Ask question

Ask question

All categories

3 years ago

6

When water waves approach an obstacle such as the supports holding up a pier, they will bend around the supports. This wave phen

omenon is called
a. reflection.
b. refraction.
c. diffraction.
d. polarization.

1 answer:

victus00 [196]3 years ago

4 0

It is called diffraction. C. The spreading out of the waves.

You might be interested in

What is one property of iodine

mamaluj [8]

Answer: I2

Explanation:

6 0

3 years ago

What happens when the mass of an object is greter

Paul [167]

The acceleration goes up.

4 0

3 years ago

A disk of radius 25 cm spinning at a rate of 30 rpm slows to a stop over 3 seconds. What is the angular acceleration? B. How man

Ne4ueva [31]

Answer:

A. α = - 1.047 rad/s²

B. θ = 14.1 rad

C. θ = 2.24 rev

Explanation:

A.

We can use the first equation of motion to find the acceleration:

$\omega_f = \omega_i + \alpha t$

where,

ωf = final angular speed = 0 rad/s

ωi = initial angular speed = (30 rpm)(2π rad/1 rev)(1 min/60 s) = 3.14 rad/s

t = time = 3 s

α = angular acceleration = ?

Therefore,

$0\ rad/s = 3.14\ rad/s + \alpha(3\ s)$

<u>α = - 1.047 rad/s²</u>

B.

We can use the second equation of motion to find the angular distance:

$\theta = \omega_it +\frac{1}{2}\alpha t^2\\\theta = (3.14\ rad/s)(3\ s) + \frac{1}{2}(1.04\ rad/s^2)(3)^2$

<u>θ = 14.1 rad</u>

C.

θ = (14.1 rad)(1 rev/2π rad)

<u>θ = 2.24 rev</u>

6 0

3 years ago

A 2.4 kg block is dropped onto a spring and platform of negligible mass. The block is released

statuscvo [17]

The speed of the block when the compression is 15 cm is 9.85 m/s.

The given parameters;

<em>mass of the block, m = 2.4 kg</em>
<em>height of the block, h = 5 m</em>
<em>compression of the spring, x = 25 cm = 0.25 m</em>

The spring constant is calculated as follows;

$F = kx\\\\mg = kx\\\\k = \frac{mg}{x} \\\\k = \frac{2.4 \times 9.8}{0.25} \\\\k = 94.08 \ N/m$

The speed of the block when the compression is 15 cm can be determined by applying the principle of conservation of energy;

$\Delta K.E = \Delta P.E\\\\\frac{1}{2} m(v^2 - v_{0 }^2 ) = mgh - \frac{1}{2} kx^2\\\\\frac{1}{2} mv^2 = mgh - \frac{1}{2} kx^2\\\\mv^2 = 2mgh - kx^2\\\\v^2 = \frac{2mgh - kx^2}{m} \\\\v = \sqrt{\frac{2mgh - kx^2}{m}} \\\\v = \sqrt{\frac{(2 \times 2.4 \times 9.8 \times 5) - (94.08 \times 0.15^2)}{2.4}} \\\\v = 9.85 \ m/s$

Thus, the speed of the block when the compression is 15 cm is 9.85 m/s.

Learn more here:brainly.com/question/14289286

8 0

3 years ago

There are competitions in which pilots fly small planes low over the ground and drop weights, trying to hit a target. Part A A p

tamaranim1 [39]

Explanation:

A pilot flying a weight; it takes t = 3.4 s to hit the ground, during which it travels a horizontal distance d = 120 m

h = g/2*t^2

h= 4.903*3.4^2

h=56.701 m

V = d/t

V= 120/3.4

V=35.3 m/sec

Now the pilot does a run at the same height but twice the speed.

How much time (t1) does it take the weight to hit the ground?

t1 = t = 3.4 sec (depending just upon height)

5 0

3 years ago