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3 years ago

Answer this question

Physics

1 answer:

ArbitrLikvidat [17]3 years ago

3 0

Answer:

i) 2

ii) No

Explanation:

i) The man will hear two sounds: one through the rail, and one in the air.

ii) The sounds will not reach the man at the same time. Sound travels faster through solids than through gases, so he will hear the sound through the rail before he hears the sound in the air.

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To calculate the velocity of an object the of the position vs time graph should be calculated

frutty [35]

<span>The vertical axis represents the velocity of the object</span>

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3 years ago

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After a plant or animal dies the 14C content decreases with a half-life of 5730 years. If an archaeologist finds an ancient fire

Shalnov [3]

Answer:

age of the site is 15411.75 years old

Explanation:

Given data

plant or animal dies = 14C

time period = 5730 year

carbon = 15.5%

to find out

age (in years) of the ancient site

solution

we know that Final value = Initial value × $0.5^{n}$

here n is half life passed

so for 15.5%

15.5% = 100% of $0.5^{n}$

0.155 = 1 × $0.5^{n}$

now take log both side

log 0.155 = log $0.5^{n}$

n = log 0.155 / log 0.5

n = 2.68966

we know here 5730 years in half life

so for 2.68966 half-lives = 2.68966 × 5730 = 15411.7518

age of the site is 15411.75 years old

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3 years ago

while flying a plane parallel to the ground a pilot releases a fuel tank in order to reduce the planes mass. what is the tanks f

stepan [7]

La velocidad vertical del tanque después de caer 10 m es 14 m/seg .

La velocidad vertical del tanque se calcula mediante la aplicación de la fórmula de velocidad , la componente vertical Vfy, del movimiento horizontal como se muestra a continuación :

Vfy=?

h = 10 m

Fórmula de Velocidad vertical Vfy:

Vfy² = 2*g*h

Vfy= √(2*9.8m/seg2* 10m )

Vfy= 14 m/seg

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3 years ago

A toy rocket, launched from the ground, rises vertically with an acceleration of 28 m/s 2 for 9.7 s until its motor stops. Disre

vredina [299]

Answer:

5080.86m

Explanation:

We will divide the problem in parts 1 and 2, and write the equation of accelerated motion with those numbers, taking the upwards direction as positive. For the first part, we have:

$y_1=y_{01}+v_{01}t+\frac{a_1t^2}{2}$

$v_1=v_{01}+a_1t$

We must consider that it's launched from the ground ( $y_{01}=0m$ ) and from rest ( $v_{01}=0m/s$ ), with an upwards acceleration $a_{1}=28m/s^2$ that lasts a time t=9.7s.

We calculate then the height achieved in part 1:

$y_1=(0m)+(0m/s)t+\frac{(28m/s^2)(9.7s)^2}{2}=1317.26m$

And the velocity achieved in part 1:

$v_1=(0m/s)+(28m/s^2)(9.7s)=271.6m/s$

We do the same for part 2, but now we must consider that the initial height is the one achieved in part 1 ( $y_{02}=1317.26m$ ) and its initial velocity is the one achieved in part 1 ( $v_{02}=271.6m/s$ ), now in free fall, which means with a downwards acceleration $a_{2}=-9,8m/s^2$ . For the data we have it's faster to use the formula $v_f^2=v_0^2+2ad$ , where d will be the displacement, or difference between maximum height and starting height of part 2, and the final velocity at maximum height we know must be 0m/s, so we have: