Use the graph to answer the following questions


1.) About how long is one half-life for Element 1? (Hint: how long does it take for half of it to decay?)

2.) A scientist finds a bone that has about 15% Element 1 left. About how old is the bone?

3.) A scientist finds a rock with about 75% of its Element 1 left. About how old is the rock?

4.) A scientist drops all his samples and is trying to figure out where to classify them. He thinks a sample is from about 9,000,000 years ago. If he tests the sample, how much of element 1 should be left in it?

5.)A paleontologist finds a skeleton that he thinks is about 1,000,000 years old. How much Element 1 should she expect to find in the sample if her guess is correct?

Use The Graph To Answer The Following Questions 1.) About How Long Is One Half-life For Element 1? (Hint:

Answers

Answer 1

1. The half-life for element 1 is 3000000 years

2. The age of the bone is 8220000 years

3. The age of the rock is 1230000 years

4. The amount of element 1 remaining is 12.5%

5. The amount of element 1 she would find is 79.55%

1. How do i determine the half-life?

The half-life of a substance is the time taken for half the substance to decay.

From the above diagram, we can see that the original amount of element 1 is 100 g. Thus, half of 100 is 50.

Looking at the diagram, the time for 50 g is 3000000 years.

Therefore, we can conclude that the half-life of element 1 is 3000000 years

2. How do i determine the age of the bone?

First, we shall determine the number of half lives that has elapsed. This is obtained as follow:

Original amount (N₀) = 100%Amount remaining (N) = 15%Number of half-lives (n) =?

2ⁿ = N₀ / N

2ⁿ = 100 / 15

2ⁿ = 6.67

Take the log of both sides

Log 2ⁿ = Log 6.67

nLog 2 = Log 6.67

Take the log of both sides

n = Log 6.67 / Log 2

n = 2.74

Finally, we shall determine the age of the bone. Details below

Half-life of element 1 (t½) = 3000000 yearsNumber of half-lives (n) = 2.74 Age of bone (t) =?

t = n × t½

t = 2.74 × 3000000

t = 8220000 years

Thus, the age of the bone is 8220000 years

3. How do i determine the age of the rock?

First, we shall determine the number of half lives that has elapsed. This is obtained as follow:

Original amount (N₀) = 100%Amount remaining (N) = 75%Number of half-lives (n) =?

2ⁿ = N₀ / N

2ⁿ = 100 / 75

2ⁿ = 1.33

Take the log of both sides

Log 2ⁿ = Log 1.33

nLog 2 = Log 6.67

Take the log of both sides

n = Log 1.33 / Log 2

n = 0.41

Finally, we shall determine the age of the rock. Details below

Half-life of element 1 (t½) = 3000000 yearsNumber of half-lives (n) = 0.41 Age of bone (t) =?

t = n × t½

t = 0.41 × 3000000

t = 1230000 years

Thus, the age of the rock is 1230000 years

4. How do i determine the amount remaining?

We shall first obtain the number of half lives that will elaspe during the time. Details below:

Half-life of element 1 (t½) = 3000000 yearsTime (t) = 9000000 yearsNumber of half-lives (n) =?

n = t / t½

n = 9000000 / 3000000

n = 3

Finally, we shall determine the amount remaining. Details below:

Original amount (N₀) = 100Number of half-lives (n) = 3Amount remaining (N) = ?

N = N₀ / 2ⁿ

N = 100 / 2³

N = 100 / 8

N = 12.5%

Thus, we can conclude that the amount remaining is 12.5%

5. How do i know how much of the element she would find?

We shall first obtain the number of half lives that will elaspe during the time. Details below:

Half-life of element 1 (t½) = 3000000 yearsTime (t) = 1000000 yearsNumber of half-lives (n) =?

n = t / t½

n = 1000000 / 3000000

n = 0.33

Finally, we shall determine the amount she would find. Details below:

Original amount (N₀) = 100Number of half-lives (n) = 0.33Amount found (N) = ?

N = N₀ / 2ⁿ

N = 100 / 2^0.33

N = 79.55%

Thus, the amount she would find is 79.55%

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Related Questions

a uniform ladder of mass m=40kg and length l=10m is leaned against a smooth vertical wall. a person of mass m=80kg stands on the ladder a distance x=7m from the bottom, as measured along the ladder. the foot of the ladder is d=1.2m from the bottom of the wall. what are the force exerted by the wall and the normal reaction exerted by the floor on the ladder?​

Answers

Answer:

according to the internet

The force exerted by the wall on the ladder is 468 N, and the normal reaction exerted by the floor on the ladder is 1152 N.

Explanation:

The force exerted by the wall on the ladder is zero, and the normal reaction exerted by the floor on the ladder is 1176 N.

What is a reaction force?

A reaction force is a force that occurs in response to an action force. According to Newton's third law of motion, "For every action, there is an equal and opposite reaction." This means that whenever an object exerts a force on another object, the second object exerts an equal and opposite force back on the first object.

For example, when you push against a wall, your body is exerting a force on the wall, but the wall is also exerting an equal and opposite force back on your body. This force is known as the reaction force. Another example is when a rocket propels itself forward by expelling exhaust gases backward. The exhaust gases exert a force on the rocket, and the rocket exerts an equal and opposite force back on the exhaust gases.

Here in the Question,

To solve this problem, we need to consider the forces acting on the ladder. There are three forces to consider: the weight of the ladder and the two forces exerted by the wall and the floor on the ladder.

Let's start by finding the weight of the ladder and the person standing on it. The weight is given by:

W = mg

where m is the mass and g is the acceleration due to gravity. For the ladder, we have:

W_ladder = 40 kg x 9.8 m/s^2 = 392 N

For the person:

W_person = 80 kg x 9.8 m/s^2 = 784 N

Next, we need to find the horizontal and vertical components of the forces exerted by the wall and the floor. Since the ladder is in equilibrium, the sum of the forces in the horizontal and vertical directions must be zero.

Let's start with the horizontal direction. The only force in this direction is the force exerted by the wall, which we'll call F_wall. Since there is no acceleration in the horizontal direction, we have:

F_wall = 0

This means that the force exerted by the wall on the ladder is zero.

Now let's look at the vertical direction. The forces in this direction are the weight of the ladder and the person (W_ladder + W_person), the normal force exerted by the floor (N_floor), and the force exerted by the wall (F_wall). Since the ladder is not accelerating in the vertical direction, the sum of the forces in this direction must be zero:

N_floor + F_wall - W_ladder - W_person = 0

Since we know that F_wall = 0, we can simplify this equation to:

N_floor - W_ladder - W_person = 0

Now we can plug in the values we found earlier:

N_floor - 392 N - 784 N = 0

Solving for N_floor, we get:

N_floor = 1176 N

So the normal reaction exerted by the floor on the ladder is 1176 N.

Therefore, the force exerted by the wall on the ladder is zero, and the normal reaction exerted by the floor on the ladder is 1176 N.

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• A helicopter's rotor speed changed from 225 rev/min to 300 rev/min in one minute. How long will it take for the rotor to have its angular velocity doubled over the initial speed. Assuming a constant angular acceleration.​

Answers

The time taken for the helicopter rotor to double its angular velocity over the initial speed is 1 minute.

What is the time of motion?

If the rotor speed changed from 225 rev/min to 300 rev/min in one minute.

This means the change in angular velocity (Δω) is:

Δω = ω2 - ω1 = 300 rev/min - 225 rev/min = 75 rev/min

We are given that the rotor doubles its angular velocity over the initial speed, so ω2 = 2ω1.

Substituting ω2 = 2ω1 into the equation for Δω, we get:

75 rev/min = 2ω1 - ω1

Simplifying, we get:

75 rev/min = ω1

Now we can solve for the time taken for the rotor to double its angular velocity over the initial speed.

Δω = αt

Substituting Δω = 75 rev/min and ω1 = 75 rev/min, we get:

75 rev/min = αt

Now we can solve for t:

t = Δω / α

We need to find α, the angular acceleration.

α = Δω / t = 75 rev/min / 1 min = 75 rev/min²

Now we can substitute the value of α back into the equation for t:

t = Δω / α = 75 rev/min / 75 rev/min² = 1 minute

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PLS HELP!!

1. Radon-222 decays with a half-life of 3.82 days. If an initial sample of the element contains 1000 atoms, it takes ___ days for that sample to decay down to 125 atoms.

a. 2 days
b. 3 days
c. 4 days
d. 5 days

Answers

Answer: 3 days

Explanation:

Work done by a force in a moving object is 100J . It was traveling at a speed of 2m/s and comes down ?

Answers

Work done by a force in a moving object is 100J . It was traveling at a speed of 2m/s and comes down to 1 m/s.

Work done is the amount energy gained (loosed) in bringing the body from initial position to final position. It is denoted by W and its SI unit is joule(J). i.e. Work(W) is force(F) times displacement(s). W=F× s When a body is displaced with 1 newton of force by 1 m, then we can say that work has been done on the body by 1 joule. Writing for it's dimension, W=F× s.

W = Fs

In this problem mass of the object is not given, consider the mass of the object is 2kg, time to speed down the object is 5s and distance is 25m then,

given,

W = 100J

M = 2kg

t = 5s

s = 25m

then,

W = m.Δv/Δt.s

100 = 2(2-v₂)/5 .25

2 - v₂= 100×5/50

2 - v₂  = 3

v₂ = 1 m/s

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Calculate the volume occupied by 1.80 mol of steam at 200 ∘C . Assume the steam is at atmospheric pressure and can be treated as an ideal gas.

Answers

The volume occupied by 1.80 mol of steam at 200°C and atmospheric pressure is approximately 40.7 L.

We can use the ideal gas law, PV = nRT, to solve for the volume of the steam. First, we need to determine the pressure of the steam at 200°C and atmospheric pressure. We can use a steam table to find that the saturation pressure of steam at 200°C is 1.013 bar.

Next, we can convert this pressure to Pascals,

1.013 bar = 101,325 Pa

We also need to convert the temperature to Kelvin 200°C + 273.15 = 473.15 K.

Now we can plug in our values,

PV = nRT

V = nRT/P

V = (1.80 mol)(8.314 J/mol•K)(473.15 K)/(101,325 Pa)

V ≈ 67.6 L

Therefore, the volume occupied by 1.80 mol of steam at 200°C and atmospheric pressure is approximately 67.6 liters.

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Question 3 of 25
Which of the following describes sound waves?

A. Mechanical waves in which the vibrations are perpendicular to the
motion of the sound

B. Electromagnetic waves in which the vibrations are perpendicular
to the motion of the sound

C. Electromagnetic waves in which the vibrations are parallel to the
motion of the sound

D. Mechanical waves in which the vibrations are parallel to the
motion of the sound

Answers

Answer:

D. Mechanical waves in which the vibrations are parallel to the motion of the sound.

Explanation:

gg

Answer: D
Mechanical waves in which the vibrations are parallel to the motion of the sound.

Explanation:
just took

The electric field component in the figure are Ēx = 3xî, Ēy = Ē₂ = 0. Calculate the flux through (1,2,3) the square surfaces of side 2 cm.​

Answers

The electric flux through a surface is calculated by the equation, φ = ∫E•dA, where E is the electric field, dA is the area vector and the integral is taken over the surface area.

What is electric flux ?

Electric flux is a measure of the amount of electric field passing through a given area. It is represented by the letter Φ (phi) and is measured in units of volt-meters (Vm). Electric flux is related to the strength of electric fields, and the amount of electric flux passing through a given area is determined by the electric field strength and the area through which it passes. Electric flux is often used in calculations of electric field strength and electric field energy, as well as in the field of electromagnetism. Electric flux can also be used to calculate the total amount of electric charge passing through a given area.

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A 89.4 kg weight-watcher wishes to climb a mountain to work off the equivalent of a large piece of chocolate cake rated at 605 (food) Calories. How high must the person climb? The acceleration due to gravity is 9.8 m/s 2 and 1 food Calorie is 103 calories. Answer in units of km.

Answers

The weight-watcher is need to climb approximately 0.298 kilometers to burn off the equivalent of a large piece of chocolate cake rated at 605 food Calories.

To determine the height that the weight-watcher needs to climb, we first need to calculate the amount of potential energy that would be required to burn off 605 food Calories.

First, we need to convert 605 food Calories into calories:

605 food Calories x 103 calories/food Calorie = 62315 calories

Next, we need to convert calories to joules:

1 calorie = 4.184 joules

62315 calories x 4.184 joules/calorie = 260808.56 joules

Now we can calculate the height that the weight-watcher must climb using the formula for potential energy:

potential energy = mass x gravity x height

where, mass = 89.4 kg, gravity = 9.8 m/s², and

potential energy = 260808.56 joules

Solving for height, we get:

height = potential energy / (mass x gravity)

height = 260808.56 joules / (89.4 kg x 9.8 m/s²)

height = 298.47 meters

Finally, we convert the height to kilometers:

298.47 meters = 0.29847 kilometers

Therefore, the weight-watcher needs to climb approximately 0.298 kilometers to burn off the equivalent of a large piece of chocolate cake rated at 605 food Calories.

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A student connects a battery to a wire and wraps the wire around an iron nail to produce an electromagnet. Which action should the student take to increase the number of paper clips the electromagnet can pick pick up?

Answers

It's worth noting that increasing any of these factors too much can cause the wire to overheat, potentially damaging the wire or the battery. So, the student should exercise caution and avoid exceeding safe limits.

To increase the number of paper clips that the electromagnet can pick up, the student can take one or more of the following actions:

Increase the number of coils: By increasing the number of times the wire is wrapped around the iron nail, the strength of the magnetic field produced by the electromagnet can be increased.

Increase the current: By increasing the voltage of the battery or using a stronger battery, the current flowing through the wire can be increased, which will increase the strength of the magnetic field.

Use a stronger iron nail: By using an iron nail with a higher magnetic permeability, the strength of the magnetic field produced by the electromagnet can be increased.

Use thicker wire: By using a wire with a larger cross-sectional area, the resistance of the wire can be reduced, which will allow more current to flow through the wire, resulting in a stronger magnetic field.

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15. A ball of mass 5 kg and a block of mass 12 kg are attached by a lightweight cord that passes over a frictionless pulley of negligible mass as shown in the figure. The block lies on a frictionless incline of angle 30o . Find the magnitude of the acceleration of the two objects and the tension in the cord. Take g = 10​

Answers

According to the question the magnitude of the acceleration of the two objects is 0 and the tension in the cord is also 0.

What is acceleration?

Acceleration is the rate of change of velocity. It is a vector quantity, meaning it has both magnitude and direction. Acceleration can be caused by a force, such as gravity or friction, or by a change in the velocity itself. In the case of a force, the acceleration is proportional to the force and inversely proportional to the mass of the object.

Let a be the acceleration of the ball and b be the acceleration of the block.

Applying Newton's second law to the ball,

T − mg sin30o = 5a

Applying Newton's second law to the block,

T + mg sin30o − 12b = 0

Solving the two equations,

T = 10a

b = 10a/12

Substituting b in the first equation,

T − mg sin30o = 5a

10a − 5(10) sin30o = 5a

5a(1 − 10 sin30o) = 0

a = 0

b = 0

T = 0

Therefore, the magnitude of the acceleration of the two objects is 0 and the tension in the cord is also 0.

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d. Write any four examples of commonly used metric prefixes.​

Answers

Answer:

1) Terahertz (Tera)

2) Kilolitre (kilo)

3) Mega

4) Gigawatt (Giga)

A basin surrounding a drain has the shape of a circular cone opening upward, having everywhere an angle of 35.0° with the horizontal. A 25.0-g ice cube is set sliding around the cone without friction in a horizontal circle of radius R. (a) Find the speed the ice cube must have as a function of R. (b) Is any piece of data unnecessary for the solution? Suppose R is made two times larger. (c) Will the required speed increase, decrease, or stay constant? If it changes, by what factor? (d) Will the time required for each revolution increase, decrease, or stay constant? If it changes, by what factor? (e) Do the answers to parts (c) and (d) seem contradictory? Explain.​

Answers

According to the information we can infer that the decrease in speed is offset by the increase in radius.

How to calculate the speed the ice cube must have as a function of R?

To calculate the speed the ice cube we have to consider that the gravitational force on the ice cube is balanced by the normal force provided by the cone. So, the speed of the ice cube must be such that the centripetal force equals the gravitational force.

In this case, to find the speed of the ice cube let M be the mass of the ice cube and r be the radius of the circular path. The gravitational force on the ice cube is given by Fg = Mg, where:

g = acceleration due to gravity.

The centripetal force is given by Fc = Mv^2/r, where:

v = speed of the ice cube

Setting Fg = Fc, we get:

Mg = Mv^2/r

Solving for v, we get:

v = sqrt(gr)

Is any piece of data unnecessary for the solution?

No piece of data is unnecessary for the solution.

Will the required speed increase, decrease, or sttay constant?

According to the information, when R is made two times larger, the required speed will decrease. For example:

From part (a), v is proportional to the square root of R. Therefore, if R is doubled, v will be multiplied by the square root of 2, which is approximately 1.414.

Will the time required for each revolution increase, decrease, or stay constant?

To know if the time required for each revolution will increase, decrease or stay constant we have to consider that shen R is made two times larger, the time required for each revolution will increase. To see this, note that the period T of the circular motion is given by:

T = 2πr/v

Do the answers to part C and D seem contradictory?

According to the information, the answers to parts (c) and (d) are not contradictory because the decrease in speed is offset by the increase in radius, resulting in a longer period of revolution. The net effect is that the ice cube will travel the same distance in each revolution, so the total time required for one complete revolution will remain constant.

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the figure shows the "before" and "after" states in an experiment where a block is sliding across a frictionless floor when an internal explosion separates it into three pieces with masses m1 = 5 kg and m2 = 12 kg and m3 = 8 kg. the block is initially moving leftward at 2 m/s. Two of the final velocity are v2 =400m/s and v3 = 9 m/s. By how much (j) does the explosion change the kenitic energy?

Answers

Change in kinetic energy after explosion is 32.9 x 10⁵J.

Mass of the first piece, m₁ = 5 kg

Mass of the second piece, m₂ = 12 kg

Mass of the third piece, m₃ = 8 kg

Velocity of the rock initially, v = 2 m/s

Velocity of the second piece, v₂ = 400 m/s

Velocity of the third piece, v₃ = 9 m/s

According to conservation of momentum,

mv = m₁v₁ + m₂v₂ + m₃v₃

Therefore velocity of the first piece,

v₁ = (mv - m₂v₂ - m₃v₃)/m₁

v₁ = [(25 x 2) - (12 x 400) - (8 x 9)]/5

v₁ = -964.4 m/s

Kinetic energy of the rock initially,

KE = 1/2 mv²

KE = 1/2 x 25 x 4

KE = 50 J

Kinetic energy after explosion,

KE' = 1/2(m₁v₁²+ m₂v₂²+ m₃v₃²)

KE' = 1/2[(5 x 93 x 10⁴) + (12 x 16 x 10⁴) + (8 x 81)]

KE' = 328.5 x 10⁴

Change in kinetic energy = KE' - KE = 32.9 x 10⁵J

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Answer: 3.26 * 10^3

Explanation:

If all of the resistors in the diagram below are equivalent, what is the voltage
across the resistor R2 in the circuit below?
5

Answers

If all of the resistors in the diagram below are equivalent, then the voltage across the resistor R2 is 2.5 V.

The value of resistor is not give take the value of resistor 5Ω

R2 and R3 are in series hence its equivalent resistor Rs is R2 + R3 = 10Ω

R1 = 5Ω and Rs = 10Ω both this resistors are in parallel hence,

I()s = V/Rs = 5/10 = 0.5 A

I(1) = V/R1 = 5 / 5 = 1 A

In order to get voltage across R2 we have to apply voltage devider rule,

voltage across R2 is

V(R2) = R2 ÷ (R1 + R2) V

Putting all the values

V(R2) = 5/10 × 5

V(R2) = 5/10 × 5 = 2.5 V

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If a student were to measure the ball's speed at each position above, at which position would
the ball be traveling the fastest?
000
A
B
C
D

Answers

Answer:

the answer is b

Explanation:

A spherical pot contains 0.75 L of hot water at an initial temperature of 95°C. The pot has an emissivity of 0.60, and the surroundings are at a temperature of 20°C. Calculate the water’s rate of heat loss by radiation.

Answers

Answer:

The rate of heat loss by radiation is given by the formula:

Q = ε * σ * A * (T^4 - T0^4)

where Q is the rate of heat loss, ε is the emissivity of the pot (0.60), σ is the Stefan-Boltzmann constant (5.67 x 10^-8 W/m^2K^4), A is the surface area of the pot (which we can assume is about the same as the area of a sphere with a diameter of 0.2 m, which is the typical size of a pot), T is the temperature of the water (95°C + 273.15K = 368.15K), and T0 is the temperature of the surroundings (20°C + 273.15K = 293.15K).

Plugging in the numbers, we get:

Q = 0.60 * 5.67 x 10^-8 * 4π(0.1)^2 * (368.15^4 - 293.15^4)

Q = 39.17 W

Therefore, the water’s rate of heat loss by radiation is 39.17 W.

Mars rotates on its axis once every 24.8 hours.
(a) What is the speed of a geosynchronous satellite orbiting Mars?
Express your answer with the appropriate units.

(b) What is the altitude of a geosynchronous satellite orbiting Mars?
Express your answer with the appropriate units.

Answers

1.44*[tex]10^{3}[/tex] m/s is the speed of a geosynchronous satellite orbiting Mars.   17.1* [tex]10^{6}[/tex]m is the altitude of a geosynchronous satellite orbiting Mars

What does geosynchronous satellite mean?

A geosynchronous satellite is positioned in an orbit with an orbital period equal to the rate of rotation of the Earth. It takes these satellites 24 hours to orbit the earth once. The equatorial plane is typically not the orbital plane for a typical geosynchronous satellite, though.

The period of rotation of the Mars is  T = 24.8 hours.

The radius of the Mars, R is 3.3895×[tex]10^{6}[/tex]

i.e. 2.05*[tex]10^{7}[/tex]m

The satellite's orbital speed is determined by; v⇒2πr/T i.e. 1.44* [tex]10^{3}[/tex]m/s

A geosynchronous satellite orbiting Mars has a height of: h⇒r-R i.e. 17.1* [tex]10^{6}[/tex]m

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an object is placed at a distance of 27.0 cm away from a thin convex lens with a focal length of 9.00 cm. how far from the lens is the image located and what type of image is formed?

Answers

According to the question the image is located 18.00 cm away from the lens and the image is a real, inverted image.

What is lens?

Lens is an optical device used to focus or disperse light, or to change the optical properties of a beam of light. It is made from transparent materials like glass, plastic or quartz and can be used in a variety of applications such as cameras, microscopes, telescopes and other optical instruments. A lens can also be used to bend light, change its direction, and magnify objects.

The image is located 18.00 cm away from the lens and the image is a real, inverted image. This can be determined using the lens equation given by 1/f = 1/d_o + 1/d_i, where f is the focal length, d_o is the object distance, and d_i is the image distance. Plugging in the given values, we get 1/9.00 = 1/27.0 + 1/d_i, which simplifies to 1/d_i = 1/9.00 - 1/27.0, or d_i = 18.00 cm.

Since the image distance is positive, this indicates that the image is real, meaning it is actually formed at the image distance. This is further confirmed by the fact that the image is inverted, as a real image is always inverted.

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(Figure 1) shows five electric charges. Four charges with the magnitude of the charge 2.0 nC form a square with the size a= 4.0 cm. Positive charge with the magnitude of q= 6.5 nC is placed in the center of the square.

Part A) What is the magnitude of the force on the 6.5 nC charge in the middle of the figure due to the four other charges?
Part B) What is the direction of the force on the 6.5 nC charge in the middle of the figure due to the four other charges?

Answers

Part A) Magnitude of force on 6.5 nC charge in the middle of figure due to four other charges is calculated as 9.89 x 10⁻³ N ; Part B) Direction of the force on 6.5 nC charge in the middle of figure due to four other charges is towards the center of square.

What is meant by electric charges?

Electric charges determine how objects interact with each other through  electromagnetic force

Let's label the four charges forming square as q₁, q₂, q₃, and q₄, and the charge in the center as q₅. The magnitude of the charges q₁-q₄ is 2.0 nC, and the magnitude of q₅ is 6.5 nC.

Part A: Force on q₅ due to each of the other charges given by Coulomb's law is : F = k * |q₁||q₅| / r²

k is Coulomb constant (k = 9.0 x 10⁹ N·m²/C²), |q₁| and |q₅| are magnitudes of charges and r is distance between charges.

Since the four charges q₁-q₄ are equidistant from q₅, then net force on q₅ due to them will be directed towards the center of square.

Distance between q₁ and q₅ is a/√2, since the diagonal of the square is √2 times the length of one side. Therefore, force on q₅ due to one of the charges is:

F₁ = k * |q₁||q₅| / (a/√2)²

F₁ = (9.0 x 10⁹ ) * (2.0 x 10⁻⁹) * (6.5 x 10⁻⁹) / (0.04 m)²

F₁ = 2.47 x 10⁻³ N

Total force on q₅ due to the four charges is: Fnet = 4F₁ = 9.89 x 10⁻³ N

Hence, magnitude of the force on 6.5 nC charge due to the four other charges is calculated as 9.89 x 10⁻³ N.

Part B: Force on q₅ due to each of the four charges is directed towards the center of square. Since the four forces are equal in magnitude and direction, then net force on q₅ due to the four charges will also be directed towards the center of the square.

So, the direction of the force on 6.5 nC charge in the middle due to four other charges is towards the center of the square.

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A rocket carrying a new 960-kg satellite into orbit misfires and places the satellite in an orbit with an altitude of 120 km, well below its operational altitude in low-Earth orbit.
(a) What would be the height of the satellite's orbit if its total energy were 550 MJ greater?

(b) What would be the difference in the system's kinetic energy? (Include the sign of the value in your answer.)

(c) What would be the difference in the system's potential energy? (Include the sign of the value in your answer.)

These answers are all in MJ. (Please answer the question instead of taking all of the points)

Answers

Answer:(a) To find the height of the satellite's orbit if its total energy were 550 MJ greater, we can use the following equation:

K2 + U2 = K1 + U1 + 550 MJ

Since the satellite is in a circular orbit, its kinetic energy is given by:

K = (1/2)mv^2

where m is the mass of the satellite, and v is its velocity.

We can use the following equation to relate the height of the satellite's orbit to its velocity:

v = sqrt(GM/R)

where G is the gravitational constant, M is the mass of the Earth, and R is the radius of the Earth plus the height of the satellite's orbit.

Therefore, we can express the kinetic energy of the satellite in terms of its height:

K = (1/2)m(GM/R)

Using these equations, we can rewrite the conservation of energy equation as:

(1/2)m(GM/(R1+h1)) - GMm/(R1+h1) = (1/2)m(GM/(R2+h2)) - GMm/(R2+h2) + 550 MJ

where R1 is the radius of the Earth, and R2 is the radius of the Earth plus h2.

Simplifying and solving for h2, we get:

h2 = [(GMm/(R1+h1)) - (GMm/(R2+h2)) - 550 MJ/(GM/(R2+h2))]^(-1) - R2

Plugging in the given values, we get:

h2 = 931 km

Therefore, the height of the satellite's orbit would be 931 km if its total energy were 550 MJ greater.

(b) To find the difference in the system's kinetic energy, we can use the following equation:

Delta K = K2 - K1

Substituting the expressions for K1 and K2, we get:

Delta K = (1/2)m(GM/(R2+h2)) - (1/2)m(GM/(R1+h1))

Plugging in the given values, we get:

Delta K = -7.5 x 10^9 J

The negative sign indicates that the system's kinetic energy has decreased.

(c) To find the difference in the system's potential energy, we can use the following equation:

Delta U = U2 - U1

Substituting the expressions for U1 and U2, we get:

Delta U = -GMm/(R2+h2) + GMm/(R1+h1)

Plugging in the given values, we get:

Delta U = 5.9 x 10^9 J

The positive sign indicates that the system's potential energy has increased.

Explanation:

At which position do you expect its speed to be one-half the speed at which it eventually hits the ground?
At which position do you expect its speed to be one-half the speed at which it eventually hits the ground?
Higher than the midpoint of the path.
Lower than the midpoint of the path.
At the midpoint of the path.

Answers

The point where the object's speed is one-half its final speed may located at 3/4 of the initial height of the object, the answer would be "lower than the midpoint of the path."

Assuming the object is subject to a constant acceleration due to gravity, the speed of the object at any given point can be described by the following equation:

v² = u² + 2as

Where:

v = final velocity

u = initial velocity

a = acceleration due to gravity (approximately 9.8 m/s² near the Earth's surface)

s = distance traveled

Let's suppose the object is dropped from rest at a height h above the ground. Then its initial velocity is u=0, and the equation simplifies to:

v² = 2gh

where g is the acceleration due to gravity and h is the initial height of the object.

To find the point at which the object's speed is one-half its final speed, we can set v = 1/2 * √(2gh) and solve for s:

1/2 * √(2gh) = √(2gh - 2gs)

Simplifying this equation, we get:

1/4 * 2gh = 2gh - 2gs

s = 3/4 * h

This means that the point where the object's speed is one-half its final speed is located at 3/4 of the initial height of the object. Therefore, the answer is "lower than the midpoint of the path."

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1. A student walks first 70m in the direction 37° north of east, and then walks 82m in the direction 20° south of east, and finally walks 28m in the direction 30° west of north. a. How far and at what angle is the student’s final position from her initial position? b. In what direction would she has to head to return to her initial position?

Answers

a. The student's final position is 100.5 meters at an angle of 25.2° north of east from her initial position and b. The student would have to head in the direction of 334.8° south of east to return to her initial position.

a. To find the student's final position, we can use vector addition. We can start by breaking each displacement vector into its x- and y-components. First displacement (70 m at 37° north of east),

x-component = 70 cos(37°) = 56.02 m

y-component = 70 sin(37°) = 42.49 m

Second displacement (82 m at 20° south of east),

x-component = 82 cos(-20°) = 78.72 m

y-component = 82 sin(-20°) = -28.12 m

Third displacement (28 m at 30° west of north),

x-component = 28 sin(30°) = 14 m

y-component = 28 cos(30°) = 24.24 m

We can add the x-components and y-components separately to get the total displacement,

x-displacement = 56.02 m + 78.72 m - 14 m = 120.74 m

y-displacement = 42.49 m - 28.12 m + 24.24 m = 38.61 m

The magnitude of the total displacement is,

|d| = √(x-displacement² + y-displacement²)

|d| = √(120.74² + 38.61²)

|d| = 126.0 m.

The direction of the total displacement can be found using the inverse tangent function,

θ = atan(y-displacement/x-displacement)

θ = atan(38.61/120.74)

θ = 17.4° north of east. Therefore, the student's final position is 126.0 m at 17.4° north of east.

b. To return to her initial position, the student would need to walk the same distance and direction as her total displacement but in the opposite direction. The opposite direction of 17.4° north of east is 162.6° south of west. Therefore, she would need to walk 126.0 m at 162.6° south of west to return to her initial position.

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Jess has a pop it powered by the sun. She holds it to the sun multiple times but it doesn't seem to work after she washed it. She tried to ma ale or work multiple times and even waited for a long period of time for the pop it to work but it didn't.


How can she fix it and why isn't it working?

Answers

If the pop was immersed in water or subjected to strong chemicals during the washing process, it may have been harmed.

Check the battery-

A dead battery can be to blame. Jess has to make sure the battery is placed and charged correctly. She might need to get help from the manufacturer if the battery cannot be changed.

Verify the solar panels-

If the pop uses solar panels for electricity, Jess has to make sure that they are clear of obstructions like debris and are functioning properly. She can use a soft cloth and some water to clean the solar panels if they are filthy.

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Calculate the angular acceleration when a force of 20N is applied tangentially to the tire of a 2.5 kg bicycle wheel. which has a radius of 50cm.

Answers

Answer:

First, we need to find the torque applied to the wheel. The torque formula is given by:

τ = r × F × sin(θ)

where

r = radius of the wheel = 50 cm = 0.5 m

F = force applied tangentially = 20 N

θ = angle between the force and the radius (which is 90 degrees for a tangential force)

So, τ = 0.5 × 20 × sin(90) = 10 Nm

The moment of inertia of a solid cylinder (which the wheel can be approximated to) is given by:

I = 0.5 × m × r^2

where

m = mass of the wheel = 2.5 kg

r = radius of the wheel = 0.5 m

So, I = 0.5 × 2.5 × 0.5^2 = 0.3125 kgm^2

The angular acceleration can be calculated using the formula:

α = τ / I

So, α = 10 / 0.3125 = 32 rad/s^2 (rounded to two decimal places)

Therefore, the angular acceleration when a force of 20N is applied tangentially to the tire of a 2.5 kg bicycle wheel with a radius of 50cm is 32 rad/s^2.

4. A car moves round a circular track at 120 mph. Give the average velocity of the car.
Explain your answer.

Answers

The average velocity of a car that is moving round a circular track at 120 mph is given as 0

How to determine the average velocity

When a car is moving around a given moves round a circular track at 120 mph, it would have a direction that is constantly changing while it is moving around the track.

After the full lap has been completed, the car is going to have to go back to its starting point. Hence the displacement is 0, average velocity is 0. This is regardless of its speed. Hnece the velocity is 0

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Questions already answered. I need to know how we get to this answer.

Question:
A basketball is launched with an initial speed of 4 m/s at an angle of 45°. The ball enters the basket .96s after it is launched. What is the distance x?

Answer:
2.7 m

Answers

The horizontal distance of the ball is determined as 2.71 m.

What is the distance x?

The horizontal distance traveled by the ball is calculated by applying the following formula as shown below;

X = Vₓt

where;

Vₓ is the horizontal component of the velocityt is the time of motion

X = 4 m/s x cos (45) x 0.96 s

X = 2.71 m

Thus, the horizontal distance of the ball is a function of the horizontal velocity and time of motion.

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HELP PLEASE I BEG! Show your work!!!!

Answers

V = f λ
V= 620Hz x 10.5
V= 6510 m/s

Answer:

C: 6510 m/s.F

Explanation:

The speed of a wave in terms of frequency and wavelength is given by :

V= Fx

V = 620 x 10.5

V = 65010 m/s

So, the speed of sound in steel is 6510 m/s.

A complete circuit with a capacitor is turned on. Charges are moved from one plate of the capacitor, through the battery, and all the way around to the other plate. Which of the following occur during that time?

Kinetic energy is decreased.
Thermal energy is decreased.
Chemical potential energy is increased.
Electrical potential energy is increased.

Answers

Answer:

Electrical potential energy is increased.

Explanation:

When a complete circuit with a capacitor is turned on, charges are moved from one plate to another through the battery, increasing the electrical potential energy.

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Jessica had a sun powered pop it. She washes it and it stops lighting up when she holds it towards the sun.

Why has it stopped working and how can it be fixed?

Answers

Based on the information provided, it seems like Jessica's sun powered pop it has stopped working because it is not receiving enough sunlight to power the device.

To fix the issue, she should try holding it in a place where it can receive direct sunlight for an extended period of time. Alternatively, if the device has a rechargeable battery, she could try charging it using a USB cable or other charging method if one is available.

If neither of these solutions work, there may be a more serious issue with the device itself and it may need to be replaced or repaired.

While this answer may provide helpful information for your inquiry, it is important to remember that using it verbatim could be seen as plagiarism. To avoid this, it is best to use your own words and properly cite any sources used. This will ensure that you are giving credit to the original author and presenting your own unique perspective on the topic.

~~~Harsha~~~

Two cars collide at an intersection. One car has a mass of 800 kg and is
moving 15 m/s to the north, while the other has a mass of 900 kg and is
moving 18 m/s to the south. What is their combined momentum?
OA. 4200 kg.m/s south
OB. 28,200 kg-m/s south
OC. 28,200 kg-m/s north
OD. 4200 kg-m/s north

Answers

Answer: option D: 4200 kg-m/s north.

Explanation: The energy of the primary car moving north is:

p1 = m1 * v1 = 800 kg * 15 m/s = 12,000 kg·m/s north

The force of the moment car moving south is:

p2 = m2 * v2 = 900 kg * (-18 m/s) = -16,200 kg·m/s north

Note that the energy of the moment car is negative because it is moving within the inverse heading to the positive course we have chosen.

To discover the combined force, we include the person momenta:

p_total = p1 + p2 = 12,000 kg·m/s north - 16,200 kg·m/s north = -4,200 kg·m/s north

In this manner, the combined momentum of the two cars is 4,200 kg·m/s within the north course.

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