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2 years ago

What is the speed over the ground of a mosquito flying 2 m/s relative to the air against a 2 m/s headwind?

Physics

2 answers:

salantis [7]2 years ago

8 0

We have that the speed over the ground of a mosquito flying 2 m/s relative to the air against a 2 m/s headwind is

X=0m/s

From the question we are told that

mosquito flying 2 m/s

against a 2 m/s headwind

Generally

The speed over the ground is the Flight Speed minus resistance speed

Generally the equation for the speed over the ground is mathematically given as

X=Flight Speed-resistance speed

Therefore

X=2-2

X=0m/s

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Katyanochek1 [597]2 years ago

7 0

Answer:

The speed of mosquito over the ground is $4m/sec$ .

Explanation:

• Relative velocity is defined as the velocity of an object B in the rest frame of another object A.

• To find the speed of mosquito over the ground, use this formula:

$$${V_{{M \mathord{\left/ {\vphantom {M A}} \right$ .

$\kern-\nulldelimiterspace} A}}} = {V_{{M \mathord{\left/ {\vphantom {M G}} \right. \kern-\nulldelimiterspace} G}}} + {V_{{A \mathord{\left/ {\vphantom {A G}} \right. \kern-\nulldelimiterspace} G}}}$$$

Where $, $A$$ represents air, $M$ represents mosquito and $G$ represents ground, thus ${V_{{A \mathord{\left/ {\vphantom {A G}} \right. \kern-\nulldelimiterspace} G}}}$ represents velocity of air with respect to ground and so on.

$\\${V_{{M \mathord{\left/ {\vphantom {M A}} \right. \kern-\nulldelimiterspace} A}}} = 2m/sec$ \\${V_{{A \mathord{\left/ {\vphantom {A G}} \right. \kern-\nulldelimiterspace} G}}} = - 2m/sec$$

• Placing the value of the given in the above formula.

$& {V_{{M \mathord{\left/ {\vphantom {M A}} \right. \kern-\nulldelimiterspace} A}}} = {V_{{M \mathord{\left/ {\vphantom {M G}} \right.$

$\kern-\nulldelimiterspace} G}}} + {V_{{A \mathord{\left/ {\vphantom {A G}} \right. \kern-\nulldelimiterspace} G}}} \\ & \Rightarrow 2 = {V_{{M \mathord{\left/ {\vphantom {M G}} \right. \kern-\nulldelimiterspace} G}}} - 2 \\$

$& \Rightarrow {V_{{M \mathord{\left/ {\vphantom {M G}} \right. \kern-\nulldelimiterspace} G}}} = 4m/sec \\\end{align}\]$$$

• Hence, the speed of mosquito over the ground is $4m/sec$ .

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1 year ago

You pull on a spring whose spring constant is 22 N/m, and stretch it from its equilibrium length of 0.3 m to a length of 0.7 m.

Liono4ka [1.6K]

Answer:

W= 4.4 J

Explanation

Elastic potential energy theory

If we have a spring of constant K to which a force F that produces a Δx deformation is applied, we apply Hooke's law:

F=K*x Formula (1): The force F applied to the spring is proportional to the deformation x of the spring.

As the force is variable to calculate the work we define an average force

$F_{a} =\frac{F_{f}+F_{i} }{2}$ Formula (2)

Ff: final force

Fi: initial force

The work done on the spring is :

W = Fa*Δx

Fa : average force

Δx : displacement

$W = F_{a} (x_{f} -x_{i} )$ :Formula (3)

$x_{f}$ : final deformation

$x_{i}$ :initial deformation

Problem development

We calculate Ff and Fi , applying formula (1) :

$F_{f} = K*x_{f} =22\frac{N}{m} *0.7m =15.4N$

$F_{i} = K*x_{i} =22\frac{N}{m} *0.3m =6.6N$

We calculate average force applying formula (2):

$F_{a} =\frac{15.4N+6.2N}{2} = 11 N$

We calculate the work done on the spring applying formula (3) : :

W= 11N*(0.7m-0.3m) = 11N*0.4m=4.4 N*m = 4.4 Joule = 4.4 J

Work done in stages

Work is the change of elastic potential energy (ΔEp)

W=ΔEp

ΔEp= Epf-Epi

Epf= final potential energy

Epi=initial potential energy

$E_{pf} =\frac{1}{2} *k*x_{f}^{2}$

$E_{pi} =\frac{1}{2} *k*x_{i}^{2}$

$E_{pf} =\frac{1}{2} *22*0.7^{2} = 5.39 J$

$E_{pf} =\frac{1}{2} *22*0.3^{2} = 0.99 J$

W=ΔEp= 5.39 J-0.99 J = 4.4J

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2 years ago

A light source radiates 60.0 W of single-wavelength sinusoidal light uniformly in all directions. What is the amplitude of the m

Anna35 [415]

To solve this problem it is necessary to take into account the concepts of Intensity as a function of Power and the definition of magnetic field.

The intensity depending on the power is defined as

$I = \frac{P}{4\pi r^2},$

Where

P = Power

r = Radius

Replacing the values that we have,

$I = \frac{60}{(4*\pi (0.7)^2)}$

$I = 9.75 Watt/m^2$

The definition of intensity tells us that,

$I = \frac{1}{2}\frac{B_o^2 c}{\mu}$

Where,

$B_0 =$ Magnetic field

$\mu =$ Permeability constant

c = Speed velocity

Then replacing with our values we have,

$9.75 = \frac{Bo^2 (3*10^8)}{(4\pi*10^{-7})}$

Re-arrange to find the magnetic Field B_0

$B_o = 2.86*10^{-7} T$

Therefore the amplitude of the magnetic field of this light is $B_o = 2.86*10^{-7} T$

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3 years ago

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2 years ago

The intensity of an earthquake wave passing through the earth is measured to be 2.5×106 j/(m2⋅s) at a distance of 43 km from the

vampirchik [111]

r₁ = distance of the point from the source = 43 km = 43000 m

I₁ = intensity of earthquake wave at distance "r₁" = 2.5 x 10⁶ W/m²

r₂ = distance of the point from the source = 1.5 km = 1500 m

I₂ = intensity of earthquake wave at distance "r₂" = ?

we know that , for a constant power , the intensity of wave is inversely proportional to the distance from the source .

I α 1/r² where I = intensity of wave , r = distance from source

hence we can write

I₁/I₂ = r₂²/r₁²

inserting the values

(2.5 x 10⁶) /I₂ = (1500/43000)²

I₂ = 2.1 x 10⁹ W/m²

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3 years ago