Answer:
A) 1.9m/s
Explanation:
Use the equation v=d/t
The time is 10 seconds. The distance is more complicated because it’s a circle so u use the equation for circumference C=2•pi•r. So you just do the circumference over time and get the answer. Hope this helps
A. The angle at which the arrow must be released to hit the bull's-eye is 20.7 °
B. The arrow will go over the branch.
<h3>A. How to determine the angle</h3>
- Range (R) = 74 m
- Initial velocity (u) = 33 m/s
- Acceleration due to gravity (g) = 9.8 m/s²
- Angle (θ) = ?
R = u²Sine(2θ) / g
74 = 33² × Sine (2θ) / 9.8
Cross multiply
74 × 9.8 = 33² × Sine (2θ)
725.2 = 1098 × Sine (2θ)
Divide both sides by 1098
Sine (2θ) = 725.2 / 1098
Sine (2θ) = 0.6605
Take the inverse of sine
2θ = Sine⁻¹ 0.6605
2θ = 41.3
Divide both sides by 2
θ = 41.3 / 2
θ = 20.7 °
<h3>B. How to determine if the arrow will go over or under the branch</h3>
To determine if the arrow will go over or under the branch situated mid way, we shall determine the maximum height attained by the arrow. This can be obtained as follow:
- Initial velocity (u) = 33 m/s
- Acceleration due to gravity (g) = 9.8 m/s²
- Angle (θ) = 20.7 °
- Maximum height (H) = ?
H = u²Sine²θ / 2g
H = [33² × (Sine 20.7)²] / (2 ×9.8)
H = 6.94 m
Thus, the maximum height attained by the arrow is 6.94 m which is greater than the height of the branch (i.e 3.50 m).
Therefore, we can conclude that the arrow will go over the branch
Learn more about projectile motion:
brainly.com/question/20326485
#SPJ1
The charge would be in equilibrium so there would be no charge in the body of the conductor.
Answer:
(b) there cannot be any charge in the body of the conductor
N2+3H2->2NH3
When 3 moles of H2 react, they produce 2 moles of NH3
3 moles of H2 have a mass of 2.02 g
2 moles of NH2 have a mass of 17.0 g
So when 2.02 g H2 react, they produce 17.0 g NH3
If 26.3g H2 react with a yield of 100%, we expect…
2.02g H2_____17.0gNH3
26.3g H2_____x=221gNH3
So now let’s calculate the percentage:
221gNH3_________100%
79.0gNH3_________x=79.0*100/221=35.7%
Answer:
Polar areas
Explanation:
From what I know Arctic, Antarctic, and polar air masses are cold. Polar (cold), Arctic (very cold), Equatorial (warm and very moist), and Tropical (warm).
- With the options you gave, you can immeditaly elimate 1. tropical areas and 2. an area where its summertime.
- Your then left with two options 1. an area where its wintertime and 2. polar areas
I then concluded that cold air masses tend to originate from POLAR AREAS.
