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VARVARA [1.3K]
3 years ago
7

Bats use a process called echolocation to find their food. This involves giving out sound waves that hit possible prey or food.

When the waves hit an object, they bounce back to the bat, allowing the bat to locate its prey. What type of wave interaction is echolocation?
Physics
2 answers:
cupoosta [38]3 years ago
8 0
The type of waves used by bats are sound waves. Most of the species use their larynx to produce ultrasound waves in the frequency range of 20 to 200 kilohertz.
These sound waves are echoed, reflected, by surroundings, in this case food or prey. These reflections are received by the specialized receptor cells in the ears of bats. The reflections are analyzed by the brain to make an image.
Fun fact: The brain cells of bats are also specialized to better analyze the frequency of ultrasound used by the bat.
Neko [114]3 years ago
4 0

reflection is the correct answer

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A 56.0kg ice skater spins about a vertical axis through her body with her arms horizontally outstretched, making 2.50 turns each
Arada [10]

Answer:

Net force = 129.4, Force as multiple of weight of her hand = 18.84

Explanation:

Given Data:

Total body weight = 56.0 kg   ;

no. of turns = 2.5/second        ;

hand to hand distance = 1.5m ;

weight of hand = 1.25% of body weight ;

Solution:

mass of hand = \frac{1.25}{100}*56 = 0.7kg ;

radius = d/2 = 1.5/2 = 0.75m     ;

Now we need to find velocity, as we know that velocity can be calculated by dividing distance by time

v = d/t = \frac{2.5*2*3.14*0.75}{1} = 11.775 m/s or 12 m/s;

a.

The formula to calculate force is given as

F = mv²/r = (0.7*11.775²)/0.75 = 129.4 N

b.

To calculate force as multiple of weight on her hand, we need to calculate the gravitational force W on her hand first.

W = gm = 9.81 * 0.7 = 6.867 N

Now the wieght on her hand can be represented by

\frac{Force_{net} }{weight of hand}  = 129.4 / 6.867 = 18.84

3 0
3 years ago
A 1.50-mm-diameter glass sphere has a charge of + 1.60 nC. What speed does an electron need to orbit the sphere 1.60 mm above th
saveliy_v [14]

Answer :

Velocity will be 3.28\times 10^{-11}m/sec

Explanation:

We have given glass surface has a diameter of 1.5 mm

And charge q = 1.60 nC

Radius of electrons orbit r = height of electron above surface + radius of sphere  = =1.6+\frac{1.5}{2}=2.35mm = 0.00235m

Force on electron is given by F=\frac{1}{4\pi \epsilon _0}\frac{qe}{r^2}, here q is charge on sphere and e is charge on electron

F=\frac{1}{4\pi \epsilon _0}\frac{qe}{r^2}=\frac{kqe}{r^2}=\frac{9\times 10^9\times 1.6\times 10^{-9}\times 1.6\times 10^{-19}}{0.00235^2}=4.172\times 10^{-13}N

This force work as centripetal force

So F=\frac{mv^2}{r}

4.172\times 10^{-13}=\frac{9.11\times 10^{-31}v^2}{0.00235}

v = =0.0328\times 10^{-9}=3.28\times 10^{-11}m/sec

   

6 0
3 years ago
Please help! the mass of an object is measured on a pan balance with a precision of 0.005 g and the recorded value of 128.01 g.
inysia [295]

Yes, these two objects have different masses.

<h3>How can we calculate that this statement is right ?</h3>

To calculate the precision of mass we are using the formula,

Precision(P) = \frac{m_1}{m_1+\triangle m}

Or,\trianglem=\frac{m_1}{P}-m₁

For the first case we are given,

m₁= The recorded value of mass

= 128.01 g

P= Precision of the mass

=0.005g

So, according to the formula, \trianglem will be,

\trianglem=  \frac{128.01}{0.005}-128.01

Or,\trianglem=25,473.99 g

Or,\trianglem=25.47 Kg

For the first case \trianglem is 25.47 Kg..

For the second case we are given,

m₁= The recorded value of mass

= 0.13 Kg

P= Precision of the mass

=0.005 Kg

So, according to the formula, \trianglem will be,

\trianglem= \frac{0.13}{0.005}-0.13

Or,\trianglem= 25.87 kg

For the second case \trianglem is 25.87 Kg.

For the two cases  \trianglem has different values, 25.47 Kg≠25.87 Kg.

Therefore we can conclude that, these two objects have different masses.

Learn more about Mass:

brainly.com/question/3187640

#SPJ4

7 0
1 year ago
Acceleration of 1.5 ms expressed in km /hr2? ​
V125BC [204]
You’re answer is 5 because !! :)
8 0
3 years ago
A 40-W lightbulb is 1.7 m from a screen. What is the intensity of light incident on the screen? Assume that a lightbulb emits ra
Sonja [21]

Answer:

Intensity, I=1.101\ W/m^2

Explanation:

Power of the light bulb, P  = 40 W

Distance from screen, r = 1.7 m

Let I is the intensity of light incident on the screen. The power acting per unit area is called the intensity of the light. Its formula is given by :

I=\dfrac{P}{A}

I=\dfrac{P}{4\pi r^2}

I=\dfrac{40\ W}{4\pi (1.7\ m)^2}

I=1.101\ W/m^2

So, the intensity of light is 1.101\ W/m^2.

6 0
3 years ago
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