave fits exactly o the string.

• Any one of $A_i$ or $\lambda_i$ or $f_i$ can be chosen to make the solution a normal mode.

Hence, the correct option is that the system can resonate at only certain resonance frequencies $f_i$ and the wavelength $\lambda_i$ must be such that $y_i(0;t) = y_i(L;t)=0
$

(C)

Expression for the wavelength of the various normal modes for a string is,

$\lambda_n=\frac{2L}{n}$ (1)

When $n=1$ , this is the longest wavelength mode.

Substitute 1 for n in equation (1).

$\lambda_n=\frac{2L}{1}\\\\2L$

When $n=2$ , this is the second longest wavelength mode.

Substitute 2 for n in equation (1).

$\lambda_n=\frac{2L}{2}\\\\L$

When $n=3$ , this is the third longest wavelength mode.

Substitute 3 for n in equation (1).

$\lambda_n=\frac{2L}{3}$

Therefore, the three longest wavelengths are $2L$ , $L$ and $\frac{2L}{3}$ .

(D)

Expression for the frequency of the various normal modes for a string is,

$f_n=\frac{v}{\lambda_n}$

For the case of frequency of the $i^{th}$ normal mode the above equation becomes.

$f_i=\frac{v}{\lambda_i}$

Here, $f_i$ is the frequency of the $i^{th}$ normal mode, v is wave speed, and $\lambda_i$ is the wavelength of $i^{th}$ normal mode.

Therefore, the frequency of $i^{th}$ normal mode is $f_i=\frac{v}{\lambda_i}$

.

6 0

3 years ago

A specific amount of water would have the same mass even if you turned it from a liquid to a solid or a gas is an example of

Elena L [17]

Answer:

b.Law of conservation of mass

7 0

3 years ago

Read 2 more answers

The direction of an electric field is always in the direction _______________ would naturally move.

Masja [62]

Answer:

Im not really sure but Id say weather .

Explanation:

6 0

3 years ago

One mole of ideal gas is slowly compressed to one-third of its original volume. in this compression, the work done on the gas ha

beks73 [17]

A boiling pot of water (the water travels in a current throughout the pot), a hot air balloon (hot air rises, making the balloon rise) , and cup of a steaming, hot liquid (hot air rises, creating steam) are all situations where convection occurs.
Read more on Brainly.com - brainly.com/question/1581851#readmore

4 0

3 years ago

In the flow past a compression corner, the upstream Mach number and pressure are 3.5 and 1 atm, respectively. Downstream of the

yulyashka [42]

Answer:

$\theta=23.7^{\circ}$

Explanation:

The ratio of pressure 2 to 1 us 5.48/1= 5.48 rounded off as 5.5.

Referring to table A.2 of modern compressible flow then $M_{\beta_1}=2.2$

Also

$M_{\beta_1}=M_1 sin \beta$ and making $sin\beta$ the subject of the formula then

$sin\beta=\frac {2.2}{3.5}\\\beta=38.94^{\circ}$

Making reference to $\theta-\beta-M$ diagram then

$\theta=23.7^{\circ}$

4 0

3 years ago