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2 years ago

5

A particle in a 799=m-long linear particle accelerator is moving at 0.875C. How long does the particle accelerator appear to the

particle?

1 answer:

barxatty [35]2 years ago

6 0

the answer is 386m because m = pi mc=2.23

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A freshly prepared sample of radioactive isotope has an activity of 10 mCi. After 4 hours, its activity is 8 mCi. Find: (a) the

Maurinko [17]

Answer:

(a). The decay constant is $1.55\times10^{-5}\ s^{-1}$

The half life is 11.3 hr.

(b). The value of N₀ is $2.38\times10^{11}\ nuclei$

(c). The sample's activity is 1.87 mCi.

Explanation:

Given that,

Activity $R_{0}=10\ mCi$

Time $t_{1}=4\ hours$

Activity R= 8 mCi

(a). We need to calculate the decay constant

Using formula of activity

$R=R_{0}e^{-\lambda t}$

$\lambda=\dfrac{1}{t}ln(\dfrac{R_{0}}{R})$

Put the value into the formula

$\lambda=\dfrac{1}{4\times3600}ln(\dfrac{10}{8})$

$\lambda=0.0000154\ s^{-1}$

$\lambda=1.55\times10^{-5}\ s^{-1}$

We need to calculate the half life

Using formula of half life

$T_{\dfrac{1}{2}}=\dfrac{ln(2)}{\lambda}$

Put the value into the formula

$T_{\dfrac{1}{2}}=\dfrac{ln(2)}{1.55\times10^{-5}}$

$T_{\dfrac{1}{2}}=44.719\times10^{3}\ s$

$T_{\dfrac{1}{2}}=11.3\ hr$

(b). We need to calculate the value of N₀

Using formula of $N_{0}$

$N_{0}=\dfrac{3.70\times10^{6}}{\lambda}$

Put the value into the formula

$N_{0}=\dfrac{3.70\times10^{6}}{1.55\times10^{-5}}$

$N_{0}=2.38\times10^{11}\ nuclei$

(c). We need to calculate the sample's activity

Using formula of activity

$R=R_{0}e^{-\lambda\times t}$

Put the value intyo the formula

$R=10e^{-(1.55\times10^{-5}\times30\times3600)}$

$R=1.87\ mCi$

Hence, (a). The decay constant is $1.55\times10^{-5}\ s^{-1}$

The half life is 11.3 hr.

(b). The value of N₀ is $2.38\times10^{11}\ nuclei$

(c). The sample's activity is 1.87 mCi.

4 0

3 years ago

Can a object have have zero velocity and nonzero acceleration?

vladimir1956 [14]

Answer:

yes

Explanation:

3 0

3 years ago

Help ASAP 100 points

Gwar [14]

Answer:

other galaxies is red shifted

Question 5 (1 point)

What does a blue shift in light from stars indicate?

Question 5 options:

The stars are moving randomly.

The stars are moving closer.

The stars are moving farther away.

The stars are stationary

7 0

3 years ago

Pepsi [2]

Answer:

Potential

Explanation:

The most accurate term is Electrostatic potential energy

It is denoted as UC

It's named like this because the force between charges or electrons is called electrostatic force .

5 0

2 years ago

A stone whirled at the end of the a rope 30cm long, makes 10 complete resolution in 2 seconds Find: A. The angular velocity in r

lord [1]

Answers:

A: Angular velocity $\omega=31.40 \frac{r a d}{s}$

B: Linear velocity $v=9.42 \frac{m}{s}$

C: Linear Distance $d=47.1 \mathrm{m}$

Given:

Radius of the rope r=30cm=0.3m

Angular distance $\Delta \theta$ =10 revolutions

Time taken t=2seconds

To find:

A: Angular velocity in radians

B: Linear speed

C: Distance covered in 5 seconds

<u>Step by Step Explanations:</u>

Solution:

A: Angular velocity in radians;

According to the formula, Angular velocity can be calculated as

Angular Velocity = angular distance/ time

$\omega=\Delta \theta / \Delta t$

Where $\omega$ =Angular velocity

$\Delta \theta$ =Angular distance=10 revolutions

Changing revolutions to radians multiply with $2 \pi$ , so that we get

$=10 \times 2 \pi$

$=10 \times 2(3.14)$

=62.80 rad/rev

$\Delta t$ =Change in time

Substitute the known values in the above equation we get

$\omega$ =62.80 / 2

$\omega=31.40 \frac{r a d}{s}$

B. Linear speed of the rope;

As per the formula

Linear speed = angular speed × radius

$v=\omega \times r$

Where $\omega$ =Angular velocity

v=Linear speed of the rope

r=Radius of the rope

Substitute the known values in the above equation we get

$v=31.40 \times 0.30$

$v=9.42 \frac{m}{s}$

C. Dsitance covered in 5 seconds;

Linear distance = linear speed × time

$d=v \times t$

Where d= Linear distance of the rope

v=Linear speed of the rope

t=Time taken

Substitute the known values in the above equation we get

$d=9.42 \times 5$

$d=47.1 \mathrm{m}$

Result:

Thus A: Angular velocity of the rope $\omega=31.40 \frac{r a d}{s}$

B Linear speed of the rope $v=9.42 \frac{m}{s}$

C: Distance covered in 5 seconds $d=47.1 \mathrm{m}$

6 0

3 years ago