Ask question

Ask question

All categories

3 years ago

14

The time it takes for a dragster to cross the finish line is unknown. The dragster accelerates from rest at 26 m/s2 for a quarte

r-mile (402 m). What is the final velocity of the dragster?

1 answer:

exis [7]3 years ago

7 0

♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️

${v}^{2} - {v(0)}^{2} = 2.a.∆x$

$v = final \: \: velocity$

$v(0) = the \: \: velocity \: \: in \: \: the \: \: rest \: \: which \: \: is \: \: zero \\$

$a = acceleration$

$∆x = covered \: \: distance$

♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️

${v}^{2} - ({0})^{2} = 2 \times 26 \times 402$

$v = \sqrt{2 \times 26 \times 402}$

$v = \sqrt{20904}$

$v = 144.58 \: \: \frac{m}{s}$

Done...

♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️

You might be interested in

Where does most star formation occur in the milky way galaxy?

k0ka [10]

Answer:

Star formation occurs most rapidly in the spiral arms, where the density of interstellar matter is highest.

4 0

2 years ago

A street light is on top of a 8 foot pole. Joe,

zubka84 [21]

8/4 = y/y-x

8y - 8x = 4y

y = 2x

y = 2 x 4

y = 8

Hope this helps

5 0

3 years ago

Please help Need good grade

ivann1987 [24]

Answer:

The other angle is 120°.

Explanation:

Given that,

Angle = 60

Speed = 5.0

We need to calculate the range

Using formula of range

$R=\dfrac{v^2\sin(2\theta)}{g}$ ...(I)

The range for the other angle is

$R=\dfrac{v^2\sin(2(\alpha-\theta))}{g}$ ....(II)

Here, distance and speed are same

On comparing both range

$\dfrac{v^2\sin(2\theta)}{g}=\dfrac{v^2\sin(2(\alpha-\theta))}{g}$

$\sin(2\theta)=\sin(2\times(\alpha-\theta))$

$\sin120=\sin2(\alpha-60)$

$120=2\alpha-120$

$\alpha=\dfrac{120+120}{2}$

$\alpha=120^{\circ}$

Hence, The other angle is 120°

6 0

4 years ago

show the trajectory of a body projected with an initial velocity vat an angle of departure thita is a paraboler

Jobisdone [24]

Answer:

Let me look up a couple of things regarding this question.

Explanation:

Then I will get back to you.

5 0

3 years ago

A horizontal pipe of diameter 1.03m has a smooth constriction to a section of diameter 0.618 m. The density of oil flowing in th

vodka [1.7K]

Velocity of the oil in the pipe: 0.76 m/s, in the constricted section: 5.87 m/s

Explanation:

We can solve this problem by using Bernoulli's equation:

$p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho v_2^2$ (1)

where

$p_1 = 7340 N/m^2$ is the pressure in the pipe

$p_2 = 5505 N/m^2$ is the pressure in the constricted section

$\rho = 821 kg/m^3$ is the density of the oil

$v_1$ is the velocity of the oil in the pipe

$v_2$ is the velocity of the oil in the constricted section

Also, according to the continuity equation,

$A_1 v_1 = A_2 v_2$

where

$A_1 = \pi r_1^2$ is the cross-sectional area of the pipe, with

$r_1 = \frac{1.03}{2}=0.515 m$ is the radius

$A_2 = \pi r_2^2$ is the cross-sectional area of the constricted section, with

$r_2=\frac{0.618}{2}=0.309 m$ is the radius

So the equation becomes

$r_1^2 v_1 = r_2^2 v_2$

So we can write

$v_2=\frac{r_1^2}{r_2^2}v_1$

Substituting into eq.(1),

$p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho (\frac{r_1^2}{r_2^2}v_1)^2$

And solving the equation for $v_1$ :

$p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho \frac{r_1^4}{r_2^4}v_1^2\\v_1=\sqrt{\frac{p_2-p_1}{\frac{1}{2}\rho-\frac{1}{2}\rho \frac{r_1^4}{r_2^4}}}=0.76 m/s$

And the velocity in the constricted section is

$v_2=\frac{r_1^2}{r_2^2}v_1=5.87 m/s$

Learn more about flow rate:

brainly.com/question/9805263

#LearnwithBrainly

7 0

3 years ago