A science student makes the following statement:

A 30-cm long string, with one end clamped and the other free to move transversely, is vibrating in its second harmonic. The wave

Ann [662]

Answer:

$\lambda$ = 40 cm

Explanation:

given data

string length = 30 cm

solution

we take here equation of length that is

L = $n \times \frac{1}{4} \lambda$ ...............1

so

total length will be here

$L = \frac{\lambda}{2} + \frac{\lambda}{4}\\$

$L = \frac{3 \lambda }{4}$

so $\lambda$ will be

$\lambda = \frac{4L}{3}\\\lambda = \frac{4\times 30}{3}$

$\lambda$ = 40 cm

5 0

3 years ago

A steel bar of rectangular cross section (1.5 in. 2 3 .0 in.) carries a tensile load P (see fig- ure). The allowable stresses in

lukranit [14]

Explanation:

Value of the cross-sectional area is as follows.

A = $1.5 \times 2.30$

= 3.45 $in^{2}$

The given data is as follows.

Allowable stress = 14,500 psi

Shear stress = 7100 psi

Now, we will calculate maximum load from allowable stress as follows.

$P_{max} = \sigma_{a}A$

= $14500 \times 3.45$

= 50025 lb

Now, maximum load from shear stress is as follows.

$P_{max} = 2 \times \tau_{a} \times A$

= $2 \times 7100 \times 3.45$

= 48990 lb

Hence, $P_{max}$ will be calculated as follows.

$P_{max} = min((P_{max})_{\sigma}, (P_{max})_{\tau})$

= 48990 lb

Thus, we can conclude that the maximum permissible load $P_{max}$ is 48990 lb.

4 0

3 years ago

A student throws a rock upwards. The rock reaches a maximum height 2.4 seconds after it was released.

Allisa [31]

Answer:

23.52 m/s

Explanation:

The following data were obtained from the question:

Time taken (t) to reach the maximum height = 2.4 s

Acceleration due to gravity (g) = 9.8 m/s²

Initial velocity (u) =..?

At the maximum height, the final velocity (v) is zero. Thus, we can obtain how fast the rock (i.e initial velocity)

was thrown as follow:

v = u – gt (since the rock is going against gravity)

0 = u – (9.8 × 2.4)

0 = u – 23.52

Collect like terms

0 + 23.52 = u

u = 23.52 m/s

Therefore, the rock was thrown at a velocity of 23.52 m/s.

7 0

3 years ago

Two black holes (the remains of exploded stars), separated by a distance of

jolli1 [7]

The largest mass is 4.7 x 10³⁰ kg and the smallest mass is 5 x 10²⁹ kg.

The given parameters;

distance between the two black holes, r = 10 AU = 1.5 x 10¹² m
gravitational force between the two black holes, F = 6.9 x 10²⁵ N.
combined mass of the two black holes = 5.20 x 10³⁰ kg

The product of the two masses is calculated from Newton's law of universal gravitational as follows;

$F = \frac{Gm_1m_2}{r^2} \\\\m_1m_2 = \frac{Fr^2}{G} \\\\m_1m_2 = \frac{(6.9\times 10^{25}) \times (1.5\times 10^{12})^2}{6.67\times 10^{-11}} \\\\m_1m_2 = 2.328 \times 10^{60} \ kg^2$

The sum of the two masses is given as;

m₁ + m₂ = 5.2 x 10³⁰ kg

m₂ = 5.2 x 10³⁰ kg - m₁

The first mass is calculated as follows;

m₁(5.2 x 10³⁰ - m₁) = 2.328 x 10⁶⁰

5.2 x 10³⁰m₁ - m₁² = 2.328 x 10⁶⁰

m₁² - 5.2 x 10³⁰m₁ + 2.328 x 10⁶⁰ = 0

solve the quadratic equation using formula method;

a = 1, b =- 5.2 x 10³⁰, c = 2.328 x 10⁶⁰

$m_1 = \frac{-b \ \ +/- \ \ \sqrt{b^2 - 4ac} }{2a} \\\\m_1 = \frac{-(-5.2\times 10^{20}) \ \ +/- \ \ \sqrt{(-5.2\times 10^{20})^2 - 4(1\times 2.328\times 10^{60})} }{2(1)} \\\\m_1 = 4.7 \times 10^{30} \ kg \ \ or \ \ 4.9 \times 10^{29} \ kg$