All categories

4 years ago

What is the answer for this question

Physics

2 answers:

Yuri [45]4 years ago

6 0

ANSWER: My sister, who is a waitress at Billy’s Big Burger Shack, is sixteen years old.

miv72 [106K]4 years ago

4 0

The correct is c. If you need help with more questions you can dm me

You might be interested in

When radioactive tracers are used for medical scans, the tracers both decay in the body via radiation, as well as being excreted

irinina [24]

Answer:

Effective half-time of the tracer is 3.6 days

Explanation:

The formula for calculating the decay due to excretion for the first process is ;

$\frac{dN_1}{dt } = - \lambda _e N_o$

here ;

$N_o$ = initial number of tracers

Then to the second process ; we have :

$\frac{dN_2}{dt } = - \lambda _e N_o$

The total decay is as a result of the overall process occurring ; we have :

$\frac{dN_{total}}{dt } = \frac{dN_1}{dt} + \frac{dN_2}{dt}$ ------ (1)

here ;

$\frac{dN_{total}}{dt } = \lambda _{total} N_o$

Putting the values in (1);we have :

$- \lambda _{Total} N_o = - \lambda_e N_o + ( -\lambda r N_o})$

$\lambda _{Total} = \lambda_e + \lambda r$

As we also know that:

$\frac{1}{t_{1/2}} = \frac{[t_{1/2}]_{radiation}+[t_{1/2}]_{excretion}}{[t_{1/2}]_{radiation}*[t_{1/2}]_{excretion}}$

$\frac{1}{t_{1/2}}_{effective}} = \frac{[t_{1/2}]_{radiation}+[t_{1/2}]_{excretion}}{[t_{1/2}]_{radiation}*[t_{1/2}]_{excretion}}$

$\frac{1}{t_{1/2}}_{effective}} = \frac{9+6}{9*6}$

$\frac{1}{t_{1/2}_{effective}}}=\frac{15}{54}$

$t_{1/2}_{effective}} = \frac{54}{15}$

= 3.6 days

5 0

4 years ago

Read 2 more answers

A 26-cm-long wire with a linear density of 20 g/m passes across the open end of an 86-cm-long open-closed tube of air. If the wi

damaskus [11]

Answer: T = 472.71 N

Explanation: The wire vibrates thus making sound waves in the tube.

The frequency of sound wave on the string equals frequency of sound wave in the tube.

L= Length of wire = 26cm = 0.26m

u=linear density of wire = 20g/m = 0.02kg/m

Length of open close tube = 86cm = 0.86m

Sound waves in the tube are generated at the second vibrational mode, hence the relationship between the length of air and and wavelength is given as

L = 3λ/4

0.86 = 3λ/4

3λ = 4 * 0.86

3λ = 3.44

λ = 3.44/3 = 1.15m.

Speed of sound in the tube = 340 m/s

Hence to get frequency of sound, we use the formulae below.

v = fλ

340 = f * 1.15

f = 340/ 1.15

f = 295.65Hz.

f = 295.65 = frequency of sound wave in pipe = frequency of sound wave in string.

The string vibrated at it fundamental frequency hence the relationship the length of string and wavelength is given as

L = λ/2

0.26 = λ/2

λ = 0.52m

The speed of sound in string is given as v = fλ

Where λ = 0.52m f = 295.65 Hz

v = 295.65 * 0.52

v = 153.738 m/s.

The velocity of sound in the string is related to tension, linear density and tension is given below as

v = √(T/u)

153.738 = √T/ 0.02

By squaring both sides

153.738² = T / 0.02

T = 153.738² * 0.02

T = 23,635.372 * 0.02

T= 472.71 N

3 0

3 years ago

_____ you remember to put the lid back on the jar of mayonnaise

Lilit [14]

Answer:

Did you remember to put the lid back on the jar of mayonnaise?

Explanation: Hope this helps :)

7 0

3 years ago

A very long stick ruled with meter markings is placed in empty space. A spaceship of rest length L = 100 m runs lengthwise along

alexira [117]

Answer:

Meter marks are on cut-off portion of stick is 100 - 60 = 40 m

Explanation:

Given data:

Spaceship length of L = 100 m

Relative velocity between the ship and stick is given as

$v = \frac{4}{5} c$

The observed length observed by the outside observer is

$L' = L\sqrt{1 -\frac{v^2}{c^2}}$

putting all value to get observe length

put $v = \frac{4}{5} c$

$L ' = 100\times \sqrt{1 - \frac{(4/5)^2c^2}{c^2}$

L' = 60 m

Meter marks are on cut-off portion of stick is 100 - 60 = 40 m

8 0

3 years ago

A missile is moving 1350 m/s at a 25.0 angle

murzikaleks [220]

I will answer both versions assuming what you want to know is the distance it travels up from and over the ground. and how long until it reaches space. 540 meters per second up and over. to reach space which is 100km above sea level, it would take about 5400 minutes

4 0

3 years ago