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3 years ago

14

A rock is launched vertically upward by a slingshot. The elastic band of the slingshot accelerates the rock from rest to 40 m/s

in a distance of 10 cm. How long will the rock take to return to its release point (time of flight)? Neglect air resistance.

1 answer:

eimsori [14]3 years ago

5 0

Answer:

0.016 s

Explanation:

Initial velocity, u = 0

use the third equation of motion to find the acceleration of the rock.

$2as=v^2-u^2\\a = \frac{40^2-0}{2\times0.1} =8000 m/s^2$

Net displacement in the vertical direction would be zero. y = 0

Use second equation:

$s=ut+\frac{1}{2}at^2\\0=0+\frac{1}{2}(8000)t^2\\t= 0.016 s$

Thus, the rock will return to its initial release point in 0.016 s.

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Answer:

A) d. (1/4)(2000m/s) = 500 m/s

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Explanation:

A)

The translational kinetic energy of a gas molecule is given as:

K.E = (3/2)KT

where,

K = Boltzman's Constant = 1.38 x 1^-23 J/K

T = Absolute Temperature

but,

K.E = (1/2) mv²

where,

v = root mean square velocity

m = mass of one mole of a gas

Comparing both equations:

(3/2)KT = (1/2) mv²

v = √(3KT)/m _____ eqn (1)

<u>FOR HYDROGEN:</u>

v = √(3KT)/m = 2000 m/s _____ eqn (2)

<u>FOR OXYGEN:</u>

velocity of oxygen = √(3KT)/(mass of oxygen)

Here,

mass of 1 mole of oxygen = 16 m

velocity of oxygen = √(3KT)/(16 m)

velocity of oxygen = (1/4) √(3KT)/m

using eqn (2)

<u>velocity of oxygen = (1/4)(2000 m/s) = 500 m/s</u>

B)

K.E = (3/2)KT

Since, the temperature is constant for both gases and K is also a constant. Therefore, the K.E of both the gases will remain same.

K.E of Oxygen = K.E of Hydrogen

<u>K.E of Oxygen = 4000 J</u>

C)

using eqn (2)

At, T = 50°C = 323 k

v = √(3KT)/m = 2000 m/s

m = 3(1.38^-23 J/k)(323 k)/(2000 m/s)²

m = 3.343 x 10^-27 kg

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Answer:

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Explanation:

Anterior or ventral - front (example, the kneecap is located on the anterior side of the leg). Posterior or dorsal - back (example, the shoulder blades are located on the posterior side of the body). Medial - toward the midline of the body (example, the middle toe is located at the medial side of the foot).

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6. State whether each of the following is an example of conduction, convection, and/or radiation. Explain if you think there is

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Points A (-5,6), B (2,-2), and C (-6,-3) are placed in three different quadrants of a Cartesian coordinate system. Convert each

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Answer: A ( $\sqrt{61}$ ,309.8°)

B (2 $\sqrt{2}$ , 315°)

C ( $3\sqrt{5}$ , 26.56°)

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For point A:

$r=\sqrt{(-5)^{2}+6^{2}}$

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$\theta=tan^{-1}(\frac{6}{-5})$

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Point A is in the II quadrant, so we substract the angle for 360° since it is in degrees:

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For point B:

$r=\sqrt{2^{2}+(-2)^{2}}$

$r=\sqrt{8}$

$r=2\sqrt{2}$

$\theta=tan^{-1}(\frac{-2}{2} )$

$\theta=tan^{-1}(1)$

$\theta=-45$ °

Point B is in IV quadrant, so:

$\theta=360-45$

$\theta=$ 315°

Polar coordinates for point B is ( $2\sqrt{2}$ , 315°)

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$r=\sqrt{(-6)^{2}+(-3)^{2}}$

$r=\sqrt{45}$

$r=3\sqrt{5}$

$\theta=tan^{-1}(\frac{-3}{-6} )$

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$\theta=$ 26.56°

Polar coordinates for point C is ( $3\sqrt{5}$ , 26.56°)

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