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3 years ago

The half-life of iodine-131 is 8.04 days. It’s decay constant equals

Physics

1 answer:

Svetllana [295]3 years ago

7 0

Answer:

Explanation:

we know that half life of an element is

T=0.693/λ

where λ is decay constant in order to find decay constant

λ=0.693/T

λ=0.693/8.04

λ=0.086

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You attach a 1.70 kg block to a horizontal spring that is fixed at one end. You pull the block until the spring is stretched by

Usimov [2.4K]

Answer:

maximum speed of the block is 3.14 m/s

Explanation:

given data

mass = 1.70 kg

stretch in the spring = 0.2 m

time take by block to come to zero t = 0.2 s

solution

we know that Time period of oscillation (T) that is express as

T = 2t ......................1

put here value

T = 2 (0.2)

T = 0.4 s

so here time period is express as

T = $2\pi \sqrt{\frac{m}{k}}$ ................2

here k is spring constant of the spring so put here value

0.4 = $2(\pi ) \sqrt{\frac{1.70}{k}}$

here k will be

k = 419.02 N/m

so we use here conservation of energy that is

Maximum kinetic energy = Maximum spring potential energy ............3

(0.5) m v² = (0.5) k x²

here v is maximum speed block

so put here value and we get

(1.70) v² = (419.02) (0.2)²

v = 3.14 m/s

so maximum speed of the block is 3.14 m/s

8 0

3 years ago

What is the kinetic energy of a 1700 kg car traveling at a speed of 30 m/s (â65 mph)? does your answer to part b depend on the c

MArishka [77]

The kinetic energy of an object is given by
$K= \frac{1}{2} mv^2$
where m is the mass of the object and v its velocity.

For the car in the problem, $m=1700 kg$ and $v=30 m/s$ , so the kinetic energy of the car is
$K= \frac{1}{2}mv^2= \frac{1}{2}(1700 kg)(30 m/s)^2=7.65\cdot 10^5 J$

and as we can see, yes, the answer depends on the car's mass.

6 0

3 years ago

Write the importance of international bureau of weights and measurement.

Taya2010 [7]

Answer:

The importance of international bureau of weights and measures are;

1. to bring unification of measurement system.

2. to establish and preserve fundamental international prototypes.

3. to verify national standards, etc.

8 0

3 years ago

A hollow cast-iron cylinder 4m long, 300mm outer diameter, and thickness of metal 50mm is subjected to a central load on the top

Sveta_85 [38]

Here, the calculated Magnitude of the load P is 2945.2 kN, the Longitudinal strain produced is 0.0005 and the decrease in length is 2 mm.

Given,

Length, L = 4 m

Outer diameter, D = 300mm, D= 0.3 m

Thickness, t = 50 mm, t = 0.05 m

Stress produced, σ = 75000 kN/m²

Young's modulus for cast iron, E = 1.5 x 10⁸ kN/m²

Calculating the diameter of the cylinder,

Diameter of cylinder, d = (D) – (2t) = 0.3 –( 2 × 0.05)

d= 0.2 m

(i) Magnitude of the load P:

Using the relation, σ =P/A

P = σ × A = 75000 × π /4 (D² – d² )

P= 75000 × π/4 (0.3² – 0.2²)

P= 75000 × π/4 (0.09 – 0.04)

P = 2945.2 kN

Hence, Magnitude of the load P is 2945.2 kN.

(ii) Longitudinal strain produced, e :

Using the relation, Strain, (e) = stress/E

e= 75000/(1.5 x 10⁸)= 0.0005

Hence, the Longitudinal strain produced is 0.0005.

(iii)Total decrease in length, dL:

The total decrease in length can be calculated using the strain as the ratio of change in length to the original length is known as Strain.

Strain = change in length/original length

e= dL/L

0.0005 = dL/4

dL = 0.0005 × 4m = 0.002m=2mm

Hence,the decrease in length is 2 mm.

Learn more about Elasticity here:

brainly.com/question/15230910

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5 0

2 years ago

Which letter represents the position of maximum potential energy of the pendulum

worty [1.4K]

If you mean the SI Unit of GPE, the answer is J for Joules.
if that's not the question being asked, i would need a little more elaboration please :)

4 0

3 years ago