Use the worked example above to help you solve this problem. The half-life of the radioactive nucleus _(88)^(226)text(Ra) is 1.6 103 yr. If a sample initially contains 4.00 1016 such nuclei, determine the following:________.
(a) the initial activity in curies µCi
(b) the number of radium nuclei remaining after 4.4 103 yr nuclei
(c) the activity at this later time µCi

Answers

Answer 1

Answer:

Explanation:

From the information given:

The half-life [tex]t_{1/2}[/tex] = 1.6103 years

The no. of the initial nuclei [tex]N_o[/tex] = [tex]4.00 \times 10^6[/tex]

Using the formula:

[tex]N = N_o exp(-\lambda t)[/tex]

where;

decay constant [tex]\lambda = \dfrac{In2}{1.6*10^3} y^{-1}[/tex]

[tex]N = N_o exp ( \dfrac{-In2}{1.6*10^3}\times 4.4 \times 10^3)[/tex]

[tex]N = N_o exp (- 1.906154747)[/tex]

The number of radium nuclei N = 5.94 × 10¹⁵

The initial activity[tex]A_o = \lambda N_o[/tex]

[tex]A_o =(\dfrac{In (2)}{1.61\times 10^3 \times 365 \times 24 \times 3600}\times 4.00 \times 10^{16})[/tex]

[tex]A_o =546075.8487 \ Bq[/tex]

Since;

1 curie = 3.7 × 10¹⁰ Bq

Then;

[tex]A_o =\dfrac{546075.8487 }{3.7\times 10^{10}}[/tex]

[tex]A_o = 1.47588 \times 10^{-5}Ci[/tex]

[tex]A_o = 14.7588 \ \mu Ci[/tex]

c) The activity at a later time is:

[tex]=5.94 \times 10^{15}( \dfrac{In (2)}{1.60 \times 10^3 \times 365\times 24 \times 3600})[/tex]

[tex]= 81599.09018 \ Bq \\ \\ = \dfrac{81599.09018}{3.7\times 10^{10}} \ Ci \\ \\ = 2.20538 \times 10^6 \ Ci \\ \\ = 2.20538 \ \mu Ci[/tex]


Related Questions

The Nardo ring is a circular test track for cars. It has a circumference of 12.5 km. Cars travel around the track at a constant speed of 100 km/h. A car starts at the easternmost point of the ring and drives for 30 minutes at this speed.

Required:
a. What distance, in km, does the car travel?
b. What is the magnitude of the car's displacement, in km, from its initial position?
c. What is the speed of the car in m/s?

Answers

Answer: 50 km, 0, 27.78 m/s

Explanation:

Given

Circumference of the track is [tex]12.5\ km[/tex]

Speed of car is [tex]100\ km/h[/tex]

Car drives for [tex]30\ \text{minute}\ or\ 0.5\ hr[/tex]

(a)Distance traveled is

[tex]\Rightarrow D=100\times 0.5\\\Rightarrow D=50\ km[/tex]

(b)displacement of the car

It can be observed that 12.5 is a multiple of 50, that is, 50 km can be interpreted as 4 complete rounds of the track.

Therefore, the displacement of the car is zero.

(c)To convert kmph to m/s, multiply the entity by [tex]\frac{5}{18}[/tex]

[tex]\Rightarrow 100\times \dfrac{5}{18}\\\\\Rightarrow 27.78\ m/s[/tex]

If a car has a force of 900 N and has a mass of 650 kg. What would be the acceleration of the
car as it moves down the street?

Answers

65 mph ;) your welcome

What is the definition of Heat Transfer

Answers

Answer:

Heat transfer is an engineering discipline that concerns the generation, use, conversion, and exchange of heat (thermal energy) between physical systems.

Explanation:

Answer:

Heat transfer is a discipline of thermal engineering that concerns the generation, use, conversion, and exchange of thermal energy between physical systems. Heat transfer is classified into various mechanisms, such as thermal conduction, thermal convection, thermal radiation, and transfer of energy by phase changes.

Explanation:

hope that helps


As a galaxy evolves and becomes more massive, what is most likely to happen?

Answers

Answer:

Explanation:

The most likely thing that would happen is that the galaxies continue becoming more and more massive, eventually becoming part of a galaxy cluster. These are massive but cluttered parts of the universe that hold many galaxies extremely close to one another. This also leads to galaxies colliding with one another, although when this happens they usually seem to pass right through each other as if they were ghosts. This is simply due to their sheer size and distance between their bodies of mass.

Answer:

It will merge with other galaxies

Explanation:

Pearson Connexus 2023

Find the magnitude of the free-fall acceleration at the orbit of the Moon (a distance of 60RE from the center of the Earth with the radius of RE). Use the gravitational acceleration on the surface of the Earth is 9.8m/s2.

Answers

Answer:

The magnitude of the free-fall acceleration at the orbit of the Moon is [tex]2.728\times 10^{-3}\,\frac{m}{s^{2}}[/tex] ([tex]\frac{2.784}{10000}\cdot g[/tex], where [tex]g = 9.8\,\frac{m}{s^{2}}[/tex]).

Explanation:

According to the Newton's Law of Gravitation, free fall acceleration ([tex]g[/tex]), in meters per square second, is directly proportional to the mass of the Earth ([tex]M[/tex]), in kilograms, and inversely proportional to the distance from the center of the Earth ([tex]r[/tex]), in meters:

[tex]g = \frac{G\cdot M}{r^{2}}[/tex] (1)

Where:

[tex]G[/tex] - Gravitational constant, in cubic meters per kilogram-square second.

[tex]M[/tex] - Mass of the Earth, in kilograms.

[tex]r[/tex] - Distance from the center of the Earth, in meters.

If we know that [tex]G = 6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}}[/tex], [tex]M = 5.972\times 10^{24}\,kg[/tex] and [tex]r = 382.26\times 10^{6}\,m[/tex], then the free-fall acceleration at the orbit of the Moon is:

[tex]g = \frac{\left(6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} \right)\cdot (5.972\times 10^{24}\,kg)}{(382.26\times 10^{6}\,m)^{2}}[/tex]

[tex]g = 2.728\times 10^{-3}\,\frac{m}{s^{2}}[/tex]

Which of the following does not represent a significant negative environmental impact of using nuclear fuel? a. air pollution b. mining c. radiation d. waste production Please select the best answer from the choices provided A B C D

Answers

the answer is a. mining.

A significant negative environmental impact of using nuclear fuel is mining. The correct option is b.

What is mining?

Mining is the process or industry of obtaining coal or other minerals from a mine.

Using nuclear fuels for mining will have no negative impact on the environment. All other terms represent negative impact of nuclear fuel Thus,  The correct option is b.

Learn more about mining.

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Two students are balancing on a 10m seesaw. The seesaw is designed so that each side of the seesaw is 5m long. The student on the left weighs 60kg and is sitting three meters away from the fulcrum at the center. The student on the right weighs 45kg. The seesaw is parallel to the ground. The mass of the board is evenly distributed so that its center of mass is over the fulcrum. What distance from the center should the student on the right be if they want the seesaw to stay parallel to the ground? 
a. 4m
b. 5m
c. 2m
d. 3m
NO LINKS. ​

Answers

Answer:

x = 4 m

Explanation:

For this exercise we must use the rotational equilibrium relationship, where we place zero at the turning point and counterclockwise rotations we will consider positive

as it indicates that the bar is in equilibrium, its center of mass coincides with the turning point, so the distance is zero and does not create torque on the system

        ∑τ  = 0

        W 3 - w x = 0

        x = 3W / w

        x = 3 Mg / mg

        x = 3 M / m

let's calculate

       x = 3 60/45

       x = 4 m

Does a ball rolling on an inclined plane have the same acceleration on the way up as it does on the way down?

Answers

Answer:

No, it does not.

Explanation:

According to the Law of Gravitation, something going down has more kinetic energy than something going up because it attracts pressure from around it when going down. When it goes up, it has less gravitational force and inertia also stops the ball from rolling upward. Therefore, without the amount of kinetic energy, it will not have the same amount of acceleration.

what is the comparison of Wind turbine and other energy sources?

Answers

Wind is a more efficient power source than solar. Compared to solar panels, wind turbines release less CO2 to the atmosphere, consume less energy, and produce more energy overall.

AnWind power or wind energy is the use of wind to provide mechanical power through wind turbines to turn electric generators for electrical power. Wind power is a popular sustainable, renewable source of power that has a much smaller impact on the environment compared to burning fossil fuels. Wind farms consist of many individual wind turbines, which are connected to the electric power transmission network. Onshore wind is an inexpensive source of electric power, competitive with or in many places cheaper than coal or gas plants. Onshore wind farms have a greater visual impact on the landscape than other power stations, as they need to be spread over more land and need to be built away from dense population. Offshore wind is steadier and stronger than on land and offshore farms have less visual impact, but construction and maintenance costs are significantly higher. Small onshore wind farms can feed some energy into the grid or provide power to isolated off-grid locations. The wind is an intermittent energy source, which cannot be dispatched on demand. Locally, it gives variable power, which is consistent from year to year but varies greatly over shorter time scales. Therefore, it must be used together with other power sources to give a reliable supply. Power-management techniques such as having dispatchable power sources (often gas-fired power plant or hydroelectric power), excess capacity, geographically distributed turbines, exporting and importing power to neighboring areas, grid storage, reducing demand when wind production is low, and curtailing occasional excess wind power, are used to overcome these problems. As the proportion of wind power in a region increases the grid may need to be upgraded. Weather forecasting permits the electric-power network to be readied for the predictable variations in production that occur. In 2019, wind supplied 1430 TWh of electricity, which was 5.3% of worldwide electrical generation, with the global installed wind power capacity reaching more than 651 G…swer:

Explanation:

How does a generator use the effect you noticed in the pickup coil to generate electrical energy? What energy transformations are taking place? Why does a generator make alternating current? What change would have to be made to make direct current?

Answers

Answer:

Movement of Electrons in opposite direction of the rotation of the pickup coil leads to the direction of Electrical energy ( Alternating current ).

Explanation:

For a Generator to generate electricity, Electrons that is found in the rotating coil of the Generator will experience some sort of force that makes them to start moving in a direction that is perpendicular to the direction of the rotating/pickup coil found in the Generator.

The conversion of mechanical energy (Rotation of the pickup coil ) to electrical energy takes place in the Generator

The generator makes an alternating current because electrons move in opposite direction of the rotating coil

In other to generate a Direct current using a generator we have to replace the slip rings with commutator.

What are used to transfer electrical energy from a power plant all the way to a home? generators turbines power transformers power lines

Answers

Answer:

Power lines

Explanation:

Generators induce that current that runs the turbines

Turbines also induce currents from water waves that is transmitted.

Transformers change A.C to D.C or vice versa.

An idea to explore the nearest Earth-sized planet in a habitable zone to our Sun is to attach a small sensor array to a 4 m by 4 m solar sail and shine a 100 GW laser on the solar sail for 10 minutes. This will get the spaceship up to speed after which it will coast the rest of the way at the speed attained after the 10 minutes of acceleration. Assume the sensor array and solar sail has a mass of 3 grams. The nearest Earth-sized planet in a habitable zone orbits the Alpha Centauri system at 4.3 Lightyears. 1 GW = 1*109 W. 1 Light year = 4*1016 m. Assume a perfectly reflecting solar sail and assume all 100 GW of power are intercepted by the solar sail.
a. What is the pressure on the Solar Sail from the laser?
b. What is the force on the Solar Sail from the laser?
c. What speed will the spaceship attain after 10 minutes assuming it started from rest and assuming a constant acceleration?
d. How long will it take the spaceship to reach Alpha Centauri in years?

Answers

Solution :

Given data:

Area of the solar sail, A = 16 [tex]$m^2$[/tex]

Mass (array + sail), m = [tex]$3 \times 10^{-3}$[/tex] kg

Power, P = 100 GW

               = [tex]$10^{11}$[/tex] W

Time, t = 10 min

           = 10 x  60 s

Distance, D = [tex]$4.3 \times 4 \times 10^{16}$[/tex] m

Kinetic energy, [tex]$KE=\frac{1}{2}mv^2=P\times t$[/tex]

[tex]$v=\sqrt{\frac{2Pt}{m}}$[/tex]

[tex]$v=\sqrt{\frac{2\times 10^{11}\times 600}{3 \times 10^{-3}}}$[/tex]

  [tex]$=2 \times 10^8$[/tex] m/s

So, the acceleration is [tex]$a=\frac{v}{t}$[/tex]

[tex]$a=\frac{2 \times 10^8}{6 \times 10^2} \ m/s^2$[/tex]

 [tex]$=333333.33 \ m/s^2$[/tex]

Therefore, force,

[tex]$F = ma$[/tex]

   [tex]$=3 \times 10^{-3}\times 333333.33$[/tex]

   = 1000 N

Pressure, [tex]$P=\frac{F}{A}$[/tex]

                   [tex]$=\frac{1000}{16}$[/tex]

                   [tex]$=62.5 \ N/m^2$[/tex]

Therefore, time taken is t

Now the distance is

[tex]$d_1=\frac{1}{2}at_1^2$[/tex]

[tex]$d_1=0.5 \times 333333.33 \times (600)^2$[/tex]

    [tex]$=6 \times 10^{10} \ m$[/tex]

Now, the distance, [tex]$d_2 = D-d_1=v.t_2$[/tex]

Now, [tex]$t_2=\frac{(17.2 \times 10^{16})-(6 \times 10^{10})}{2 \times 10^8}$[/tex]

            = 859999700 s

Therefore, total time is

[tex]$T=t_1+t_2$[/tex]

  = 86000300 s

  = 27.27 years

write the dimensional formula of gravitational constant and specific heat capacity.​

Answers

Answer:

SEE EXPLANATION

Explanation:

The dimensional formula of gravitational constant is given by,

M^-1 L^3 T^-2  

Where,  

M = Mass

L = Length

T = Time

Dimensional formula of Specific Heat Capacity =[ M^0L^2T^-2K^-1]

A book is sitting on a table. The book is
applying a 65 N down on the table.
How much force is the table exerting on the
book?

Answers

Answer:

65 Newtons

Explanation:

A book is kept on the table. The book is applying 65 N of force down on the table. So the table will also apply 65 N force to the book.

What is Force?

According to physics, a force is an effect that has the power to change an object's motion. A force can cause a massed item to accelerate or modify its velocity. A push or a pull is a straightforward method to explain force. Considering that a force has both a magnitude and a direction, it is a vector quantity.

The pace at which an object's direction changes when it is traveling is referred to as its velocity, and is measured by a particular unit of time and from a particular point of view.

When the book is applying force on the table that is mg, and it is 65 N the table will also generate a normal reaction force which will be equal to the force applied by the book which is 65 N.

To get more information about Force :

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Although we have discussed single-slit diffraction only for a slit, a similar result holds when light bends around a straight, thin object, such as a strand of hair. In that case, a is the width of the strand. From actual laboratory measurements on a human hair, it was found that when a beam of light of wavelength 631.8 nm was shone on a single strand of hair, and the diffracted light was viewed on a screen 1.20 m away, the first dark fringes on either side of the central bright spot were 5.02 cm apart.

Required:
How thick was this strand of hair?

Answers

Answer:

d = 1.51 x 10⁻⁵ m = 15.1 μm

Explanation:

We will use young's Double Slit formula here:

[tex]Y = \frac{\lambda L}{d}\\\\d = \frac{\lambda L}{Y}[/tex]

where,

d = width of strand = ?

λ = wavelength = 631.8 nm = 6.318 x 10⁻⁷ m

L = Screen to hair distance = 1.2 m

Y = fringe spacing = 5.02 cm = 0.0502 m

Therefore,

[tex]d = \frac{(6.318\ x\ 10^{-7}\ m)(1.2\ m)}{0.0502\ m}[/tex]

d = 1.51 x 10⁻⁵ m = 15.1 μm

The yearly whole-body dose caused by radioactive 40K absorbed in our tissues is 17 mrem. Note: 40K also emits gamma rays, many of which leave the body before being absorbed. Because fatty tissue has low potassium concentration and muscle has a higher concentration, gamma ray emissions indicate indirectly a person's fat content. Assuming that 40K undergoes beta decay with an RBE of 1.4 , determine the absorbed dose in rads. How much beta ray energy does an 75-kg person absorb in one year? Use that 1rad=10^−2J/kg .

Answers

Answer:

The beta ray Energy (E) absorbed = 0.009105 Joules

Explanation:

The absorbed dose rate in the tissue can be determined by using the known formula:

Dose equivalent = absorbed dose × relative biological effective (RBE)

[tex]absorbed \ dose = \dfrac{Dose \ equivalent}{RBE}[/tex]

[tex]absorbed \ dose = \dfrac{17 \times 10^{-3} \ rem}{1.4}[/tex]

absorbed dose = 12.14 × 10⁻³ rad

absorbed dose = 12.14 m rad

However, since the absorbed dose is determined, the energy absorbed in one year is can be expressed as:

Energy (E) = mass × absorbed rate

Energy (E) = 75 kg × 12.14 × 10⁻³ rad

Energy (E) = 75 kg × 12.14 × 10⁻³× 10⁻² J/kg

Energy (E) = 75 × 12.14 × 10⁻³× 10⁻² Joules

Energy (E) = 0.009105 Joules

in a simple machine the input work is 187.5 N and output work is 125N. If load distance is 25 cm and effort distance is 75 cm calculate value of load and effort ​

Answers

Answer:

1. Laod = 500 N

2. Effort = 250 N

Explanation:

From the question given above, the following data were obtained:

Input work = 187.5 J

Output work = 125 J

Load distance = 25 cm

Effort distance = 75 cm

1. Determination of the load.

Output work = 125 J

Load distance = 25 cm = 25/100 = 0.25 m

Load =?

Output work = Load × Load distance

125 = Load × 0.25

Divide both side by 0.25

Load = 125 / 0.25

Laod = 500 N

2. Determination of the effort.

Input work = 187.5 J

Effort distance = 75 cm = 75/100 = 0.75 m

Effort =?

Input work = Effort × Effort distance

187.5 = Effort × 0.75

Divide both side by 0.75

Effort = 187.5 / 0.75

Effort = 250 N

On a day when the water is flowing relatively gently, water in the Niagara River is moving horizontally at 4.5 m/s before shooting over Niagara Falls. After moving over the edge, the water drops 53 m to the water below. If we ignore air resistance, how much time does it take for the water to go from the top of the falls to the bottom? How far does the water move horizontally during this time?

Answers

Answer: 3.28 s, 14.76 m

Explanation:

Given

Water moves horizontally with a speed of [tex]v=4.5\ m/s[/tex]

It falls from a height of [tex]h=53\ m[/tex]

Time is taken to fall this much height is

[tex]\Rightarrow t=\sqrt{\dfrac{2h}{g}}\quad [\text{As there is no initial vertical velocity}]\\\\\Rightarrow t=\sqrt{\dfrac{2\times 53}{9.8}}\\\\\Rightarrow t=3.28\ s[/tex]

In this time, water moves horizontally by

[tex]\Rightarrow d=4.5\times 3.28\\\Rightarrow d=14.76\ m[/tex]

A 2.4 tonne boulder falls from the top of a 125 m cliff onto the ground below
What speed does it hit the ground with?
(Ignore air resistance.)
Give your answer to the nearest 0.1 m/s

Use g = 9.8 m/s2 for the acceleration due to gravity.

Answers

Answer:

49.5 m/s

Explanation:

Let's use the energy approach to solve this problem.

We are given the mass and displacement of the boulder.

The formula for potential energy is PE = mgh.

The formula for kinetic energy is KE = 1/2mv².

If we ignore air resistance, we can use the conservation of energy to solve this problem by setting PE and KE equal to each other. By doing so, we realize that the mass of the boulder does not affect the speed that it hits the ground (final velocity).

mgh = 1/2mv²

Divide both sides of the equation by m.

gh = 1/2v²

Let's take the downwards direction as positive. The displacement of the boulder is 125 m and g = 9.8 m/s².

(9.8)(125) = 1/2v²

Now we can solve for v.

1225 = 1/2v²

Multiply both sides by 2.

2450 = v²

Take the square root of both sides.

v = 49.49747468

Rounded to the nearest 0.1 m/s, the final velocity of the boulder is 49.5 m/s.

A scientist wants to create a new material using nanotechnology. In which size range will she be working? 1 nm to 100 nm 1 to 10,000 nm 100 to 1000 nm 100 to 10,000 nm

Answers

Answer:

I believe it’s A

Explanation:

Answer:

A is correct (1 nm to 100 nm)

Explanation:

I just took the exam

1. What kind of force is the force of gravity? (1 Point) contact force
non-contact force electromagnetic force imaginary force

Answers

Answer:

the answer is electromagnetic force because gravity is a force of attraction

please select all that would be categorized as vectors

Answers

Answer:

1. 10 ft/s² straight up

2. 4 Km north-west

3. 20 m/s South

Explanation:

To successfully categorize the quantities given above as vectors, we must understand what vector quantity is.

A vector quantity is a quantity that has both magnitude and direction.

Considering the question given above, the vector quantities are:

1. 10 ft/s² straight up

2. 4 Km north-west

3. 20 m/s South

The remaining quantities in the question are not considered as vectors because they only have magnitude but no direction. They are therefore referred to as scalar quantities.

74Be decays with a half-life of about 53 d. It is produced in the upper atmosphere, and filters down onto the Earth's surface.

If a plant leaf is detected to have 350 decays/s of 74Be, how long do we have to wait for the decay rate to drop to 25 per second?

Answers

Answer:

201.8 days

Explanation:

The activity of a radioactive sample as a function of times is :

[tex]$R=R_0e^{\frac{0.693t}{T_{1/2}}}$[/tex]

Here, [tex]$R_0$[/tex] = the initial activity

          [tex]$T_{1/2}$[/tex] = half life

          t = elapsed time

Now rearranging the equation for time, t, we get:

[tex]$\frac{R}{R_0}=e^{-\frac{0.693t}{T_{1/2}}}$[/tex]

[tex]$\ln\left(\frac{R}{R_0}\right)=-\frac{0.693t}{T_{1/2}}$[/tex]

[tex]$t=\frac{-\ln\left(\frac{R}{R_0}\right)T_{1/2}}{0.693}$[/tex]

[tex]$t=\frac{-\ln\left(\frac{25}{350}\right)\times 53}{0.693}$[/tex]

 = 201.8 days

Therefore, the required time is 201.8 days

Magnets on the tracks
How can levitating trains support the claim that magnetic fields exist between objects exerting magnetic force on each other even when there is no physical contact?

Answers

Answer:

Explanation: They levitate sort of because when to opposite forces of magnets cant go together they repel

The wings of the blue-throated hummingbird, which inhabits Mexico and the southwestern United States, beat at a rate of up to 900 times per minute.

Required:
a. Calculate the period of vibration of the bird’s wings.
b. Calculate the frequency of the wings’ vibratio
c. Calculate the angular frequency of the bird’s wingbeats.

Answers

Answer:

a) the period of vibration of the bird’s wings is 0.0667s

b) the frequency of the wings’ vibration is 15 Hz

c) the angular frequency of the bird’s wingbeats is 94.2 rad/s

Explanation:

Given the data in the question;

beat at a rate of up to 900 times per minute.

The number of vibrations per unit time can be be determined as follows;

frequency f = number of vibration / time ---------- let this be equation 1

Also Time period T is;

T = 1/f { time period and frequency are reciprocal to each other }

T = 1/f ---------- Let this be equation 2

Then, the angular frequency ω can be determined using the formular;

ω = 2π × f -------------- Let this be equation 3.

Now, given that; The wings of the blue-throated hummingbird beat at a rate of up to 900 times per minute; Hence,

from equation 1, frequency f = number of vibration / time

f = 900 / 1 min

f = 900 / 60 sec

f = 15 Hz

Therefore, the frequency of the wings’ vibration is 15 Hz

a) the period of vibration of the bird’s wings

from equation 2, T = 1/f ,

we substitute in the value of f

T = 1 / 15 Hz

T = 0.0667s

Therefore, the period of vibration of the bird’s wings is 0.0667s

c) the angular frequency of the bird’s wingbeats;

from equation 3, ω = 2π × f

we substitute

ω = 2π × 15 Hz

ω = 94.2 rad/s

Therefore, the angular frequency of the bird’s wingbeats is 94.2 rad/s

Plane-polarized light is incident on a single polarizing disk, with the direction of E0 parallel to the direction of the transmission axis. Through what angle should the disk be rotated so that the intensity in the transmitted beam is reduced by a factor of 6.00?

Answers

Answer: [tex]65.9^{\circ}[/tex]

Explanation:

Given

Intensity must be reduced by a factor of 6

Intensity is given by [tex]I=I_o\cos ^2\theta\\[/tex]

Substitute [tex]I[/tex] by [tex]\frac{I_o}{6}[/tex]

[tex]\Rightarrow \dfrac{I_o}{6}=I_o\cos ^2\theta\\\\\Rightarrow \cos^2\theta =\dfrac{1}{6}\\\\\Rightarrow \cos \theta=\dfrac{1}{\sqrt{6}}\\\\\Rightarrow \theta=65.9^{\circ}[/tex]

So, the disk must be rotated by an angle of [tex]65.9^{\circ}[/tex] .

A light bulb has 240 Ω of resistance when connected to a 120 V electrical socket. What is the current through the light bulb? (AKS 8b)

Question 36 options:

2.0 A


1.0 A


0.50 A


0.20 A

Answers

Answer:

0.50 A

Explanation:

We can find the current from this through Ohm's Law:

V = IR

Now, in it's original form, we're only looking for Voltage. However, we can switch this formula around to find each part of it:

I = V / R (current)

R = V / I (resistance)

In this case, we'll use the formula for current:

V = 120 V

R = 240 Ω

I = ?

I = 120 / 240

I = 0.5 A

135g of an unknown substance gains 9133 J of heat as it is heated from 25⁰C to 100⁰C. Using the chart below, determine the identity of the unknown substance.

Answers

Answer:

The unknown substance is Aluminum.

Explanation:

We'll begin by calculating the change in the temperature of substance. This can be obtained as follow:

Initial temperature (T₁) = 25 ⁰C

Final temperature (T₂) = 100 ⁰C

Change in temperature (ΔT) =?

ΔT = T₂ – T₁

ΔT = 100 – 25

ΔT = 75 ⁰C

Finally, we shall determine the specific heat capacity of the substance. This can be obtained as follow:

Change in temperature (ΔT) = 75 ⁰C

Mass of the substance (M) = 135 g

Heat (Q) gained = 9133 J

Specific heat capacity (C) of substance =?

Q = MCΔT

9133 = 135 × C × 75

9133 = 10125 × C

Divide both side by 10125

C = 9133 / 10125

C = 0.902 J/gºC

Thus, the specific heat capacity of substance is 0.902 J/gºC

Comparing the specific heat capacity (i.e 0.902 J/gºC) of substance to those given in the table above, we can see clearly that the unknown substance is aluminum.

A tension of 35.0N is applied downward on a 2.50kg mass. What is the acceleration of the mass?
A. -59.5 m/s2
B. -14.0 m/s2
C. -4.20 m/s2
D. -23.8 m/s2​

Answers

Answer:

D. -23.8 m/s2​

Explanation:

Answer:

[tex]\text{B. }-14.0\:\mathrm{m/s^2}[/tex]

Explanation:

From Newton's 2nd Law, we have [tex]\Sigma F=ma[/tex].

Substituting given values, we get:

[tex]-35.0=2.50a,\\a=\frac{-35.0}{2.50},\\a=\boxed{-14.0\:\mathrm{m/s^2}}[/tex]

*Note the negative sign simply implies that the mass is travelling downwards (direction).

4. How do ordinary objects behave in microgravity?
Bubbles rose faster
Bubbles paused
Bubbles kept rising

Answers

Answer:

Bubbles paused

Explanation:

the air bubble doesn't rise because it is no lighter than the water around it—there's no buoyancy. The droplet doesn't fall from the leaf because there's no force to pull it off. It's stuck there by molecular adhesion.

for instance, onto the International Space Station, gravity becomes negligible, and the laws of physics act differently than here on Earth

On Earth, the buoyancy of the air bubbles causes them to rise to the top together, creating a segregation between air and water. However, in microgravity, nothing forces the air bubbles to interact and thus rise together, Green said.

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