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3 years ago

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As a wave travels through a medium, it displaces particles in a direction parallel to the motion of the wave. We can conclude th

is is a

1 answer:

Gre4nikov [31]3 years ago

5 0

Transvere wave because the direction which the particles are being displaced

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A 210 g block is dropped onto a relaxed vertical spring that has a spring constant of k = 2.0 N/cm. The block becomes attached t

Yuliya22 [10]

Answer:

a) $W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J$

b) $W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J$

c) $V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s$

d) $d_1 =0.183m$ or 18.3 cm

Explanation:

For this case we have the following system with the forces on the figure attached.

We know that the spring compresses a total distance of x=0.10 m

Part a

The gravitational force is defined as mg so on this case the work donde by the gravity is:

$W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J$

Part b

For this case first we can convert the spring constant to N/m like this:

$2 \frac{N}{cm} \frac{100cm}{1m}=200 \frac{N}{m}$

And the work donde by the spring on this case is given by:

$W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J$

Part c

We can assume that the initial velocity for the block is Vi and is at rest from the end of the movement. If we use balance of energy we got:

$W_{g} +W_{spring} = K_{f} -K_{i}=0- \frac{1}{2} m v^2_i$

And if we solve for the initial velocity we got:

$V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s$

Part d

Let d1 represent the new maximum distance, in order to find it we know that :

$-1/2mV^2_i = W_g + W_{spring}$

And replacing we got:

$-1/2mV^2_i =mg d_1 -1/2 k d^2_1$

And we can put the terms like this:

$\frac{1}{2} k d^2_1 -mg d_1 -1/2 m V^2_i =0$

If we multiply all the equation by 2 we got:

$k d^2_1 -2 mg d_1 -m V^2_i =0$

Now we can replace the values and we got:

$200N/m d^2_1 -0.21kg(9.8m/s^2)d_1 -0.21 kg(5.50 m/s)^2) =0$

$200 d^2_1 -2.058 d_1 -6.3525=0$

And solving the quadratic equation we got that the solution for $d_1 =0.183m$ or 18.3 cm because the negative solution not make sense.

5 0

2 years ago

The standard normal curve is not symmetrical. <br> a. True<br> b. False

FrozenT [24]

Answer;

The above statement is false

Explanation;

Symmetrical distribution, commonly known as symmetric distribution or normal distribution, is typically unimodal, meaning it shows only one peak in graph form.

It is a type of distribution where the left side of the distribution mirrors the right side. By definition, a symmetric distribution is never a skewed distribution.

All normal distributions are symmetric and have bell-shaped density curves with a single peak.

7 0

3 years ago

Read 2 more answers

Water (density = 1 ´ 103 kg/m3) flows at 10 m/s through a pipe with radius 0.030 m. the pipe goes up to the second floor of the

Hatshy [7]

density of water = $1000 kg/m^3$

velocity of flow = $10 m/s$

radius of pipe = $0.030 m$

Height of second floor = $2 m$

Now we can use here Bernuoli's Equation to find the speed of water flow at second floor

$P_1 + 1/2\rho v_1^2 + \rho g h_1= P_2 + 1/2 \rho v_2^2 + \rho g h_2$

$P + 1/2 * 1000 * 10^2 + 1000* 9.8 * 0 = P + 1/2 * 1000 * v^2 + 1000*9.8*2$

$v = 7.8 m/s$

Now in order to find the radius of pipe we can use equation of continuity

$A_1 v_1 = A_2 v_2$

$\pi *0.030^2 * 10 = \pi * r^2 * 7.8$

$r = 0.034 m$

So radius of pipe at second floor is 0.034 meter

3 0

2 years ago

A speeding Thunderbird left skid marks on the road that were 76.7 m long when it came to a stop. If the acceleration of the car

FrozenT [24]

Answer: 39.2 m/s

Explanation:

You can use the kinematic equation:

$v_f^2=v_i^2+2a*d$

We know the final velocity because it says it came to a stop. So now all we gotta do is plug in.

$0^2=v_i^2+2(-10)(76.7)\\v_i^2=1,534\\v_i=\sqrt{1534} \\=39.166 m/s$

4 0

3 years ago

Which electromagnetic waves have the shortest wavelengths and highest frequencies?

Usimov [2.4K]

Answer: Gamma rays

Explanation: The given waves belong to the electromagnetic spectrum which consists of different electromagnetic radiations arranged in terms of increasing wavelengths or decreasing frequencies.

$E= h\times \nu$

$\nu=\frac{c}{\lambda}$

Thus $E=\frac{h\times c}{\lambda}$

E= energy

$\nu$ = frequency

c = speed of light

$\lambda$ = wavelength

Thus frequency and wavelength are inversely related. The waves having high energies ave high frequencies and have shorter wavelengths.

Thus gamma rays having highest energy have highest frequency and shortest wavelength.

3 0

2 years ago

Read 2 more answers