Guest #7253071

3 weeks ago

**Answer:**

25g

**Explanation:**

According to a quick internet search, the half-life of I-131 is 8 days.

The amount left after 16 days can be calculated with the radioactive exponential decay formula:

[tex]A_(_t_)=A_0e^(^\frac{\ln(.5)}{T}^)^t[/tex]

Where:

[tex]A_(_t_)[/tex] = the amount left as a function of time

[tex]A_0[/tex] = the original amount (100g)

T = the half-life of the isotope (8d)

t = time (16d)

So:

[tex]A_(_t_)=A_0e^(^\frac{\ln(0.5)}{T}^)^t\\A_(_t_)=(100g)e^(^\frac{\ln(0.5)}{8d}^)^(^1^6^d^)\\A_(_t_)=(100g)e^-^2^(^\ln(2))[/tex]

[tex]A_(_t_)=25g[/tex]

This answer is intuitive because the isotope has been through two half-lives:

[tex]100g(\frac{1}{2})(\frac{1}{2})=25g[/tex]

Question

We all know the trick of pretending to lose a cell phone connection, but the receiving person (in reality) can usually tell you have a good connection. This time, you are in luck- you are using a cell phone at 850 MHz, on one side of you is the cell phone tower, and on the other side, a tall building.

1) what is the wavelength of your cell phone transmission?

2) How far from the building should you stand to lose reception?

3) Since the phone companies employ good engineers, that trick won't actually work. However, you happen to see a chain-link fence nearby, and (out of desperation) try to use that as a diffraction grating to scatter the radiation. Estimating the link spacing as 5 cm, will this work?

1) what is the wavelength of your cell phone transmission?

2) How far from the building should you stand to lose reception?

3) Since the phone companies employ good engineers, that trick won't actually work. However, you happen to see a chain-link fence nearby, and (out of desperation) try to use that as a diffraction grating to scatter the radiation. Estimating the link spacing as 5 cm, will this work?

Solution 1

The **transmission** **waves** from a smartphone, which range in **frequency** from 800 to 2400 MHz and belong within the category of the **electromagnetic** **spectrum**, have a **wavelength** of around one foot.

**define frequency ?**

The **frequency** of a repeated event is its number of instances per unit of time. It differs from **angular** **frequency** and is sometimes referred to as temporal **frequency** for clarification. The unit of **frequency** is (Hz), or one occurrence per second. The time **lapsed** between **occurrences** is measured by the time, which is the **reciprocal** of the **frequency**.

For instance, if a ** bets **120 times per minute , the period, Tâ€”the space between betsâ€”is half a second. The **frequency** of oscillatory and **vibrtory** **phenomena**, such as **mechanical vibrions**, sound, and **radio waves**, is a crucial characteristic in science and **engineering**.

The **transmission** **waves** from a smartphone, which range in **frequency** from 800 to 2400 MHz and belong within the RF category of the **electromagnetic** **spectrum**, have a **wavelength** of around one feet.

Standing **8 feet **away is advised, according to **FEMA**.

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Question

A bus slows down from 8.22 m/s to a stop in 13.3 s. Its wheels have a radius of 0.530 m. Find the initial angular speed of the wheels.

Solution 1

**Answer:**

**The initial angular velocity of the bus wheel:**

**Ï‰â‚€ Â â‰ˆ 15.6 rad / s**

**Explanation:**

Given:

Vâ‚€ = 8.22 m/s - **Starting speed of the bus**

V= 0 m/s - **Final speed of the bus (the bus has stopped)**

Ï‰ = 0 rad/c - **Final angular velocity of the bus wheel**

t = 13.3 s - **Bus deceleration time**

R = 0.530 m - **Bus wheel radius**

____________________

Ï‰â‚€ - ? Â - **The initial angular speed of the wheels.**

1)** Bus linear acceleration module:**

a = ( Vâ‚€ - V) / t = (8.22 - 0) / 13.3 â‰ˆ 0.62 m/sÂ²

2) **Bus angular accelerations module:**

Îµ = a / R = 0.62 / 0.530 â‰ˆ 1.17 rad / sÂ²

3) **The initial angular velocity of the bus wheel:**

Ï‰â‚€ = ÎµÂ·t Â = 1.17Â·13.3 â‰ˆ **15.6 rad / s**

Question

A sound pulse emitted underwater reflects off a school of fish and is detected at the same place 0.10 s later. How far away are the fish? (Assume the water temperature is 20 degrees C.)

Solution 1

I'll assume that the **properties** of the water are that sound waves in fresh water move at a speed of 1497 m/s at a temperature of 25 Â°C. If the speed is 0.01s, you must divide it by 100 to get 14.97; but, since it travels there and back in 0.01s, you must split it in half i.e 7.485m.

**Without a medium, what kind of wave can travel?**

**Electromagnetic** waves

In contrast to mechanical waves, electromagnetic waves can travel without a medium. This implies that electromagnetic waves can pass not only through solid objects like air and rock but also through empty space.

Answer and **justification** Since electromagnetic waves are not conveyed by the vibration of particles, they do not need a medium to travel. Electric and magnetic fields, which can exist anywhere, oscillate, creating waves. Waves can therefore exist in solids, liquids, gases, or empty space.

Know more about **reflection**:

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Question

Which of the position versus time graphs below matches the following description: a speed versus time graph has a horizontal line at 5 m/s from zero to five seconds?

Solution 1

The **position versus time graph** that matches the description: a **speed** versus time graph has a horizontal line at 5 m/s from zero to five seconds is Graph A; The correct option is A.

A Â **position versus time graph **is a graph that plots the position of an object against the time it takes for the object to move from one point to another.

The slope of a position versus time graph represents the **speed** or velocity of the object.

The slope of a position time graph = distance or position/time

A speed versus time graph is a graph that plots the speed of an object against the time it takes for the object to move to that speed.

The slope of a position versus time graph represents the **acceleration** of the object.

Considering the given **speed versus time graphs**:

Graph A: the speed versus time graph has a horizontal line at 5 m/s from zero to five seconds.

Graph B: the speed versus time graph shows variation is speed over time

Graph C: the speed versus time graph shows a constant increase in speed over time

Graph D; the speed versus time graph shows a constant decrease in speed over time

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Question

6. Mikey can kick the ball to Sage with a force of 20 N, making it travel 60m. Give the work done on the ball.

Solution 1

**Answer:**

**Explanation:**

Given:

F =20 N

D = 60 m

________

W - ?

W = FÂ·D = 20Â·60 = 1 200 J

Question

.

A 2155 kg demolition ball swings at the end of a 17.5m cable on the arc of a vertical

circle. At the lowest point of the swing, the ball is moving at 8.6m/s. Determine the

tension in the cable at the lowest point. (Hint: T-mg =mv^2/r)

Solution 1

**Answer:**

**Thread tension:**

**T =** F + P = 9 108 + 21 120 â‰ˆ **30 230 N**

**Explanation:**

Given:

m = 2155 kg

R = 17.5 m

V = 8.6 m/s

g = 9.8 m/sÂ²

_______________

T - ?

**Modulus of centripetal force (force is directed towards the center of the circular arc):**

[tex]F = \frac{m\cdot V^2}{R} = \frac{2155\cdot 8.6^2}{17.5} = 9108 N[/tex]

**Ball weight (directed vertically down):**

*W = mÂ·g = 2155Â·9.8 â‰ˆ 21 120 N*

**Let's use the hint. Thread tension:**

**T =** F + W = 9 108 + 21 120 â‰ˆ **30 230 N**

Solution 2

The **tension** in the cable at the lowest point of the swing is [tex]30248.196 N[/tex].

To determine the **tension **in the cable at the lowest point of the** swing**, we can use the hint provided:

[tex]T-mg=\frac{mv^{2}}{r}[/tex]

where T represents** tension**,

m is the mass of the **demolition ball** (2155 kg),

g is the **acceleration** due to gravity (approximately 9.8 m/s^2),

v is the **velocity** of the ball (8.6 m/s), and

r is the** radius** of the circular path (17.5 m).

Plugging in the given values, we have:

[tex]T-2155\times 9.81=\frac{2155\times 8.6^{2}}{17.5}[/tex]

[tex]T=30248.196 N[/tex]

Therefore, the **tension** in the cable at the lowest point of the** swing** is approximately [tex]30248.196 N[/tex].

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Question

Baseballs arenâ€™t especially bouncy. A baseball fired at 20 m/s

at a solid wall rebounds at 11 m/s. What fraction of the initial kinetic energy is transformed into thermal energy?

at a solid wall rebounds at 11 m/s. What fraction of the initial kinetic energy is transformed into thermal energy?

Solution 1

The fraction of the **initial kinetic energy **is transformed into **thermal energy **is 3/4.

The fraction of the** initial kinetic energy** is transformed into **thermal energy **is calculated by applying the principle of conservation of energy as shown below.

Mathematically, this is given as;

Î”K.E = E

where;

- Î”K.E is the change in kinetic energy
- E is the thermal energy

E = K.Ef - K.Ei

where;

- K.Ef is the final kinetic energy
- K.Ei is the initial kinetic energy

E = (Â¹/â‚‚muÂ² - Â¹/â‚‚mvÂ²) / (Â¹/â‚‚muÂ²)

where;

- m is the mass of the baseball
- v is the final velocity of the baseball
- u is the initial velocity of the baseball

E = (uÂ² - vÂ²) / (uÂ²)

E = (20Â² - 10Â²) / (20Â²)

E = 0.75 = 3/4

Thus, the fraction of the **initial kinetic energy **is transformed into **thermal energy **is a function of the change in speed of the baseball.

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Question

Polaris is also called the North Star because Earth's North

Pole points toward it. As Earth rotates, other stars seem to

spin around Polaris in a counterclockwise direction. In one.

day, the stars will make one complete circle (360Â°) around

Polaris. Compare the positions of the constellations in the

two diagrams. Calculate how many hours have passed

between the first and second diagrams.

Pole points toward it. As Earth rotates, other stars seem to

spin around Polaris in a counterclockwise direction. In one.

day, the stars will make one complete circle (360Â°) around

Polaris. Compare the positions of the constellations in the

two diagrams. Calculate how many hours have passed

between the first and second diagrams.

Solution 1

**Answer:****.****.****.****.****.****.****.****.****.****.****.****.****.****.****.****.**

**Explanation:****.****.****.****.****.****.****.****.****.****.****.****.****.**

Question

what is the acceleration of a 38 kg crate being pushed horizontally by an 83 N force?

Solution 1

The magnitude of **acceleration **of the crate be **7.14 m/sÂ².**

**Acceleration **is rate of change of velocity with time. Due to having both direction and magnitude, it is a vector quantity. Si unit of **acceleration **is meter/secondÂ² (m/sÂ²).

**Mass **of the create be: m = 38kg.

**Force **applied on the create : F = 83N.

From Newton's** second law of motion**, we can write:

**force **= mass Ã— acceleration

â‡’ **acceleration **= force/mass = 38/83m/sÂ² = 0.45 m/sÂ².

Hence, **acceleration **of the create be 0.45 m/sÂ².

Learn more about **acceleration **here:

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Question

Using the right hand rule, what do the three numbered vectors represent in the diagram below?

Solution 1

Using the right-hand rule, the correct answer is:

A) Magnetic field

B) Electric current

C) Force

The equation corresponding to this graphical representation is

F = IBL

where I is the current, B the magnetic field, L the length of the wire and F the magnetic force.

A) Magnetic field

B) Electric current

C) Force

The equation corresponding to this graphical representation is

F = IBL

where I is the current, B the magnetic field, L the length of the wire and F the magnetic force.

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