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satela [25.4K]
3 years ago
15

What is one way in which radioactive decay is used to benefit society?

Physics
2 answers:
snow_lady [41]3 years ago
7 0
<span>They are used to measure and map effluent and pollution discharges from factories and sewerage plants, and the movement of sand around harbours, rivers and bays. Radioactive materials used for such purposes have short half-lives and decay to background levels within days.</span>
Fudgin [204]3 years ago
4 0

Answer:

Sample Response: Radioisotopes are used in medicine as tracers to detect and diagnose medical problems. Radioisotopes are also used in radiation therapy to treat cancer.

On Edge

Explanation:

You might be interested in
A small drop of water is suspended motionless in air by a uniform electric field that is directed upward and has a magnitude of
Gennadij [26K]

To solve this problem we will apply the concepts related to the electric field such as the smelting of the Force and the load (In this case the force is equivalent to the weight). Later we will apply the ratio of the total charge as a function of the multiplication of the number of electrons and their individual charge.

E = \frac{mg}{q}

Here,

m = mass

g = Acceleration due to gravity

Rearranging to find the charge,

q = \frac{mg}{E}

Replacing,

q = \frac{(3.37*10^{-9})(9.8)}{11000}

q = 3.002*10^{-12}C

Since the field is acting upwards the charge on the drop should be negative to balance it in air. The equation to find the number of electrons then is

q = ne

Here,

n = Number of electrons

e = Charge of each electron

n = \frac{q}{e}

Replacing,

n = \frac{3.002*10^{-12}}{1.6*10^{-19}}

n = 2.44*10^7

Therefore the number of electrons that reside on the drop is 2.44*10^7

5 0
3 years ago
If a receiver is overly selective:
VMariaS [17]

Answer:

C) only part of the bandwidth of the AM signal is amplified, causing some of the sideband information to be lost and distortion results.

Explanation:

Selectivity is the ability of a receiver to respond only to a specific signal on a wanted frequency and reject other signals nearby in frequency.

If a receiver is overly selective, only part of the bandwidth of the AM signal is amplified, causing some of the sideband information to be lost and distortion results. Whereas, if a receiver is underselective, the receiver can pick different signals on different frequencies at the same time.

7 0
3 years ago
A capacitor with a capacitance of 50µf when connected to a battery of 400 V. The charge and energy stored on it is? a. 0.05 C an
nekit [7.7K]

Answer:

c. 0.02 C and 4 J

Explanation:

Applying,

Q = CV................ Equation 1

Where Q = Charge, C = Capacitance of the capacitor, V = Voltage.

From the question,

Given: C = 50 μF = 50×10⁻⁶ F, V = 400 V

Substitute these values into equation 1

Q = (50×10⁻⁶)(400)

Q = 0.02 C.

Also Applying

E = CV²/2............. Equation 2

Where E = Energy stored.

Therefore,

E = (50×10⁻⁶ )(400²)/2

E = 4 J

Hence the right option is c. 0.02 C and 4 J

3 0
3 years ago
1)Determine, in terms of unit vectors, the resultant of the five forces illustrated in the figure, Consider F1=20 N, F2= 12 N, F
LiRa [457]

Explanation:

1) F₁ lies in a plane perpendicular to the xy plane, 60° from the x axis.  The angle between F₁ and the +z axis is 30°.  Therefore, the vector is:

<F₁> = 20 (sin 30° cos 60° i + sin 30° sin 60° j + cos 30° k)

<F₁> = 20 (¼ i + ¼√3 j + ½√3 k)

<F₁> = 5 i + 5√3 j + 10√3 k

F₂ is in the xy plane.  Its slope is -24/7.  The vector is:

<F₂> = 12 (-⁷/₂₅ i + ²⁴/₂₅ j + 0 k)

<F₂> = -3.36 i + 11.52 j

F₃ is parallel to the +x axis.  The vector is:

<F₃> = 17 (i + 0 j + 0 k)

<F₃> = 17 i

F₄ is parallel to the -z axis.  The vector is:

<F₄> = 15 (0 i + 0 j − k)

<F₄> = -15 k

F₅ is in the xy plane.  It forms a 15° angle with the -y axis.  The vector is:

<F₅> = 9 (-sin 15° i − cos 15° j + 0 k)

<F₅> = -9 sin 15° i − 9 cos 15° j

The resultant vector is therefore:

<F> = (5 − 3.36 + 17 − 9 sin 15°) i + (5√3 + 11.52 − 9 cos 15°) j + (10√3 − 15) k

<F> = 16.31 i + 11.49 j + 2.32 k

2) Sum of forces at point B in the x direction:

∑F = ma

Tbc cos 40° − ¹⁵/₁₇ Tab = 0

Tbc cos 40° = ¹⁵/₁₇ Tab

Tbc = 1.15 Tab

Sum of forces at point B in the y direction:

∑F = ma

Tbc sin 40° + ⁸/₁₇ Tab − mAg = 0

Tbc sin 40° + ⁸/₁₇ Tab = (2 kg) (10 m/s²)

(1.15 Tab) sin 40° + ⁸/₁₇ Tab = 20 N

1.21 Tab = 20 N

Tab = 16.52 N

Tbc = 19.02 N

Sum of forces at point C in the x direction:

∑F = ma

Tcd sin 25° − Tbc cos 40° = 0

Tcd sin 25° = Tbc cos 40°

Tcd = 1.81 Tbc

Tcd = 34.48 N

3(a) When the crane is on the verge of tipping, the center of gravity is directly over point F.  Relative to point A:

3.7 m = [ (390 kg) (0.9 m) + (90 kg) (9 m cos θ + 1.7 m) + (80 kg) (9 m cos θ + 2.9 m) ] / (390 kg + 90 kg + 80 kg)

2072 kgm = 351 kgm + 810 kgm cos θ + 153 kgm + 720 kgm cos θ + 232 kgm

1336 kgm = 1530 kgm cos θ

θ = 29.17°

3(b) 3.7 m = [ (390 kg) (0.9 m) + (90 kg) (x + 1.7 m) + (80 kg) (x + 2.9 m) ] / (390 kg + 90 kg + 80 kg)

2072 kgm = 351 kgm + (90 kg) x + 153 kgm + (80 kg) x + 232 kgm

1336 kgm = (170 kg) x

x = 7.86 m

4) Find the lengths of the cables.

Lab = √((2 m)² + (3 m)² + (5 m)²)

Lab = √38 m

Lac = √((2 m)² + (3 m)² + (5 m)²)

Lac = √38 m

Lde = √((2 m)² + (3 m)²)

Lde = √13 m

Sum of forces in the x direction:

∑F = ma

-5/√38 Fab − 5/√38 Fac − 2/√13 Fde + Rx = 0

Sum of forces in the y direction:

∑F = ma

2/√38 Fab − 2/√38 Fac = 0

Fab = Fac

Sum of forces in the z direction:

∑F = ma

3/√38 Fab + 3/√38 Fac + 3/√13 Fde − mg = 0

Sum of moments about the y-axis:

∑τ = Iα

(3/√38 Fab) (5 m) + (3/√38 Fac) (5 m) + (3/√13 Fde) (2 m) − (mg) (2 m) = 0

Substitute Fab = Fac and simplify:

6/√38 Fab + 3/√13 Fde − mg = 0

30/√38 Fab + 6/√13 Fde − 2mg = 0

Double first equation:

12/√38 Fab + 6/√13 Fde − 2mg = 0

Subtract from the second equation:

28/√38 Fab = 0

Fab = 0

Fac = 0

Solve for Fde:

3/√38 Fab + 3/√38 Fac + 3/√13 Fde − mg = 0

3/√13 Fde = mg

3/√13 Fde = (1.7 kg) (10 m/s²)

Fde = 20.43 N

Solve for Rx:

-5/√38 Fab − 5/√38 Fac − 2/√13 Fde + Rx = 0

Rx = 2/√13 Fde

Rx = 11.33 N

8 0
3 years ago
A 1000.kg car moving at north at 100. km/hr brakes to a stop in 50. m within a uniform acceleration. what are magnitude and dire
Marianna [84]

Answer:

Force=7.71KN in the opposite direction

Explanation:

Given the mass of the car(M)=1000Kg

The initial speed of the car(u)=100km/hr

we know that 1km/hr=5/18 m/sec

100km/hr=100*5/18 m/sec=27.77m/sec

the distance travelled before it stops (s)=50m

let the acceleration of the car be a

The final velocity of the car is 0.

we know that  v^{2}-u^{2}=2as

0^{2}-27.77^{2}=2(-a)(50)

a=7.712m/sec^{2}

We know that F=Ma

F=1000* 7.712=7712N=7.71KN

6 0
3 years ago
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