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3 years ago

What is the difference between em waves and mechanical waves?

Physics

1 answer:

Anastaziya [24]3 years ago

7 0

Answer:

Electromagnetic waves differ from mechanical waves in that they do not require a medium to propagate. This means that electromagnetic waves can travel not only through air and solid materials, but also through the vacuum of space.

Explanation:

Since electromagnetic waves do not require a medium to pass through, they are faster in matter that has fewer particles. ... ~ Sound cannot travel in a vacuum because mechanical waves, such as sound waves, require a medium to propagate through.

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I need to know how to solve this problem for an upcoming exam

Nataly_w [17]

W=f*d = 5000N * 10 stories * each 4m = 200 000 J

7 0

3 years ago

The maximum current output of a 60 ω circuit is 11 A. What is the rms voltage of the circuit?

bezimeni [28]

Answer:

660V

Explanation:

V=IR

V=11×60

=660V

hope this helps

please like and Mark as brainliest

8 0

4 years ago

An 89.0 kg fullback moving east with a speed of 5.6 m/s is tackled by an 85.0 kg opponent running west at 2.84 m/s, and the coll

Korolek [52]

Answer:

a. $v_f=1.477m/s$

b. Δ $K=1558.3J$

c. $E_k=1034.7 J$

Explanation:

a).

Momentum conserved

$p_{ix}=p_{fx}$

$m_1*v_1+m_2*v_2=(m_1+m_2)*v_f$

$v_f=\frac{m_1*v_1+m_2*v_2}{m_1+m_2}$

$v_f=\frac{89.0kg*5.6m/s+85.0kg*-2.84m/s}{(89.0+85.0)kg}$

$v_f=1.477m/s$

b).

Δ $K=K_i-K_f$

$\frac{1}{2}*m_1*v_1^2+\frac{1}{2}*m_2*v_2^2=\frac{1}{2}*(m_1+m_2)*v_f^2$

$\frac{1}{2}*89.0kg*(5.6m/s)^2+\frac{1}{2}*85.0kg*(2.84m/s)^2=\frac{1}{2}*(89.0+85.0)kg*(1.447m/s)^2$

Δ $K=1558.3J$

c).

$E_k=\frac{1}{2}*89kg*(5.8m/s)^2-\frac{1}{2}*(85+89)kg*(1.44m/s)^2$

$E_k=1034.7 J$

d).

All of which has been lost as mechanical energy, and is now thermal energy warmer football players, noise a loud crunch for example.

8 0

3 years ago

At the end of cylindrical rod of length l = 1 m and mass M = 1 kg rotating horizontaly along the vertical axis in its center wit

matrenka [14]

Answer:

w = 0.943 rad / s

Explanation:

For this problem we can use the law of conservation of angular momentum

Starting point. With the mouse in the center

L₀ = I w₀

Where The moment of inertia (I) of a rod that rotates at one end is

I = 1/3 M L²

Final point. When the mouse is at the end of the rod

$L_{f}$ = I w + m L² w

As the system is formed by the rod and the mouse, the forces during the movement are internal, therefore the angular momentum is conserved

L₀ = L_{f}

I w₀ = (I + m L²) w

w = I / I + m L²) w₀

We substitute the moment of inertia

w = 1/3 M L² / (1/3 M + m) L² w₀

w = 1 / 3M / (M / 3 + m) w₀

We substitute the values

w = 1/3 / (1/3 + 0.02) w₀

w = 0.943 w₀

To finish the calculation the initial angular velocity value is needed, if we assume that this value is w₀ = 1 rad / s

w = 0.943 rad / s

3 0

3 years ago

Johnson is dragging a bag on a ice surface. He pulls on the strap with a force of 112 N at an angle of 45° to the horizontal to

Nady [450]

Answer with Explanation:

We are given that

Force=F=112 N

$\theta=45^{\circ}$

Distance, $s=84 m$

Time, t=3.33 minutes

We have to find the work done by Johnson on the bag and the power generated by Johnson.

Work done, W= $Fscos\theta$

Using the formula

Work done, W= $112\times 84cos45=6652.46 J$

Power, P= $\frac{W}{t}=\frac{6652.46}{3.33\times 60}=33.3 watt$

Using 1 minute=60 s

Hence, the power generated by Johnson=33.3 watt

5 0

3 years ago