The energy needed to get a reaction started is the

A skier moving at 4.75 m/s encounters a long, rough, horizontal patch of snow having a coefficient of kinetic friction of 0.220

disa [49]

First we need to find the acceleration of the skier on the rough patch of snow.
We are only concerned with the horizontal direction, since the skier is moving in this direction, so we can neglect forces that do not act in this direction. So we have only one horizontal force acting on the skier: the frictional force, $\mu m g$ . For Newton's second law, the resultant of the forces acting on the skier must be equal to ma (mass per acceleration), so we can write:
$ma=-\mu m g$
Where the negative sign is due to the fact the friction is directed against the motion of the skier.
Simplifying and solving, we find the value of the acceleration:
$a=-(0.220)(9.81 m/s^2)=-2.16 m/s^2$

Now we can use the following relationship to find the distance covered by the skier before stopping, S:
$2aS=v_f^2-v_i^2$
where $v_f=0$ is the final speed of the skier and $v_i=4.75 m/s$ is the initial speed. Substituting numbers, we find:
$S=- \frac{v_i^2}{2a}=- \frac{(4.75 m/s)^2}{2(-2.16 m/s^2)}=5.23 m$

5 0

3 years ago

Starting from zero, the electric current takes 2 seconds to reach half its maximum possible value in an RL circuit with a resist

Leno4ka [110]

Answer:

time=4s

Explanation:

we know that in a RL circuit with a resistance R, an inductance L and a battery of emf E, the current (i) will vary in following fashion

$i(t)=\frac{E}{R}(1-e^\frac{-t}{\frac{L}{R}})$ , where $i$ max= $\frac{E}{R}$

Given that, at i(2)= $\frac{imax}{2} =\frac{E}{2R}$

⇒ $\frac{E}{2R}=\frac{E}{R}(1-e^\frac{-2}{\frac{L}{R}})$

⇒ $\frac{1}{2}=1-e^\frac{-2}{\frac{L}{R}}$

⇒ $\frac{1}{2}=e^\frac{-2}{\frac{L}{R}}$

Applying logarithm on both sides,

⇒ $log(\frac{1}{2})=\frac{-2}{\frac{L}{R}}$

⇒ $log(2)=\frac{2}{\frac{L}{R}}$

⇒ $\frac{L}{R}=\frac{2}{log2}$

Now substitute $i(t)=\frac{3}{4}imax=\frac{3E}{4R}$

⇒ $\frac{3E}{4R}=\frac{E}{R}(1-e^\frac{-t}{\frac{L}{R}})$

⇒ $\frac{3}{4}=1-e^\frac{-t}{\frac{L}{R}}$

⇒ $\frac{1}{4}=e^\frac{-t}{\frac{L}{R}}$

Applying logarithm on both sides,

⇒ $log(\frac{1}{4})=\frac{-t}{\frac{L}{R}}$

⇒ $log(4)=\frac{t}{\frac{L}{R}}$

⇒ $t=log4\frac{L}{R}$

now subs. $\frac{L}{R}=\frac{2}{log2}$

⇒ $t=log4\frac{2}{log2}$

also $log4=log2^{2}=2log2$

⇒ $t=2log2\frac{2}{log2}$

⇒ $t=4$

5 0

3 years ago

Please answer this. Science 7th grade.

nikdorinn [45]

Answer:

the answer would be 2

Explanation:

it would be 2 because if u look at the diagram the darkest arrow is pointsin towards earth and the moon and when the moon is infront of the sun it cause's an eclispe

6 0

3 years ago

Massive stars terminate in a brilliant explosion called a

BigorU [14]

super nova is named for such explosion

6 0

3 years ago

which planet detection method would work best for a massive planet that's close to its host star but doesn't pass between us and

Alona [7]

The Doppler method is most frequently employed to find extrasolar planets, however it works best for finding extremely large planets that are near to their host star.

<h3>Which planet is the hottest?</h3>

Venus is the hottest planet within our solar system, with average global temperatures hot enough even to melt lead due to the runaway greenhouse effect caused by its thick atmosphere, which traps heat. Venus is around 700°F (390°C) hotter that it would be in the absence of the greenhouse effect.

<h3>Is it 8 or 9 planets?</h3>

Our solar system is made up of ten planetary systems, 146 moons, countless comets, meteorites, and other space objects, as well as Pluto and several dwarf planets. In the system, there is just the Sun. The eight planets are Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, and finally Neptune.

To know more about planet visit:

brainly.com/question/15268075

#SPJ4

3 0

1 year ago