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3 years ago

What is a transverse wave?

Physics

1 answer:

Tema [17]3 years ago

5 0

The answer is C. they travel perpendicular to the direction of energy

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4. A hockey puck with a momentum of -17 kg x m/s when it

Nonamiya [84]

Answer:

-2 kg x m/s

Explanation:

-17 + 54 = 35 + x

x = -2

7 0

3 years ago

Ocean waves pass through two small openings, 20.0 m apart, in a breakwater. You're in a boat 70.0 m from the breakwater and init

Klio2033 [76]

Answer:

λ = 5.65m

Explanation:

The Path Difference Condition is given as:

δ= $(m+\frac{1}{2})\frac{lamda}{n}$ ;

where lamda is represent by the symbol (λ) and is the wavelength we are meant to calculate.

m = no of openings which is 2

∴δ= $\frac{3*lamda}{2}$

n is the index of refraction of the medium in which the wave is traveling

To find δ we have;

δ= $\sqrt{70^2+(33+\frac{20}{2})^2 }-\sqrt{70^2+(33-\frac{20}{2})^2 }$

δ= $\sqrt{4900+(\frac{66+20}{2})^2}-\sqrt{4900+(\frac{66-20}{2})^2}$

δ= $\sqrt{4900+(\frac{86}{2})^2 }-\sqrt{4900+(\frac{46}{2})^2 }$

δ= $\sqrt{4900+43^2}-\sqrt{4900+23^2}$

δ= $\sqrt{4900+1849}-\sqrt{4900+529}$

δ= $\sqrt{6749}-\sqrt{5429}$

δ= 82.15 -73.68

δ= 8.47

Again remember; to calculate the wavelength of the ocean waves; we have:

δ= $\frac{3*lamda}{2}$

δ= 8.47

8.47 = $\frac{3*lamda}{2}$

λ = $\frac{8.47*2}{3}$

λ = 5.65m

3 0

3 years ago

How do theories differ from Laws and Principles?

g100num [7]

A scientific law is the simple mathematical expression of the relationship involved. A principle is the same relationship expressed in words. A theory is the explanation of the facts that make up the relationship.

8 0

3 years ago

What does the atomic number of an atom tell us?

blondinia [14]

Answer:

The number of protons in the atom!

Explanation:

8 0

2 years ago

Read 2 more answers

A race car has a centripetal acceleration of 15.625 m/s2 as it goes around a curve. If the curve is a circle with radius 40 m, w

myrzilka [38]

The centripetal acceleration is given by
$a_c = \frac{v^2}{r}$
where v is the tangential speed and r the radius of the circular orbit.

For the car in this problem, $a_c = 15.625 m/s^2$ and r=40 m, so we can re-arrange the previous equation to find the velocity of the car:
$v= \sqrt{a_c r}= \sqrt{(15.625 m/s^2)(40 m)}=25 m/s$

8 0

2 years ago