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Leno4ka [110]
2 years ago
10

A wave is a disturbance that carries energy through space. Which of these characterizes the fundamental difference between mecha

nical waves and electromagnetic waves
Physics
2 answers:
d1i1m1o1n [39]2 years ago
4 0

Explanation:

Electromagnetic waves are waves that have no medium to travel or propagate whilr mechanical waves need a medium for its transmission or propagation.

Electromagnetic wave can travel in a vacuum whereas mechanical waves do not. These media are water, air etc. Examples of electromagnetic waves include light, radio signals etc.

shusha [124]2 years ago
3 0

Answer:

Explanation:

Mechanical waves and electromagnetic waves are the types of waves we have. The major difference between this type of waves is that, while mechanical waves requires material medium for its propagation, electromagnetic waves does not require material medium for its propagation.

Example of mechanical waves are waves produced by a ripple tank.

Example of an electromagnetic waves are the electromagnetic spectrum which includes radio waves, infrared waves, visible light, ultraviolet rays etc... This waves are produced naturally without any medium compare to mechanical wave.

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Engineers who design battery-operated devices such as cell phones and MP3 players try to make them as efficient as possible. An
Svetradugi [14.3K]


efficiency= [useful energy transferred ÷ total energy supply]×100%

So, [5500÷10000]×100%=0.55×100

                                        =55%

5 0
2 years ago
The Bernoulli equation is valid for steady, inviscid, incompressible flows with a constant acceleration of gravity. Consider flo
irina1246 [14]

Answer:

p+\frac{1}{2}ρV^{2}+ρg_{0}z-\frac{1}{2}ρcz^{2}=constant

Explanation:

first write the newtons second law:

F_{s}=δma_{s}

Applying bernoulli,s equation as follows:

∑δp+\frac{1}{2} ρδV^{2} +δγz=0\\

Where, δp is the pressure change across the streamline and V is the fluid particle velocity

substitute ρg for {tex]γ[/tex] and g_{0}-cz for g

dp+d(\frac{1}{2}V^{2}+ρ(g_{0}-cz)dz=0

integrating the above equation using limits 1 and 2.

\int\limits^2_1  \, dp +\int\limits^2_1 {(\frac{1}{2}ρV^{2} )} \, +ρ \int\limits^2_1 {(g_{0}-cz )} \,dz=0\\p_{1}^{2}+\frac{1}{2}ρ(V^{2})_{1}^{2}+ρg_{0}z_{1}^{2}-ρc(\frac{z^{2}}{2})_{1}^{2}=0\\p_{2}-p_{1}+\frac{1}{2}ρ(V^{2}_{2}-V^{2}_{1})+ρg_{0}(z_{2}-z_{1})-\frac{1}{2}ρc(z^{2}_{2}-z^{2}_{1})=0\\p+\frac{1}{2}ρV^{2}+ρg_{0}z-\frac{1}{2}ρcz^{2}=constant

there the bernoulli equation for this flow is p+\frac{1}{2}ρV^{2}+ρg_{0}z-\frac{1}{2}ρcz^{2}=constant

note: ρ=density(ρ) in some parts and change(δ) in other parts of this equation. it just doesn't show up as that in formular

4 0
2 years ago
To practice Problem-Solving Strategy 25.1 Power and Energy in Circuits. A device for heating a cup of water in a car connects to
jeyben [28]

Answer:

The time to boil the water is 877 s

Explanation:

The first thing we must do is calculate the external resistance (R) of the circuit, from the description we notice that it is a series circuit, by which the resistors are added

    V = i (r + R)

We replace we calculate

     r + R = V / i

     R = v / i - r

     R = 10/12 -0.04

     R = 0.793 Ω

We calculate the power supplied

     P = V i = I² R

     P = 12² 0.793

     P = 114 W

This is the power dissipated in the external resistance

We use the relationship, that power is work per unit of time and that work is the variation of energy

     P = E / t

     t = E / P

     t = 100 10³/114

     t = 877 s

The time to boil the water is 877 s

4 0
2 years ago
Imagine that you replace the block in the video with a happy or sad ball identical to the one used as a pendulum, so that the sa
gizmo_the_mogwai [7]

Answer: a) The Answer to the question is option a) None of it.

Explanation:

The reason is because according to the law of conservation of energy Energy can neither be created nor destroyed but can be transformed from one form to the other. Therefore none of the kinetic energy was dissipated, rather it was transformed to another form of energy.

6 0
2 years ago
At 5000-kg freight car runs into 10000-kg freight car at rest. they couple upon collision and move with a speed of 2 m/s. what w
aliina [53]

Solution for the problem is:

Total momentum before collision is always equal to total momentum after collision. So note that:
Momentum of car A = 5000 x Xm/s 
Momentum of car A + B = 15,000 x 2m/s 

So combining the two, will give us the equation:
15,000/5,000 = 3 
3 x 2 =6m/s

6 0
2 years ago
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