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tiny-mole [99]
2 years ago
5

Imagine that a loudspeaker is producing a quiet note with a low pitch. How will its vibrations change:

Physics
1 answer:
borishaifa [10]2 years ago
3 0

Answer:

See below

Explanation:

Higher pitch mens higher frequecy of vibrations

 louder means more amplitude of vibrations (stronger vibrations)

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_ acceleration occurs when an object speeds up.<br> Answer<br> Positive
Nina [5.8K]

_ acceleration occurs when an object speeds up.

Answer

Positive

6 0
4 years ago
Read 2 more answers
A parallel-plate vacuum capacitor has 8.38 J of energy stored in it. The separation between the plates is 2.30 mm. If the separa
Elanso [62]

Answer:

Explanation:

plate separation = 2.3 x 10⁻³ m

capacity C₁ = ε A / d

= ε A / 2.3 x 10⁻³

C₂ = ε A / 1.15 x 10⁻³

\frac{C_2}{C_1} = \frac{2.3}{1.15}

a ) when charge remains constant

energy = \frac{q^2}{2C}

q is charge and C is capacity

energy stored initially E₁= \frac{q^2}{2C_1}

energy stored finally E₂ = \frac{q^2}{2C_2}

\frac{E_1}{E_2} = \frac{C_2}{C_1} = \frac{2.3}{1.15}

E_2 = \frac{1.15}{2.3 } \times E_1

= \frac{1.15}{2.3 } \times 8.38

= 4.19 J

b )

In this case potential diff remains constant

energy of capacitor = 1/2 C V²

energy is proportional to capacity as V is constant .

\frac{E_2}{E_1} = \frac{C_2}{C_1}

\frac{E_2}{8.38} = \frac{2.3}{1.15}

E_2 = 16.76 .

8 0
3 years ago
Ernest Rutherford (the first New Zealander to be awarded the Nobel Prize in Chemistry) demonstrated that nuclei were very small
Vaselesa [24]

Answer:

The final kinetic energy of the Helium nucleus (alpha particle) after been scattered through an angle of 120° is

8.00 x 10-13J

Explanation:

In Rutherford Scattering experiment, the collision of the helium nucleus with the gold nucleus is an ELASTIC COLLISION. This means that the kinetic energy is conserved ( The same before and after the collision).

Thus, the final kinetic energy of the helium nucleus is the same as initial kinetic energy (8.00 x 10^-13Joules)

Although, the kinetic energy is converted to potential energy in Coulomb's law equation.

That is,

1/2(mv^2) = (K* q1q2)/r

Where m is the mass of helium nucleus, v is its colliding velocity, k is electrostatic constant, q1 is the charge on helium nucleus, q2 is the charge on gold nucleus, r is impact parameter

6 0
3 years ago
A 5000-kg freight car runs into a 10,000-kg freight car at rest. They couple upon collision and move with a speed of 2 m/s. What
Vitek1552 [10]

Answer:

C) 6 m/s

Explanation:

Given that

m₁=5000 kg

The initial velocity of 5000 kg car =u₁

m₂=10,000 kg

The initial velocity of 10000 kg car =u₂ = 0 m/s

After collision the final speed of the both car,v = 2 m/s

There is no any external force on the system that is why linear momentum will be conserved.

Linear momentum P = m v

m₁u₁ + m₂u₂ = (m₂ + m₁) v

5000 x u₁ + 10000 x 0 = (5000 + 10000) x 2

5000 x u₁ = 15000 x 2

5 x u₁ = 15 x 2

u₁ = 6 m/s

Therefore the answer is C.

C) 6 m/s

4 0
3 years ago
You are trying to climb a castle wall so, from the ground, you throw a hook with a rope attached to it at 24.1 m/s at an angle o
Serhud [2]

Answer:

The value is  h  =  13.2 \  m

Explanation:

From the question we are told that

    The speed of the rope with hook is u =  24 .1 \  m/s

     The angle is  \theta = 65.0^o

      The speed at which it hits top of the wall is  v  =  16.3 m/s

Generally from kinematic equation we have that

      v_y^2  =  u_y ^2 + *  2 (-g)* h

Here h is the height of the wall so

      [16.3 sin (65)]^2 =  [24.1 sin (65)] ^2+   2 (-9.8)* h

=>    h  =  13.2 \  m

4 0
3 years ago
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