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Oksana_A [137]
1 year ago
11

The graph shows the range of frequencies commonly heard by some animals.

Physics
2 answers:
Vsevolod [243]1 year ago
3 0

Concept:

Frequency- It is defined as the number of oscillations occur in one second.

Its SI unit is Hertz (Hz)

Given: Produced sound vibrations is 18,500 cycles in 0.75 seconds

∵ In 0.75 second, produced sound has oscillations = 18,500 cycles

∴ In 1.0 second, produced sound has oscillations = (18,500 ÷ 0.75) Hz

The frequency of the sound will be ≈ 24,667 Hz

From the study of the given graph, only the animals (c) Cats, (b) Moths and (a) Bats can hear the produced sound because their upper audible frequency range is greater than 24,667 Hz.

nignag [31]1 year ago
3 0

Answer:

A, B, C

bats, moths, cats

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An electron is accelerated by a potential difference of 3000V and enters a region of a uniform magnetic field. As a result the e
valkas [14]

<span>The election has energy of 3000 eV entering the magnetic field. </span>

converting to J, energy = qeV =1.6E-19*3000 where q=1.6E-19

<span>Kinetic energy = 1/2mv^2 with m is mass of electron</span>

1/2mv^2 = qeV

<span>v = sqrt[1.6E-19*3000*2/9.1E-31] = 3.25E6</span>

<span>Magnetic force = eVB where B is magnetic field strength</span>

<span>Force = ma where a=radial acceleration=v^2/r </span>

evB = mv^2/r

B =  2*3000/0.26/3.25E6 = 7.1E-3T = 71Gauss


5 0
1 year ago
Gibbons, small Asian apes, move by brachiation, swinging below a handhold to move forward to thenext handhold. A 9.3kggibbon has
Mumz [18]

Upward force provided by the branch is 260 N

<u>Explanation:</u>

Given -

Mass of Gibbon, m = 9.3 kg

Length of the branch, r = 0.6 m

Speed of the movement, v = 3.3 m/s

Upward force, T = ?

The tension force in the rod must be greater than the weight at the bottom of the swing in order to provide an upward centripetal acceleration.

Therefore,

F net = T - mg

F net = ma = mv²/r

Thus,

T = mv²/r + mg

T = m ( v²/r + g)

T = 9.3 [ ( 3.3)² / 0.6 + 9.8]

T = 259.9 N ≈ 260 N

Therefore, upward force provided by the branch is 260 N

5 0
1 year ago
A 146N force is needed to pull a 350 N block across a horizontal surface at a constant speed by a rope making an angle of 50 deg
vfiekz [6]

Answer:

F = force

f = friction

u = coefficient of friction

R = normal reaction force

a = Acceleration

m = mass of block

g = gravity

f = uR

F = Ma

Say the block is moving to the right.

The 146N force thus acts to the right, and the friction force to the left, since it resists movement.  

The 146N force acts to the right, but the horizontal component of it is 146 cos 50  = 93.84: So this is the force to the right.

Since F = uR and we're trying to find u, we need both F and R. R is easy to get since it is just m x g. This is in fact already given as the weight 350N. So R = 350.

The block is moving at a constant speed, so the force to the right must = the force to the left.

F = ma, so 93.84 - f = (350/g) x 0

This means f must be 93.84 also.  

so  we have f = uR,  

      93.84 = u x 350

so u = 0.268 or  

0.27 to 2dp.  

Hope you understand this.

Explanation:

7 0
1 year ago
Hurry plz!!!
lora16 [44]
The statement is true
4 0
1 year ago
Two technicians are discussing what could be the cause of slow cranking and excessive current draw. Technician A says that an en
svet-max [94.6K]

Answer:

Both the technician are correct.

Explanation:

The cause of slow cracking  and excessive current draw in the engine can be caused because of mechanical fault as well as defect in the starter motor or binding in it. Some more reasons can be defective battery, faulty electrical connections, a charging system problem or starter rundown when parked.

So, both the technician are correct here.

8 0
1 year ago
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