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3 years ago

8

A flare is launched from a boat. The height, , in meters, of the flare above the water is approximately modelled by the function

, where is the number of seconds after the flare is launched. How many seconds will it take for the flare to return to the water?

1 answer:

Tpy6a [65]3 years ago

6 0

Answer:

10 seconds

Explanation:

If the height is modeled by the function $h(t)=-15t^{2} +150t$ , then the seconds it takes to reach the water (when the height equals 0) is modeled by the following.

$0=-15t^{2} +150t\\0=-15t*(t-10)\\\\t=0\\t=10$

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How long does it take to travel a distance of 672km at a speed of 95km/h?

Brilliant_brown [7]

Answer:

7.07 hours

Explanation:

divide the distance by the speed

so in this case, divide 672 by 95

6 0

3 years ago

You toss a ball straight up with an initial speed of 30m/s. How high does it go, and how long is it in the air (neglecting air r

Brut [27]

Explanation:

Given that,

A ball is tossed straight up with an initial speed of 30 m/s

We need to find the height it will go and the time it takes in the air.

At its maximum height, its final speed, v = 0 and it will move under the action of gravity. Using equation of motion :

v = u +at

Here, a = -g

v = u -gt

i.e. u = gt

$t=\dfrac{u}{g}\\\\t=\dfrac{30\ m/s}{9.8\ m/s^2}\\\\t=3.06\ s$

So, the time for upward motion is 3.06 seconds. It means that it will in air for 3.06×2 = 6.12 seconds

Let d is the maximum distance covered by it.

$d=ut-\dfrac{1}{2}gt^2$

Putting all values

$d=30(3.06)-\dfrac{1}{2}\times 9.8\times (3.06)^2\\\\d=45.91\ m$

Hence, it will go to a height of 45.91 m and it will in the air for 6.12 seconds.

8 0

3 years ago

1. A heat engine operates between two reservoirs at T2 = 600 K and T1 = 350 K. It takes in 1.00 x 103 J of energy from the highe

Alex73 [517]

Answer:

$\Delta S_u=2.1429\ J.K^{-1}$

$W_c=416.67\ J$

Explanation:

Given:

temperature of source reservoir, $T_H=600\ K$

temperature of sink reservoir, $T_L=350\ K$

energy absorbed from the source, $Q_{in}=1000\ J$

work done, $W=250\ J$

a.

<u>Now change in entropy of the surrounding:</u>

$\Delta S_u=\frac{dQ_L}{T_L}$

<em>Since heat engine is a device that absorbs heat from a high temperature reservoir and does some work giving out heat in the universe as the byproduct.</em>

$\Delta S_u=\frac{Q_H-W}{T_L}$

$\Delta S_u=\frac{1000-250}{350}$

$\Delta S_u=2.1429\ J.K^{-1}$

b.

<u>We know Carnot efficiency is given as:</u>

$\eta_c=1-\frac{T_L}{T_H}$

$\eta_c=1-\frac{350}{600}$

$\eta_c=0.4167$

<u>Now the Carnot work done:</u>

$W_c=Q_H\times \eta_c$

$W_c=1000\times 0.4167$

$W_c=416.67\ J$ .......................(1)

c.

From eq. (1) we have the Carnot work, so the difference:

$\Delta W=W_c-W$

$\Delta W=416.67-250$

$\Delta W=166.67\ J$

Now, we find:

$T_L.\Delta S_u=350\times 2.1429$

5 0

3 years ago

a football player starts from rest and speeds up to a final velocity of 12 m/s. the speeding up takes a total of 6 seconds. what

gizmo_the_mogwai [7]

Answer:

2 m/s²

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 0 m/s

Final velocity (v) = 12 m/s

Time (t) = 6 s

Acceleration (a) =?

The acceleration of the player can be obtained as follow:

v = u + at

12 = 0 + (a × 6)

12 = 6a

Divide both side by 6

a = 12 / 6

a = 2 m/s²

Thus, the acceleration of the player is 2 m/s²

8 0

3 years ago

What types of energy does a ball rolling down a hill have?

Yuki888 [10]

Kinetic energy is the answer to your question.

7 0

3 years ago

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