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3 years ago

An athlet starting from stationary moves with an acceleration 2.2m/s^2 for 5 seconds, then for other 5 seconds

Physics

1 answer:

ycow [4]3 years ago

4 0

Answer:

The graph is shown below.

The athlete's speed at the finish line is 11 m/s

Total distance covered is 82.5 m.

Explanation:

The graph of speed versus time is drawn below.

Acceleration in the first 5 seconds, $a=2.2$ m/s²

From the graph, the slope of line OA is the acceleration in the first 5 seconds.

Slope of line OA is given as:

$slope, a=\frac{AD}{OD}\\2.2=\frac{AD}{5}\\AD =2.2\times 5=11\textrm{ m/s}$

Now, the length AD is nothing but speed at point A or B as AB is a straight line.

Therefore, the speed when crossing the finish line is the speed at B which is equal to 11 m/s.

Distance covered is given by the total area under the graph.

The total area can be divided into two shapes; a triangle and a rectangle.

The area under the graph is the sum of areas of triangle OAD and rectangle ABCD.

Area of triangle OAD is, $A_{tri}=\frac{1}{2}\times OD\times AD=\frac{1}{2}\times 5\times 11=27.5$

Area of rectangle ABCD is, $A_{rec}=AB\times AD=5\times 11=55$

Therefore, the total distance covered till the finish line is given as:

$d_{total}=27.5+55=82.5\textrm{ m}$

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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

4 years ago

Which processes transfer energy from the core to the photosphere?

nikklg [1K]

<span>-radiation and convection

Hope this helps :)
</span>

7 0

3 years ago

Read 2 more answers

A series circuit is set up with an AA battery along with an mystery material and ammeter; however, there’s no current passing th

Ahat [919]

Answer:

A. Insulator

Explanation:

Since there is no current passing through at all.

7 0

3 years ago

Jumping up before the elevator hits. After the cable snaps and the safety system fails, an elevator cab free-falls from a height

noname [10]

Answer:

a) I = 2279.5 N s , b) F = 3.80 10⁵ N, c) I = 3125.5 N s and d) F = 5.21 10⁵ N

Explanation:

The impulse is equal to the variation in the amount of movement.

I =∫ F dt = Δp

I = m $v_{f}$ - m v₀

Let's calculate the final speed using kinematics, as the cable breaks the initial speed is zero

$v_{f}$ ² = V₀² - 2g y

$v_{f}$ ² = 0 - 2 9.8 30.0

$v_{f}$ = √588

$v_{f}$ = 24.25 m/s

a) We calculate the impulse

I = 94 24.25 - 0

I = 2279.5 N s

b) Let's join the other expression of the impulse to calculate the average force

I = F t

F = I / t

F = 2279.5 / 6 10⁻³

F = 3.80 10⁵ N

just before the crash the passenger jumps up with v = 8 m / s, let's take the moments of interest just when the elevator arrives with a speed of 24.25m/s down and as an end point the jump up to vf = 8 m / n

c) I = m $v_{f}$ - m v₀

I = 94 8 - 94 (-24.25)

I = 3125.5 N s

d) F = I / t

F = 3125.5 / 6 10⁻³

F = 5.21 10⁵ N

7 0

3 years ago

Two identical isolated conducting spheres are picked for an experiment. Only one of them is charged. If the spheres are now brie

aev [14]

Because they are conducting, when you bring them together the charge is split equally among the two spheres (because they have the same radius the amount of charge is also equal). Now they will repel each other because of the net charge on each with the same polarity.

5 0

3 years ago