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2 years ago

Give ten (10) words related to health on the grid. Write your answer on a separate sheet.

Physics

1 answer:

Vesna [10]2 years ago

7 0

Answer:

Answer

Explanation:

Answer:

1. Fitness 6. Health

2. Medicine 7. Diet

3. Diseases 8. Treatment

4. Exercises 9. Activities

5. Nutrition 10. Workouts

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A ball is dropped from rest from the top of a building of height h. At the same instant, a second ball is projected vertically u

uranmaximum [27]

Answer:

a) $t = \sqrt{\frac{h}{2g}}$

b) Ball 1 has a greater speed than ball 2 when they are passing.

c) The height is the same for both balls = 3h/4.

Explanation:

a) We can find the time when the two balls meet by equating the distances as follows:

$y = y_{0_{1}} + v_{0_{1}}t - \frac{1}{2}gt^{2}$

Where:

$y_{0_{1}}$ : is the initial height = h

$v_{0_{1}}$ : is the initial speed of ball 1 = 0 (it is dropped from rest)

$y = h - \frac{1}{2}gt^{2}$ (1)

Now, for ball 2 we have:

$y = y_{0_{2}} + v_{0_{2}}t - \frac{1}{2}gt^{2}$

Where:

$y_{0_{2}}$ : is the initial height of ball 2 = 0

$y = v_{0_{2}}t - \frac{1}{2}gt^{2}$ (2)

By equating equation (1) and (2) we have:

$h - \frac{1}{2}gt^{2} = v_{0_{2}}t - \frac{1}{2}gt^{2}$

$t=\frac{h}{v_{0_{2}}}$

Where the initial velocity of the ball 2 is:

$v_{f_{2}}^{2} = v_{0_{2}}^{2} - 2gh$

Since $v_{f_{2}}^{2}$ = 0 at the maximum height (h):

$v_{0_{2}} = \sqrt{2gh}$

Hence, the time when they pass each other is:

$t = \frac{h}{\sqrt{2gh}} = \sqrt{\frac{h}{2g}}$

b) When they are passing the speed of each one is:

For ball 1:

$v_{f_{1}} = - gt = -g*\sqrt{\frac{h}{2g}} = - 0.71\sqrt{gh}$

The minus sign is because ball 1 is going down.

For ball 2:

$v_{f_{2}} = v_{0_{2}} - gt = \sqrt{2gh} - g*\sqrt{\frac{h}{2g}} = (\sqrt{1} - \frac{1}{\sqrt{2}})*\sqrt{gh} = 0.41\sqrt{gh}$

Therefore, taking the magnitude of ball 1 we can see that it has a greater speed than ball 2 when they are passing.

c) The height of the ball is:

For ball 1:

$y_{1} = h - \frac{1}{2}gt^{2} = h - \frac{1}{2}g(\sqrt{\frac{h}{2g}})^{2} = \frac{3}{4}h$

For ball 2:

$y_{2} = v_{0_{2}}t - \frac{1}{2}gt^{2} = \sqrt{2gh}*\sqrt{\frac{h}{2g}} - \frac{1}{2}g(\sqrt{\frac{h}{2g}})^{2} = \frac{3}{4}h$

Then, when they are passing the height is the same for both = 3h/4.

I hope it helps you!

7 0

3 years ago

ou have just moved into a new apartment and are trying to arrange your bedroom. You would like to move your dresser of weight 3,

Dimas [21]

So far, since you moved into the apartment until the end of this much of the story, you haven't done ANY work on the dresser yet.

I'll admit that you pushed, groaned and grunted, sweated and strained plenty. You're physically and mentally exhausted, you're not interested in the dresser at the moment, and right now you just want to snappa cappa brew, crash on the couch, and watch cartoons on TV. But if you've done your Physics homework, you know you haven't technically done any work yet.

In Physics, "Work" is the product of Force times Distance.

Since the dresser hasn't budged yet, the Distahce is zero. So no matter how great the Force may be, it's multiplied by zero, so the Work is zero.

5 0

3 years ago

What comes before Action-reaction?

Anarel [89]

These two forces are called action and reaction forces and are the subject of Newton's third law of motion. Formally stated, Newton's third law is: For every action, there is an equal and opposite reaction. The statement means that in every interaction, there is a pair of forces acting on the two interacting objects.

4 0

3 years ago

A wheel of radius 25cm has eight spokes. It is mounted on a fixed axle and is rotating at a constant angular speed w. You shoot

spayn [35]

Explanation:

We will assume that the rim of the wheel is also very thin, like the spokes. The distance s between the spokes along the rim is

$s = \frac{1}{8}C = \frac{1}{8}(2\pi)(0.25\:\text{m}) = 0.196\:\text{m}$

The 20-cm arrow, traveling at 6 m/s, will travel its length in

$t = \dfrac{0.2\:\text{m}}{6\:\text{m/s}} = \dfrac{1}{30}\:\text{s}$

The fastest speed that the wheel can spin without clipping the arrow is

$v = \dfrac{s}{t} = \dfrac{0.196\:\text{m}}{\left(\dfrac{1}{30\:\text{s}}\right)} = 5.9\:\text{m/s}$

The angular velocity $\omega$ of the wheel is given by

$\omega = \dfrac{v}{r} = \dfrac{5.9\:\text{m/s}}{0.25\:\text{m}} = 23.6\:\text{rad/s}$

In terms of rev/s, we can convert the answer above as follows:

$23.6\:\dfrac{\text{rad}}{\text{s}}×\dfrac{1\:\text{rev}}{2\pi\:\text{rad}} = 3.8\:\text{rev/s}$

As you probably noticed, I did the calculations based on the assumption that I'm aiming for the edge of the wheel because this is the part of the wheel where a point travels a longer linear distance compared to ones closer to the axle, thus giving the arrow a better chance to pass through the wheel without getting clipped by the spokes. If you aim closer to the axle, then the wheel needs to spin slower to allow the arrow to get through without hitting the spokes.

3 0

3 years ago

Why is the pipe that carries the heated glycol from the box to the tank at an angle?

Alik [6]

Answer:

Bok.............................

5 0

2 years ago