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abruzzese [7]
2 years ago
3

Male and female mosquitoes emit different sound frequencies while in flight, and their antennae are sensitive to different frequ

encies. The sound made by male mosquitoes in flight is close to 650 Hz, whereas the sound made by females is close to 400 Hz. The natural frequency of males' antennae is about 400 Hz, whereas that of females is about 200 Hz. Biologists suspect that some mosquitoes can detect the presence of other mosquitoes based on nerve signals generated when their antennae vibrate strongly.
Assuming that the biologists are correct, the emitted frequencies of flying mosquitoes and the vibrational frequencies of their antennae suggest it is likely that:

A) both sexes can detect flying females.
B) female mosquitoes can detect flying females.
C) female mosquitoes can detect flying males.
D) male mosquitoes can detect flying females.
E) male mosquitoes can detect flying males.
F) female mosquitoes can detect flying members of both sexes.
Physics
1 answer:
Phoenix [80]2 years ago
4 0

Answer:

A) both sexes can detect flying females

Explanation:

Sinve both sexes can detect flying members, and flying females and males emits different frequencies.

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What is the speed of a beam of electrons when the simultaneous influence of an electric field of 1.56×104v/m and a magnetic fiel
sashaice [31]

1) 3.38\cdot 10^6 m/s

When both the electric field and the magnetic field are acting on the electron normal to the beam and normal to each other, the electric force and the magnetic force on the electron have opposite directions: in order to produce no deflection on the electron beam, the two forces must be equal in magnitude

F_E = F_B\\qE = qvB

where

q is the electron charge

E is the magnitude of the electric field

v is the electron speed

B is the magnitude of the magnetic field

Solving the formula for v, we find

v=\frac{E}{B}=\frac{1.56\cdot 10^4 V/m}{4.62\cdot 10^{-3} T}=3.38\cdot 10^6 m/s

2) 4.1 mm

When the electric field is removed, only the magnetic force acts on the electron, providing the centripetal force that keeps the electron in a circular path:

qvB=m\frac{v^2}{r}

where m is the mass of the electron and r is the radius of the trajectory. Solving the formula for r, we find

r=\frac{mv}{qB}=\frac{(9.1 \cdot 10^{-31} kg)(3.38\cdot 10^6 m/s)}{(1.6\cdot 10^{-19} C)(4.62\cdot 10^{-3}T)}=4.2\cdot 10^{-3} m=4.1 mm

3) 7.6\cdot 10^{-9}s

The speed of the electron in the circular trajectory is equal to the ratio between the circumference of the orbit, 2 \pi r, and the period, T:

v=\frac{2\pi r}{T}

Solving the equation for T and using the results found in 1) and 2), we find the period of the orbit:

T=\frac{2\pi r}{v}=\frac{2\pi (4.1\cdot 10^{-3} m)}{3.38\cdot 10^6 m/s}=7.6\cdot 10^{-9}s

7 0
2 years ago
Two identical carts travel at the same speed toward each other, and then a collision occurs. The graphs show the momentum of eac
madam [21]

Explanation :

The interaction between two objects is termed as the collision. The collision can be of two types i.e. elastic collision and inelastic collision.

In this case, two identical carts travel at the same speed toward each other, and then a collision occurs. In an inelastic collision, the momentum before and after the collision remains the same but its kinetic energy gets lost.

After the collision, both the object sticks over each other and moves with one velocity.

Out of the given graph, the graph that shows a perfectly inelastic collision is attached. It shows that after the collision both the carts move with the same velocity.

5 0
2 years ago
A Turtle and a Snail are 360 meters apart, and they start to move towards each other at 3 p.m. If the Turtle is 11 times as fast
Neko [114]

Answer:

Snail's speed = \frac{30m}{2400s} = 0.0125m/s

Turtle's speed =  \frac{330m}{2400s} = 0.1375m/s

Explanation:

Let the snail's speed be x m/s

The turtle's speed then is 11x m/s

Speed = Distance ÷ Time

Since speed and distance are directly proportional;

The ratio of the distances snail and turtle cover before they meet is x:11x respectively.

Simplified, the ratio of snail distance : turtle distance = 1:11

So snail covers a distance of \frac{1}{12} × 360 = 30m

And turtle covers a distance of \frac{11}{12} × 360 = 330m

The time each took before they met is 40 × 60 = 2400 seconds

Snail's speed = \frac{30m}{2400s} = 0.0125m/s

Turtle's speed =  \frac{330m}{2400s} = 0.1375m/s

8 0
2 years ago
Read 2 more answers
Ron fills a beaker with glycerin (n = 1.473) to a depth of 5.0 cm. if he looks straight down through the glycerin surface, he wi
Tju [1.3M]

By law of refraction we know that image position and object positions are related to each other by following relation

\frac{\mu_1}{h_o} = \frac{\mu_2}{h_i}

here we know that

\mu_1 = 1.473

h_o = 5 cm

\mu_2 = 1

now by above formula

\frac{1.473}{5} = \frac{1}{h_i}

h_i = 3.39 cm

so apparent depth of the bottom is seen by the observer as h = 3.39 cm

7 0
2 years ago
The rotational speeds of four generators are listed in RPM (revolutions per minute). Arrange the generators in order based on th
artcher [175]

with the same generator, so the only factor for producing the slectric field is only the speed. The faster the rotational speed of the generator the greater it produce electric field. So the sequence is 3000 rpm < 3200 rpm < 3400 rpm < 3600 rpm

4 0
2 years ago
Read 2 more answers
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