Answer:
D.
a control group
Explanation:
In a scientific experiment such as the one above, there is an experimental group and a control group. The experimental group is the group that receives the treatment while the control group does not receive any treatment. The control group helps the researcher to observe if the treatment had any significant effect.
In this case, it will help Alan and Monica to determine if fertilizer X actually had an effect on the plant. Therefore, the pot with o grams of fertilizer in it is the control group.
It deals with skin diseases
Answer:
R = 6.3456 10⁴ mile
Explanation:
For this exercise we will use Newton's second law where force is gravitational force
F = m a
The satellite is in a circular orbit therefore the acceleration is centripetal
a = v² / r
Where the distance is taken from the center of the Earth
G m M / r² = m v² / r
G M / r = v²
The speed module is constant, let's use the uniform motion relationships, with the length of the circle is
d = 2π r
v = d / t
The time for a full turn is called period (T)
Let's replace
G M / r = (2π r / T)²
r³ = G M T²²2 / 4π²
r = ∛ (G M T² / 4π²)
We have the magnitudes in several types of units
T = 88.59 h (3600 s / 1h) = 3.189 10⁵ s
Re = 6.37 10⁶ m
Let's calculate
r = ∛ (6.67 10⁻¹¹ 5.98 10²⁴ (3,189 10⁵)²/4π²)
r = ∛ (1.027487 10²⁴)
r = 1.0847 10⁸ m
This is the distance from the center of the Earth, the distance you want the surface is
R = r - Re
R = 108.47 10⁶ - 6.37 10⁶
R = 102.1 10⁶ m
Let's reduce to miles
R = 102.1 10⁶ m (1 mile / 1609 m)
R = 6.3456 10⁴ mile
Answer:
a) 725.5 m
b) 630 m
Explanation:
Given data:
acceleration of Helicopter = 7.0 m/s^2
time spent upwards by helicopter = 11.0 seconds
a) Determine the maximum height above ground reached by the helicopter
h1 = at^2 /2
= 7 * 11^2 / 2
= ( 7 * 121 ) / 2 = 423.5 m
also v = a*t = 7 * 11 = 77 m/s
also we calculate h2
h2 = v^2 / 2g
= (77^2) / 2 * 9.81
= 302 m
therefore the maximum height = 302 + 423.5 = 725.5 m
b) Given that ; power deploys a jet pack strapped on his back at 7.0 s and with a downward acceleration of ; 1.0 m/s^2
<u>Determine distance Power reaches before helicopter crashes </u>
s = ut + 1/2 at^2
h.gt^2 - 77t - 423.5 m = 0
h.gt^2 - 77t = 423.5
t = 17. 66 secs
Yf = 423.5 + 77 *7 - 4.9 *7
Yf = 928.2
Vf = u + at
= 77 - 9.8*7 = 8.4 m/secs
t' = 17.66 - 7 = 10.66 secs
hence
Yf = 725.5 - 8.4 * 10.66 + 1/2 * -1 * 10.66
= 630 meters
