Answer:
Savannas have a fairly constant temperature all year; temperate grasslands have a greater seasonal temperature variation.
Explanation:
For example, the African Savanna has an almost constant temperature all year (see the first figure below).
The difference between summer and winter temperatures is only about 5 °C, and the rate of temperature change is quite slow.
The temperature of a temperate grassland (see the second figure below) has a much greater seasonal variation.
The summers are hot, and the winters are cold. The difference between summer and winter temperatures is about 30 °C, with a rapid rate of temperature change from one season to the next.
V ( initial ) = 20 m/s
h = 2.30 m
h = v y * t + g t ² / 2
d = v x * t
1 ) At α = 18°:
v y = 20 * sin 18° = 6.18 m/s
v x = 20 * cos 18° = 19.02 m/ s
2.30 = 6.18 t + 4.9 t²
4.9 t² + 6.18 t - 2.30 = 0
After solving the quadratic equation ( a = 4.9, b = 6.18, c = - 2.3 ):
t 1/2 = (- 6.18 +/- √( 6.18² - 4 * 4.9 * (-2.3)) ) / ( 2 * 4.9 )
t = 0.3 s
d 1 = 19.02 m/s * 0.3 s = 5.706 m
2 ) At α = 8°:
v y = 20* sin 8° = 2.78 m/s
v x = 20* cos 8° = 19.81 m/s
2.3 = 2.78 t + 4.9 t²
4.9 t² + 2.78 t - 2.3 = 0
t = 0.46 s
d 2 = 19.81 * 0.46 = 9.113 m
The distance is:
d 2 - d 1 = 9.113 m - 5.706 m = 3.407 m
GOOD LUCK AND HOPE IT HELPS U
Answer:
Isothermal : P2 = ( P1V1 / V2 ) , work-done 
Adiabatic : : P2 = 
, work-done =
W = 
Explanation:
initial temperature : T
Pressure : P
initial volume : V1
Final volume : V2
A) If expansion was isothermal calculate final pressure and work-done
we use the gas laws
= PIVI = P2V2
Hence : P2 = ( P1V1 / V2 )
work-done :
B) If the expansion was Adiabatic show the Final pressure and work-done
final pressure
where y = 5/3
hence : P2 =
Work-done
W =
Where 
A. Formula: F=ma or F/m=a
10,000N/1,267kg≈7.9m/
B. Formula: a=
and s=d/t
speed= 394.6/15
s=26.3m/s
a=
a=1.75m/
C. 7.9-1.75=difference of 6.15m/
D. The force that most likely caused this difference is friction forces
Answer:
A) θ = 13.1º , B) E
Explanation:
A) For this exercise, let's use Newton's second law, let's set a reference frame where the axis ax is in the radial direction and is horizontal, the axis y is vertical.
In this reference system the only force that we must decompose is the Normal one, let's use trigonometry
sin θ = Nₓ / N
cos θ = 
/ N
Nₓ = N sin θ
Ny = N cos θ
x-axis (radial)
Nₓ = m a
where the acceleration is centripetal
a = v² / R
we substitute
-N sin θ = -m v² / R (1)
the negative sign indicates that the force and acceleration towards the center of the circle
y-axis (Vertical)
Ny - W = 0
N cos θ = mg
N = mg / cos θ
we substitute in 1
mg / cos θ sin θ = m v² / R
g tan θ = v² / R
θ = tan⁻¹ (v² / gR)
we calculate
θ = tan⁻¹ (25² / 9.8 274)
θ = 13.1º
B) when comparing the equations the correct one is E
