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3 years ago

What is the average velocity of a car if it travels from position 25m to a position of -7m in 34 seconds?

Physics

2 answers:

zubka84 [21]3 years ago

6 0

Answer:

vp = 0.94 m/s

Explanation

Formula

Vp = position/ time

position: Initial position - Final position

Position = 25 m - (-7 m) = 25 m + 7 m = 32 m

Then

Vp = 32 m / 34 seconds

Vp = 0.94 m/s

CaHeK987 [17]3 years ago

4 0

Answer:

Average velocity = 0.94 m/s

Explanation:

Given

Initial position A = 25 m

Final position B = -7 m

Duration t = 34 s

Solution

Average Velocity

$v_{avg} =\frac{Displacement}{time} \\\\v_{avg} =\frac{A-B}{t} \\\\v_{avg} =\frac{25-(-7)}{34} \\\\v_{avg} =0.94 m/s$

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Expectant mothers many times see their unborn child for the first time during an ultrasonic examination. In ultrasonic imaging,

Rzqust [24]

A) A. 380 kHz

To clerly see the image of the fetus, the wavelength of the ultrasound must be 1/4 of the size of the fetus, therefore

$\lambda=\frac{1}{4}(1.6 cm)=0.4 cm=0.004 m$

The frequency of a wave is given by

$f=\frac{v}{\lambda}$

where

v is the speed of the wave

$\lambda$ is the wavelength

For the ultrasound wave in this problem, we have

v = 1500 m/s is the wave speed

$\lambda=0.004 m$ is the wavelength

So, the frequency is

$f=\frac{1500 m/s}{0.004 m}=3.75\cdot 10^5 Hz=375 kHz \sim 380 kHz$

B) B. f(c+v)/c−v

The formula for the Doppler effect is:

$f'=\frac{v\pm v_r}{v\pm v_s}f$

where

f' is the apparent frequency

v is the speed of the wave

$v_r$ is the velocity of the receiver (positive if the receiver is moving towards the source, negative if it is moving away from the source)

$v_s$ is the speed of the source (positive if the source is moving away from the receiver, negative if it is moving towards the receiver)

f is the original frequency

In this problem, we have two situations:

- at first, the ultrasound waves reach the blood cells (the receiver) which are moving towards the source with speed

$v_r = +v$ (positive)

- then, the reflected waves is "emitted" by the blood cells (the source) which are moving towards the source with speed

$v_s = -v$

also

v = c = speed of sound in the blood

So the formula becomes

$f'=\frac{c + v}{v - v_s}f$

C. A. The gel has a density similar to that of skin, so very little of the incident ultrasonic wave is lost by reflection

The reflection coefficient is

$R=\frac{(Z_1 -Z_2)^2}{(Z_1+Z_2)^2}$

where Z1 and Z2 are the acoustic impedances of the two mediums, and R represents the fraction of the wave that is reflected back. The acoustic impedance Z is directly proportional to the density of the medium, $\rho$ .

In order for the ultrasound to pass through the skin, Z1 and Z2 must be as close as possible: therefore, a gel with density similar to that of skin is applied, in order to make the two acoustic impedances Z1 and Z2 as close as possible, so that R becomes close to zero.

3 0

3 years ago

Two protons move with uniform circular motion in the presence of uniform magnetic fields. Proton one moves twice as fast as prot

lubasha [3.4K]

Answer:

r₂ = 4 r

Explanation:

For this exercise let's use Newton's second law with the magnetic force

F = q v x B

bold letters indicate vectors, the magnitude of this expression is

F = q v B sin θ

in this case we assume that the angle is 90º between the speed and the magnetic field.

If we use the rule of the right hand with the positive charge, the thumb in the direction of the speed, the fingers extended in the direction of the magnetic field, the palm points in the direction of the force, which is towards the center of the circle, therefore the force is radial and the acceleration is centripetal

a = v² / r

let's use Newton's second law

F = ma

q v B = m v² / r

r = $\frac{qB}{mv}$

Let's apply this expression to our case.

Proton 1

r = \frac{qB_1}{mv_1}

Proton 2

r₂ = $\frac{q \ B_2}{m \ v_2}$

in the exercise indicate some relationships between the two protons

* v₁ = 2 v₂

v₂ = v₁ / 2

* B₂ = 2B₁

we substitute

r₂ = $\frac{q \ 2B_1}{m \ \frac{v_1}{2} }$

r₂ = 4 $\frac{qB_1}{mv_1}$

r₂ = 4 r

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3 years ago

Determine the minimum angle at which a roadbedshould be banked

poizon [28]

To solve this problem, apply the concepts related to the relationship given between the centripetal Force and the Weight.

The horizontal force component is equivalent to the weight of the car, while the vertical component is linked to the centripetal force exerted on the car, therefore,

$T cos\theta = mg \rightarrow T = \frac{mg}{cos\theta}$

$Tsin\theta = \frac{mv^2}{r} \rightarrow T = \frac{mv^2/r}{sin\theta}$

Equating both equation we have that,

$\frac{mv^2}{r} = mgtan\theta$

$tan\theta = \frac{v^2}{rg}$

Rearranging to find the angle we have that,

$\theta = tan^{-1} (\frac{v^2}{rg})$

Our values are given as,

$r = 2.00*10^2m$

$v = 20m/s$

$\theta = tan^{-1}(\frac{20^2}{(2*10^{2})(9.8)})$

$\theta = 11.53 \°$

Therefore the minimum angle will be 11.53°

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3 years ago

Astronomers study radio waves to learn about

il63 [147K]

Answer:

Radio astronomy began in 1933 when an engineer named Karl Jansky accidentally discovered that radio waves come not just from inventions we create but also from natural stuff in space.

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4 years ago

An electric pump has 2kW power . How much water will it lift every minute if the height is 10 m ?

valkas [14]

Answer:

Given that,

Power = 2000 W
time = 60 seconds
distance= 10m

Power = work done ÷ time

Here, since the movement is vertical, w = mgh

So,

Power = mgh÷t

2000 = (m × 9.8 ×10) ÷ 60

m = (2000 ×60) ÷98

m = 1224.5kg

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3 years ago