Answer:
The correct answer is: waxing gibbous, 3 days
Explanation:
Waning quarter moon: hair removal time and bangs cuts.
The growing quarter as a moment of growth, development and evolution. On the contrary, the waning moon is associated with a time of completion, debugging or liquidation of pending issues.
We must take advantage of the influence of the lunar cycle in our favor according to the action we are going to take. If you have trouble growing your hair, try to go to the hairdresser in a crescent moon: it will grow faster. It is no nonsense. Since I cut my bangs to the Cleopatra, the touch-ups last me for another 1-1.5 weeks. As I reviewed the bangs in a growing room, in just a couple of weeks I was returning to the hairdresser.
That affects hair removal. There are many people who take appointments to the beautician to shave by consulting the lunar calendar. The hair removal done as soon as the dwindling is the best because it lasts longer, lasts for another week until the next appointment.
4. Grass - Caterpillar - Hedgehog - Fox
5. Caterpillar, Rabbit, Mouse.
6. Cougar and Fox.
7. Bacteria
8. The bird, hedgehog, Fox and cougar would be effected since the Hedgehogs and birds would soon die out due to the loss of their food. Once they die out, the cougar and Fox would have no predators left to eat.
Answer:
time taken with speed 23 km/h will be 1.8 hours or 1 hour 48 minutes
Explanation:
Given:
Time is inversely proportional to the speed
mathematically,
t ∝ (1/r)
let the proportionality constant be 'k'
thus,
t = k/r
therefore, for case 1
time = 3 hr
speed = 14 km/hr
3 = k/14
also,
for case 2
let the time be = t
r = 23 km/h
thus,
we have
t = k/23
on dividing equation 2 by 1
we get
or
or
t = 1.8 hr = or 1 hour 48 minutes ( 0.8 hours × 60 minutes/hour = 48 minutes)
Answer:
The shortest braking distance is 35.8 m
Explanation:
To solve this problem we must use Newton's second law applied to the boxes, on the vertical axis we have the norm up and the weight vertically down
On the horizontal axis we fear the force of friction (fr) that opposes the movement and acceleration of the train, write the equation for each axis
Y axis
N- W = 0
N = W = mg
X axis
-Fr = m a
-μ N = m a
-μ mg = ma
a = μ g
a = - 0.32 9.8
a = - 3.14 m/s²
We calculate the distance using the kinematics equations
Vf² = Vo² + 2 a x
x = (Vf² - Vo²) / 2 a
When the train stops the speed is zero (Vf = 0)
Vo = 54 km/h (1000m/1km) (1 h/3600s)= 15 m/s
x = ( 0 - 15²) / 2 (-3.14)
x= 35.8 m
The shortest braking distance is 35.8 m
