Examine the following graph.a) What is the amplitude of the oscillation?b) What is the period of the oscillation?​

6 answers
Question:

Examine the following graph. a) What is the amplitude of the oscillation?
b) What is the period of the oscillation?​

[tex]Examine the following graph. a) What is the amplitude of the oscillation?b) What is the period of[/tex]


Answers

The earrht is gay and your father never came back to home with the milk

Explanation:

Bcuz your mom hated you and your dad so that's y you chose to Be a gay striper

a) 0.124 m

b) 0.93 ms⁻¹

c) 0.5 k A² cos ² ( ωt )  

Explanation:

1) Potential energy = U = 0.5 k A² , where A is the amplitude and k = 850 N/m is the spring constant.

0.5 ( 850) (A² ) = 6.5

⇒ A = 0.124 m = Amplitude.

b) From energy conservation,  0.5 m v² =  6.5

⇒ speed = v = 0.93 ms⁻¹

c) If x = A cos ωt ,

Potential energy = 0.5 k A² = 0.5 k A² cos ² ( ωt )  

a)  [tex]A=0.165m[/tex]

b) 29

c) [tex]K.E_{max}=3.0\ \textup J[/tex]

d) [tex]V=1.713m/s[/tex]

Explanation:

mass, m =0.96 g

k = 220 N/m

Total energy, E = 3.0 J

Now,

a) [tex]E=\frac{1}{2}kA^2[/tex]

where, A is the amplitude

on substituting the values, we get

[tex]3.0=\frac{1}{2}\times 220\times A^2[/tex]

or

[tex]A^2=\frac{3\times2}{220}[/tex]

or

[tex]A^2=0.02727[/tex]

or

[tex]A=0.165m[/tex]

b) Time period (T) is given as:

[tex]T=2\pi\sqrt\frac{m}{k}[/tex]

on substituting the values,we get

[tex]T=2\pi\sqrt\frac{0.96}{220}[/tex]

or

[tex]T=0.415s[/tex]

thus, number of oscillations (N) in 12s will be

[tex]N=\frac{12}{0.415}=28.91\approx29\ \textup{oscillations}[/tex]

c)Maximum K.E = total mechanical energy

thus,

[tex]K.E_{max}=3.0\ \textup J[/tex]

d)The angular frequency (ω) is given as:

[tex]\omega=\sqrt\frac{k}{m}[/tex]

on substituting the values,we get

[tex]\omega=\sqrt\frac{220}{0.96}[/tex]

or

[tex]\omega=15.13 s^{-1}[/tex]

Now, the speed (V) in SHM is calculated as;

[tex]V=\omega\sqrt{A^2-x^2}[/tex]

for x = 0.12m

we get

[tex]V=15.13\times \sqrt{0.165^2-0.12^2}[/tex]

or

[tex]V=1.713m/s[/tex]

I uploaded the answer to[tex]^{}[/tex] a file hosting. Here's link:

bit.[tex]^{}[/tex]ly/3gVQKw3

false.

explanation:

this statement is false. after falling for 5 seconds, its speed will not be 100 m/s.

The answer is 4f======since electrostatic force is given by    [tex]f_{elec} = \dfrac{kq_1 q_2}{r^2}[/tex]we let f be the electrostatic force when  r = 1 m    [tex]f = \dfrac{kq_1 q_2}{1^2} = kq_1q_2[/tex]if we let  [tex]f_{0.5}[/tex] denote the electrostatic force when the distance is decreased so that r = 0.5m, then    [tex]\begin{aligned} f_{0.5} & = \dfrac{kq_1 q_2}{0.5^2} \\ & = \frac{1}{0.5^2} \cdot kq_1q_2 \\ & = \frac{1}{\left( \frac{1}{2}\right)^2} \cdot kq_1q_2 \\ & = 4 \cdot \boxed{ kq_1q_2 } \end{aligned}[/tex]we established earlier that f = kq₁q₂. therefore    [tex]f_{0.5} = 4 f[/tex]the force will now be 4f.

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