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allochka39001 [22]
2 years ago
13

Phases of the moon. Help!???????

Physics
1 answer:
Veseljchak [2.6K]2 years ago
6 0

Answer:

The Lunar Month.

New Moon.

Waxing Crescent Moon.

First Quarter Moon.

Waxing Gibbous Moon.

Full Moon.

Waning Gibbous Moon.

Third Quarter Moon.

Explanation:

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The force on an object use Socratic glad to help ..
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When resting, a person has a metabolic rate of about 3.0 105 joules per hour. The person is submerged neck-deep into a tub conta
jek_recluse [69]

Answer:

The temperature after half an hour is 19.3002^{\circ}

Solution:

As per the question;

Metabolic rate of the person is 3.0105 J/h

Temperature, T = 19.30^{\circ}

Mass of the water, m_{w} = 1.2103 kg

Time duration, t = 0.5 h = 30 min = 180 s

Now,

Heat, Q = ms\Delta t

Thus heat transfer in half an hour:

Q = 3.0105\times 0.5 = 1.505 J

Now, the temperature of water after half an hour, T' is given by:

Q = m_{w}s\Delta T = ms(T' - T)

where

s = 4186 J

1.505 = 1.21103\times 4186\times (T' - 19.30^{\circ})

T' = 19.3002 ^{\circ}

4 0
3 years ago
What happens to momentum when two objects collide
Lostsunrise [7]

Answer:

The sum of momenta of each object before and after collision is equal . Thus when they collide , their momentum will change but its sum will be sane as the sum of the momentum before collision

3 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
Where is heat transferred by conduction in a lava lamp?
Semmy [17]
It is transferred by direct contact, say you touched it, you would feel the heat
8 0
2 years ago
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