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ZanzabumX [31]
3 years ago
12

Please help, i’ll give brainliest!!

Physics
1 answer:
mrs_skeptik [129]3 years ago
5 0

Answer:

duty h gucuuvu h just hc i oicuxp o cut o icucj x uc jo 8cuc8c

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A 16 kg mass suspended from a light spring is replaced by a 4 kg mass. What factor changes the frequency of the oscillation? (a)
AnnZ [28]

Answer:

Frequency change by a factor of 2.

(b) is correct option.

Explanation:

Given that,

Mass = 16 kg

Replaced mass = 4 kg

We need to calculate the frequency

Using formula of frequency  

f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}

Put the value into the formula

f_{1}=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{16}}

f_{1}=\dfrac{1}{2\pi}\dfrac{\sqrt{k}}{4}}....(I)

f_{2}=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{4}}

f_{2}=\dfrac{1}{2\pi}\dfrac{\sqrt{k}}{2}}...(II)

f_{2}=2f_{1}

Hence, Frequency change by a factor of 2.

5 0
3 years ago
Which two statements are true about energy transformations?
Ghella [55]
Energy is never created and energy never is destroyed
5 0
3 years ago
Read 2 more answers
A cork on the surface of a pond bobs up and down two times per second on ripples having a wavelength of 7.60 cm. If the cork is
mario62 [17]

The time taken for the cork to reach the shore is 92.1 s.

The given parameters:

  • <em>Frequency of the oscillation, f = 2 Hz</em>
  • <em>Wavelength of the oscillation, λ = 7.6 cm</em>
  • <em>Distance of the cork from the shore, d = 14.0 m</em>

The speed of the wave is calculated as follows;

v = f \lambda \\\\v = 2 \times 0.076 \\\\v = 0.152 \ m/s

The time taken for the cork to reach the shore is calculated as follows;

time = \frac{distance}{speed} \\\\time = \frac{14}{0.152} \\\\time = 92.1 \ s

Learn more about equation of wave here: brainly.com/question/4692600

6 0
3 years ago
Two horses are pulling a barge with a mass 2000 kg along a canal. The cable connected to the first horse makes an angle of θ1= 4
Shkiper50 [21]

Answer:

a = 0.5195 m/s²

θ = 9.997º ≈ 10º

Explanation:

We apply Newton's 2nd Law as follows:

∑ Fx = m*ax

∑ Fy = m*ay

Then we have

∑ Fx = F₁x + F₂x = m*ax    ⇒ 600*Cos 40º + 600*Cos (-20º) = 2000*ax

⇒   ax = 0.5117 m/s²

∑ Fy = F₁y + F₂y = m*ay    ⇒ 600*Sin 40º + 600*Sin (-20º) = 2000*ay

⇒   ay = 0.0902 m/s²

the magnitude of the acceleration of the barge is

a = √(ax² + ay²) = √((0.5117 m/s²)² + (0.0902 m/s²))= 0.5196 m/s²

and the direction is

θ = Arctan (ay / ax) = Arctan (0.0902 / 0.5117) = 9.997º ≈ 10º

8 0
3 years ago
Which part of the comet exists when it is away from the Sun? A. tail B. coma C. nucleus D. meteor E. core
Katena32 [7]
The answer is E I think.. don’t trust me 100%
6 0
4 years ago
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