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3 years ago

This chart shows characteristics of three different types of waves.

Physics

1 answer:

prohojiy [21]3 years ago

7 0

Answer:

1. Primary or P waves are push and pull waves

2. Secondary, S or Shear Waves are also called transverse wave

3. L or surface waves reach the earth's surface after P and S waves

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A car slams on its brakes, coming to a complete stop in 4.0 s. The car was traveling south at 60.0 mph. Calculate the accelerati

lutik1710 [3]

We need to apply the following expression

Vf = Vo + a.t

Where a is the acceleration
t is the time it takes to achieve Speed Vf
And Vo the initial speed

In this problem Vo = 60 miles/hour
Vf = 0 (comes to a stop)
t = 4 seconds

a = -60 [miles/hours] / 4 [s] = -15 miles/hour/second (negative as it goes in the opposite direction)

6 0

3 years ago

Read 2 more answers

How much thermal energy does it take to raise the temperature of 2.5 kg of

KonstantinChe [14]

Answer:

390kJ

Explanation:

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2 years ago

Một ô tô đi với vận tốc 60 km/h trên nửa phần đầu của đoạn đường AB. Trong nửa đoạnđường còn lại ô tô đi nửa thời gian đầu với v

melamori03 [73]

Answer:

Huhfgyyeyvvucvyehukfsdjbjz/////???

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3 years ago

A 326 g object is attached to a spring and executes simple harmonic motion with a period of 0.250 s. If the total energy of the

Basile [38]

Answer:

a) $v_{max}=5.98\frac{m}{s}$

b) $k=205.92\frac{N}{m}$

c) $A=0.24m$

Explanation:

a) In the equilibrium position of the system, that is when the spring is not elongated, the potential energy is zero. Therefore, the total energy of the system is the maximum kinetic energy:

$E=K_{max}\\E=\frac{mv_{max}^2}{2}\\v_{max}=\sqrt{\frac{2E}{m}}\\v_{max}=\sqrt{\frac{2(5.83J)}{0.326kg}}\\v_{max}=5.98\frac{m}{s}$

b) The force constant of the spring can be calculated from the natural frequency of the system:

$\omega^2=\frac{k}{m}\\k=m\omega^2$

Recall that $\omega=\frac{2\pi}{T}\\$ , that is the distance traveled in one revolution divided into the time of one revolution. Replacing and solving for k:

$k=\frac{m4\pi^2}{T^2}\\k=\frac{(0.326kg)4\pi^2}{(0.25s)^2}\\k=205.92\frac{N}{m}$

c) The maximum speed is directly proportional to the amplitude of the motion:

$v_{max}=A\omega\\A=\frac{v_{max}}{\omega}\\A=\frac{(v_{max})T}{2\pi}\\A=\frac{(5.98\frac{m}{s})0.25s}{2\pi}\\A=0.24m$

3 0

3 years ago

A fish swimming in a horizontal plane has velocity i = (4.00 + 1.00 ) m/s at a point in the ocean where the position relative to

ryzh [129]

Answer:

a. The horizontal component of acceleration a₁ = 0.68 m/s²

The vertical component of acceleration a₂ = -0.11 m/s²

b. -9.19° = 350.81° from the the positive x-axis

Explanation:

The initial velocity v₁ of the fish is v₁ = 4.00i + 1.00j m/s. Its final velocity after accelerating for t = 19.0 s is v₂ = 17.0i - 1.00j m/s

a. The acceleration a = (v₂ - v₁)/t = [17.0i - 1.00j - (4.00i + 1.00j)]/19 = [(17.0 -4.0)i - (-1.0 -1.0)j]/19 = (13.0i - 2.0j)/19 = 0.68i - 0.11j m/s²

The horizontal component of acceleration a₁ = 0.68 m/s²

The vertical component of acceleration a₂ = -0.11 m/s²

b. The direction of the acceleration relative to the unit vector i,

tanθ = a₂/a₁ = -0.11/0.68 = -0.1618

θ = tan⁻¹(-0.1618) = -9.19° ⇒ 360 + (-9.19) = 350.81° from the the positive x-axis

6 0

3 years ago