Answer:
the ball travelled approximately 60 m towards north before stopping
Explanation:
Given the data in the question;
First course : 
= 0.75 m/s², 
= 20 m, 
= 10 m/s
now, form the third equation of motion;
v² = u² + 2as
we substitute
² = (10)² + (2 × 0.75 × 20)
² = 100 + 30
² = 130
= √130
= 11.4 m/s
for the Second Course:
= 11.4 m/s, 
= -1.15 m/s², 
= 0
Also, form the third equation of motion;
v² = u² + 2as
we substitute
0² = (11.4)² + (2 × (-1.15) × 
)
0 = 129.96 - 2.3
2.3 = 129.96
= 129.96 / 2.3
= 56.5 m
so;
|d| = √( ² + ² )
we substitute
|d| = √( (20)² + (56.5)² )
|d| = √( 400 + 3192.25 )
|d| = √( 3592.25 )
|d| = 59.9 m ≈ 60 m
Therefore, the ball travelled approximately 60 m towards north before stopping
To solve this we assume
that the gas inside is an ideal gas. Then, we can use the ideal gas
equation which is expressed as PV = nRT. At a constant pressure and number of
moles of the gas the ratio T/V is equal to some constant. At another set of
condition of temperature, the constant is still the same. Calculations are as
follows:
T1 / V1 = T2 / V2
V2 = T2 x V1 / T1
V2 = 659.7 x 28 / 504.7
<span>V2 = 36.60 in^3</span>
Answer:
The potential energy increases if the orbital radius increases.
Explanation:
The orbital radius of the electron increases means the distance from the nucleus of the hydrogen atom increase.
The nucleus is positively charged .
The potential energy is given by P.E = -
where Z is the atomic number
r is the radius
The negative sign indicates that the electron which is revolving is bound to nucleus.
As the radius and potential energy are inversely proportional it is clear that when <em>radius increase</em> the<em> potential energy become less negative </em>which means the potential energy increases when the orbital radius increase.
