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3 years ago

A small object is attached to a horizontal spring and set in simple harmonic motion with amplitude A and period T .

Physics

1 answer:

kupik [55]3 years ago

4 0

Answer:

t = 3/2T

To find how long it takes to cover a total distance of 6A, we need to find the time it takes to cover a distance A then multiply by 6.

The step to the solution is given below in the attachment.

Explanation:

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MakcuM [25]

Answer:

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Explanation:

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3 years ago

As an object moves along the x axis, many measurements are made of its position, enough to generate a smooth, accurate graph of

Bas_tet [7]

The best and most correct answers among the choices provided by your question are he second and third choices.
The velocity at any instant the average velocity during some time interval cannot be obtained from the graph alone.

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

6 0

3 years ago

Read 2 more answers

A lens of focal length 10.0 cm

fgiga [73]

Answer:

30cm

Explanation:

7 0

3 years ago

Suppose we measure the energy stored in some inductor to be E when there is a current I running through it. If I double the curr

slavikrds [6]

Answer:

If I double the current in the inductor, the new total energy will become 4E (option f).

Explanation:

The coil or inductor is a passive component made of an insulated wire that stores energy in the form of a magnetic field due to its form of coiled turns of wire, through a phenomenon called self-induction. In other words, inductors store energy in the form of a magnetic field. The energy stored in the space where there is a magnetic field in the inductor is:

$E=\frac{1}{2} *L*I^{2}$

where E is Energy [J], L is Inductance [H] and I is Current [A].

If you double the current in the inductor, then the new value of the current is I'= 2*I. So replacing the new total energy is:

$E'=\frac{1}{2} *L*I'^{2}=\frac{1}{2} *L*(2*I)^{2}=\frac{1}{2} *L*4*I^{2}=4*\frac{1}{2} *L*I^{2}$

Then:

$E'=4*E$

If I double the current in the inductor, the new total energy will become 4E (option f).

3 0

3 years ago

A bat moving at 3.7 m/s is chasing a ying insect. The bat emits a 36 kHz chirp and receives back an echo at 36.79 kHz. At what s

3241004551 [841]

Answer:

The speed the bat is gaining on its prey is 0.03m/s

Explanation:

Given;

speed of the bat, v₀ = 3.7 m/s

frequency of the bat, F₀ = 36 kHz

frequency of the source, Fs = 36.79

This is relative motion between a source of the sound and the observer. The phenomenon is known as Doppler effect.

Apply the following equation to determine the speed of the insect which is the source;

$F_0 = F_s[\frac{v+v_0}{v-v_s} ]\\\\\frac{F_0}{F_s} = [\frac{v+v_0}{v-v_s} ]\\\\\frac{36.79}{36} = \frac{340+3.7}{340-v_s}\\\\1.0219 = \frac{343.7}{340-v_s}\\\\ 340-v_s = \frac{343.7}{1.0219}\\\\340-v_s = 336.33\\\\v_s = 340-336.33\\\\v_s = 3.67 \ m/s$

The speed the bat is gaining on its prey = 3.7m/s - 3.67m/s = 0.03 m/s

Therefore, the speed the bat is gaining on its prey is 0.03m/s

8 0

3 years ago