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2 years ago

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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What best describes an impulse acting on an object?

Triss [41]

The force on an object use Socratic glad to help ..

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2 years ago

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: