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2 years ago

"Younger teens often brag about a lack of sleep, but that usually stops when they get older. Why do you think that is?"

1 answer:

3 0

as we grow up we kinda experience some age staging so a teen got a lotta stuff going on they wanna be social active they be working on their popularity. so i guess thats it

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A rock is rolling down a hill. At position 1, its velocity is 2.0 m/s. Twelve seconds later, as it passes position 2, its veloci

alukav5142 [94]

As this happens over twelve seconds, you would take the total difference in velocities and divide it by twelve to find the change per second

44.0 m/s - 2.0 m/s = 42.0 m/s

42.0 m/s / 12 s = 3.5 m/s2

the acceleration of the rock would be 3.5 m/s2

4 0

3 years ago

Read 2 more answers

Starting from rest, a 2.1x10^-4 kg flea springs straight upward. While the flea is pushing off from the ground, the ground exert

Rainbow [258]

Answer:

a) The flea's speed when it leaves the ground is $v=1.88m/s$

b) The flea move $18cm$ upward while it is pushing off

Explanation:

Knwons

Mass $m=2.1\times 10^{-4} kg$ , Work $W=3.7\times 10^{-4} J$ and Force $F=0.41N$

a) Here we are going to use $W=\frac{1}{2} mv^{2}$ , so $v=\sqrt{\frac{2W}{m} }= \sqrt{\frac{2(3.7\times 10^{-4} J)}{2.1\times 10^{-4} kg} }=1.88m/s$

a) Here we are going to use $W=mgh$ , so $h=\frac{W}{mg}= \frac{3.7\times 10^{-4} J}{(2.1\times 10^{-4} kg)(9.8m/s^{2})} =0,1797m$ or $18cm$ approx.

7 0

3 years ago

If the speed of sound in air is 343 m/s and the wavelength of this note is 2.62 m, what is the frequency of this C3 note? (Round

snow_lady [41]

Answer:

The frequency of of the note = 131 Hz.

Explanation:

Frequency: Frequency can be defined as the number of complete oscillation completed by a wave in one seconds. The S.I unit of frequency is Hertz ( Hz)

v = λf ............................ Equation 1

Making f the subject of the equation,

f = v/λ .......................... Equation 2

Where v = Speed, λ = wavelength, f = frequency

Given: v = 343 m/s, λ = 2.62 m.

Substituting these values into equation 2

f = 343/2.62

f = 131 Hz

Thus the frequency of of the note = 131 Hz.

8 0

2 years ago

Ballon volume of 3200ml of xenon gas is at a gauge pressure of 122kPa and a temperature of 27c. What is the volume when the ball

N76 [4]

Given:

The initial volume of the gas, V₁=3200 ml=3.2×10⁻³ m³

The initial pressure of the gas, P₁=122 kPa

The initial temperature of the gas, T₁=27 °C=300 K

The final temperature, T₂=65 °C=338 K

The final pressure, P₂=112 kPa

To find:

The final volume of xenon gas.

Explanation:

From the combined gas law,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

Where V₂ is the volume after it is heated.

On rearranging the above equation,

$V_2=\frac{T_2P_1V_1}{T_1P_2}$

On substituting the known values,

$\begin{gathered} V_2=\frac{338\times112\times10^3\times3.2\times10^{-3}}{300\times112\times10^3} \\ =3.61\text{ m}^3 \end{gathered}$

Final answer:

The volume of the balloon when it is heated is 3.61 m³

4 0

1 year ago

How many revolutions per minute would a 23 m -diameter Ferris wheel need to make for the passengers to feel "weightless" at the

kirza4 [7]

Answer:

Approximately $6.2\; {\rm rpm}$ , assuming that the gravitational field strength is $g = 9.81\; {\rm m\cdot s^{-2}}$ .

Explanation:

Let $\omega$ denote the required angular velocity of this Ferris wheel. Let $m$ denote the mass of a particular passenger on this Ferris wheel.

At the topmost point of the Ferris wheel, there would be at most two forces acting on this passenger:

Weight of the passenger (downwards), $m\, g$ , and possibly
Normal force $F_\text{normal}$ that the Ferris wheel exerts on this passenger (upwards.)

This passenger would feel "weightless" if the normal force on them is $0$ - that is, $F_\text{normal} = 0$ .

The net force on this passenger is $(m\, g - F_\text{normal})$ . Hence, when $F_\text{normal} = 0$ , the net force on this passenger would be equal to $m\, g$ .

Passengers on this Ferris wheel are in a centripetal motion of angular velocity $\omega$ around a circle of radius $r$ . Thus, the centripetal acceleration of these passengers would be $a = \omega^{2}\, r$ . The net force on a passenger of mass $m$ would be $m\, a = m\, \omega^{2}\, r$ .

Notice that $m\, \omega^{2} \, r = (\text{Net Force}) = m\, g$ . Solve this equation for $\omega$ , the angular speed of this Ferris wheel. Since $g = 9.81\; {\rm m\cdot s^{-2}}$ and $r = 23\; {\rm m}$ :

$\begin{aligned} \omega^{2} = \frac{g}{r}\end{aligned}$ .

$\begin{aligned} \omega &= \sqrt{\frac{g}{r}} \\ &= \sqrt{\frac{9.81\; {\rm m \cdot s^{-2}}}{23\; {\rm m}}} \\ &\approx 0.653\; {\rm rad \cdot s^{-1}} \end{aligned}$ .

The question is asking for the angular velocity of this Ferris wheel in the unit ${\rm rpm}$ , where $1\; {\rm rpm} = (2\, \pi\; {\rm rad}) / (60\; {\rm s})$ . Apply unit conversion:

$\begin{aligned} \omega &\approx 0.653\; {\rm rad \cdot s^{-1}} \\ &= 0.653\; {\rm rad \cdot s^{-1}} \times \frac{1\; {\rm rpm}}{(2\, \pi\; {\rm rad}) / (60\; {\rm s})} \\ &= 0.653\; {\rm rad \cdot s^{-1} \times \frac{60\; {\rm s}}{2\, \pi\; {\rm rad}} \times 1\; {\rm rpm} \\ &\approx 6.2\; {\rm rpm} \end{aligned}$ .

3 0

1 year ago