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Bingel [31]
2 years ago
9

The half life of uranium-235 is 4. 5 billion years. If 0. 5 half-lives have elapsed, how many years have gone by?

Physics
1 answer:
igor_vitrenko [27]2 years ago
5 0

Answer:

2.25 billion years

Explanation:

i am smort

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50200 J of heat are removed from
Dmitry_Shevchenko [17]

Correct Answer:

3.1375

Explanation:

Use equation Q=mcΔT to find m

Plug in all variables -50200=x\cdot 2000\cdot -8

Answer: 3.1375

4 0
3 years ago
These questions !plz !! i need help!!!
Nataly_w [17]

(6) Wagon B is at rest so it has no momentum at the start. If <em>v</em> is the velocity of the wagons locked together, then

(140 kg) (15 m/s) = (140 kg + 200 kg) <em>v</em>

==>   <em>v</em> ≈ 6.2 m/s

(7) False. If you double the time it takes to perform the same amount of work, then you <u>halve</u> the power output:

<em>E</em> <em>/</em> (2<em>t </em>) = 1/2 × <em>E/t</em> = 1/2 <em>P</em>

<em />

3 0
3 years ago
Using the definition of moment of inertia, calculate icm, the moment of inertia about the center of mass, for this object.
Step2247 [10]
<span> The masses have no inertia about their own CM, and "the object" is the two masses. </span>
<span>1. Icm (at point A) = 2mr^2 
hope this helps</span>
3 0
3 years ago
Find the time t2 that it would take the charge of the capacitor to reach 99.99% of its maximum value given that r=12.0ω and c=50
defon

Answer:

Explanation:

Given that, .

R = 12 ohms

C = 500μf.

Time t =? When the charge reaches 99.99% of maximum

The charge on a RC circuit is given as

A discharging circuit

Q = Qo•exp(-t/RC)

Where RC is the time constant

τ = RC = 12 × 500 ×10^-6

τ = 0.006 sec

The maximum charge is Qo,

Therefore Q = 99.99% of Qo

Then, Q = 99.99/100 × Qo

Q = 0.9999Qo

So, substituting this into the equation above

Q = Qo•exp(-t/RC)

0.9999Qo = Qo•exp(-t / 0.006)

Divide both side by Qo

0.9999 = exp(-t / 0.006)

Take In of both sodes

In(0.9999) = In(exp(-t / 0.006))

-1 × 10^-4 = -t / 0.006

t = -1 × 10^-4 × - 0.006

t = 6 × 10^-7 second

So it will take 6 × 10^-7 a for charge to reached 99.99% of it's maximum charge

8 0
3 years ago
An open pipe of length 0.58 m vibrates in the third harmonic with a frequency of 939Hz. What is the speed of sound through the a
vaieri [72.5K]

363 m/s is the speed of sound through the air in the pipe.

Answer: Option B

<u>Explanation:</u>

The formula used to calculate the wavelength given as below,

      Wavelength (\lambda)=\frac{\text { wave velocity }(v)}{\text { frequency }(f)}

      v=\lambda \times f   --------> eq. 1

In power system, harmonics define by positive integers of the fundamental frequency. So the third order harmonic is a multiple of the third fundamental frequency. Each harmonic creates an additional node and an opposite node, as well as an additional half wave within the string.

If the number of waves in the circuit is known, the comparison between standing wavelength and circuit length can be calculated algebraically. The general expression for this given as,

         L=\frac{n \lambda}{2}

For first harmonic, n =1

         L=\frac{\lambda}{2}

For second harmonic, n =2

         L=\frac{2 \lambda}{2}=\lambda

For third harmonic, n =3

         L=\frac{3 \lambda}{2}

         \lambda=\frac{2 L}{3}   -------> eq. 2

Here given f = 939 Hz, L = 0.58 m...And, substitute eq 2 in eq 1 and values, we get

   v=\frac{2 \times 0.58 \times 989}{3}=\frac{1089.24}{3}=363.08 \mathrm{m} / \mathrm{s}

3 0
3 years ago
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