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
Current flow depends on other things in addition to the circuit configuration.
If the SAME voltage is applied to some arrangement of the SAME components, the greatest current will occur when they are all in parallel.
Answer:
v₂ = 0.56 m / s
Explanation:
This exercise can be done using Bernoulli's equation
P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂
Where points 1 and 2 are on the surface of the glass and the top of the straw
The pressure at the two points is the same because they are open to the atmosphere, if we assume that the surface of the vessel is much sea that the area of the straw the velocity of the surface of the vessel is almost zero v₁ = 0
The difference in height between the level of the glass and the straw is constant and equal to 1.6 cm = 1.6 10⁻² m
We substitute in the equation
+ ρ g y₁ =
+ ½ ρ v₂² + ρ g y₂
½ v₂² = g (y₂-y₁)
v₂ = √ 2 g (y₂-y₁)
Let's calculate
v₂ = √ (2 9.8 1.6 10⁻²)
v₂ = 0.56 m / s
40 seconds I’m pretty sure sorry if I’m wrong
<span>Range = 88.5 Km/h - 94.5 Km/h</span><span>
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