1. They have evolved their leaves into spikes for minimum water loss through transpiration.
2. They have a waxy layer for minimum water loss.
3. They have thick walls for minimum water loss.
4. They can take water from atmosphere.
5. They change the photo energy from Sun into an intermediate stage and store it, so that they can make food even in night.
Answer:
The coefficient of kinetic friction between the sled and the snow is 0.0134
Explanation:
Given that:
M = mass of person = 52 kg
m = mass of sled = 15.2 kg
U = initial velocity of person = 3.63 m/s
u = initial velocity of sled = 0 m/s
After collision, the person and the sled would move with the same velocity V.
a) According to law of momentum conservation:
Total momentum before collision = Total momentum after collision
MU + mu = (M + m)V

Substituting values:

The velocity of the sled and person as they move away is 2.81 m/s
b) acceleration due to gravity (g) = 9.8 m/s²
d = 30 m
Using the formula:

The coefficient of kinetic friction between the sled and the snow is 0.0134
Answer:
a) 0.138J
b) 3.58m/S
c) (1.52J)(I)
Explanation:
a) to find the increase in the translational kinetic energy you can use the relation

where Wp is the work done by the person and Wg is the work done by the gravitational force
By replacing Wp=Fh1 and Wg=mgh2, being h1 the distance of the motion of the hand and h2 the distance of the yo-yo, m is the mass of the yo-yo, then you obtain:

the change in the translational kinetic energy is 0.138J
b) the new speed of the yo-yo is obtained by using the previous result and the formula for the kinetic energy of an object:

where vf is the final speed, vo is the initial speed. By doing vf the subject of the formula and replacing you get:

the new speed is 3.58m/s
c) in this case what you can compute is the quotient between the initial rotational energy and the final rotational energy

hence, the change in Er is about 1.52J times the initial rotational energy
The distance between Mars and the Sun in the scale model would be 1140 m
Explanation:
In this scale model, we have:
represents an actual distance of

The actual distance between Mars and the Sun is 228 million km, therefore

On the scale model, this would corresponds to a distance of
.
Therefore, we can write the following proportion:

And solving for
, we find:

Learn more about distance:
brainly.com/question/3969582
#LearnwithBrainly
Answer:
s = 30330.7 m = 30.33 km
Explanation:
First we need to calculate the speed of sound at the given temperature. For this purpose we use the following formula:
v = v₀√[T/273 k]
where,
v = speed of sound at given temperature = ?
v₀ = speed of sound at 0°C = 331 m/s
T = Given Temperature = 10°C + 273 = 283 k
Therefore,
v = (331 m/s)√[283 k/273 k]
v = 337 m/s
Now, we use the following formula to calculate the distance traveled by sound:
s = vt
where,
s = distance traveled = ?
t = time taken = 90 s
Therefore,
s = (337 m/s)(90 s)
<u>s = 30330.7 m = 30.33 km</u>