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Fantom [35]
3 years ago
5

Alex climbs to the top of a tall tree while his friend Gary waits on the ground below. Alex throws down a ball at 8 m/s from 50

m above the ground at the same time Gary throws a ball up. At what speed must Gary throw a ball up in order for the two balls to cross paths 25 m above the ground? The starting height of the ball thrown upward is 1.5 m above the ground. Ignore the effects of air resistance. whats the answer in m/s?
Physics
1 answer:
BaLLatris [955]3 years ago
5 0

Answer:22.62 m/s

Explanation:

Given

two balls are separated by a distance of 50 m

Alex throws  the ball from a height of 50 m with a velocity of 8 m/s and Gary launches a ball with some velocity  exactly at the same time.

ball  from ground travels a distance of 25 m in t sec

For Person on tree  

25=ut+\frac{1}{2}gt^2

25=8t+\frac{1}{2}\times 9.81\times t^2--------1

For person at ground

23.5=ut-\frac{1}{2}gt^2---------2

Solve equation (1)

50=16t+9.81t^2

9.81t^2+16t-50=0

t=\frac{-16\pm\sqrt{256+4\times 50\times 9.81}}{2\times 9.81}=\frac{47.1-16}{19.62}=1.58 s

put the value of t in equation 2

23.5=u\times 1.58-\frac{9.81\times 1.58^2}{2}

u=\frac{35.744}{1.58}=22.62 m/s

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Thats a personal question

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It asks about you

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Which is another term for the free enterprise system
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Answer:

Another term for free enterprise system would be capitalism

6 0
3 years ago
In a flying ski jump, the skier acquires a speed of 110 km/h by racing down a steep hill and then lifts off into the air from a
matrenka [14]

Answer:

Approximately \displaystyle\rm \left[ \begin{array}{c}\rm191\; m\\\rm-191\; m\end{array}\right].

Explanation:

Consider this 45^{\circ} slope and the trajectory of the skier in a cartesian plane. Since the problem is asking for the displacement vector relative to the point of "lift off", let that particular point be the origin (0, 0).

Assume that the skier is running in the positive x-direction. The line that represents the slope shall point downwards at 45^{\circ} to the x-axis. Since this slope is connected to the ramp, it should also go through the origin. Based on these conditions, this line should be represented as y = -x.

Convert the initial speed of this diver to SI units:

\displaystyle v = \rm 110\; km\cdot h^{-1} = 110 \times \frac{1}{3.6} = 30.556\; m\cdot s^{-1}.

The question assumes that the skier is in a free-fall motion. In other words, the skier travels with a constant horizontal velocity and accelerates downwards at g (g \approx \rm -9.81\; m\cdot s^{-2} near the surface of the earth.) At t seconds after the skier goes beyond the edge of the ramp, the position of the skier will be:

  • x-coordinate: 30.556t meters (constant velocity;)
  • y-coordinate: \displaystyle -\frac{1}{2}g\cdot t^{2} = -\frac{9.81}{2}\cdot t^{2} meters (constant acceleration with an initial vertical velocity of zero.)

To eliminate t from this expression, solve the equation between t and x for t. That is: express t as a function of x.

x = 30.556\;t\implies \displaystyle t = \frac{x}{30.556}.

Replace the t in the equation of y with this expression:

\begin{aligned} y = &-\frac{9.81}{2}\cdot t^{2}\\ &= -\frac{9.81}{2} \cdot \left(\frac{x}{30.556}\right)^{2}\\&= -0.0052535\;x^{2}\end{aligned}.

Plot the two functions:

  • y = -x,
  • \displaystyle y= -0.0052535\;x^{2},

and look for their intersection. Refer to the diagram attached.

Alternatively, equate the two expressions of y (right-hand side of the equation, the part where y is expressed as a function of x.)

-0.0052535\;x^{2} = -x,

\implies x = 190.35.

The value of y can be found by evaluating either equation at this particular x-value: x = 190.35.

y = -190.35.

The position vector of a point (x, y) on a cartesian plane is \displaystyle \left[\begin{array}{l}x \\ y\end{array}\right]. The coordinates of this skier is approximately (190.35, -190.35). The position vector of this skier will be \displaystyle\rm \left[ \begin{array}{c}\rm191\\\rm-191\end{array}\right]. Keep in mind that both numbers in this vectors are in meters.

4 0
3 years ago
A suitcase has a width of 55 cm. Which unit conversation fraction should you use to find the width in inches?
Crank

Answer:

2,54 cm are equal to 1 inch

Explanation:

Doing the conversion:

55[cm]*\frac{1[inch]}{2,54[cm]} =21,65[inch]

6 0
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A bodybuilder deadlifts a 215 kg weight to a height of 0.90 m above the ground. If he deadlifts this weight 10 times in a span o
wariber [46]

A bodybuilder deadlifts 215 kg to a height of 0.90 m. If he deadlifts this weight 10 times in 45 s, the power exerted is 421 W (b.)

<h3>What is power?</h3>

In physics, power (P) is the work (W) done over a period of time.

  • Step 1. Calculate the work done by the bodybuilder each time.

The bodybuilder lifts a 215 kg (m) weight to a height of 0.90 m (h). Being the gravity (g) of 9.81 m/s², we can calculate the work done in each lift using the following expression.

W = m × g × h = 215 kg × 9.81 m/s² × 0.90 m = 1.9 × 10³ N

  • Step 2. Calculate the work done by the bodybuilder over 10 times.

W = 10 × 1.9 × 10³ N = 1.9 × 10⁴ N

  • Step 3. Calculate the power exerted by the bodybuilder.

The bodybuilder does a work of 1.9 × 10⁴ N in a 45-s span.

P = 1.9 × 10⁴ N/45 s = 421 W

A bodybuilder deadlifts 215 kg to a height of 0.90 m. If he deadlifts this weight 10 times in 45 s, the power exerted is 421 W (b.)

Learn more about power here: brainly.com/question/911620

#SPJ1

4 0
1 year ago
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