Answer:
An object decreases in size due to the collision of materials. An object increases in size due to the addition of materials. Gas particles are formed from solar nebula materials.
Answer:
15.3 s and 332 m
Explanation:
With the launch of projectiles expressions we can solve this problem, with the acceleration of the moon
gm = 1/6 ge
gm = 1/6 9.8 m/s² = 1.63 m/s²
We calculate the range
R = Vo² sin 2θ / g
R = 25² sin (2 30) / 1.63
R= 332 m
We will calculate the time of flight,
Y = Voy t – ½ g t2
Voy = Vo sin θ
When the ball reaches the end point has the same initial height Y=0
0 = Vo sin t – ½ g t2
0 = 25 sin (30) t – ½ 1.63 t2
0= 12.5 t – 0.815 t2
We solve the equation
0= t ( 12.5 -0.815 t)
t=0 s
t= 15.3 s
The value of zero corresponds to the departure point and the flight time is 15.3 s
Let's calculate the reach on earth
R2 = 25² sin (2 30) / 9.8
R2 = 55.2 m
R/R2 = 332/55.2
R/R2 = 6
Therefore the ball travels a distance six times greater on the moon than on Earth
Answer:
Electric current is defined as the rate of flow of electric charge in a circuit from point one point to another. This is carried by electrically charged particles within the circuit. Current is represented by the symbol I and its unit measured in Amperes. It is therefore related to the voltage and resistance of the circuit. If the current in the circuit reduces, the rate at which the charge and current on the capacitor reduces also proportionally in an exponential manner.
Explanation:
Since a decrease in the flow of current in the circuit is observed, the implication for the rate at which the charge and voltage on the capacitor is also an exponential decrease in the rate of flow with time. This is because the electric current is directly proportional to the electric charge and the time.
A trough is the lowest point in a wave.
<u>Answer:</u>
<em>The initial distance between the trains is 1450 m.
</em>
<u>Explanation:</u>
In the question two trains are of equal length 400 m and moves at a uniform speed of 72 km/h. train A is moving ahead of train B. If the train B has to overtake train A it should accelerate.
Train B’s acceleration is 
and it accelerated for 50 seconds.
<em>
</em>
<em>t=50 s
</em>
<em>initial speed u=72km/h
</em>
<em>we have to convert this speed into m/s </em>
<em>
</em>
<em>Distance covered in accelerating phase 
</em>
<em>
</em>
<em>
</em>
If a train is just behind another, the distance covered by the train located behind during overtaking phase will be equal to the sum of the lengths of the trains.
<em>Here length of train A+length of train 
</em>
<em>Hence the initial distance between the trains = 
</em>
