PLEASE HELP ME WITH THIS ONE QUESTION
The half-life of Barium-139 is 4.96 x 10^3 seconds. A sample contains 3.21 x 10^17 nuclei. What is the decay constant for this decay?

A) 1.67 x 10^-4 s^-1

B) 5.43 x 10^-4 s^-1

C) 1.40 x 10^-4 s^-1

D) 2.22 x 10^-4 s^-1

Answer: θ would equal approximately 28.7°

This is a kinematics problem, where one is only given the theta value 13.0° in regards to the range; thus, the problem is testing one's understanding of the relationships between the variables.

Range (aka x) = (v₀ sin (2θ₀))/g, where θ₀ = 13.0°

Now if we multiply the range by 2, we get:

2x = 2((v₀ sin (2θ₀))/g) → to verbalize, if range equates to (v₀ sin (2θ₀))/g, and doubling the range equals twice the product value, then:

2θ = sin⁻¹(2sin(2(13.0° )) = sin⁻¹(2(0.76255845048)) = sin⁻¹ (1.52511690096) = 57.35560850015109°/2 = θ

Thus, θ = 28.67780425

It's been awhile since I did this; though I hope it helped!

Answer: 31.89seconds

Explanation:

Based on the information given, we are meant to calculate deceleration which will be:

t = V/a

where, a = mg

Therefore, t = V/mg

t = 25/0.08 × 9.8

t = 25/0.784

t = 31.89seconds

Therefore, the time that it will take before the hockey puck slides to a stop is 31.89seconds.

Answer:

(c) at rest

Explanation:

Given

See attachment for the distance time graph

Required

What does the graph illustrate?

From the graph, we can see that the line of distance is a horizontal line.

This suggests that a time increases, the distance remains unchanged

When distance remains unchanged over time, then it means the object is at rest.

Hence, (c) is correct

Answer:

The angle is 4.1 rad.

Explanation:

The centripetal acceleration (α) is given by:

[tex] \alpha = \omega^{2} r [/tex]    (1)                  

Where:

ω: is the angular velocity  

r: is the radius

And the tangential acceleration (a) is:                      

[tex] a = \alpha r [/tex]      (2)

Since the magnitude of "α" is 8.2 times the magnitude of "a" (equating (2) and (1)) we have:

[tex] \omega^{2} r = 8.2\alpha r   [/tex]

[tex] \omega^{2} = 8.2\alpha [/tex]    (3)      

Now, we can find the angle with the following equation:

[tex] \omega_{f}^{2} = \omega_{0}^{2} + 2\alpha \Delta \theta [/tex]

Where:

[tex] \omega_{f}[/tex]: is the final angular velocity                                                                              [tex] \omega_{0}[/tex]: is the initial angular velocity = 0 (it starts from rest)

[tex]\Delta \theta[/tex]: is the angle

[tex] \omega^{2} = 2\alpha \Delta \theta [/tex]     (4)    

By entering equation (3) into (4) we can calculate the angle:

[tex] 8.2\alpha = 2\alpha \Delta \theta [/tex]

[tex] \Delta \theta = 4.1 rad [/tex]

Therefore, the angle is 4.1 rad.

I hope it helps you!                  

Answer:

499.7 J

Explanation:

Since total mechanical energy is conserved,

U₁ + K₁ + W₁ = U₂ + K₂ + W₂ where U₁ = potential energy at bottom of incline = mgh₁, K₁ = kinetic energy at bottom of incline = 1/2mv₁² and W₁ = work done by friction at bottom of incline, and U₂ = potential energy at top of incline = mgh₂, K₁ = kinetic energy at top of incline = 1/2mv₂² and W₂ = work done by friction at top of incline. m = mass = 4 kg, h₁ = 0 m, v₁ = 21.3 m/s, W₁ = 0 J, h₂ = radius of circular ramp = 10 m, v₂ = 2.8 m/s, W₂ = unknown.

So, U₁ + K₁ + W₁ = U₂ + K₂ + W₂

mgh₁ + 1/2mv₁²  + W₁ = mgh₂ + 1/2mv₂²  + W₂

Substituting the values of the variables into the equation, we have

mgh₁ + 1/2mv₁²  + W₁ = mgh₂ + 1/2mv₂²  + W₂

4 kg × 9.8 m/s²(0) + 1/2 × 4 kg × (21.3 m/s)²  + 0 = 4 kg × 9.8 m/s² × 10 m + 1/2 × 4 kg × (2.8 m/s)²  + W₂

0 + 2 kg × 453.69 m²/s² = 392 kgm²/s² + 2 kg × 7.84 m²/s²  + W₂

907.38 kgm²/s² = 392 kgm²/s² + 15.68 kgm²/s²  + W₂

907.38 kgm²/s² = 407.68 kgm²/s² + W₂

W₂ = 907.38 kgm²/s² - 407.68 kgm²/s²

W₂ = 499.7 kgm²/s²

W₂ = 499.7 J

Since friction is a non-conservative force, the work done by all the non-conservative forces is thus W₂ = 499.7 J

Answer:

66,7 seconds

Explanation:

the formula for height/distance is : S=v.t

Answer: 4

Explanation:

MA = output force / input force

MA = 20 / 5

MA = 4

Hope this helps.  Please mark brainliest.

I can not even read this question.

What are you trying to even say?

The acceleration of the car with two (2) washers added is equal to 0.33 [tex]m/s^2[/tex].

Given the following data:

An acceleration can be defined as the rate of change of velocity of an object with respect to time and it is measured in meter per seconds square.

In Science, the average acceleration of an object is calculated by subtracting its initial velocity from the final velocity and dividing by the change in time for the given interval.

Mathematically, average acceleration is given by this formula:

[tex]a = \frac{V\;-\;U}{t_f-t_i}[/tex]

Where:  

Substituting the given parameters into the formula, we have;

[tex]a = \frac{0.36\;-\;0.13}{2.61\;-\;1.92}\\\\a=\frac{0.23}{0.69}[/tex]

a = 0.33 [tex]m/s^2[/tex]

Read more on acceleration here: brainly.com/question/24728358

Answer: The energy released as thermal energy is 6.5 J

Explanation:

Energy stored by the spider when it relaxes is given by:

[tex]E_o=\text{Resilience}\times \text{Work}[/tex]

We are given:

Resilience = 0.35

Work done = 10.0 J

Putting values in above equation, we get:

[tex]E_o=0.35\times 10\\\\E_o=3.5J[/tex]

Energy released at thermal energy is the difference between the work done and the energy it takes to relaxes, which is given by the equation:

[tex]E_T=\text{Work done}-E_o[/tex]

Putting values in above equation, we get:

[tex]E_T=(10-3.5)=6.5J[/tex]

Hence, the energy released as thermal energy is 6.5 J

The energy released as thermal energy when 10 J of work is done to stretch silk will be 6.5 J

Thermal energy refers to the energy contained within a system that is responsible for its temperature. Heat is the flow of thermal energy.

Energy stored by the spider when it relaxes is given by:

[tex]\rm E_o=Resilience \ \times Work[/tex]

We are given:

Resilience = 0.35

Work done = 10.0 J

Putting values in above equation, we get:

[tex]\rm E_o=0.35\times 10[/tex]

[tex]E_o=3.5\ J[/tex]

Energy released at thermal energy is the difference between the work done and the energy it takes to relaxes, which is given by the equation:

[tex]E_T=\rm Work done -E_o[/tex]

Putting values in above equation, we get:

[tex]E_T=(10-3.5)=6.5\ J[/tex]

Hence, the energy released as thermal energy is 6.5 J

To know more about thermal energy follow

https://brainly.com/question/19666326

Answer:

The period the field must be reduced to zero is 9.81 x 10⁻⁵ s

Explanation:

Given;

initial value of the magnetic field, B₁ = 0.276 T

number of turns of the solenoid, N = 517 turns

diameter of the solenoid, d = 10.5 cm = 0.105 m

induced emf, = 12.6 kV = 12,600 V

when the field becomes zero, then the final magnetic field value, B₂ = 0

The induced emf is given by Faraday's law;

[tex]emf = -\frac{NA\Delta B}{t} \\\\emf = -\frac{NA (B_2 -B_1)}{t} \\\\t = -\frac{NA (B_2 -B_1)}{emf}\\\\t = \frac{NA (B_1 -B_2)}{emf}\\\\where;\\\\t \ is \ the \ time \ when \ B = 0 \ \ (i.e\ B_2 = 0)\\\\A \ is \ the \ area \ of \ the \ coil\\\\A = \frac{\pi d^2}{4} = \frac{\pi (0.105)^2}{4} = 0.00866 \ m^2\\\\t= \frac{(517) \times (0.00866)\times (0.276 -0)}{12,600}\\\\t = 9.81 \times 10^{-5} \ s[/tex]

Therefore, the period the field must be reduced to zero is 9.81 x 10⁻⁵ s

Answer:

The cup is acted upon by an unbalanced force which is the acceleration of the car, but before it was an object at rest that stayed at rest.

Explanation:

Newton's first law of motion states, "if a body is at rest or moving at a constant speed in a straight line, it will remain at rest or keep moving in a straight line at constant speed unless it is acted upon by a force."

Since the cup is at rest while sitting on top of the car, it stays at rest as the car begins to move. Since the car is accelerating and the cup is not, the cup falls off of the car.

Answer:

Explanation:

From the given information:

The initial PE [tex](PE)_i[/tex] = m×g×h

= 5 kg × 9.81 m/s² × 10 m

= 490.5 J

The change in Potential energy P.E of the box is:

ΔP.E = [tex]P.E_f -P.E_i[/tex]

ΔP.E = 0 - [tex](PE)_i[/tex]

ΔP.E = [tex]-P.E_i[/tex]

If we take a look at conservation of total energy for determining the change in the internal energy of the box;

[tex]\Delta P.E + \Delta K.E + \Delta U = 0[/tex]

[tex]\Delta U = -\Delta P.E - \Delta K.E[/tex]

this can be re-written as:

[tex]\Delta U =- (-\Delta P.E_i) - \Delta K.E[/tex]

Here, K.E = 0

Also, 70% goes into raising the internal energy for the box;

Thus,

[tex]\Delta U =(70\%) \Delta P.E_i-0[/tex]

[tex]\Delta U =(0.70) (490.5)[/tex]

ΔU = 343.35  J

Thus, the magnitude of the increase is = 343.35 J

Answer:

The number of photons in the pulse is 1.04 x 10²⁴

Explanation:

Given;

frequency of the emitted photons, f = 2.83 x 10¹¹ Hz

duration of the incident light, t = 3 ms = 3 x 10⁻³ s

power of the incident light, P = 65,000 W

The energy of each photon emitted is calculated as;

E = hf

where;

h is Planck's constant, = 6.626 x 10⁻³⁴ Js

E =  6.626 x 10⁻³⁴ x  2.83 x 10¹¹

E = 1.875 x 10⁻²² J

let the number of photons in the pulse = n

n(E)= Power x time

[tex]n = \frac{Pt}{E} \\\\n = \frac{65,000 \times 3\times 10^{-3}}{1.875 \times 10^{-22}} \\\\n = 1.04 \times 10^{24} \ photons[/tex]

I= 0.39 A

OPTION B is the correct answer.

Answer:

9 joules of heat energy was produced

Explanation: there is no acceleration therefore its not a kinetic energy

Energy= force × distance

= 3×3

=9