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4 years ago

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James Stewart, 2002 Motocross/Supercross Rookie of the Year, is leading a race when he runs out of gas near the finish line. He

is moving at 16 m/s when he enters a section of the course covered with sand where the effective coefficient of friction is 0.90.
Determine the distance the racer travels before coming to a stop.

1 answer:

Elza [17]4 years ago

6 0

Explanation:

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The smallest unit of charge is − 1.6 × 10 − 19 C, which is the charge in coulombs of a single electron. Robert Millikan was able

vovangra [49]

Answer:

$-8.0 \times 10 ^{-19 }\ C,\ -3.2 \times 10 ^{-19 }\ C, -4.8 \times 10 ^{-19 }\ C$

Explanation:

<u>Charge of an Electron</u>

Since Robert Millikan determined the charge of a single electron is

$q_e=-1.6\cdot 10^{-19}\ C$

Every possible charged particle must have a charge that is an exact multiple of that elemental charge. For example, if a particle has 5 electrons in excess, thus its charge is $5\times -1.6\cdot 10^{-19}\ C=-8 \cdot 10^{-19}\ C$

Let's test the possible charges listed in the question:

$-8.0 \times 10 ^{-19 }$ . We have just found it's a possible charge of a particle

$-3.2 \times 10 ^{-19 }$ . Since 3.2 is an exact multiple of 1.6, this is also a possible charge of the oil droplets

$-1.2 \times 10 ^{-19 }$ this is not a possible charge for an oil droplet since it's smaller than the charge of the electron, the smallest unit of charge

$-5.6 \times 10 ^{-19 },\ -9.4 \times 10 ^{-19 }$ cannot be a possible charge for an oil droplet because they are not exact multiples of 1.6

Finally, the charge $-4.8 \times 10 ^{-19 }\ C$ is four times the charge of the electron, so it is a possible value for the charge of an oil droplet

Summarizing, the following are the possible values for the charge of an oil droplet:

$-8.0 \times 10 ^{-19 }\ C,\ -3.2 \times 10 ^{-19 }\ C, -4.8 \times 10 ^{-19 }\ C$

5 0

3 years ago

Can somebody google or pls if u know the correct answer help!!!

love history [14]

Answer:

Nutrients

Oxygen

Water

Stable body temperature

Atmospheric pressure

Explanation:

8 0

3 years ago

Read 2 more answers

imagine that 501 people are present in a movie theater of volume 8.00 x10^3 that is sealed shut so no air can escape. Each perso

Semenov [28]

Answer:

The temperature of air will increase by $\Delta T=41044.967\ K$

Explanation:

Given:

no. of person in a theater, $n=501$

volume of air in the theater, $V=8\times 10^3\ m^3$

rate of heat given off by each person, $P=110\ J.s^{-1}$

duration of movie, $t=2\ hr=7200\ s$

initial pressure in the theater, $p_i=1.01\times 10^5\ Pa$

initial temperature in the theater, $T_i=20+273=293\ K$

specific heat capacity of air at the given conditions, $c=1.0061\ J.kg^{-1}.K^{-1}$

<u>The total quantity of heat released by the total people in the theater during the movie:</u>

$Q=n.P.t$

$Q=501\times 110\times 7200$

$Q=396792000\ J$

<u>Form the relation of heat capacity:</u>

$Q=m.c.\Delta T$

∵ $p_i.V=m.R.T$

$Q=(\frac{p_i.V}{R.T}) \times c\times (T_f-T_i)$

$396792000=(\frac{1.01\times 10^5\times 8\times 10^3}{287\times 293}) \times 1.0061\times (T_f-293)$

$T_f=41337.967\ K$

Change in temperature of air:

$\Delta T=41044.967\ K$

8 0

3 years ago

FrozenT [24]

I would think the answer is c.

5 0

4 years ago

Read 2 more answers

The objective lens and the eyepiece of a telescope are spaced 85 cm apart. If the eyepiece is123 D what is the total magnificati

dedylja [7]

Answer:

The magnification would be "103.55". A further explanation is given below.

Explanation:

The given values are:

Distance between lens and eyepiece,

L = 85 cm

Eyepiece is,

= 123 D

Now,

The refractive power of eye piece will be:

⇒ $\frac{1}{f_e}=123D$

$f_e=\frac{1}{123D}$

$f_e=0.813 \ cm$

The length of the telescope will be:

⇒ $L=f_0+f_e$

⇒ $f_0=L-f_e$

On substituting the values, we get

⇒ $=85-0.813$

⇒ $=84.187 \ cm$

Now,

The magnification of the telescope will be:

⇒ $M=\frac{f_0}{f_e}$

⇒ $=\frac{84.187}{0.813}$

⇒ $=103.55$

5 0

3 years ago