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Dmitry_Shevchenko [17]

3 years ago

14

on june 20, 1969, two american astronauts named Buss Aldrin and Neil Armstrong became the first humans to contact the surface of

the moon? True or False

1 answer:

xxTIMURxx [149]3 years ago

7 0

Nope.
Pretty sneaky.
It was JULY 20.
I remember it well.

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A gas expands against a constant external pressure of 2.00 atm until its volume has increased from 6.00 to 10.00 L. During this

mars1129 [50]

Answer:

ΔU = - 310.6 J (negative sign indicates decrease in internal energy)

W = 810.6 J

Explanation:

a.

Using first law of thermodynamics:

Q = ΔU + W

where,

Q = Heat Absorbed = 500 J

ΔU = Change in Internal Energy of Gas = ?

W = Work Done = PΔV =

P = Pressure = 2 atm = 202650 Pa

ΔV = Change in Volume = 10 L - 6 L = 4 L = 0.004 m³

Therefore,

Q = ΔU + PΔV

500 J = ΔU + (202650 Pa)(0.004 m³)

ΔU = 500 J - 810.6 J

<u>ΔU = - 310.6 J (negative sign indicates decrease in internal energy)</u>

<u></u>

b.

The work done can be simply calculated as:

W = PΔV

W = (202650 Pa)(0.004 m³)

<u>W = 810.6 J</u>

7 0

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:

$v_{02}^2+2a_2(y_2-y_{02})=v_2^2=0m/s$

Then, to get $y_2$ , we do:

$2a_2(y_2-y_{02})=-v_{02}^2$

$y_2-y_{02}=-\frac{v_{02}^2}{2a_2}$

$y_2=y_{02}-\frac{v_{02}^2}{2a_2}$

And we substitute the values:

$y_2=y_{02}-\frac{v_{02}^2}{2a_2}=(1317.26m)-\frac{(271.6m/s)^2}{2(-9.8m/s^2)}=5080.86m$

3 0

3 years ago

If a stone is dropped from a height of 400 feet, its height after t seconds is given by s = 400 − 16t2. Find its instantaneous v

soldier1979 [14.2K]

Answer:

$v = -32 t$

Explanation:

given,

s = 400- 16 t²

we know,

Velocity of an object is defined as the change in displacement per unit change in time.

velocity an also be return as

$v = \dfrac{ds}{dt}$

$v = \dfrac{d}{dt}(400-16t^2)$

$v= 0 -2\times 16 t$

$v = -32 t$

Hence, instantaneous velocity function given by $v = -32 t$

To calculate instantaneous velocity, you need to insert value of time.

ex, instantaneous velocity at t = 4 s

v = -32 x 4 = -128 m/s.

5 0

3 years ago

The weight of a body of certain mass becomes zero in space.why?write with reasons

Anvisha [2.4K]

Answer:

Weight is what you get when a certain amount of gravity is acting on that mass, and something, like the surface of a planet, is resisting that action. In space, when falling freely, there's nothing resisting the pull of gravity so weight disappears. Mass however stays.

hope this helps u

Explanation:

7 0

3 years ago

PLEASE HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! IM DESPERATE!!!!!!!!!!!!!! EVEN IF YOU DONT KNOW , GIVE A GUESS. What would

dezoksy [38]

Well, A = T or U C = G G = C T or U = A So it would be like this; DNA Sequence: GCTAATTGCATCCGA The Complementary Sequence: CGATTAACGTAGGCT Hope this helped :)

3 0

3 years ago