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4 years ago

Please help with this

Physics

1 answer:

frosja888 [35]4 years ago

8 0

912.

outer ear:
pinna
ear canal

middle ear:
ossicles and ear drum

inner ear:
semcircular canals
cochlea
auditory nerve

13.
frequency = wavespeed ÷ wavelength

14.
if frequency increases you would experience a higher pitch in sound

15.
humans can hear 20Hz to 20kHz

16.
The Doppler effect is the change in frequency or wavelength of a wave for an observer who is moving relative to the wave source. Can be used for machines measuring speed via doppler effect.

17.
Doppler in hospitals can be used for ultrasound to provide images for diagnosis and monitoring.

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Miswer

quester [9]

Answer: 4.0

Explanation:

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3 years ago

The voltage supplied to a circuit is 17 V and the current running through is 10 A. What is the power generated?

Lelu [443]

Answer:

170 W

Explanation:

Applying

P = VI.................... Equation 1

Where P = Power generated in watt, V = Voltage supplied to the circuit, I = Current running through the circuit.

From the question,

Given: V = 17 V, I = 10 A

Substitute these values into equation 1

P = (17×10)

P = 170 Watt.

Hence the power generated is 170 W.

The right option is A. 170 W

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3 years ago

A football punker attempts to kick the football so that it lands on the ground 67.0 m from where it is kicked and stays in the a

Flauer [41]

To solve this problem we will apply the linear motion kinematic equations. We will find the two components of velocity and finally by geometric and vector relations we will find both the angle and the magnitude of the vector. In the case of horizontal speed we have to

$v_x = \frac{x}{t}$

$v_x = \frac{67}{4.5}$

$v_x = 14.89m/s$

The vertical component of velocity is

$-h = v_y t -\frac{1}{2} gt^2$

Here,

h = Height

g = Gravitational acceleration

t = Time

$v_y$ = Vertical component of velocity

$-1.23 = v_y(4.5)-\frac{1}{2} (9.8)(4.5)^2$

$-1.23= 4.5v_y - 99.225$

$v_y = 21.77m/s$

The direction of the velocity will be given by the tangent of the components, then

$tan\theta = \frac{v_y}{v_x}$

$\theta = tan^{-1} (\frac{21.77}{14.89})$

$\theta = 55.59\°$

The magnitude is given vectorially as,

$|V| = \sqrt{v_x^2+v_y^2}$

$|V| = \sqrt{14.89^2 +21.77^2}$

$|V| = 26.37m/s$

Therefore the angle is 55.59° and the velocity is 26.37m/s

6 0

3 years ago

Two identical copper blocks are connected by a weightless, unstretchable cord through a frictionless pulley at the top of a thin

likoan [24]

Answer:

13.6 N

Explanation:

Since one side of the wedge is vertical and the wedge makes and angle of 33 with the horizontal, the angle between the weight of the copper block on the incline and the incline is thus 90 - 33 = 57.

Let M be the mass of the block that hangs, m be the mass of the block on the incline and T be the tension in the weightless unstretchable cord.

We assume the motion is downwards in the direction of the hanging block, M.

We now write equations of motion for each block.

Mg - T = Ma (1) and T - mgcos57 - F = ma where F is the frictional force on the block on the incline and a is their acceleration.

Now, since both blocks do not move, a = 0.

So, Mg - T = M(0) = 0 and T - mgcos57 - F = m(0) = 0

Mg - T = 0 (3) and T - mgcos57 - F = 0 (4)

From (3), T = Mg

Substituting T into (4), we have

T - mgcos57 - F = 0

Mg - mgcos57 - F = 0

So, Mg - mgcos57 = F

F = Mg - mgcos57

F = (M - mcos57)g

Since g = acceleration due to gravity = 9.8 m/s², and M = 2.94 kg and m = 2.85 kg.

We find F, thus

F = (2.94 kg - 2.85 kgcos57)9.8 m/s²

F = (2.94 kg - 2.85 kg × 0.5446)9.8 m/s²

F = (2.94 kg - 1.552 kg)9.8 m/s²

F = (1.388 kg)9.8 m/s²

F = 13.6024 kgm/s²

F ≅ 13.6 N

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3 years ago

Clois what is the weight of a body in the earth, if its weig is 5Nin moon? <br>

Debora [2.8K]

Explanation:

because the moon has less mass than earth, the force due to gravity at the lunar surface is only about 1/6 that on earthso,the weight of a body on earth is 6×5N =30N

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3 years ago