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lorasvet [3.4K]

3 years ago

12

A water slide of length l has a vertical drop of h. Abby's mass is m. An average friction force of magnitude f opposes her motio

n. She starts down the slide at initial speed v_i Use work-energy ideas to develop an expression for her speed at the bottom of the slide. Then evaluate your result using unit analysis and limiting case analysis.

1 answer:

OLEGan [10]3 years ago

8 0

Answer:

The solution is in the attached file below

Explanation:

You might be interested in

The greatest speed with which an athlete can jump vertically is around 5 m/sec. Determine the speed at which Earth would move do

katrin2010 [14]

Answer:

Approximately $2.0 \times 10^{-23}\; \rm m \cdot s^{-1}$ if that athlete jumped up at $1.8\; \rm m \cdot s^{-1}$ . (Assuming that $g = 9.81\; \rm m\cdot s^{-1}$ .)

Explanation:

The momentum $p$ of an object is the product of its mass $m$ and its velocity $v$ . That is: $p = m \cdot v$ .

Before the jump, the speed of the athlete and the earth would be zero (relative to each other.) That is: $v(\text{athlete, before}) = 0$ and $v(\text{earth, before}) = 0$ . Therefore:

$\begin{aligned}& p(\text{athlete, before}) = 0\end{aligned}$ and $p(\text{earth, before}) = 0$ .

Assume that there is no force from outside of the earth (and the athlete) acting on the two. Momentum should be conserved at the instant that the athlete jumped up from the earth.

Before the jump, the sum of the momentum of the athlete and the earth was zero. Because momentum is conserved, the sum of the momentum of the two objects after the jump should also be zero. That is:

$\begin{aligned}& p(\text{athlete, after}) + p(\text{earth, after}) \\ & =p(\text{athlete, before}) + p(\text{earth, before}) \\ &= 0\end{aligned}$ .

Therefore:

$p(\text{athelete, after}) = - p(\text{earth, after})$ .

$\begin{aligned}& m(\text{athlete}) \cdot v(\text{athelete, after}) \\ &= - m(\text{earth}) \cdot v(\text{earth, after})\end{aligned}$ .

Rewrite this equation to find an expression for $v(\text{earth, after})$ , the speed of the earth after the jump:

$\begin{aligned} &v(\text{earth, after}) \\ &= -\frac{m(\text{athlete}) \cdot v(\text{athlete, after})}{m(\text{earth})} \end{aligned}$ .

The mass of the athlete needs to be calculated from the weight of this athlete. Assume that the gravitational field strength is $g = 9.81\; \rm N \cdot kg^{-1}$ .

$\begin{aligned}& m(\text{athlete}) = \frac{664\; \rm N}{9.81\; \rm N \cdot kg^{-1}} \approx 67.686\; \rm N\end{aligned}$ .

Calculate $v(\text{earth, after})$ using $m(\text{earth})$ and $v(\text{athlete, after})$ values from the question:

$\begin{aligned} &v(\text{earth, after}) \\ &= -\frac{m(\text{athlete}) \cdot v(\text{athlete, after})}{m(\text{earth})} \\ &\approx -2.0 \times 10^{-23}\; \rm m \cdot s^{-1}\end{aligned}$ .

The negative sign suggests that the earth would move downwards after the jump. The speed of the motion would be approximately $2.0 \times 10^{-23}\; \rm m \cdot s^{-1}$ .

3 0

3 years ago

CH4 + 2O2 → CO2 + 2H2O The number in front of a compound in a balanced chemical equation is known as a(n) A) charge. B) coeffici

Tamiku [17]

The numbers in front of a compound is called a subscript. It shows how many atoms of that specific element it is next two that is involved in the reaction or is in that compound.

The answer is then D.

Coefficient is found behind the compounds, the charges are written above.

6 0

3 years ago

If April launched a water balloon directly upwards with a speed of 40 m/s, how high would it be after 6 seconds? Use 10 m/s

Volgvan

Answer:

495m

Explanation:

ANSWER

u

stone

=10m/s (initial velocity of stone)

t=11s

∴H=−ut+

2

1

gt

2

(H=height of baloon)

H=−10×11+

2

1

×10×121=605−110=495m

3 0

4 years ago

A potter's wheel is rotating around a vertical axis through its center at a frequency of 2.0 rev/srev/s . The wheel can be consi

vagabundo [1.1K]

Answer:1.7 rev/s

Explanation:

Given

Frequency of wheel $N_1=2\ rev/s$

angular speed $\omega_1=2\pi N_1=4\pi\ rad/s$

mass of wheel $m_1=4.5\ kg$

diameter of wheel $d_1=0.30\ m=30\ cm$

radius of wheel $r_1=\frac{d_1}{2}=\frac{30}{2}=15\ cm$

mass of clay $m_2=2.8\ kg$

the radius of the chunk of clay $r_2=8\ cm$

Moment of inertia of Wheel

$I_1=\dfrac{m_1r_1^2}{2}=\dfrac{4.5\times 15^2}{2}\ kg-cm^2$

Combined moment of inertia of wheel and clay chunk

$I_2=\dfrac{m_1r_1^2}{2}+\dfrac{m_2r_2^2}{2}=\dfrac{4.5\times 15^2}{2}+\dfrac{2.8\times 8^2}{2}\ kg-cm^2$

Conserving angular momentum

$\Rightarrow I_1\omega_1=I_2\omega_2\\\Rightarrow \dfrac{4.5\times 15^2}{2}\cdot 4\pi=(\dfrac{4.5\times 15^2}{2}+\dfrac{2.8\times 8^2}{2})\omega_2\\\\\Rightarrow \omega _2=\dfrac{4\pi }{1+\dfrac{2.8}{4.5}\times (\dfrac{8}{15})^2}=\dfrac{4\pi}{1+0.1769}=0.849\times 4\pi$

Common frequency of wheel and chunk of clay is

$\Rightarrow N_2=\dfrac{4\pi \times 0.849}{2\pi}=1.698\approx 1.7\ rev/s$

5 0

3 years ago

In order to measure current in a circuit, the multi-meter needs to be placed in __A__ , while when measuring voltage, the multi-

lakkis [162]

Answer:

A- series B- parallel

Explanation:

In order to measure current in a circuit, the multi-meter needs to be placed in series with the circuit while when measuring voltage, the multi-meter needs to be placed in parallel with the circuit.

It should be however noted that the same current flows in a series connected circuit and same voltage flows through loads connected in parallel. The ammeter is placed in series with the load to ensure that same value of currents is flowing in both the ammeter and loads(since same current flows in series connected circuit elements and all the amount of voltage must be made to appear on the load for the current to be measured accurately.

Voltmeter is connected in parallel to the load due to high value of current possessed by the voltmeter. The parallel connection will cause the current flowing through the voltmeter to reduce to zero so that it won't have effect (increase) on the amount of current initially on the resistor thereby measuring the exact amount of voltage on the load.

3 0

4 years ago