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3 years ago

What is the frequency of a sound wave that has a wave length of 3.5 m?

Physics

1 answer:

polet [3.4K]3 years ago

3 0

Answer:

85.654988MHz

Explanation:

f = C/λ

λ (Lambda) = Wavelength in meters

c = Speed of Light (299,792,458 m/s)

f = Frequency (MHz)

299,792,458 m/s. ÷ 3.5. = 85.654988MHz

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Why can the atmosphere hold on to heat

Naddika [18.5K]

Answer:

A. Air

Explanation:

4 0

3 years ago

Compute the dot product of the vectors u and v, and find the angle between the vectors. Bold v equals 7 Bold i minus Bold j and

OLga [1]

Answer:

$\theta = 106.3 degree$

Explanation:

As we know that

$\vec w = -\hat i + 7\hat j$

$\vec v = 7\hat i - \hat j$

also we know that

$\vec v. \vec w = -14$

it is given as

$\vec v. \vec w = (-\hat i + 7\hat j).(7\hat i - \hat j)$

$\vec v. \vec w = - 7 - 7 = -14$

also we can find the magnitude of two vectors as

$|v| = \sqrt{(-1)^2 + (7)^2}$

$|v| = \sqrt{50}$

similarly we have

$|w| = \sqrt{(7^2) + (-1)^2}$

$|w| = \sqrt{50}$

now we know the formula of dot product as

$\vec v. \vec w = |v||w| cos\theta$

$-14 = (\sqrt{50})^2cos\theta$

$\theta = cos^{-1}(\frac{-14}{50})$

$\theta = 106.3 degree$

3 0

3 years ago

An unbalanced force of 20N acts on a 4.0kg mass what is it's acceleration

tankabanditka [31]

Hi there!

According to Newton's second law:

∑F = m · a, where:

∑F = net force (N = kgm/s²)

m = mass (kg)

a = acceleration (m/s²)

Rearrange to solve for acceleration:

F/m = a

20N / 4.0kg = 5 m/s²

4 0

3 years ago

A ball bearing of radius of 1.5 mm made of iron of density

Serjik [45]

Answer:

$\boxed{\sf Viscosity \ of \ glycerine \ (\eta) = 14.382 \ poise}$

Given:

Radius of ball bearing (r) = 1.5 mm = 0.15 cm

Density of iron (ρ) = 7.85 g/cm³

Density of glycerine (σ) = 1.25 g/cm³

Terminal velocity (v) = 2.25 cm/s

Acceleration due to gravity (g) = 980.6 cm/s²

To Find:

Viscosity of glycerine ( $\sf \eta$ )

Explanation:

$\boxed{ \bold{v = \frac{2}{9} \frac{( {r}^{2} ( \rho - \sigma)g)}{ \eta} }}$

$\sf \implies \eta = \frac{2}{9} \frac{( {r}^{2}( \rho - \sigma)g )}{v}$

Substituting values of r, ρ, σ, v & g in the equation:

$\sf \implies \eta = \frac{2}{9} \frac{( {(0.15)}^{2} \times (7.85 - 1.25) \times 980.6)}{2.25}$

$\sf \implies \eta = \frac{2}{9} \frac{(0.0225 \times 6.6 \times 980.6)}{2.25}$

$\sf \implies \eta = \frac{2}{9} \times \frac{145.6191}{2.25}$

$\sf \implies \eta = \frac{2}{9} \times 64.7196$

$\sf \implies \eta = 2 \times 7.191$

$\sf \implies \eta = 14.382 \: poise$

6 0

3 years ago

A 4.30 g bullet moving at 943 m/s strikes a 730 g wooden block at rest on a frictionless surface. The bullet emerges, traveling

Ymorist [56]

Answer:

(a)2.7 m/s

(b) 5.52 m/s

Explanation:

The total of the system would be conserved as no external force is acting on it.

Initial momentum = final momentum

⇒(4.30 g × 943 m/s) + (730 g × 0) = (4.30 g × 484 m/s) + (730 g × v)

⇒ 730 ×v = (4054.9 - 2081.2) =1973.7

⇒v=2.7 m/s

Thus, the resulting speed of the block is 2.7 m/s.

(b) since, the momentum is conserved, the speed of the bullet-block center of mass would be constant.

$V_{COM} = \frac{m_b}{m_b+m_{bl}}v_{bi}=\frac{4.30}{4.30+730}\times 943 m/s = 5.52 m/s$

Thus, the speed of the bullet-block center of mass is 5.52 m/s.

4 0

4 years ago