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11111nata11111 [884]

3 years ago

13

(a) A submarine descends to a depth of 70 m below the surface of water. The density of the water is 1050 kg/m3. Atmospheric pres

sure is 1.0 × 10^5 Pa (1.0 × 10⁵ Pa). Calculate

1 answer:

timurjin [86]3 years ago

5 0

Answer:

sttouyietETwe2e664yrwtwwteuwtrwruwuuwwuwtwuw7w7w5w7w772253536464647

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A train increases its speed steadily from 10 m/s to 20 m/s in

rewona [7]

Answer:

15m/s

Explanation:

add the two speeds and divide by 2

10+20=30

30/2=15

3 0

3 years ago

If a ball is thrown straight up into the air with an initial velocity of 65 ft/s, its height in feet after t seconds is given by

fgiga [73]

Answer:

a) $v_{1}=\frac{(62.5-66)ft}{(2.5-2)s}=-7ft/s$

$v_{2}=\frac{(65.94-66)ft}{(2.1-2)s}=-0.6ft/s$

$v_{3}=\frac{(66.0084-66)ft}{(2.01-2)s}=0.84ft/s$

$v_{4}=\frac{(66.001-66)ft}{(2.001-2)s}=1ft/s$

b) $v=65-32(2)=1ft/s$

Explanation:

From the exercise we got the ball's equation of position:

$y=65t-16t^{2}$

a) To find the average velocity at the given time we need to use the following formula:

$v=\frac{y_{2}-y_{1} }{t_{2}-t_{1} }$

Being said that, we need to find the ball's position at t=2, t=2.5, t=2.1, t=2.01, t=2.001

$y_{t=2}=65(2)-16(2)^{2} =66ft$

$y_{t=2.5}=65(2.5)-16(2.5)^{2} =62.5ft$

$v_{1}=\frac{(62.5-66)ft}{(2.5-2)s}=-7ft/s$

--

$y_{t=2.1}=65(2.1)-16(2.1)^{2} =65.94ft$

$v_{2}=\frac{(65.94-66)ft}{(2.1-2)s}=-0.6ft/s$

--

$y_{t=2.01}=65(2.01)-16(2.01)^{2} =66.0084ft$

$v_{3}=\frac{(66.0084-66)ft}{(2.01-2)s}=0.84ft/s$

--

$y_{t=2.001}=65(2.001)-16(2.001)^{2} =66.001ft$

$v_{4}=\frac{(66.001-66)ft}{(2.001-2)s}=1ft/s$

b) To find the instantaneous velocity we need to derivate the equation

$v=\frac{df}{dt}=65-32t$

$v=65-32(2)=1ft/s$

7 0

3 years ago

For a given initial projectile speed Vo, calculate what launch angle A gives the longest range R. Show your work, don't just quo

pickupchik [31]

The optimal angle of 45° for maximum horizontal range is only valid when initial height is the same as final height.

<span>In that particular situation, you can prove it like this: </span>

<span>initial velocity is Vo </span>
<span>launch angle is α </span>

<span>initial vertical velocity is </span>
<span>Vv = Vo×sin(α) </span>

<span>horizontal velocity is </span>
<span>Vh = Vo×cos(α) </span>

<span>total time in the air is the the time it needs to fall back to a height of 0 m, so </span>
<span>d = v×t + a×t²/2 </span>
<span>where </span>
<span>d = distance = 0 m </span>
<span>v = initial vertical velocity = Vv = Vo×sin(α) </span>
<span>t = time = ? </span>
<span>a = acceleration by gravity = g (= -9.8 m/s²) </span>
<span>so </span>
<span>0 = Vo×sin(α)×t + g×t²/2 </span>
<span>0 = (Vo×sin(α) + g×t/2)×t </span>
<span>t = 0 (obviously, the projectile is at height 0 m at time = 0s) </span>
<span>or </span>
<span>Vo×sin(α) + g×t/2 = 0 </span>
<span>t = -2×Vo×sin(α)/g </span>

<span>Now look at the horizontal range. </span>
<span>r = v × t </span>
<span>where </span>
<span>r = horizontal range = ? </span>
<span>v = horizontal velocity = Vh = Vo×cos(α) </span>
<span>t = time = -2×Vo×sin(α)/g </span>
<span>so </span>
<span>r = (Vo×cos(α)) × (-2×Vo×sin(α)/g) </span>
<span>r = -(Vo)²×sin(2α)/g </span>

<span>To find the extreme values of r (minimum or maximum) with variable α, you must find the first derivative of r with respect to α, and set it equal to 0. </span>

<span>dr/dα = d[-(Vo)²×sin(2α)/g] / dα </span>
<span>dr/dα = -(Vo)²/g × d[sin(2α)] / dα </span>
<span>dr/dα = -(Vo)²/g × cos(2α) × d(2α) / dα </span>
<span>dr/dα = -2 × (Vo)² × cos(2α) / g </span>

<span>Vo and g are constants ≠ 0, so the only way for dr/dα to become 0 is when </span>
<span>cos(2α) = 0 </span>
<span>2α = 90° </span>
<span>α = 45° </span>

4 0

3 years ago

To Calculate Velocity and Average Velocity you can use the same formula as both.

Elanso [62]

Answer:

I believe it's True! Brainliest??

Explanation: Hope you have a great day :)

4 0

3 years ago

Stars form from clouds of gas and dust. As a protostar gravitationally contracts within its parent cloud, "conservation of angul

vredina [299]

Answer:

D) True. the protostar rotates more quickly.

Explanation:

If the system is isolated, the angular momentum must be retained.

Initial

L₀ = I w₀

Final

$L_{f}$ = $I_{f}$ $w_{f}$

L₀ = $L_{f}$

I w₀ = $I_{f}$ $w_{f}$

$w_{f}$ = I / $I_{f}$ w₀

In general, the radius of the cloud decreases significantly to form the star, the moment of inertia must decrease, so the angular velocity must increase

Let's examine the answers

A) False. The opposite happens

B) False. Speed changes

C) False. For this there must be an external force, which does not exist

D) True. You agree with the above

5 0

3 years ago