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marysya [2.9K]
3 years ago
14

The froghopper Philaenus spumarius is supposedly the best jumper in the animal kingdom. To start a jump, this insect can acceler

ate at 4.00 km/s2 over a distance of 2.0 mm as it straightens its specially designed "jumping legs."
(a) Find the upward velocity with which the insect takes off.
(b) In what time interval does it reach this velocity?
(c) How high would the insect jump if air resistance were negligible? The actual height it reaches is about 70 cm, so air resistance must be a noticeable force on the leaping froghopper.
Physics
1 answer:
Kay [80]3 years ago
5 0

Answer:

Part a)

v_f = 4 m/s

Part b)

t = 0.001 s

Part c)

d = 0.815 m

Explanation:

Part a)

As we know that initially the grass hopper is at rest at the ground position

Now the acceleration is given as

a = 4000 m/s^2

distance of the legs that it stretched is given as

s = 2.0 mm

so we have

v_f^2 - v_i^2 = 2 a d

v_f^2 - 0 = 2(4000)(0.002)

v_f = 4 m/s

Part b)

time taken to reach this speed is given as

v_f - v_i = at

4 - 0 = 4000 t

t = 0.001 s

Part c)

as the grass hopper reach the maximum height its final speed would be zero

so we will have

v_f^2 - v_i^2 = 2 a d

0 - 4^2 = 2(-9.81) d

d = 0.815 m

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Answer:

14 m/s

Explanation:

Using the principle of conservation of energy, the potential energy is converted to kinetic energy, assuming any losses.

Kinetic energy is given by ½mv²

Potential energy is given by mgh

Where m is the mass, v is the velocity, g is acceleration due to gravity and h is the height.

Equating kinetic energy to be equal to potential energy then

½mv²=mgh

V

Making v the subject of the formula

v=√(2gh)

Substituting 9.81 m/s² for g and 10 m for h then

v=√(2*9.81*10)=14.0071410359145 m/s

Rounding off, v is approximately 14 m/s

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3 years ago
A 78.5-kg man floats in freshwater with 3.2% of his volume above water when his lungs are empty, and 4.85% of his volume above w
Dima020 [189]

Answer:

A) V_air = 1.295 L

B) Volume is not reasonable

Explanation:

A) Let;

m be total mass of the man

m_p be the mass of the man that pulled out of the water because of the buoyant force that pulled out of the lung

m_3 be the mass above the water with the empty lung

m_5 be the mass above the water with full lung

F_b be the buoyant force due to the air in the lung

V_a be the volume of air inside man's lungs

w_p be the weight that the buoyant force opposes as a result of the air.

Now, we are given;

m = 78.5 kg

m_3 = 3.2% × 78.5 = 2.512 kg

m_5 = 4.85% × 78.5 = 3.80725 kg

Now, m_p = m_5 - m_3

m_p = 3.80725 - 2.512

m_p = 1.29525 kg

From archimedes principle, we have the formula for buoyant force as;

F_b = (m_displaced water)g = (ρ_water × V_air × g)

Where ρ_water is density of water = 1000 kg/m³

Thus;

F_b = w_p = 1.29525 × 9.81

F_b = 12.7064 N

As earlier said,

F_b = (ρ_water × V_air × g)

Thus;

V_air = F_b/(ρ_water × × g)

V_air = 12.7064/(1000 × 9.81)

V_air = 1.295 × 10^(-3) m³

We want to convert to litres;

1 m³ = 1000 L

Thus;

V_air = 1.295 × 10^(-3) × 1000

V_air = 1.295 L

B) From research, the average lung capacity of an adult human being is 6 litres of air.

Thus, the calculated lung volume is not reasonable

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A person of mass 70 kg stands at the center of a rotating merry-go-round platform of radius 2.9 m and moment of inertia 900 kg⋅m
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Explanation:

It is given that,

Mass of person, m = 70 kg

Radius of merry go round, r = 2.9 m

The moment of inertia, I_1=900\ kg.m^2

Initial angular velocity of the platform, \omega=0.95\ rad/s

Part A,

Let \omega_2 is the angular velocity when the person reaches the edge. We need to find it. It can be calculated using the conservation of angular momentum as :

I_1\omega_1=I_2\omega_2

Here, I_2=I_1+mr^2

I_1\omega_1=(I_1+mr^2)\omega_2

900\times 0.95=(900+70\times (2.9)^2)\omega_2

Solving the above equation, we get the value as :

\omega_2=0.574\ rad/s

Part B,

The initial rotational kinetic energy is given by :

k_i=\dfrac{1}{2}I_1\omega_1^2

k_i=\dfrac{1}{2}\times 900\times (0.95)^2

k_i=406.12\ rad/s

The final rotational kinetic energy is given by :

k_f=\dfrac{1}{2}(I_1+mr^2)\omega_1^2

k_f=\dfrac{1}{2}\times (900+70\times (2.9)^2)(0.574)^2

k_f=245.24\ rad/s

Hence, this is the required solution.

5 0
3 years ago
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