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3 years ago

If a sound wave travels 680 m in 2 seconds and has a frequency of 400 hz, what is the wavelength of the wave?

Physics

1 answer:

Kipish [7]3 years ago

5 0

Answer: 0.85 meters (with and without sigfigs)

Explanation: To find the wavelength, you just have to switch around the equation for wave speed: v (wave speed) = λ (wavelength)*f (frequency) so λ (wavelength) = v (wave speed)/f (frequency). You don't have the wave speed but you can calculate it. Since wave speed is measured in meters/second or m/s, you just have to divide the amount of meters you were given by the amount of seconds. You will get 340 m/s. Next, you have to plug the values into the equation: λ (wavelength) = 340 m/s (wave speed)/400 Hz (frequency). The answer is 0.85 meters (seconds cancel) and has the correct number of significant figures.

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An object is formed by attaching a uniform, thin rod with a mass of mr = 6.9 kg and length L = 4.88 m to a uniform sphere with m

Bumek [7]

Answer:

a) total moment of inertia is 1359.05 kg m^2

b) angular acceleratio is 0.854rad/sec^2

Explanation:

Given data:

m1=6.9 kg

L=4.88 m

m2=34.5 kg

R=1.22 m

we klnow that moment of inertia for rod is given as

J1=(1/12) ×m×L^2

$J1 = (1/12) \times 6.9 \times 4.88^2 = 13.693 kg m^2$

moment of inertia for sphere is given as

J1=(2/5) ×m×r^2

$J1 = (2/5) \times 34.5 \times 1.22^2 = 20.539 kg m^2$

As object rotates around free end of rod then for sphere the axis around what it rotates is at a distance of d2=L+R

For rod distance is d1=0.5*L

By Steiner theorem

for the rod we get $J_1'=J_1 + m_1\times d_1^2$

$J_1' = 13.693 + 2.44^2\times 6.9 = 54.77 kg m^2$

for the sphere we get $J_2' = J_2 + m_2\times d_2^2$

$J2' = 20.539 + 34.5\times 6.1^2 = 1304.28 kg^m2$

And the total moment of inertia for the first case is

$J_{t1} = J_1'+J_2' = 54.77 + 1304.28 = 1359.05kg.m^2$

b) F=476 N

The torque for system is given as

$M = F\times d\times sin(a)$

where a is angle between Force and distance d

and where d represent distance from rotating axis.

In this case a = 90 degree

$M = F\times L/2$

M=476*2.44 = 1161.44 Nm

The acceleration is calculated as

$a_1 = \frac{M}{J_{t1}}$

$= \frac{1161.44}{1359.05}$

= 0.854 rad/sec^2

4 0

3 years ago

Question 6 of 10

sergejj [24]

The answer is C.89,866. I multiplied 262x343.

3 0

2 years ago

The age of the universe is around 100,000,000,000,000,000s. A top quark has a lifetime of roughly 0.000000000000000000000001s. W

Maslowich

Answer:

10⁴¹ s quark top lives have been in the history of the universe.

Explanation:

You need to determine how many quark top lives there have been in the history of the universe, that is, what is the age of the universe divided by the lifetime of a top quark. Expressed in a formula, this is:

$t\frac{Age of the universe}{Lifetime of a top quark}$

Yo know that the "Age of the universe" is 100,000,000,000,000,000 which can also be expressed as 10¹⁷ s .

You also know that the "Lifetime of a top quark" is 0.000000000000000000000001 which can also be expressed as 10⁻²⁴ s.

Then $t=\frac{10^{17} }{10^{-24} }$

Recalling that the result of dividing two powers of the same base is another power with the same base where the exponent is the subtraction of the initial exponents, it is possible to calculate this division as follows:

$t=10^{17-(-24)}$