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Paha777 [63]
4 years ago
8

Which of the following statements about electromagnetic radiation is true?

Physics
1 answer:
victus00 [196]4 years ago
3 0
B beause beashsusisiisisisizi I jxoddiidid
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Specific heats

Explanation:

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Before starting the simulation and having the waves encounter the barrier, the wavelength is manipulated. This is the (1. BLANK)
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1.) independent

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3.) gap width

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Anna [14]
Pitch is the sensation of certain frequencies to the ear. High frequency = high pitch, low frequency = low pitch. 

f = c(speed of the wave) /  <span>λ (wavelength)

1. 343m/s / 0.77955m = 439.99 Hz   
     This corresponds to pitch A 

2. 343m/s / 0.52028m = 659.26 Hz
</span>     This corresponds to pitch E 
<span>
3. 343m/s / 0.65552m = 523.349 Hz
    </span>This corresponds to pitch C

4. using f = c /  λ
  λ = c / f<span>
     = 343m/s / 587.33 = 0.583999 m = 0.584 m

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3 0
3 years ago
Two horizontal rods are each held up by vertical strings tied to their ends. Rod 1 has length L and mass M; rod 2 has length 2L
antiseptic1488 [7]

Answer:

Rod 1 has greater initial angular acceleration; The initial angular acceleration for rod 1 is greater than for rod 2.

Explanation:

For the rod 1 the angular acceleration is

\tau_1 = I_1\alpha _1 \\\\\alpha_1 = \dfrac{\tau_1}{I_1}

Similarly, for rod 2

\alpha_2 = \dfrac{\tau_2}{I_2}.

Now, the moment of inertia for rod 1 is

I_1 = \dfrac{1}{3}ML^2,

and the torque acting on it is (about the center of mass)

\tau_1 = Mg\dfrac{L}{2};

therefore, the angular acceleration of rod 1 is  

\alpha_1 = \dfrac{Mg\dfrac{L}{2}}{\dfrac{1}{3}ML^2},

\boxed{\alpha_1 = \dfrac{3g}{2L} }

Now, for rod 2 the moment of inertia is

I_2 = \dfrac{1}{3}(2M)(2L)^2

I_2 = \dfrac{8}{3} ML^2,

and the torque acting is (about the center of mass)

\tau _2 = (2M)g \dfrac{(2L)}{2}

\tau _2 = 2MgL;

therefore, the angular acceleration \alpha_2 is

\alpha_2 = \dfrac{2MgL;}{\dfrac{8}{3} ML^2,}.

\boxed{\alpha_2 = \dfrac{3g}{4L}}

We see here that

\dfrac{3g}{2L} > \dfrac{3g}{4L}

therefore

\boxed{\alpha_1 > \alpha_2.}

In other words , the initial angular acceleration for rod 1 is greater than for rod 2.

7 0
3 years ago
A standard basketball (mass = 624 grams; 24.3 cm in diameter) is held fully under water. a. Calculate the buoyant force and weig
galina1969 [7]

Answer:

Explanation:

Mass = 624 gm = .624 kg

weight = .624 x 9.8

= 6.11 N

Radius of ball = 12.15 x 10⁻² cm

volume of ball

= 4/3 x 3.14 x ( 12.15 x 10⁻²)³

= 7509.26 x 10⁻⁶ m³

Buoyant force = weight of displaced water

= 7509.26 x 10⁻⁶ x 10³ x 9.8

= 73.59 N

b ) Since buoyant force exceeds the weight of the ball , it will float .

c )

Let volume v sticks out while floating .

Volume under water

= 7509.26 x 10⁻⁶ - v

its weight

= (7509.26 x 10⁻⁶ - v ) x 10³ x 9.8

For floating

(7509.26 x 10⁻⁶ - v ) x 10³ x 9.8  =  .624 x 9.8 ( weight of ball )

(7509.26 x 10⁻⁶ - v ) x 10³ = .624

7.509 - v x 10³ = .624

v x 10³ = 7.509 - .624

v x 10³ = 6.885

v = 6.885 x 10⁻³ m³

fraction

= v / total volume

=  6.885 x 10⁻³ / 7.51 x 10⁻³

91.67 %

8 0
3 years ago
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