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kirill [66]
3 years ago
8

A sample of a radioisotope shows an activity of 999 disintegrations per minute due to beta decay. If after 1.10 years the activi

ty is 952 disintegrations per minute, what is the half-life of this radioisotope?
Physics
1 answer:
galina1969 [7]3 years ago
3 0

Answer:

15.8 year

Explanation:

This this a first order decay reaction first order decay reaction is that reaction in which the rate of reaction is linearly depend on the concentration of reactants. for the first order decay the equation is given as

ln\frac{A}{A_0}= -Kt

where A_0=initial activity and A is the activity at time t K is a rate constant

using above equation ln\frac{952}{999}= -K×1.10

K=0.043809 /year

Half life=\frac{ln2}{K}

=\frac{ln2}{0.043809}

=15.8 year

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The magnitude of a velocity vector is called speed. Suppose that a wind is blowing from the direction N45°W at a speed of 50 km/
velikii [3]

To solve this problem, you must figure out (in vector form) both the wind vector and plane vector

w⃗ = wind vector

P⃗ = plane vector

 To get the  true course of the plane, you need to add the plane and wind vectors, the formula would be

w⃗ +P⃗ ,

 which will result to the ground speed.

ground speed=||w⃗ +P⃗ ||

 

Using the planar representation of your situation, this will help you understand the equation, use this to make the equation more understandable.

w⃗ =AB¯¯¯¯¯¯¯¯,   P⃗ =AC¯¯¯¯¯¯¯¯

 

the smaller circle is of radius 50 (similar to the wind speed) and the larger circle is of radius 200 (similar to the plane vector.  To get the coordinates of these two vectors,  use polar coordinates.

Let East be 0 degrees, so since the wind vector is on the circle of radius 50, we have:

w⃗ =⟨50cos(135),50sin(135)⟩=⟨−252√, 252√⟩.

P⃗ =⟨200cos(60),200sin(60)⟩=⟨100,\1003√⟩.

w⃗ +P⃗ =⟨100−252√ , 1003√+252√⟩

||w⃗ +P⃗ ||=(100−252√)2+(1003√+252√)2

√≈218.349218.

 

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8 0
2 years ago
One solution to minimize resonance with buildings is to ______ the width to span ratio.
Shtirlitz [24]

Answer:

Increase

Explanation:

Resonance is a phenomenon which occurs when a body A in motion set another body B into motion of it own natural frequency. So for resonance to be minimize in a body is to increase the width to span ratio. So as to reduce the overall vibration which affects directly building resonance, the stiffness or trusses and girders should be increase. The increase in this aspect helps to reinforce building structure and support.

3 0
2 years ago
13
gladu [14]

v = u + at

50 = 0 + a*10

50 = 10a

a = 5 m/s^2

4 0
2 years ago
A bug splats against the windshield of a car traveling at high speeds down a backcountry road. Which statement correctly compare
zvonat [6]

Answer:

C. The bug's change in momentum is equal to the car's change in momentum.

Explanation:

As we know by Newton's 2nd law

F = \frac{\Delta P}{\Delta t}

here we have also know that when car hits the bug then force applied by wind shield on the bug is same as the force applied by the bug on the car's wind shield as per Newton's III law

F_{12} = F_{21}

so we know that

\frac{\Delta P_{12}}{\Delta t} = \frac{\Delta P_{21}}{\Delta t}

so we have

\Delta P_{12} = \Delta P_{21}

so correct answer will be

C. The bug's change in momentum is equal to the car's change in momentum.

6 0
3 years ago
A 10-kg package drops from chute into a 25-kg cart with a velocity of 3 m/s. The cart is initially at rest and can roll freely w
amid [387]

Answer:

(a) the final velocity of the cart is 0.857 m/s

(b) the impulse experienced by the package is 21.43 kg.m/s

(c) the fraction of the initial energy lost is 0.71

Explanation:

Given;

mass of the package, m₁ = 10 kg

mass of the cart, m₂ = 25 kg

initial velocity of the package, u₁ = 3 m/s

initial velocity of the cart, u₂ = 0

let the final velocity of the cart = v

(a) Apply the principle of conservation of linear momentum to determine common final velocity for ineleastic collision;

m₁u₁  + m₂u₂ = v(m₁  +  m₂)

10 x 3   + 25 x 0   = v(10  +  25)

30  = 35v

v = 30 / 35

v = 0.857 m/s

(b) the impulse experienced by the package;

The impulse = change in momentum of the package

J = ΔP = m₁v - m₁u₁

J = m₁(v - u₁)

J = 10(0.857 - 3)

J = -21.43 kg.m/s

the magnitude of the impulse experienced by the package = 21.43 kg.m/s

(c)

the initial kinetic energy of the package is calculated as;

K.E_i = \frac{1}{2} mu_1^2\\\\K.E_i = \frac{1}{2} \times 10 \times (3)^2\\\\K.E_i = 45 \ J\\\\

the final kinetic energy of the package;

K.E_f = \frac{1}{2} (m_1 + m_2)v^2\\\\K.E_f = \frac{1}{2} \times (10 + 25) \times 0.857^2\\\\K.E_f = 12.85 \ J

the fraction of the initial energy lost;

= \frac{\Delta K.E}{K.E_i} = \frac{45 -12.85}{45} = 0.71

7 0
2 years ago
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