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4 years ago

8

Which of the following statements about electromagnetic radiation is true?

1 answer:

victus00 [196]4 years ago

3 0

B beause beashsusisiisisisizi I jxoddiidid

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14. A spoon is placed in a cup of hot

sweet [91]

Answer:

Specific heats

Explanation:

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3 years ago

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

AnnZ [28]

1.) independent

2.) dependent

3.) gap width

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3 years ago

Read 2 more answers

Pls help i begg youuuuu

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