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
If the liquid is denser than the coin, then the coin will eventually
come to rest floating, with part of it above the surface of the liquid.
That's exactly the situation if you drop the coin into mercury.
Density of copper . . . 8.96 gm/cm³
iron . . . 7.87
zinc . . . 7.13
silver . . 10.5
nickel . . .8.91
lead . . . 11.4
Density of mercury . . . 13.53 gm/cm³ !
Heat must be added to raise the temperature from 20.0 •C to 82 •C is
Q= (4200 x 1.8 + 880 x 1.3) x (82 - 20) = 539648J
Answer: c. 1.89 x 10^14 J
Explanation:
By Einstein's equation, we know that:
E = m*c^2
Where m is the mass-consumed in this case:
m = 2.10g
And we must rewrite this in Kg, knowing that:
1kg = 1000g
Then:
m = 2.10g = (2.10/1000) kg = 0.0021 kg
And c is the speed of light:
c = 3*10^8 m/s.
Then the energy will be:
E = 0.0021 kg*(3*10^8 m/s)^2 = 1.89*10^14 Joules.
The correct option is:
c. 1.89 x 10^14J
speed of point A is given as
![v_a = 20.9 m/s](https://tex.z-dn.net/?f=v_a%20%3D%2020.9%20m%2Fs)
![v_a = r*\omega](https://tex.z-dn.net/?f=v_a%20%3D%20r%2A%5Comega)
here r = distance from the axis
so here we have
![20.9 = r*23](https://tex.z-dn.net/?f=20.9%20%3D%20r%2A23)
![r = 0.91 m](https://tex.z-dn.net/?f=r%20%20%3D%200.91%20m)
now the distance of point A and B is 0.71 m
while the distance of axis from point A is 0.91 m
so distance of axis from point B will be given as
![r = 0.91 + 0.71 = 1.62 m](https://tex.z-dn.net/?f=%20r%20%3D%200.91%20%2B%200.71%20%3D%201.62%20m)
now for the speed of point b is given as
![v = r \omega](https://tex.z-dn.net/?f=v%20%3D%20r%20%5Comega)
![v = 1.62 * 23](https://tex.z-dn.net/?f=v%20%3D%201.62%20%2A%2023)
![v = 37.26 m/s](https://tex.z-dn.net/?f=v%20%3D%2037.26%20m%2Fs)
so end point B will move with speed 37.26 m/s