The frog's launch speed and the time spends in the air are 22.5m/s and 2.73s respectively.
To find the answer, we need to know about the time of flight and range of projectile motion.
<h3>What's the expression of range of a projectile motion?</h3>
- Range = U²× sin(2θ)/g
- U= initial velocity, θ= angle of projectile and g= acceleration due to gravity
- U=√{Range×g/sin(2θ)}
- Here, range= 2.20m, = 36.5°
- U= √{2.20×9.8/sin(73)}
U= √{2.20×9.8/sin(73)} = 22.5m/s
<h3>What's the expression of time of flight in projectile motion?</h3>
- Time of flight= (2×U×sinθ)/g
- So, T= (2×22.5×sin36.5°)/9.8
= 2.73 s
Thus, we can conclude that the frog's launch speed and the time spends in the air are 22.5m/s and 2.73s respectively.
Learn more about the range and time period of projectile motion here:
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D, 0.140 liters! Hang on a sec and I'll show you a trick I use.
A 300-kg bear grasping a vertical tree slides down at constant velocity. The friction force between the
tree and the bear is
<span>I'll tell you how to do it but you must crunch the numbers.
Use Kepler's 3rd Law
T^2 = k R^3
where k = 4(pi)^2/ GM
G =gravitational constant = 6.67300 × 10-11 m3 kg-1 s-2
M = mass of this new planet
pi = 3.14159265
T =3.09 days = 266976 seconds
R = (579,000,000km)/9 = 64333333.3 km
a)
Solve Kepler's 3rd Law for M. Your answer will be in kg
b)
mass of the sun = 1.98892 × 10^30 kilograms
Form the ratio
M(planet)/M(sun) </span>
Answer:
b and d
a, c, e, and f
Explanation:
Ideal gas law:
PV = nRT
Solving for temperature:
T = PV / (nR)
Therefore, temperature is directly proportional to pressure and volume, and inversely proportional to the number of molecules.
T = k PV / N
Let's say that T₀ is the temperature when P = 100 kPa, V = 4 L, and N = 6×10²³.
a) T = k PV / N = T₀
b) T = k (2P) V / N = 2T₀
c) T = k (P/2) (2V) / N = T₀
d) T = k PV / (N/2) = 2T₀
e) T = k P (V/2) / (N/2) = T₀
f) T = k (P/2) V / (N/2) = T₀
b and d have the highest temperature,
a, c, e, and f have the lowest temperature.
