The froghopper (Philaenus spumarius), the champion leaper of the insect world, has a mass of 12.3mg and leaves the ground (in the most energetic jumps) at 4.00m/s from a vertical start. The jump itself lasts a mere 1.00ms before the insect is clear of the ground. (a) find the force that the ground exerts on the froghopper during its jump. (b) express the force of (a) in terms of the frog hoppers weight.
F=ma=mv/t =12.3•10⁻⁶•4/1•10⁻³ =49.2•10⁻³ N
F/mg = 49.2•10⁻³/12.3•10⁻⁶•9.8 =408.2
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did partial reflection and refraction still occur when light was travelling more slowly in the first medium than in second? explain?
Elena's answer is actually right
To find the force that the ground exerts on the froghopper during its jump, we can use Newton's second law, which states that force (F) is equal to mass (m) multiplied by acceleration (a). In this case, the acceleration can be calculated using the initial velocity (u), final velocity (v), and time (t) using the formula a = (v - u) / t.
Given:
Mass (m) = 12.3 mg = 12.3 x 10^-6 kg
Initial velocity (u) = 0 m/s (since the jump starts from rest)
Final velocity (v) = 4.00 m/s
Time (t) = 1.00 ms = 1.00 x 10^-3 s
(a) Calculation of acceleration:
a = (v - u) / t
= (4.00 - 0) / (1.00 x 10^-3)
= 4.00 x 10^3 m/s^2
Now, we can use Newton's second law to find the force:
F = m * a
= 12.3 x 10^-6 kg * 4.00 x 10^3 m/s^2
= 0.0492 N
Therefore, the force that the ground exerts on the froghopper during its jump is 0.0492 Newtons.
(b) To express the force in terms of the frog hopper's weight, we can use the equation W = m * g, where W is the weight, m is the mass, and g is the acceleration due to gravity.
Weight (W) = m * g
The force exerted by the ground is equal to the weight of the froghopper, so we can write:
Force (F) = Weight (W)
Dividing both sides of the equation by m, we get:
F / m = W / m
Since W = m * g, we can substitute:
F / m = m * g / m
The mass (m) cancels out, and we're left with:
F / m = g
Therefore, the force (F) can be expressed in terms of the froghopper's weight (W) as:
F = W = m * g
So, the force of (a) in terms of the frog hopper's weight is simply the weight of the froghopper, which is 12.3 x 10^-6 kg multiplied by the acceleration due to gravity (g).