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3 years ago

The height of a transverse wave from the midpoint to the crest or trough is the .

Physics

2 answers:

Ne4ueva [31]3 years ago

7 0

From the midpoint to the crest or trough, the height of a transverse wave is known as amplitude.

<u>Explanation: </u>

The crest’s or trough’s midpoint in the transverse wave are amplitudes in the height. The transverse waves which propagate to the perpendicular of the line in it. The waves have three features, namely, frequency, amplitude and wavelength.

Amplitude defines as the vertical height from the equilibrium line or position in the mid-point of the wave formation. There are two formation in transverse waves which are crest and trough.

Sedaia [141]3 years ago

4 0

Answer:Amplitude

Explanation:

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Answer:

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4 years ago

3. You are flying 2586 miles from San Francisco to New York. An hour into the flight, you are 600

Citrus2011 [14]

Answer:

268.2 m/s (1dp)

Explanation:

600 miles = 965,606m (1mile = 1609.34m)

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3 years ago

Read 2 more answers

How do i find stretch? The problem in questioning has already given me the elastic energy and k-value, but I have no idea how to

finlep [7]

Answer:

Stretch can be obtained using the Elastic potential energy formula.

The expression to find the stretch (x) is $x=\sqrt{\frac{2\times EPE}{k}}$

Explanation:

Given:

Elastic potential energy (EPE) of the spring mass system and the spring constant (k) are given.

To find: Elongation in the spring (x).

We can find the elongation or stretch of the spring using the formula for Elastic Potential Energy (EPE).

The formula to find EPE is given as:

$EPE=\frac{1}{2}kx^2$

Rewriting the above expression in terms of 'x', we get:

$x=\sqrt{\frac{2\times EPE}{k}}$

Example:

If EPE = 100 J and spring constant, k = 2 N/m.

Elongation or stretch is given as:

$x=\sqrt{\frac{2\times EPE}{k}}\\\\x=\sqrt{\frac{2\times 100}{2}}\\\\x=\sqrt{100}=10\ m$

Therefore, the stretch in the spring is 10 m.

So, stretch in the spring can be calculated using the formula for Elastic Potential Energy.

6 0

3 years ago

A rock is lifted by a machine to a height of 10m. If it has a mass of 22 kilograms

zmey [24]

Answer:

2156J

Explanation:

Given parameters:

Height of lift = 10m

Mass = 22kg

Unknown:

Work done by the machine = ?

Solution:

Work done is the force applied to move a body through a certain distance.

So;

Work done = Force x distance

Here;

Work done = mass x acceleration due to gravity x height

Work done = 22 x 9.8 x 10 = 2156J

5 0

3 years ago

A sports car is advertised to be able to stop in a distance of 50.0 m from a speed of 80 km. What is its acceleration and how ma

Flauer [41]

Explanation:

Given that,

Initial speed of the sports car, u = 80 km/h = 22.22 m/s

Final speed of the runner, v = 0

Distance covered by the sports car, d = 80 km = 80000 m

Let a is the acceleration of the sports car. It can be calculated using third equation of motion as :

$v^2-u^2=2ad$

$a=\dfrac{v^2-u^2}{2d}$

$a=\dfrac{0-(22.22)^2}{2\times 80000}$

$a=-0.00308\ m/s^2$

Value of g, $g=9.8\ m/s^2$

$a=\dfrac{-0.00308}{9.8}\ m/s^2$

$a=(-0.000314)\ g\ m/s^2$

Hence, this is required solution.

8 0

3 years ago