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marta [7]
3 years ago
7

It takes 5.0 seconds for a wave with a wavelength of 2.5 m to travel past your location. What’s the period of the wave. What is

the frequency of the wave and what’s the speed of the wave
Physics
1 answer:
Anna35 [415]3 years ago
8 0
(a) Period of the wave
The period of a wave is the time needed for a complete cycle of the wave to pass through a certain point.
So, if an entire cycle of the wave passes through the given location in 5.0 seconds, this means that the period is equal to 5.0 s: T=5.0 s.

(b) Frequency of the wave
The frequency of a wave is defined as
f= \frac{1}{T}
since in our problem the period is T=5.0 s, the frequency is
f= \frac{1}{5.0 s}=0.2 Hz

(c) Speed of the wave
The speed of a wave is given by the following relationship between frequency f and wavelength \lambda:
v=f \lambda = (0.2 Hz)(2.5 m)=0.5 m/s

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How do you write 0.345 in scientific notation?
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3.45 × 10^-1

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Read 2 more answers
Consider a rifle, which has a mass of 2.44 kg and a bullet which has a mass of 150 grams and is loaded in the firing chamber.Whe
svetlana [45]

Answer:

a. 0 kgm/s

b. 0 kgm/s

c. 66 kgm/s

d. -66 kgm/s

e. 0 kgm/s

f. -27.05 m/s

g. 173.68 N

h. 12.58 m/s

i. 0.772 m

j. 14487 J

Explanation:

150 g = 0.15 kg

a. Before the bullet is fired, both rifle and the bullet has no motion. Therefore, their velocity are 0, and so are their momentum.

b. The combination is also 0 here since the bullet and the rifle have no velocity before firing.

c. After the bullet is fired, the momentum is:

0.15*440 = 66 kgm/s

d. As there's no external force, momentum should be conserved. That means the total momentum of both the bullet and the rifle is 0 after firing. That means the momentum of the rifle is equal and opposite of the bullet, which is -66kgm/s

e. 0 according to law of momentum conservation.

f. Velocity of the rifle is its momentum divided by mass

v = -66 / 2.44 = -27.05 m/s

g. The average force would be the momentum divided by the time

f = -66 / 0.38 = 173.68 N

h. According the momentum conservation the bullet-block system would have the same momentum as before, which is 66kgm/s. As the total mass of the bullet-block is 5.1 + 0.15 = 5.25. The velocity of the combination right after the impact is

66 / 5.25 = 12.58 m/s

i. The normal force and also friction force due to sliding is

F_f = N\mu = Mg\mu = 5.25*9.81*0.83 = 42.75N

According to law of energy conservation, the initial kinetic energy will soon be transformed to work done by this friction force along a distance d:

W = K_e

dF_f = 0.5Mv_0^2

d = \frac{Mv_0^2}{2F_f} = \frac{5.25*12.58^2}{2*42.75} = 0.772 m

j.Kinetic energy of the bullet before the impact:

K_b = 0.5*m_bv_b^2 = 0.5*0.15*440^2 = 14520 J

Kinetic energy of the block-bullet system after the impact:

K_e = 0.5Mv_0^2 = 0.5 * 5.25*12.58^2 = 33 J

So 14520 - 33 = 14487 J was lost during the lodging process.

3 0
3 years ago
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