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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Answer:
Explanation:
The formula to determine the size of a capillary tube is
h = 2•T•Cos θ / r•ρ•g
Where
h = height of liquid level
T = surface tension
r = radius of capillary tube
ρ = density of liquid
θ = angle of contact = 0°
g =acceleration due to gravity=9.81m/s²
The liquid is water then,
ρ = 1000 kg / m³
Given that,
T = 0.0735 N/m
h = 0.25mm = 0.25 × 10^-3m
Then,
r = 2•T•Cos θ / h•ρ•g
r = 2 × 0.0735 × Cos0 / 2.5 × 10^-3 × 1000 × 9.81
r = 5.99 × 10^-3m
Then, r ≈ 6mm
The radius of the capillary tube is 6mm
So, the minimum size is
Volume = πr²h
Volume = π × 6² × 0.25
V = 2.83 mm³
The minimum size of the capillary tube is 2.83mm³
Answer:
m = 2.01[kg]
Explanation:
This problem can be solved using Newton's second law which tells us that the force applied on a body is equal to the product of mass by acceleration.

where:
F = force = 12.5 [N]
m = mass [kg]
a = acceleration = 6.2 [m/s²]
![12.5=m*6.2\\m = 2.01[kg]](https://tex.z-dn.net/?f=12.5%3Dm%2A6.2%5C%5Cm%20%3D%202.01%5Bkg%5D)