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4 years ago

Waves on a pond are an example of which kind of wave?

Physics

1 answer:

KATRIN_1 [288]4 years ago

6 0

Waves on a pond are an example of which kind of wave?

B. surface waves

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Which of the following would be an example of basic research

abruzzese [7]

Basic research, also called pure research or fundamental research, is scientific research aimed to improve scientific theories for improved understanding or prediction of natural or other phenomena.

(source: wikipedia)

5 0

3 years ago

Read 2 more answers

A hockey puck oscillates on a frictionless, horizontal track while attached to a horizontal spring. The puck has mass 0.160 kg a

marshall27 [118]

Explanation:

The given data is as follows.

mass (m) = 0.160 kg, spring constant (k) = 8 n/m,

Maximum speed ( $v_{m}$ ) = 0.350 m/s

Formula for angular frequency is as follows.

$\omega = \sqrt{\frac{{k}{m}}$

$\omega = \sqrt{\frac{{8}{0.160}}$

$\omega$ = 7.07 rad/sec

(a) Formula to calculate the amplitude is as follows.

$\nu_{max} = A \omega$

A = $\frac{\nu}{\omega}$

= $\frac{0.35}{7.07}$

= 0.05 m

Hence, value of amplitude is 0.05 m.

(b) Displacement = 0.030 m

Formula for mechanical energy is as follows.

M.E = $\frac{1}{2}kA^{2}$

Putting the values into the above formula as follows.

M.E = $\frac{1}{2}kA^{2}$

= $\frac{1}{2} \times 8 \times (0.05)^{2}$

= $9.8 \times 10^{-3}$ Joule

For x = 0.03,

As, P.E = $\frac{1}{2} \times kx^{2}$

= $\frac{1}{2} \times 8 \times (0.03)$

= $3.6 \times 10^{-3}$

Hence, calculate the kinetic energy as follows.

K.E = M.E - P.E

= ( $9.8 \times 10^{-3}$ - $3.6 \times 10^{-3}$ ) J

= $6.2 \times 10^{-3}$ J

Thus, we can conclude that kinetic energy of the puck when the displacement of the glider is 0.0300 m is $6.2 \times 10^{-3}$ J.

7 0

3 years ago

At the local swimming pool, the diving board is elevated h = 8.5 m above the pool surface and overhangs the pool edge by L = 2 m

Tasya [4]

Answer:

Part a)

$t_w = \sqrt{\frac{2h}{g}}$

Part b)

$t_w = 1.32 s$

Part c)

$d_w = 6.88 m$

Explanation:

Part a)

As we know that the diver will have zero velocity in vertical direction

so here we can say that

$\Delta y = v_y t + \frac{1}{2}a_y t^2$

$h = \frac{1}{2}gt^2$

$t_w = \sqrt{\frac{2h}{g}}$

Part b)

as we know that

$h = 8.5 m$

$g = 9.81 m/s^2$

so we will have

$t_w = \sqrt{\frac{2(8.5)}{9.81}}$

$t_w = 1.32 s$

Part c)

Distance covered by diver from the edge of the pool is given as

$d_w = L + v_w t_w$

$d_w = 2 + (3.7)(1.32)$

$d_w = 6.88 m$

7 0

4 years ago

What aspects of his fitness program could keep him from achieving his goal?

Flura [38]

Idk maybe excersice everyday

4 0

3 years ago

if the vessel in the sample problem accelerates fir 1.00 min, what will its speed be after that minute ?

LUCKY_DIMON [66]

Answers:

a) 154.08 m/s=554.68 km/h

b) 108 m/s=388.8 km/h

Explanation:

The complete question is written below:

In 1977 off the coast of Australia, the fastest speed by a vessel on the water was achieved. If this vessel were to undergo an average acceleration of $1.80 m/s^{2}$ , it would go from rest to its top speed in 85.6 s.

a) What was the speed of the vessel?

b) If the vessel in the sample problem accelerates for 1.00 min, what will its speed be after that minute?

Calculate the answers in both meters per second and kilometers per hour

a) The average acceleration $a_{av}$ is expressed as:

$a_{av}=\frac{\Delta V}{\Delta t}=\frac{V-V_{o}}{\Delta t}$ (1)

Where:

$a_{av}=1.80 m/s^{2}$

$\Delta V$ is the variation of velocity in a given time $\Delta t$ , which is the difference between the final velocity $V$ and the initial velocity $V_{o}=0$ (because it starts from rest).