Ask question

Ask question

All categories

2 years ago

12

on a recent adventure trip, Anita Break went rock climbing. Anita was able to perform 1370 J of work in 100 seconds. Determine A

nita's power rating during this portion of the climb

1 answer:

Zanzabum2 years ago

5 0

      Power = (energy) / (time)

   =    (1370 joules) / (100 seconds)

   = 13.7 joules/second

   = 13.7 watts .

That's not an awful lot of power, especially for a strenuous activity like
rock-climbing. Shoot ! Even I could probably perform at that level.

Compare 13.7 watts to the light power coming out of a 20-watt night light.

   13.7 watts = 0.018 horsepower. (rounded)

You might be interested in

A rock is suspended from a string, and it accelerates downward. Which one of the following statements is correct? Group of answe

Tems11 [23]

Answer:

a. The magnitude of the tension in the string is greater than the magnitude of the weight of the rock.

Explanation:

During the motion of the rock while it is in downward motion we can say

$T - mg cos\theta = ma_c$

since it is performing circular motion so we will have its acceleration towards its center

$T = mgcos\theta + ma_c$

$a_c = \frac{v^2}{L}$

$T = mgcos\theta + \frac{mv^2}{L}$

So at the lowest point of the path we can say

$T = mg + \frac{mv^2}{L}$

so correct answer is

a. The magnitude of the tension in the string is greater than the magnitude of the weight of the rock.

4 0

2 years ago

A rocket is fired straight upward, starting from rest with an acceleration of 25.0 m/s^2. It runs out of fuel at the end of 4.00

svetoff [14.1K]

Answer:

a) 200 m

b) 100 m/s

c) 709 m

d) -118.2 m/s

e) 26.24 s

Explanation:

The rocket flies upward with constant acceleretion.

The equation for position under constant acceleration is:

Y(t) = Y0 + Vy0 * t + 1/2 * a * t^2

Y0 = 0

V0 = 0

Y(t) = 1/2 * 25 * t^2

Y(t) = 12.5 * t^2

And speed under constant acceleration:

Vy(t) = Vy0 + a * t

Vy(t) = 25 * t

It burns for 4 s and runs out of fuel

Y(4) = 12.5 * 4^2 = 200 m

V(4) = 25 * 4 = 100 m/s

Form t = 4 the rocket will coast, it will be in free fall, affected only by gravity

It will be under constant acceleration. These new equations will have different starting constants.

Y(t) = Y4 + Vy4 * (t - 4) + 1/2 * g * (t - 4)^2

Vy(t) = Vy4 + g * (t - 4)

When it reaches its highest point it will have a speed of zero.

0 = Vy4 + g * (t - 4)

0 = 100 - 9.81 * (t - 4)

100 = 9.81 * (t - 4)

t - 4 = 100 / 9.81

t = 10.2 + 4 = 14.2 s

At that moment it will have a height of:

Y(14.2) = 200 + 100 * (14.2 - 4) - 1/2 * 9.81 * (14.2 - 4)^2 = 709 m

The rocket will fall and hit the ground:

Y(t) = 0 = 200 + 100 * (t - 4) - 1/2 * 9.81 * (t - 4)^2

0 = 200 + 100 * t - 400 - 4.9 * (t^2 - 8 * t +16)

0 = -4.9 * t^2 + 139.2 * t -278.4

Solving this equation electronically:

t = 26.24 s

At that time the speed will be:

Vy(t) = 100 - 9.81 * (26.24 - 4) = -118.2 m/s

6 0

2 years ago

Reactive elements such as alkali metals and halogens are found in nature only as

bonufazy [111]

Compounds

Explanation:

Reactive elements such as alkali metals and halogens are found in nature only as compounds. Such elements are too unstable to remain as stable atoms, therefore they readily combine and form compounds.

Compounds are formed when two atoms combines together to share electrons.
They either lose, gain, or share electrons between themselves.
In the end, they end up becoming more stable.
This is the reason why atoms combine.
Unstable elements are very reactive especially alkali metals and halogens.
On their own, they are unstable and prefers to bond with other atoms in order to gain a measure of stability.
This is why they are found in combined state in nature.

Learn more:

Compound brainly.com/question/10585691

Noble gases brainly.com/question/1781595

#learnwithBrainly

4 0

3 years ago

A charge is divided q1 and (q-q1)what will be the ratio of q/q1 so that force between the two parts placed at a given distance i

Arturiano [62]

Answer:

$q / q_{1} = 2$ , assuming that $q_{1}$ and $(q - q_{1})$ are point charges.

Explanation:

Let $k$ denote the coulomb constant. Let $r$ denote the distance between the two point charges. In this question, neither $k$ and $r$ depend on the value of $q_{1}$ .

By Coulomb's Law, the magnitude of electrostatic force between $q_{1}$ and $(q - q_{1})$ would be:

$\begin{aligned}F &= \frac{k\, q_{1}\, (q - q_{1})}{r^{2}} \\ &= \frac{k}{r^{2}}\, (q\, q_{1} - {q_{1}}^{2})\end{aligned}$ .

Find the first and second derivative of $F$ with respect to $q_{1}$ . (Note that $0 < q_{1} < q$ .)

First derivative:

$\begin{aligned}\frac{d}{d q_{1}}[F] &= \frac{d}{d q_{1}} \left[\frac{k}{r^{2}}\, (q\, q_{1} - {q_{1}}^{2})\right] \\ &= \frac{k}{r^{2}}\, \left[\frac{d}{d q_{1}} [q\, q_{1}] - \frac{d}{d q_{1}}[{q_{1}}^{2}]\right]\\ &= \frac{k}{r^{2}}\, (q - 2\, q_{1})\end{aligned}$ .

Second derivative:

$\begin{aligned}\frac{d^{2}}{{d q_{1}}^{2}}[F] &= \frac{d}{d q_{1}} \left[\frac{k}{r^{2}}\, (q - 2\, q_{1})\right] \\ &= \frac{(-2)\, k}{r^{2}}\end{aligned}$ .

The value of the coulomb constant $k$ is greater than $0$ . Thus, the value of the second derivative of $F$ with respect to $q_{1}$ would be negative for all real $r$ . $F\!$ would be convex over all $q_{1}$ .

By the convexity of $\! F$ with respect to $\! q_{1} \!$ , there would be a unique $q_{1}$ that globally maximizes $F$ . The first derivative of $F\!$ with respect to $q_{1}\!$ should be $0$ for that particular $\! q_{1}$ . In other words:

$\displaystyle \frac{k}{r^{2}}\, (q - 2\, q_{1}) = 0$ <em>.</em>

$2\, q_{1} = q$ .

$q_{1} = q / 2$ .

In other words, the force between the two point charges would be maximized when the charge is evenly split:

$\begin{aligned} \frac{q}{q_{1}} &= \frac{q}{q / 2} = 2\end{aligned}$ .

3 0

2 years ago

a 5.0 kg ball is dropped from a 2.5 m high window. what is the velocity of the ball just before it hits the ground?

julsineya [31]

Answer:

Approximately $7.0\; \rm m \cdot s^{-1}$ .

Assumption: the ball dropped with no initial velocity, and that the air resistance on this ball is negligible.

Explanation:

Assume the air resistance on the ball is negligible. Because of gravity, the ball should accelerate downwards at a constant $g = 9.81 \; \rm m \cdot s^{-2}$ near the surface of the earth.

For an object that is accelerating constantly,

$v^2 - u^2 = 2\, a \, x$ ,

where

$u$ is the initial velocity of the object,
$v$ is the final velocity of the object.
$a$ is its acceleration, and
$x$ is its displacement.

In this case, $x$ is the same as the change in the ball's height: $x = 2.5\; \rm m$ . By assumption, this ball was dropped with no initial velocity. As a result, $u = 0$ . Since the ball is accelerating due to gravity, $a = 9.81\; \rm m \cdot s^{-2}$ .

$v^2 - u^2 = 2\, g \cdot h$ .

In this case, $v$ would be the velocity of the ball just before it hits the ground. Solve for

$v^2 = 2\, a\, x + u^2$ .

$\begin{aligned}v &= \sqrt{2\, a\, x + u^2} \\ &= \sqrt{2\times 9.81 \times 2.5 + 0} \\ &\approx 7.0\; \rm m\cdot s^{-1}\end{aligned}$ .

3 0

2 years ago