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3 years ago

HELP

Physics

1 answer:

Mice21 [21]3 years ago

4 0

Answer:

Lesson 4.1: Introducing the Liquid Oxygen Problem. 102–103. Warm-Up ... chemists to understand why a machine used to make fuel for Titan's lander mission.

Explanation:

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4. A ball is thrown with an initial speed vi at an angle θi with the horizontal. The

GalinKa [24]

Disclaimer: I ended up finding what's asked for in the reverse order (e)-(a).

At time $t$ , the horizontal position $x$ and vertical position $y$ of the ball are given respectively by

$x = v_i \cos(\theta_i) t$

$y = v_i \sin(\theta_i) t - \dfrac g2 t^2$

and the horizontal velocity $v_x$ and vertical velocity $v_y$ are

$v_x = v_i \cos(\theta_i)$

$v_y = v_i \sin(\theta_i) - gt$

The ball reaches its maximum height with $v_y=0$ . At this point, the ball has zero vertical velocity. This happens when

$v_i \sin(\theta_i) - gt = 0 \implies t = \dfrac{v_i \sin(\theta_i)}g$

which means

$y = \dfrac R6 = v_i \sin(\theta_i) \times \dfrac{v_i \sin(\theta_i)}g - \dfrac g2 \left(\dfrac{v_i \sin(\theta_i)}g\right)^2 \\\\ \implies R = \dfrac{6{v_i}^2 \sin^2(\theta_i)}g - \dfrac{3{v_i}^2 \sin^2(\theta_i)}g \\\\ \implies R = \dfrac{3{v_i}^2 \sin^2(\theta_i)}g$

At the same time, the ball will have traveled half its horizontal range, so

$x = \dfrac R2 = v_i \cos(\theta_i) \times \dfrac{v_i \sin(\theta_i)}g \\\\ \implies R = \dfrac{2{v_i}^2 \cos(\theta_i) \sin(\theta_i)}g$

Solve for $v_i$ and $\theta_i$ :

$\dfrac{3{v_i}^2 \sin^2(\theta_i)}g = \dfrac{2{v_i}^2 \cos(\theta_i) \sin(\theta_i)}g \\\\ \implies 3 \sin^2(\theta_i) = 2 \cos(\theta_i) \sin(\theta_i) \\\\ \sin(\theta_i) (3\sin(\theta_i) - 2 \cos(\theta_i)) = 0$

Since $0^\circ$ , we cannot have $\sin(\theta_i)=0$ , so we're left with (e)

$3 \sin(\theta_i) - 2\cos(\theta_i) = 0 \\\\ \implies 3 \sin(\theta_i) = 2\cos(\theta_i) \\\\ \implies \tan(\theta_i) = \dfrac23 \\\\ \implies \boxed{\theta_i = \tan^{-1}\left(\dfrac23\right) \approx 33.7^\circ}$

Now,

$\cos\left(\tan^{-1}\left(\dfrac23\right)\right) = \dfrac3{\sqrt{13}}$

$\sin\left(\tan^{-1}\left(\dfrac23\right)\right) = \dfrac2{\sqrt{13}}$

so it follows that (d)

$R = \dfrac{2{v_i}^2 \times\frac3{\sqrt{13}} \times \frac2{\sqrt{13}}}g \\\\ \implies {v_i}^2 = \dfrac{13Rg}{12} \\\\ \implies \boxed{v_i = \sqrt{\dfrac{13Rg}{12}}}$

Knowing the initial speed and angle, the initial vertical component of velocity is (c)

$v_y = \sqrt{\dfrac{13Rg}{12}} \sin\left(\tan^{-1}\left(\dfrac23\right)\right) \\\\ \implies v_y = \sqrt{\dfrac{13Rg}{12}} \times \dfrac2{\sqrt{13}} \\\\ \implies \boxed{v_y = \sqrt{\dfrac{Rg}3}}$

We mentioned earlier that the vertical velocity is zero at maximum height, so the speed of the ball is entirely determined by the horizontal component. (b)

$v_x = \sqrt{\dfrac{13Rg}{12}} \times \dfrac3{\sqrt{13}} \\\\ \implies v_x = \dfrac{\sqrt{3Rg}}{2}$

Then with $v_y=0$ , the ball's speed $v$ is

$v = \sqrt{{v_x}^2 + {v_y}^2} \\\\ \implies v = v_x \\\\ \implies \boxed{v = \dfrac{\sqrt{3Rg}}2}$

Finally, in the work leading up to part (e), we showed the time to maximum height is

$t = \dfrac{v_i \sin(\theta_i)}g$

but this is just half the total time the ball spends in the air. The total airtime is then

$2t = \dfrac{2 \times \sqrt{\frac{13Rg}{12}} \times \frac2{\sqrt{13}}}g \\\\ \implies 2t = 2\sqrt{\dfrac R{3g}}$

and the ball is in the air over the interval (a)

$\boxed{0 < t < 2\sqrt{\frac R{3g}}}$

7 0

2 years ago

A go cart with a mass of 60 kg is moving at a rate of 10 m/s. How much Kinetic Energy does the go cart have?

maria [59]

Answer:

Explanation:

The kinetic energy of an object can be found by using the formula

$k = \frac{1}{2} m {v}^{2} \\$

m is the mass in kg

v is the velocity in m/s

From the question

m = 60 kg

v = 10 m/s

We have

$k = \frac{1}{2} \times 60 \times {10}^{2} \\ = 30 \times 100 \\ = 3000$

We have the final answer as

Hope this helps you

6 0

3 years ago

Read 2 more answers

Which researcher first realized that certain psychic secretions pointed to a simple but important form of learning

AnnZ [28]

Ivan Pavlov was the researcher that first realized that certain psychic secretions pointed to a simple but important form of learning.

The way he figured this out was when dogs paired certain things with food, and because of this, the dogs started salivating.

8 0

4 years ago

HELP WILL MEDAL !!! if you give a inpproper statement you will be repoted

ehidna [41]

1: Weather has a somewhat unpredictable pattern.
2: They can misinterpret the pattern of weather that looks like snow.
3: If they predict it early then the weather can change making there interpretation wrong.

Hope that helps a little bit. Please don’t report me I tried.

8 0

3 years ago

A ball is thrown horizontally from the top of a building 54 m high. The ball strikes the ground at a point 35 m horizontally awa

Ivanshal [37]

Answer:

V=34.2 m/s

Explanation:

Given that

Height , h= 54 m

Horizontal distance , x = 35 m

Given that , the ball is thrown horizontally , therefore the initial vertical velocity will be zero.

In vertical direction :