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lara31 [8.8K]
3 years ago
12

How does half-life apply to real life?

Physics
2 answers:
Lyrx [107]3 years ago
6 0
The term half-life is defined as the time it takes for one-half of the atoms of a radioactive material to disintegrate."
wlad13 [49]3 years ago
3 0
half-life? what do you mean
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Which of the following is known as the following cycle because it's reservoir is a rock.
inessss [21]

Letra a obrigadoaijahahavqvqgqgqg

3 0
3 years ago
An object that is dropped straight down from a height of 100 m has a vertical change in position that is less than that of an id
sergey [27]
Objects dropped straight or thrown horizontally from the same height
change their vertical velocity at the same rate, and fall through equal
vertical distances in equal time intervals.

The statement is false. 
5 0
3 years ago
ANSWER ASAP Which factor determines the amount of gas that can dissolve in ocean water? (A) the motion of the air above the ocea
Andru [333]

Answer:

The amount of each gas that can dissolve in the ocean depends on the solubility and saturation of the gas in water. Solubility refers to the amount of a dissolved gas that the water can hold under a particular set of conditions, which are usually defined as 0o C and 1 atmosphere of pressure.

Explanation:

hope this helps

6 0
3 years ago
Read 2 more answers
The spring of a spring gun has force constant k = 400 N/m and negligible mass. The spring is compressed 6.00 cm and a ball with
nikdorinn [45]

Answer:

A) v = 6.93 m/s

B) v = 4.9 m/s

C) x_m = 0.015m

D) v_max = 5.2 m/s

Explanation:

We are given;

x = 6 cm = 0.06 m

k = 400 N

m = 0.03 kg

F = 6N

A) from work energy law, work dome by the spring on ball which now became a kinetic energy is;

Ws = K.E = ½kx²

Similarly, kinetic energy of ball is;

K.E = ½mv²

So, equating both equations, we have;

½kx² = ½mv²

Making v the subject gives;

v = √(kx²/m)

Plugging in the relevant values to give;

v = √((400 × 0.06²)/0.03)

v = √48

v = 6.93 m/s

B) If there is friction, the total work is;

Ws = ½kx² - - - (1)

Work of the ball is;

Wb = KE + Wf

So, Wb = ½mv² + fx - - - (2)

Combining both equations, we have;

½mv² + fx = ½kx²

Plugging in the relevant values, we have;

(½ × 0.03 × v²) + (6 × 0.06) = ½ × 400 × 0.06²

0.015v² + 0.36 = 0.72

0.015v² = 0.72 - 0.36

v² = 0.36/0.015

v = √24

v = 4.9 m/s

C) The speed is greatest where the acceleration stops i.e. where the net force on the ball is zero. (ie spring force matches 6.0N friction)

So, from F = Kx;

(x is measured into barrel from end where F = 0)

Thus; 6.0 = 400x

x_m = 6/400

x_m = 0.015m from the end after traveling 0.045m

D) Initial force on ball = (Kx - F) =

[(400 x 0.06) - 6.0] = 18N

Final force on ball = 0N

Mean Net force on ball = ½(18 + 0)

Mean met force, F_m = 9N

Net Work Done on ball = KE = 9N x 0.045m = 0.405 J

Thus;

½m(v_max)² = 0.405J

(v_max)² = 2 x 0.405/0.03

(v_max)² = 27

v(max) = √27

v_max = 5.2 m/s

6 0
3 years ago
A charge feels a 7.89 x 10-7 N force when it moves 2090 m/s at a 29.4° angle to a 4.23 x 10-3 T magnetic field. What is the amou
tamaranim1 [39]

We have that the amount of the charge q is

q=1.8*10^{-7}

From the Question we are told that

Force F=7.89 x*10^{-7}

Velocity V=2090m/s

Angle \theta=29.4

Magnetic field B=4.23 * 10^{-3} T

Generally, the equation for Force F is mathematically given by

F=qVBsin\theta\\\\q=\frac{F}{VBsin\theta}

q=\frac{7.89 x*10^{-7}}{4.23 * 10^{-3} T*sin29.4*2090m/s}

q=1.8*10^{-7}

In conclusion

The amount of the charge q is

q=1.8*10^{-7}

For more information on this visit

brainly.com/question/1470439

6 0
3 years ago
Read 2 more answers
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