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2 years ago

10

the (speed) of an object is a vector quantity that tells us the speed and direction of an objects momentum is this correct? If n

ot what is the correct answer

2 answers:

Ivenika [448]2 years ago

7 0

Answer:

sorry kailangan points wh

irina [24]2 years ago

5 0

The statement is incorrect.

The second word in the statement ... the word in (parentheses) ... should be (velocity), not (speed).

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A whistle you use to call your hunting dog has a frequency of 21 kHz, but your dog is ignoring it. You suspect the whistle may n

uranmaximum [27]

The Doppler effect is the right concept to solve this problem. The Doppler effect is understood as the change in apparent frequency of a wave produced by the relative movement of the source with respect to its observer. Mathematically it can be described as,

$f = (1-\frac{v_0}{v})f_0$

Here,

$f_0$ = Frequency of the sound from the Whistle

f = Frequency of sound heard

v = Speed of the sound in the Air

Replacing we have that

$1- \frac{v_0}{343} = \frac{20kHz}{21kHz}$

$\frac{v_0}{343} = 1-\frac{20}{21}$

$\frac{v_0}{343} = \frac{1}{21}$

$v_0 = \frac{1}{21}(343)$

$v_0 = 16.33m/s$

Therefore the minimum speed to know if the whistle is working is 16.33m/s

3 0

3 years ago

A 10 g particle undergoes SHM with an amplitude of 2.0 mm and a maximum acceleration of magnitude 8.0 multiplied by 103 m/s2, an

Nat2105 [25]

Answer:

a) $T=0.0031416s$

b) $v_{max}=6.283\frac{m}{s}$

c) $E=0.1974J$

d) $F=80N$

e) $F=40N$

Explanation:

1) Important concepts

Simple harmonic motion is defined as "the motion of a mass on a spring when it is subject to the linear elastic restoring force given by Hooke's Law (F=-Kx). The motion experimented by the particle is sinusoidal in time and demonstrates a single resonant frequency".

2) Part a

The equation that describes the simple armonic motion is given by $X=Acos(\omega t +\phi)$ (1)

And taking the first and second derivate of the equation (1) we obtain the velocity and acceleration function respectively.

For the velocity:

$\frac{dX}{dt}=v(t)=-A\omega sin(\omega t +\phi)$ (2)

For the acceleration

$\frac{d^2 X}{dt}=a(t)=-A\omega^2 cos(\omega t+\phi)$ (3)

As we can see in equation (3) the acceleration would be maximum when the cosine term would be -1 and on this case:

$A\omega^2=8x10^{3}\frac{m}{s^2}$

Since we know the amplitude A=0.002m we can solve for $\omega$ like this:

$\omega =\sqrt{\frac{8000\frac{m}{s^2}}{0.002m}}=2000\frac{rad}{s}$

And we with this value we can find the period with the following formula

$T=\frac{2\pi}{\omega}=\frac{2 \pi}{2000\frac{rad}{s}}=0.0031416s$

3) Part b

From equation (2) we see that the maximum velocity occurs when the sine function is euqal to -1 and on this case we have that:

$v_{max}=A\omega =0.002mx2000\frac{rad}{s}=4\frac{m rad}{s}=4\frac{m}{s}$

4) Part c

In order to find the total mechanical energy of the oscillator we can use this formula:

$E=\frac{1}{2}mv^2_{max}=\frac{1}{2}(0.01kg)(6.283\frac{m}{s})^2=0.1974J$

5) Part d

When we want to find the force from the 2nd Law of Newton we know that F=ma.

At the maximum displacement we know that X=A, and in order to that happens $cos(\omega t +\phi)=1$ , and we also know that the maximum acceleration is given by::

$|\frac{d^2X}{dt^2}|=A\omega^2$

So then we have that:

$F=ma=mA\omega^2$

And since we have everything we can find the force

$F=ma=0.01Kg(0.002m)(2000\frac{rad}{s})^2 =80N$

6) Part e

When the mass it's at the half of it's maximum displacement the term $cos(\omega t +\phi)=1/2$ and on this case the acceleration would be given by;

$|\frac{d^2X}{dt^2}|=A\omega^2 cos(\omega t +\phi)=A\omega^2 \frac{1}{2}$

And the force would be given by:

$F=ma=\frac{1}{2}mA\omega^2$

And replacing we have:

$F=\frac{1}{2}(0.01Kg)(0.002m)(2000\frac{rad}{s})^2 =40N$

8 0

2 years ago

The green bars are called "error bars." They indicate the range of uncertainty that scientists have about the data on the graph.

vovangra [49]

This graph shows data up to about 2010. So it couldn't have been drawn before 2010. OF COURSE the data from only 10 years earlier was more reliable than the data that was 120 years old ! It wasn't even measured the same way back then as it is now.

4 0

3 years ago

Read 2 more answers

A uniform solid cylinder of radius r and mass m can roll inside a hollow cylinder of radius R > r without slipping. A pendulu

iogann1982 [59]

Answer: The two answers are in explanation.

Explanation: Please find the attached files for the solution

8 0

3 years ago

A 1400 kg car drives up a 16 m high hill. The car’s speeds at the bottom and top are 21 m/s and 13 m/s, respectively. If the car

Roman55 [17]

Answer:

Explanation:

mass, m = 1400 kg

height, h = 16 m

initial velocity, u = 21 m/s

final velocity, v = 13 m/s

Work done by engine, We = 80 kJ

Let the work done by the friction force is Wf.

Use the work energy theorem

net work done = change in kinetic energy

work done by engine + work done by friction force + work done by the gravitational force = Change in kinetic energy

80000 + Wf - m x g x h = 0.5 m ( v² - u²)

80000 + Wf - 1400 x 9.8 x 16 = 0.5 x 1400 x ( 169 - 441 )

- 139520 + Wf = - 190400

Wf = 50880 J

4 0

3 years ago