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lilavasa [31]
3 years ago
9

A string has a mass of 0.8Kg. The length of the string is 2m. The tension in the string is 5N. The string is stretched and fixed

at both ends such that when struck it forms the second harmonic. Calculate the frequency of the wave which results on the string.
Physics
1 answer:
viktelen [127]3 years ago
3 0

Answer:

1.25Hz

Explanation:

For waves on a string, the second harmonic is obtained from;

2fo = 1/l √T/M

Where;

l = length of the string

M= mass in kilograms

T = tension in the string

2fo = 1/2√5/0.8

2f0 = 1.25Hz

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which of the numbers on this figure indicates typical continental conditions (regional metamorphism)?
Tpy6a [65]

The number that indicates typical continental conditions (regional metamorphism) is that showing schist and gneiss rocks.

<h3>What is regional metamorphism?</h3>

Regional metamorphism occurs when rocks undergoes changes as a result of high temperatures and pressure deep within the earth's crust.

Regional metamorphic rocks are usually foliated or squashed in appearance.

Examples of regional metamorphism rocks are schist and gneiss rocks.

Therefore, the figure that indicates typical continental conditions (regional metamorphism) is that showing schist and gneiss rocks.

Learn more about regional metamorphism at: brainly.com/question/14678538

#SPJ11

4 0
2 years ago
Read 2 more answers
Which device is based on the expansion of matter as temperature increases?
Nat2105 [25]
Thaattttttt would be mass vs volume im in science now. hope it helps
3 0
3 years ago
Read 2 more answers
Big Ben, a large artifact in England, has a mass of 1x10^8 kilograms and the Empire State Building 1x10^9 kilograms. The distanc
TiliK225 [7]

Answer:

The force, exerted by Big Ben on the Empire State Building is 2.66972 × 10⁻⁷ N

Explanation:

The question relates to the force of gravity experienced between two bodies

The given parameters are;

The mass of Big Ben, M₁ = 1 × 10⁸ kg

The mass of the Empire State Building, M₂ = 1 × 10⁹ kg

The distance between the two Big Ben and the Empire State Building, r = 5,000,000 meters

By Newton's Law of gravitation, we have;

F=G \times \dfrac{M_{1} \times M_{2}}{r^{2}}

Where;

F = The force exerted by Big Ben on the Empire State Building and vice versa

G = The universal gravitational constant = 6.67430 × 10⁻¹¹ N·m²/kg²

M₁, M₂, and r are the given parameters

By plugging in the values of the parameters and the constant into the equation for Newton's Law of gravitation, we have;

F=6.67430 \times 10^{-11} \times \dfrac{1 \times 10^8 \times 1 \times 10^9}{(5,000,000)^{2}} = 2.66972 \times 10^{-7}

The force, 'F', exerted by Big Ben on the Empire State Building is F = 2.66972 × 10⁻⁷ N.

3 0
3 years ago
Uniformly charged ring with 180 nC/m and radius R= 58 cm. Find the magnitude of the electric field in KN/C at a point P on the a
raketka [301]

Answer:

3.135 kN/C

Explanation:

The electric field on the axis of a charged ring with radius R and distance z from the axis is E = qz/{4πε₀[√(z² + R²)]³}

Given that R = 58 cm = 0.58 m, z = 116 cm = 1.16m, q = total charge on ring = λl where λ = charge density on ring = 180 nC/m = 180 × 10⁻⁹ C/m and l = length of ring = 2πR. So q = λl = λ2πR = 180 × 10⁻⁹ C/m × 2π(0.58 m) = 208.8π × 10⁻⁹ C and ε₀ = permittivity of free space = 8.854 × 10⁻¹² F/m

So, E = qz/{4πε₀[√(z² + R²)]³}

E = 208.8π × 10⁻⁹ C × 1.16 m/{4π8.854 × 10⁻¹² F/m[√((1.16 m)² + (0.58 m)²)]³}

E = 242.208 × 10⁻⁹ Cm/{35.416 × 10⁻¹² F/m[√(1.3456 m² + 0.3364 m²)]³}

E = 242.208 × 10⁻⁹ Cm/35.416 × 10⁻¹² F/m[√(1.682 m²)]³}

E = 6.839 × 10³ Cm²/[1.297 m]³F

E = 6.839 × 10³ Cm²/2.182 m³F

E = 3.135 × 10³ V/m

E = 3.135 × 10³ N/C

E = 3.135 kN/C

3 0
3 years ago
A parallel plate capacitor is constructed using two square metal sheets, each of side L = 10 cm. The plates are separated by a d
Snowcat [4.5K]

Answer:

The energy stored is 1.4 x 10^-9 J.

Explanation:

Side of square, L = 10 cm = 0.1 m

Distance, d = 2 mm = 0.002 m

Electric field, E = 4000 V/m

The energy stored in the capacitor is

U = 0.5 C V^2

The capacitance is given by

C = \frac{\varepsilon o A}{d}\\\\So \\\\U = 0.5\frac{\varepsilon o A}{d}\times E^2 d^2\\\\U = 0.5\times 8.85\times 10^{-12}\times 0.1\times 0.1\times 4000\times 4000\times 0.002\\\\U = 1.4\times10^{-9} J

7 0
3 years ago
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