Waves Please need help fast

A catamaran with a mass of 5.44×10^3 kg is moving at 12 knots. How much work is required to increase the speed to 16 knots? (One

Andre45 [30]

The work that is required to increase the speed to 16 knots is 14,176.47 Joules

If a catamaran with a mass of 5.44×10^3 kg is moving at 12 knots, hence;

5.44×10^3 kg = 12 knots

For an increased speed to 16knots, we will have:

x = 16knots

Divide both expressions

$\frac{5.44 \times 10^3}{x} = \frac{12}{16}\\12x = 16 \times 5.44 \times 10^3\\x = 7.23\times 10^3kg\\$

To get the required work done, we will divide the mass by the speed of one knot to have:

$w=\frac{7230}{0.51}\\w= 14,176.47Joules$

Hence the work that is required to increase the speed to 16 knots is 14,176.47 Joules

Learn more here: brainly.com/question/25573786

8 0

2 years ago

Write an expression for a transverse harmonic wave that has a wavelength of 2.5 m and propagates to the right with a speed of 13

lyudmila [28]

Answer:

$y = 0.14 Cos\left ( 2.512x-34.66t \right )$

Explanation:

wavelength, λ = 2.5 m

speed, v = 13.8 m/s

Amplitude, A = 0.14 m

The general equation of the transverse harmonic wave which is travelling right is given by

$y = A Sin\left ( \frac{2\pi }{\lambda } (x - vt)+\phi \right )$

where, Ф is phase

At t = 0, x = 0 , y = 0.14 m

0.14 = 0.14 Sin Ф

Ф = π/2

So, the equation is

$y = 0.14Sin\left ( \frac{2\pi }{2.5 } (x - 13.8t)+\frac{\pi }{2} \right )$

$y = 0.14 Cos\left ( 2.512x-34.66t \right )$

3 0

3 years ago

A 570 kg elevator accelerates downwards at 1.5 m/s2 for the first 13 m of its motion.

jeka94

Mass of the elevator (m) = 570 Kg
Acceleration = 1.5 m/s^2
Distance (s) = 13 m
Let the force be F.
We know, F = ma,
Therefore, F = (570 × 1.5) N = 855 N
Angle between distance and force (θ) = 0°
We know, work done = F s Cos θ
Therefore, work done by the cable during this part
= (855 × 13 × Cos 0°) J
= (855 × 13 × 1) J
= 11115 J

Answer:

11115 J

Hope you could get an idea from here.

Doubt clarification - use comment section.

6 0

2 years ago

An alternator consists of a coil of area A with N turns that rotates in a uniform field B around a diameter perpendicular to the

svet-max [94.6K]

Answer:

(a): emf = $\rm 2\pi f NBA\sin(2\pi ft).$

(b): Amplitude of alternating voltage = 20.942 Volts.

Explanation:

Given:

Area of the coil = A.
Number of turns of coil = N.
Magnetic field = B
Rotation frequency = f.

(a):

The magnetic flux through the coil is given by

$\phi = \vec B \cdot \vec A=BA\cos\theta$

where,

$\vec A$ = area vector of the coil directed along the normal to the plane of the coil.

$\theta$ = angle between $\vec B$ and $\vec A$ .

Assuming, the direction of magnetic field is along the normal to the plane of the coil initially.

At any time t, the angle which magnetic field makes with the normal to the plane of the coil is $2\pi ft$

Therefore, the magnetic flux linked with the coil at any time t is given by

$\phi(t) = NBA\cos(2\pi ft)$

According to Faraday's law of electromagnetic induction, the emf induced in the coil is given by

$e=-\dfrac{d\phi}{dt}\\=-\dfrac{d(NBA\cos(2\pi ft))}{dt}\\=-NBA(-2\pi f\sin(2\pi ft))\\=2\pi f NBA\sin(2\pi ft).$

(b):

The amplitude of the alternating voltage is the maximum value of the emf and emf is maximum when $\sin(2\pi ft)=1.$

Therefore, the amplitude of the alternating voltage is given by

$e_o = 2\pi ft NBA.$

We have,

$N=100\\A=10^{-2}\ m^2\\B=0.1\ T\\f=2000\ rev/ min = 2000\times \dfrac{1}{60}\ rev/sec=33.33\ rev/sec.$

Putting all these values,

$e_o = 2\pi \times 33.33\times 100\times 0.1\times 10^{-2}=20.942\ Volts.$

7 0

3 years ago

An electric heater containing two heating wires X and Y is connected to a power supply of electromotive force(emf) 9.0V and negl

mina [271]

Answer:

0.4 ohms.

Explanation:

From the circuit,

The voltage reading in the voltmeter = voltage drop across each of the parallel resistance.

1/R' = 1/R1+1/R2

R' = (R1×R2)/(R1+R2)

R' = (2.4×1.2)/(2.4+1.2)

R' = 2.88/3.6

R' = 0.8 ohms.

Hence the current flowing through the circuit is

I = V'/R'................ Equation 1

Where V' = voltmeter reading

I = 6/0.8

I = 7.5 A

This is the same current that flows through the variable resistor.

Voltage drop across the variable resistor = 9-6 = 3 V

Therefore, the resistance of the variable resistor = 3/7.5

Resistance = 0.4 ohms.

7 0

3 years ago