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1 year ago

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A longitudinal wave is observed to be moving along a slinky. Adjacent crests are 2.4 m apart. Exactly 6 crests are observed to m

ove past a given point in 9.1 s. Determine the wavelength, frequency, and speed of this wave.

1 answer:

Umnica [9.8K]1 year ago

8 0

Answer:

5

Explanation:

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When the boy crashes his bumper car into the girl's bumper car, the momentum from his car is transferred to hers. What evidence

Darya [45]

Answer:

Where is the text?

Explanation:

If you refer to the short sentence you wrote as text, I believe the answer is probably the word "crashes" because it shows how the momentum was transferred.

8 0

2 years ago

Read 2 more answers

You are using a hydrogen discharge tube and high quality red and blue light filters as the light source for a Michelson interfer

boyakko [2]

Answer:

final displacement = +24484.5 nm

Explanation:

The path difference when 158 bright spots were observed with red light (λ1 = 656.3 nm) is given as;

Δr = 2d2 - 2d1 = 150λ1

So, 2d2 - 2d1 = 150λ1

Dividing both sides by 2 to get;

d2 - d1 = 75λ1 - - - - eq1

Where;

d1 = distance between the fixed mirror and the beam splitter

d2 = position of moveable mirror from splitter when 158 bright spots are observed

Now, the path difference between the two waves when 114 bright spots were observed is;

Δr = 2d'2 - 2d1 = 114λ1

2d'2 - 2d1 = 114λ1

Divide both sides by 2 to get;

d'2 - d1 = 57λ1

Where;

d'2 is the new position of the movable mirror from the splitter

Now, the displacement of the moveable mirror is (d2 - d'2). To get this, we will subtract eq2 from eq1.

(d2 - d1) - (d'2 - d1) = 75λ1 - 57λ2

d2 - d1 - d'2 + d1 = 75λ1 - 57λ2

d2 - d'2 = 75λ1 - 57λ2

We are given;

(λ1 = 656.3 nm) and λ2 = 434.0 nm.

Thus;

d2 - d'2 = 75(656.3) - 57(434)

d2 - d'2 = +24484.5 nm

5 0

2 years ago

A truck is traveling down a road with a 4-percent grade at a speed of 75 mi/h when its brakes are applied to slow it down to 22.

kvasek [131]

Answer:

3.964 s

Explanation:

Metric unit conversion:

1 miles = 1.6 km = 1600 m.

1 hour = 60 minutes = 3600 seconds

75 mph = 75 * 1600 / 3600 = 33.3 m/s

22.5 mph = 22.5 * 1600/3600 = 10 m/s

Let g = 9.81 m/s2

Friction is the product of coefficient and normal force, which equals to the gravity

$F_f = \mu N = \mu mg$

The deceleration caused by friction is friction divided by mass according to Newton 2nd law.

$a = F_f / m = \mu mg / m = \mu g = 0.6 *9.81 = 5.886 m/s^2$

So the time required to decelerate from 33.3 m/s to 10 m/s so the wheels don't slide, with the rate of 5.886 m/s2 is

$t = \frac{\Delta v}{a} = \frac{33.3 - 10}{5.886} = 3.964 s$

3 0

1 year ago

Which is found farthest from the center of an atom?

xxTIMURxx [149]

Electron

Hope this helps

5 0

1 year ago

What is the acceleration during the pushing-off phase, in m/27 Express your answer in meters per second squared. A bush baby, an

Sindrei [870]

Answer:

$a = 12.78 g's$

Explanation:

Height reached by the object after push off is given as

$H = 2.3 m$

$v_f^2 - v_i^2 = 2 a s$

now we have

$0 - v^2 = 2(-9.81)(2.3)$

$v = 6.72 m/s$

now we know that this push last for total distance of 0.18 m

so during the push we will have

$v_f^2 - v_i^2 = 2 a d$

$6.72^2 - 0 = 2a(0.18)$

$a = 125.35 m/s^2$

now in terms of g = 9.81 m/s/s we have

$a = \frac{125.35}{9.81} gs$

$a = 12.78 g's$

6 0

1 year ago