Answer:
What is the best description for the volume of air volume of air provided in a high quality rescue breath?
Explanation:
A) only Enough air to create a visible rise of the chest.
B) Until you can no longer force air in.
C) plenty of air make sure it is adequate to sustain life
D) clear and obvious rise of the chest, sustained over a few seconds
Complete Question
A football coach walks 18 meters westward, then 12 meters
eastward, then 28 meters westward, and finally 14 meters
eastward.
a
From this motion what is the distance covered
b
What is the magnitude and direction of the displacement
Answer:
a
b
Magnitude
Direction
West
Explanation:
From the question we are told that
The first distance covered westward is ![d_w_1 = 18 \ m /tex] The first distance covered eastward is [tex]d_e1 = 12 \ m /tex] The second distance covered westward is [tex]d_w_2 = 28 \ m /tex] The second distance covered eastward is [tex]d_e2 = 14 \ m /tex] Generally the distance covered is mathematically represented as [tex]D = d_w1 + d_w2 + d_e1 + d_e2](https://tex.z-dn.net/?f=d_w_1%20%20%3D%20%2018%20%5C%20%20m%20%2Ftex%5D%3C%2Fp%3E%3Cp%3E%20%20%20%20%20The%20%20first%20distance%20covered%20eastward%20is%20%5Btex%5Dd_e1%20%3D%20%2012%20%5C%20%20m%20%2Ftex%5D%3C%2Fp%3E%3Cp%3E%20%20%20%20%20%20The%20second%20distance%20covered%20westward%20is%20%5Btex%5Dd_w_2%20%20%3D%20%2028%20%5C%20%20m%20%2Ftex%5D%3C%2Fp%3E%3Cp%3E%20%20%20%20%20%20The%20%20second%20distance%20covered%20eastward%20is%20%5Btex%5Dd_e2%20%3D%20%2014%20%5C%20%20m%20%2Ftex%5D%3C%2Fp%3E%3Cp%3E%20%20%20%3C%2Fp%3E%3Cp%3EGenerally%20the%20distance%20covered%20is%20mathematically%20represented%20as%20%3C%2Fp%3E%3Cp%3E%20%20%20%20%20%20%5Btex%5DD%20%3D%20%20d_w1%20%2B%20d_w2%20%2B%20d_e1%20%2B%20d_e2)
=> 
=>
For the second question eastward is in the direction of the positive x-axis so it would be positive and westward is in the direction of the negative x-axis so it would be negative
The magnitude of the displacement is
=> 
=>
The direction is west
The solution for this problem is:
For 1st minimum, let m be equal to 1.
d = slit width
D = screen distance.
Θ = arcsin (m * lambda/ (d))
= 0.13934 rad, 7.9836 deg
y = D*tan (Θ)
y = 6.50 * tan (7.9836)
= 0.91161 m is the distance from the central maximum to the first-order minimum
Answer:
Maximum altitude above the ground = 1,540,224 m = 1540.2 km
Explanation:
Using the equations of motion
u = initial velocity of the projectile = 5.5 km/s = 5500 m/s
v = final velocity of the projectile at maximum height reached = 0 m/s
g = acceleration due to gravity = (GM/R²) (from the gravitational law)
g = (6.674 × 10⁻¹¹ × 5.97 × 10²⁴)/(6370000²)
g = -9.82 m/s² (minus because of the direction in which it is directed)
y = vertical distance covered by the projectile = ?
v² = u² + 2gy
0² = 5500² + 2(-9.82)(y)
19.64y = 5500²
y = 1,540,224 m = 1540.2 km
Hope this Helps!!!
