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alexgriva [62]
4 years ago
15

A heavy object moving [E] collides with a much lighter object moving [W]. If the collision is completely inelastic, is it possib

le for both objects to be at rest after the collision? Explain.
Physics
1 answer:
Gekata [30.6K]4 years ago
5 0

It is possible for both objects to be resting and not paying attention to what's coming their way, and they will remain at rest because even their alarm can't wake them up.

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Look in the comments. I d a r e y o u
Alika [10]

Answer:

Where is the comments??

Explanation:

3 0
3 years ago
Read 2 more answers
A wave is travelling at 3000 m/s has a wavelength of 1 m.
Levart [38]

Answer:

a] 3000hz

b]3.33 × 10⁻⁴

ci]300

ii] 3000

iii]60,000

Explanation:

3 0
3 years ago
Some superconductors are capable of carrying a very large quantity of current. If the measured current is 1.00 ´ 105 A, how many
Zielflug [23.3K]

Answer:

The 6.25 \times 10^{23} electrons are moving through the superconductor per second.

Explanation:

Given :

Current I = 1 \times 10^{5} A

Charge of electron e = 1.6 \times 10^{-19} C

Time t = 1 sec

From the formula of current,

Current is the number of charges flowing per unit time.

   I = \frac{ne}{t}

Where n = number of charges means in our case number of electrons

   n = \frac{It}{e}

   n = \frac{1 \times 10^{5} }{1.6 \times 10^{-19} }

   n = 6.25 \times 10^{23}

Therefore, 6.25 \times 10^{23} electrons are moving through the superconductor per second.

5 0
4 years ago
What are the answers to the question
Vitek1552 [10]

Answer: the link isnt loading

7 0
3 years ago
A piano tuner sounds two strings simultaneously. One has been previously tuned to vibrate at 293.0 Hz. The tuner hears 3.0 beats
ololo11 [35]

Answer:

Part a)

f_B = 290 Hz

Part B)

percentage increase is

percentage = 1.38%

Explanation:

Part a)

As we know that the beat frequency is

f_A - f_B = 3

after increasing the tension the beat frequency is decreased and hence the tension in string B will increase

So we have

293 - f_B = 3

f_B = 290 Hz

Part B)

percentage increase in the tension of the string will be given as

f_A - f_B' = 1

f_B' = 292 Hz

now we have

f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}

so we have

T_1 = C (290)^2

T_2 = C(292)^2

so we have

\frac{\Delta T}{T} = \frac{292^2 - 290^2}{290^2}

percentage increase is

percentage = 1.38

4 0
4 years ago
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