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• ¿Qué rapidez tendrá un sonido de 150 Hz con una longitud de onda de 2 metros?

kobusy [5.1K]

Answer:

Speed = 300 m/s

Explanation:

Given the following data;

Frequency = 150 Hz

Wavelength = 2 meters

To find the speed of the wave;

Mathematically, the speed of a wave is given by the formula:

$Speed = wavelength * frequency$

Substituting into the formula, we have;

$Speed = 2 * 150$

Speed = 300 m/s

7 0

3 years ago

A student obtained a clean dry glass-stoppered flask. she weighed the flask and stopper on an analytical balance and found the t

svetoff [14.1K]

29.213 cm3

First, calculate the mass of the water used. You do this by subtracting the original mass of the flask from the combined mass of the water and flask, giving:60.735 g - 31.601 g = 29.134 g
So we now know we have 29.134 g of water. To calculate the volume of the flask, simply divide by the density of the water, giving:29.134 g / (0.9973 g/cm3) = 29.213 cm3

6 0

3 years ago

A cylindrical Nickel rod (9 mm diameter, 50 m long) is pulled in tension with a load of 6,283 N. What would the elongation of th

Norma-Jean [14]

Answer:

0.29 m

Explanation:

9 mm = 0.009 m in diameter

Cross-sectional area $A = \pi d^2/4 = \pi * 0.009^2/4 = 6.36\times 10^{-5} m^2$

Let the tensile modulus of Nickel $E = 170 \times 10^9Pa$ .

The elongation of the rod can be calculated using the following formula:

$\Delta L = \frac{F L}{A E} = \frac{6283*50}{6.36\times 10^{-5} * 170 \times 10^9} = \frac{314150}{1081200} = 0.29 m$

6 0

3 years ago

A hanging weight, with a mass of m1 = 0.365 kg, is attached by a string to a block with mass m2 = 0.825 kg as shown in the figur

morpeh [17]

The speed of the block after it has moved the given distance away from the initial position is 1.1 m/s.

<h3>Angular Speed of the pulley </h3>

The angular speed of the pulley after the block m1 fall through a distance, d, is obatined from conservation of energy and it is given as;

K.E = P.E

$\frac{1}{2} mv^2 + \frac{1}{2} I\omega^2 = mgh\\\\\frac{1}{2} m_2v_0^2 + \frac{1}{2} \omega^2(m_1R^2_2 + m_2R_2^2) + \frac{1}{2} \omega^2( \frac{1}{2} MR_1^2 + \frac{1}{2} MR_2^2) = m_1gd- \mu_km_2gd\\\\\frac{1}{2} m_2v_0^2 + \frac{1}{2} \omega^2[R_2^2(m_1 + m_2)+ \frac{1}{2} M(R_1^2 + R_2^2)] = gd(m_1 - \mu_k m_2)\\\\$

$\frac{1}{2} m_2v_0 + \frac{1}{4} \omega^2[2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] = gd(m_1 - \mu_k m_2)\\\\2m_2v_0 + \omega^2 [2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] = 4gd(m_1 - \mu_k m_2)\\\\\omega^2 [2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] = 4gd(m_1 - \mu_k m_2) - 2m_2v_0^2\\\\\omega^2 = \frac{ 4gd(m_1 - \mu_k m_2) - 2m_2v_0^2}{2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)} \\\\\omega = \sqrt{\frac{ 4gd(m_1 - \mu_k m_2) - 2m_2v_0^2}{2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)}} \\\\$

Substitute the given parameters and solve for the angular speed;

$\omega = \sqrt{\frac{ 4(9.8)(0.7)(0.365 \ - \ 0.25\times 0.825) - 2(0.825)(0.82)^2}{2(0.03)^2(0.365 \ + \ 0.825)\ \ +\ \ 0.35(0.02^2\ + \ 0.03^2)}} \\\\\omega = \sqrt{\frac{3.25}{0.00214\ + \ 0.000455 } } \\\\\omega = 35.39 \ rad/s$

<h3>Linear speed of the block</h3>

The linear speed of the block after travelling 0.7 m;

v = ωR₂

v = 35.39 x 0.03

v = 1.1 m/s

Thus, the speed of the block after it has moved the given distance away from the initial position is 1.1 m/s.

Learn more about conservation of energy here: brainly.com/question/24772394

5 0

2 years ago

Which of the following items has the most inertia while at rest?

netineya [11]

Answer:

airplane

Explanation:

as greater mass greater inertia

8 0

3 years ago