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2 years ago

What is the wavelength of a wave that has a speed of 160 m/s and a period of 3.7 ms?

Physics

1 answer:

Elza [17]2 years ago

3 0

Answer:

160m/s

Explanation:

The speed of a wave is related to its frequency and wavelength, according to this equation:

v=f ×λ

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Zanzabum

Days and nights are equal in length everywhere.(gradpoint)

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3 years ago

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Match the following.

Eddi Din [679]

Answer:

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Explanation:

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3 years ago

What is the mass of a ball that has 29j of potential energy and is lifted 2.0m?

Salsk061 [2.6K]

Answer:

1.48kg

Explanation:

Here,

potential energy (P.E) = 29j

height (h) = 2m

acceleration due to gravity(g) =

$9.8m {s}^{ - 2}$

mass(m) = ?

we know,

P.E = mgh

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2 years ago

Two cannonballs are dropped from a second-floor physics lab at height h above the ground. Ball B has four times the mass of ball

denpristay [2]

Answer:

1:4

Explanation:

The formula for calculating kinetic energy is:

$KE=\dfrac{1}{2}mv^2$

If the mass is multiplied by 4, then, the kinetic energy must be increased by 4 as well. Since they will be travelling at the same speed when they are at the same point, the relation between KA and KB must be 1:4 or 1/4. Hope this helps!

3 0

3 years ago

A silver wire has a cross sectional area a = 2.0 mm2. a total of 9.4 × 1018 electrons pass through the wire in 3.0 s. the conduc

marta [7]

This problem uses the relationships among current I, current density J, and drift speed vd. We are given the total of electrons that pass through the wire in t = 3s and the area A, so we use the following equation to to find vd, from J and the known electron density n, so:

$v_{d} = \frac{J}{n\left | q \right |}$

The current I is any motion of charge from one region to another, so this is given by:

 $I = \frac{\Delta Q}{\Delta t} = \frac{9.4x1018electrons}{3s} = 3189.73(A)$

The magnitude of the current density is:

$J = \frac{I}{A} = \frac{3189.73}{2x10^{-6}} = 1594.86(A/m^{2})$

Being:

$A=2mm^{2} = 2x10^{-6}m^{2}$

Finally, for the drift velocity magnitude vd, we find:

 $v_{d} = \frac{1594.86}{5.8x1028\left |1.60x10^{-19}|\right } = 1.67x10^{18}(m/s)$

Notice: The current I is very high for this wire. The given values of the variables are a little bit odd

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3 years ago