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3 years ago

Find the velocity of a wave of period 2 s and wavelength 0.5m. *

Physics

1 answer:

Helen [10]3 years ago

4 0

Answer:

The velocity of wave is 0.25m/s

Explanation:

First, you have to find the fequency using the formula, F = 1/Time :

$f = \frac{1}{t}$

Let t = 2s,

$f = \frac{1}{2}$

$f = 0.5Hz$

Next, you have to use speed formula, V = f×λ :

Let f = 0.5Hz,

Let λ = 0.5m,

$v = 0.5 \times 0.5$

$v = 0.25$

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Explanation:

<h2>Speed is the rate of change of distance with time.</h2>

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3 years ago

A student finds an unlabeled liquid container in his lab. He notices that the container has two liquids. Since the two liquids h

raketka [301]

Answer:

$\rho = 1848.03 kg m^{-3}$

Explanation:

given data:

density of water \rho = 1 gm/cm^3 = 1000 kg/m^3

height of water = 20 cm =0.2 m

Pressure p = 1.01300*10^5 Pa

pressure at bottom

$P = P_{fluid} + P_{h_2 o}$

$P = P_{fluid} + \rho g h$

$P_{fluid} = P - \rho g h$

= 1.01300*10^5 - 1000*0.2*9.8

= 99340 Pa

$p_{fluid} = P_{atm} + \rho g h_{fluid}$ h_[fluid} = 0.307m

$99340 = 104900 + \rho *9.8*0.307$

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3 years ago

Assume you are going to race the three objects (hollow sphere, disk and ring used in Experiment 8.2) by releasing them from rest

Ann [662]

Answer:

a. The disk

b. Because it has the smallest rotational inertia

Explanation:

a. Which object do you expect to reach the bottom of the inclined plan first?

I would expect the disk to reach the bottom first.

b. Why?

This is because the disk has the smallest rotational inertia.

The rotational inertial of the hollow sphere, disk and ring are 2/3MR², 1/2MR² and MR² respectively.

Since the three objects are rolling from the same height, they have the same mechanical energy.

But, since the disk has the smallest rotational inertia, it would have the smallest rotational kinetic energy and largest translational kinetic energy. The disk's smaller rotational kinetic energy will cause to rotate less but translate more than the other objects and thus reach the bottom first.

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3 years ago

_____Is defined as the conductors and equipment for delivering electric energy from the serving utility to the wiring system of

insens350 [35]

Answer:

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3 years ago

Calculate the percentage increase in length of a wire of diameter 2.2 mm stretched by a load of

vesna_86 [32]

Answer:

0.21%

Explanation:

We are given;

Mass; m = 100 kg

Diameter; d = 2.2 mm = 2.2 × 10^(-3) m

Young's modulus; E = 12.5 x 10^(10) N/m².

Formula for area is;

A = πd²/4

A = (π/4) x (2.2 x 10^(-3))²

A = 3.8 x 10^(-6) m²

Force; F = mg

g is acceleration due to gravity and has a constant value of 9.8 m/s²

F = 100 × 9.8

F = 980 N

Formula for young's modulus is;

E = Stress/strain

Formula for stress = F/A

Formula for strain = ΔL/L

Thus;

E = (F/A)/(ΔL/L)

Making ΔL/L the subject, we have;

ΔL/L = (F/A)/E

Plugging in the relevant values;

ΔL/L = 980/(3.8 x 10^(-6) × 12.5 × 10^(10))

ΔL/L = 0.0021

Then percentage increase in length of a wire = 0.0021 × 100% = 0.21%

8 0

3 years ago