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sdas [7]
3 years ago
8

Are longitudinal waves mechanical or electromagnetic?

Physics
2 answers:
irakobra [83]3 years ago
7 0
Longitudinal waves r mechanical
OlgaM077 [116]3 years ago
5 0

Answer:n contrast, an electromagnetic wave, such as light or radio waves, doesn't need a medium and can travel through empty space. While all electromagnetic waves are transverse, mechanical waves can be transverse or longitudinal, which brings us to our next type of wave.

Explanation:

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A 7600 kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.35 m/s2 and feels no appreci
ollegr [7]

Answer:

a) The rocket reaches a maximum height of 737.577 meters.

b) The rocket will come crashing down approximately 17.655 seconds after engine failure.

Explanation:

a) Let suppose that rocket accelerates uniformly in the two stages. First, rocket is accelerates due to engine and second, it is decelerated by gravity.

1st Stage - Engine

Given that initial velocity, acceleration and travelled distance are known, we determine final velocity (v), measured in meters per second, by using this kinematic equation:

v = \sqrt{v_{o}^{2} +2\cdot a\cdot \Delta s} (1)

Where:

a - Acceleration, measured in meters per square second.

\Delta s - Travelled distance, measured in meters.

v_{o} - Initial velocity, measured in meters per second.

If we know that v_{o} = 0\,\frac{m}{s}, a = 2.35\,\frac{m}{s^{2}} and \Delta s = 595\,m, the final velocity of the rocket is:

v = \sqrt{\left(0\,\frac{m}{s} \right)^{2}+2\cdot \left(2.35\,\frac{m}{s^{2}} \right)\cdot (595\,m)}

v\approx 52.882\,\frac{m}{s}

The time associated with this launch (t), measured in seconds, is:

t = \frac{v-v_{o}}{a}

t = \frac{52.882\,\frac{m}{s}-0\,\frac{m}{s}}{2.35\,\frac{m}{s} }

t = 22.503\,s

2nd Stage - Gravity

The rocket reaches its maximum height when final velocity is zero:

v^{2} = v_{o}^{2} + 2\cdot a\cdot (s-s_{o}) (2)

Where:

v_{o} - Initial speed, measured in meters per second.

v - Final speed, measured in meters per second.

a - Gravitational acceleration, measured in meters per square second.

s_{o} - Initial height, measured in meters.

s - Final height, measured in meters.

If we know that v_{o} = 52.882\,\frac{m}{s}, v = 0\,\frac{m}{s}, a = -9.807\,\frac{m}{s^{2}} and s_{o} = 595\,m, then the maximum height reached by the rocket is:

v^{2} -v_{o}^{2} = 2\cdot a\cdot (s-s_{o})

s-s_{o} = \frac{v^{2}-v_{o}^{2}}{2\cdot a}

s = s_{o} + \frac{v^{2}-v_{o}^{2}}{2\cdot a}

s = 595\,m + \frac{\left(0\,\frac{m}{s} \right)^{2}-\left(52.882\,\frac{m}{s} \right)^{2}}{2\cdot \left(-9.807\,\frac{m}{s^{2}} \right)}

s = 737.577\,m

The rocket reaches a maximum height of 737.577 meters.

b) The time needed for the rocket to crash down to the launch pad is determined by the following kinematic equation:

s = s_{o} + v_{o}\cdot t +\frac{1}{2}\cdot a \cdot t^{2} (2)

Where:

s_{o} - Initial height, measured in meters.

s - Final height, measured in meters.

v_{o} - Initial speed, measured in meters per second.

a - Gravitational acceleration, measured in meters per square second.

t - Time, measured in seconds.

If we know that s_{o} = 595\,m, v_{o} = 52.882\,\frac{m}{s}, s = 0\,m and a = -9.807\,\frac{m}{s^{2}}, then the time needed by the rocket is:

0\,m = 595\,m + \left(52.882\,\frac{m}{s} \right)\cdot t + \frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right)\cdot t^{2}

-4.904\cdot t^{2}+52.882\cdot t +595 = 0

Then, we solve this polynomial by Quadratic Formula:

t_{1}\approx 17.655\,s, t_{2} \approx -6.872\,s

Only the first root is solution that is physically reasonable. Hence, the rocket will come crashing down approximately 17.655 seconds after engine failure.

7 0
3 years ago
At speeds over 30 mph, you should maintain a following distance of at least ________ behind the vehicle ahead of you.
ICE Princess25 [194]

At speeds over 30 mph, you should maintain a following distance of at least <u>three full seconds</u> behind the vehicle ahead of you.

As a general rule and common sense at a speed of 30 mph you can leave three full seconds so that you can achieve a prudent distance between the car you are driving and the car in front in order to be able to perform some kind of maneuver if an accident or unforeseen event occurs.

To count the full three seconds you can use the technique of counting the Mississippis as follows: Mississippi one, Mississippi two, Mississippi three.

<h3>What is an accident?</h3>

An accident is an unexpected event that generally causes damage, injury or negative consequences.

Learn more about accident at: brainly.com/question/28070413

#SPJ4

7 0
2 years ago
A 15-kg block at rest on a horizontal frictionless surface is attached to a very light ideal spring of force constant 450 N/m. T
den301095 [7]

Answer:

0.266 m

Explanation:

Assuming the lump of patty is 3 Kg then applying the principal of conservation of linear momentum,

P= mv where p is momentum, m is mass and v is the speed of an object. In this case

m_pv_p=v_c(m_p+m_b) where sunscripts p and b represent putty and block respectively, c is common velocity.

Substituting the given values then

3*8=v(15+3)

V=24/18=1.33 m/s

The resultant kinetic energy is transferred to spring hence we apply the law of conservation of energy

0.5(m_p+m_b)v_c^{2}=0.5kx^{2} where k is spring constant and x is the compression of spring. Substituting the given values then

(3+15)*1.33^{2}=450*x^{2}\\x\approx 0.266 m

7 0
3 years ago
Light is shone on a diffraction grating
Pani-rosa [81]

Answer:

    λ = 482.05 nm

Explanation:

The diffraction phenomenon and the diffraction grating is described by the expression

         d sin θ = m λ

where d is the distance between two consecutive slits, λ the wavelength and m an integer representing the order of diffraction

in this case they indicate the distance between slits, the angle and the order of diffraction

         λ = \frac{d sin \theta }{m}d sin θ / m

let's calculate

         λ = 1.00 10⁻⁶ sin 74.6 / 2

         λ = 4.82048 10⁻⁷ m

Let's reduce to nm

         λ = 4.82048 10⁻⁷ m (10⁹ nm / 1 m)

         λ = 482.05 nm

3 0
3 years ago
If a horse is running at a speed of 35 m/s when it's rider pulls up on the reigns. After 5 seconds the horse slows down to a pac
Nataliya [291]
The answer is simply 5.51
7 0
3 years ago
Read 2 more answers
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