PHYSICS COLLEGE

g A particular guitar string has a length of 60.0 cm, and a mass per unit length of 2.00 grams/meter. You hear a pure tone of 660 Hz when a particular standing wave, represented by the animation above, is excited on the string. Calculate the wavelength of this standing wave.

Answers

Answer 1

Answer:

Answer:

wavelength of the standing wave will be equal to 30 cm

Explanation:

We have given length of the guitar string L = 60 cm

Mass per unit length $\mu =2gram/m$

Frequency is given f = 660 Hz

We have to find the wavelength of the standing wave

Length of the string will be 2 times of the wavelength of the wave

So $L=2\lambda$

$\lambda =(L)/(2)=(60)/(2)=30cm$

So wavelength of the standing wave will be equal to 30 cm

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You are standing on the edge of a ravine that is 17.5 m wide. You notice a cave on the opposite wall whose ceiling is 7.3 m below your feet. The cave is 4.2 m deep, and has a vertical back wall. You decide to kick a rock across the ravine into the cave. What initial horizontal velocity must you give the rock so that the rock barely misses the overhang? The acceleration of gravity is 9.8 m/s 2 .

Answers

Answer:

v₀ₓ = 14.34 m / s

Explanation:

We can solve this problem using the projectile launch equations.

Let's look for the time it takes to descend to the height of the cave

y = $v_(oy)$ t - ½ g t²

As it rises horizontally the initial vertical speed is zero

y = 0 - ½ gt²

t = √2 y / g

t = √2 7.3 / 9-8

t = 1.22 s

This is the same time to cross the ravine

x = v₀ₓ t

v₀ₓ = x / t

v₀ₓ = 17.5 / 1.22

v₀ₓ = 14.34 m / s

This is the minimum speed.

A circular loop of wire has radius 7.80cm . A sinusoidal electromagnetic plane wave traveling in air passes through the loop, with the direction of the magnetic field of the wave perpendicular to the plane of the loop. The intensity of the wave at the location of the loop is 2.03�10?2W/m2 , and the wavelength of the wave is 6.20m .What is the maximum emf induced in the loop? Express your answer with the appropriate units.I stumbled through the formulas I do know for EMF but cant seem to figure out how to get the right answer. Please help and provide explanation! Thanks

Answers

Answer:

fem = - 4.50 10²² V

Explanation:

For the solution of this problem we must use the equation of the induced electromotive force or Faraday's law

E = - d Φ $._(B)$ / dt = d (BA cos θ) dt

In this case they tell us that the magnetic field is perpendicular to the plane of the loop, as the normal to the surface of the loop is in the direction of the radius, the angle between the field and this normal is zero, so cos 0º = 1. The area of the loop is constant, with this the equation is

E = - A dB / dt (1)

To find field B, we have the relationships of electromagnetic waves

E = c B

The intensity or poynting vector for the wave is described by the equation

S = I = 1 / μ₀ E x B = 1 /μ₀ E B

We replace

I = 1 /μ₀ (cB) B = c /μ₀ B²

This is the instantaneous intensity.

B = √ (μ₀ I /c)

We substitute in equation 1

E = - A μ₀/c d I / dt

With the maximum value we are asked to change it derived from variations

E = -A c/μ₀ ΔI / Δt

It remains to find the time of the variation. Let's use the equation

c = λ f = λ / T

T = λ / c

T = 6.20 / 3 10⁸

T = 2.06 10⁻⁸ s

We already have all the values to calculate the fem

fem = - π r² c/μ₀ ΔI/Δt

fem = - (π 0.078²) (3 10⁸/(4π 10⁻⁷) (2.03 10² -0) / (2.06 10⁻⁸ - 0)

fem = - 4.50 10²² V

A firefighting crew uses a water cannon that shoots water at 27.0 m/s at a fixed angle of 50.0 ∘ above the horizontal. The firefighters want to direct the water at a blaze that is 12.0 m above ground level. How far from the building should they position their cannon? There are two possibilities (d1Part A:
d1=_____m
Part B:
d2=______m

Answers

Answer:

Explanation:

In projectile motion , range of projectile is given by the expressions

R = u²sin2θ / g

where u is velocity of projectile.

u = 27 m/s θ = 50

12 = 27² sin 2θ / 9.8

sin 2θ = .16

θ = 9.2 / 2

= 4.6

When we place 90- θ in place of θ , in the formula of range , we get the same value of projectile. hence at 85.4 ° , the range will be same.

If a ball leaves the ground with a vertical velocity of 5.46 m/s, how long does it takethe ball to reach the highest point?

Answers

Answer:

0.557 s

Explanation:

Given:

v₀ = 5.46 m/s

v = 0 m/s

a = -9.8 m/s²

Find: t

v = at + v₀

0 m/s = (-9.8 m/s²) t + 5.46 m/s

t = 0.557 s

A metal block of mass 3 kg is falling downward and has velocity of 0.44 m/s when it is 0.8 m above the floor. It strikes the top of a relaxed vertical spring of length 0.4 m. The spring constant is 2000 N/m. After striking the spring, the block rebounds. What is the maximum height above the floor that the block reaches after the impact

Answers

Answer:

$y_(max) = 0.829\,m$

Explanation:

Let assume that one end of the spring is attached to the ground. The speed of the metal block when hits the relaxed vertical spring is:

$v = \sqrt{(0.8\,(m)/(s))^(2) + 2\cdot (9.807\,(m)/(s^(2)) )\cdot (0.4\,m)}$

$v = 2.913\,(m)/(s)$

The maximum compression of the spring is calculated by using the Principle of Energy Conservation:

$(3\,kg)\cdot (9.807\,(m)/(s^(2)))\cdot (0.4\,m) + (1)/(2)\cdot (3\,kg)\cdot (2.913\,(m)/(s) )^(2) = (3\,kg) \cdot (9.807\,(m)/(s^(2)))\cdot (0.4\,m-\Delta s) + (1)/(2)\cdot (2000\,(N)/(m))\cdot (\Delta s) ^(2)$

After some algebraic handling, a second-order polynomial is formed:

$12.728\,J = (1)/(2)\cdot (2000\,(N)/(m) )\cdot (\Delta s)^(2) - (3\,kg)\cdot (9.807\,(m)/(s^(2)) )\cdot \Delta s$

$1000\cdot (\Delta s)^(2)-29.421\cdot \Delta s - 12.728 = 0$

The roots of the polynomial are, respectively:

$\Delta s_(1) \approx 0.128\,m$

$\Delta s_(2) \approx -0.099\,m$

The first root is the only solution that is physically reasonable. Then, the elongation of the spring is:

$\Delta s \approx 0.128\,m$

The maximum height that the block reaches after rebound is:

$(3\,kg) \cdot (9.807\,(m)/(s^(2)) )\cdot (0.4\,m-\Delta s) + (1)/(2)\cdot (2000\,(N)/(m))\cdot (\Delta s)^(2) = (3\,kg)\cdot (9.807\,(m)/(s^(2)) )\cdot y_(max)$

$y_(max) = 0.829\,m$

Answer:

0.81 m

Explanation:

In all moment, the total energy is constant:

Energy of sistem = kinetics energy + potencial energy = CONSTANT

So, it doesn't matter what happens when the block hit the spring, what matters are the (1) and (2) states:

(1): metal block to 0.8 m above the floor

(2): metal block above the floor, with zero velocity ( how high, is the X)

Then:

$E_(kb1) + E_(gb1) = E_(kb2) + E_(gs2)$

$E_(kb1) + E_(gb1) = 0 + E_(gs2)$

$(1)/(2)*m*V_(b1) ^(2) + m*g*H_(b1) = m*g*H_(b2)$

$H_(b2) = (V_(b1) ^(2) )/(2g) + H_(b1)$

Replacing data:

$H_(b2) = (0.44^(2) )/(2*9.81) + 0.8$

$H_(b2) = (0.44^(2) )/(2*9.81) + 0.4$

HB2 ≈ 0.81 m

A spring has a spring constant of 1350 N/m. You place the spring vertically with one end on the floor. You then drop a 1.3 kg book onto it from a height of 0.8 m above the top of the spring. Find the maximum distance the spring will be compressed. Express your answer with the appropriate mks units.

Answers

Answer:

0.123 m.

Explanation:

From Hook's law,

The potential energy of the book = the energy stored in the spring.

mgh = 1/2ke².................. Equation 1

Where m= mass of the book, g = acceleration due to gravity, h = height, k = spring constant of the spring, e = distance of compression.

make e the subject of the equation

e = √(2mgh/k).................. Equation 2

Given: m = 1.3 kg, h = 0.8 m, k = 1350 N/m

Constant: g = 9.8 m/s²

Substitute into equation 2

e = √(2×1.3×0.8×9.8/1350)

e = √(20.384/1350)

e = √(0.0151)

e = 0.123 m.

Answer:

0.015m (downwards)

Explanation:

When the book is dropped on the top of the spring at that height, the potential energy ( $E_(P)$ ) of the book is converted to elastic energy ( $E_(E)$ ) on the spring thereby causing a compression on the spring. i.e

$E_(P)$ = $E_(E)$

But;

The potential energy $E_(P)$ of the mass (book), is the product of the mass(m) of the book, the height(h) from which it was dropped and the acceleration due to gravity (g). i.e

$E_(P)$ = - m x g x h [the -ve sign shows a decrease in height as the mass (book) drops]

Also;

The elastic energy ( $E_(E)$ ) of compression of the spring is given by

$E_(E)$ = $(1)/(2)$ x k x c