In the figure, two identical spheres of radius R are set up a distance d away from each other (d is the distance from the center of one to the center of the other). One sphere has charge +Q and is centered at x = 0. The other sphere has a charge -Q and is centered at x = d. You are using this set-up to accelerate a small particle with charge q and mass m. The particle starts at a position xi and ends up at a position xf = d - xi.

(The distance from the positive sphere to xi is equal to the distance from xf to the negative sphere.)

The values in this problem are:

d = 4.70 m, R = 0.16 m, Q = 8.65x10-6 C, q = 6.37x10-8 C
m = 1.87x10-5 kg, xi = 0.78 m

If the small charge starts at rest, what is the final speed of the charge when it reaches xf?

Give your answer in m/s to at least three digits, to avoid being counted incorrect due to rounding.

Note: You can assume that the spheres are far enough apart that the charge on the sphere is uniformly distributed on their surfaces.

Answers

Answer 1

The final speed of the charge, when it reaches xf, is approximately 1.31 m/s.

To find the final speed of the charge when it reaches xf, we can use the principle of conservation of energy. The initial kinetic energy of the charge is zero since it starts at rest, and the final kinetic energy is given by: Kf = (1/2)mvf^2

The potential energy of the charge is due to the electric potential created by the charged spheres. The potential energy at xi is Ui = k * (|Q| * |q|) / xi

where k is the Coulomb constant (8.99x10^9 N m^2/C^2).

The potential energy at xf is:

Uf = k * (|Q| * |q|) / xf

The change in potential energy as the charge moves from xi to xf is:

ΔU = Uf - Ui

According to the conservation of energy, the change in potential energy is equal to the change in kinetic energy:

ΔU = Kf - Ki

Since the initial kinetic energy is zero, we have:

Kf = ΔU

Substituting the expressions for ΔU, Ui, and Uf, we get:

(1/2)mvf^2 = k * (|Q| * |q|) * (1/xi - 1/xf)

Simplifying the equation and solving for vf, we have:

vf = sqrt(2 * k * (|Q| * |q|) * (1/xi - 1/xf) / m)

Plugging in the given values, we get:

vf = sqrt(2 * (8.99x10^9 N m^2/C^2) * (8.65x10^-6 C * 6.37x10^-8 C) * (1/0.78 m - 1/(4.70 - 0.78) m) / (1.87x10^-5 kg))

vf ≈ 1.31 m/s

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Related Questions

You have an object with a charge of negative 3 micro-Coulombs. How many extra electrons are on this object? 1.875×10
13
electrons B 5.33×10
14
electrons (C) 1.6×10
19
electrons D 6.25×10
18
electrons

Answers

The no. of extra electrons are on this object 1.875 x 10^13 electrons. The correct answer is A.

To determine the number of extra electrons on an object with a charge of -3 micro-Coulombs, we need to use the elementary charge (e) as a conversion factor.

The elementary charge represents the charge of a single electron and is approximately equal to 1.6 x 10^-19 Coulombs.

Given:

Charge of the object = -3 micro-Coulombs = -3 x 10^-6 Coulombs

To calculate the number of extra electrons, we divide the total charge by the elementary charge:

Number of extra electrons = (Charge of the object) / (Elementary charge)

Number of extra electrons = (-3 x 10^-6 C) / (1.6 x 10^-19 C)

Number of extra electrons ≈ -1.875 x 10^13 electrons

Since the charge of the object is negative, it indicates an excess of electrons. However, the magnitude of the charge is what determines the number of extra electrons, so we take the absolute value.

Therefore, the correct answer is A) 1.875 x 10^13 electrons.

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A 1.0μF capacitor is being charged by a 6.0 V battery through a 10 MΩ resistor. Determine the potential across the capacitor at time t=1.0 s. Express your answer in volts to two significant figures. - Part B Determine the potential across the capacitor at time t=5.0 s. Express your answer in volts to two significant figures. Part C Determine the potential across the capacitor at time t=20 s. Express your answer in volts to two significant figures.

Answers

The potential across the capacitor at t = 20s is approximately 6.0 V.

To determine the potential across the capacitor at different times, we can use the formula for the charging of a capacitor in an RC circuit:

Vc = V₀ × (1 - e^(-t/RC))

where: Vc is the potential across the capacitor at time t,

V₀ is the initial potential (battery voltage),

t is the time,

R is the resistance, and

C is the capacitance.

Given:

V₀ = 6.0 V

t₁ = 1.0 s

t₂ = 5.0 s

t₃ = 20 s

R = 10 MΩ = 10 × 10⁶ Ω

C = 1.0 μF = 1.0 × 10⁻⁶ F

Let's calculate the potentials across the capacitor at each time:

Part A: t = 1.0 s

Vc₁ = V₀ (1 - e^(-t₁/RC))

Substituting the values:

Vc₁ = 6.0 V(1 - e^(-1.0 s / (10 * 10⁶ Ω * 1.0 * 10⁻⁶ F)))

Calculating:

Vc₁ ≈ 5.9 V

Therefore, the potential across the capacitor at t = 1.0 s is approximately 5.9 V.

Part B: t = 5.0 s

Vc₂ = V₀(1 - e^(-t₂/RC))

Substituting the values:

Vc₂ = 6.0 V (1 - e^(-5.0 s / (10 * 10⁶ Ω * 1.0 * 10⁻⁶ F)))

Calculating:

Vc₂ ≈ 5.8 V

Therefore, the potential across the capacitor at t = 5.0 s is approximately 5.8 V.

Part C: t = 20 s

Vc₃ = V₀(1 - e^(-t₃/RC))

Substituting the values:

Vc₃ = 6.0 V (1 - e^(-20 s / (10 × 10⁶ Ω × 1.0 × 10⁻⁶ F)))

Calculating:

Vc₃ ≈ 6.0 V

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Final answer:

Using the formula for charging a capacitor, the potential across the 1.0μF capacitor charged by a 6.0 V battery through a 10 MΩ resistor is approximately 0.63 volts at 1 second, 2.99 volts at 5 seconds, and 5.26 volts at 20 seconds.

Explanation:

The problem involves a charging capacitor with a given resistance and capacitance. This situation is governed by the formula for charging a capacitor: V = emf(1 - e-t/RC), where V is the voltage across the capacitor, t is the time, R is the resistance, C is the capacitance, and e is the base of natural logarithm.

With the provided values, we can determine the potential of the capacitor at given times:

Part A: At t = 1.0 s, V = 6.0V * (1 - e^(-1.0s / (10 * 10^6 Ω* 1*10^-6 µF) = 0.63 V <-> rounded to two significant figures.

Part B: At t = 5.0 s, V = 6.0V * (1 - e^(-5.0s / (10 * 10^6 Ω * 1*10^-6 µF)) = 2.99 V <-> rounded to two significant figures.

Part C: At t = 20.0 s, V = 6.0V * (1 - e^(-5.0s / (10 * 10^6 Ω * 1*10^-6 µF)) = 5.26 V <-> rounded to two significant figures.

Therefore, the potential across the capacitor reaches asymptotically to the battery voltage and takes several multiples of the time constant (RC) to reach this voltage.

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1. A 2.08 m tall man stands 1.78 m from a lens with focal length 28.3 cm.

How tall (in m) is his image formed by the lens? Be sure to include the sign to indicate orientation!

2. When laser light of some unknown wavelength hits a diffraction grating with 20 slits per millimeter, it produces bright fringes separated by 27.2 mm on a screen that is 2.41 m away.

Given the pattern formed, what must be the wavelength of the light (in nm)?

Answers

1. The height of the image is 0.287 m.

2. The wavelength of the light is 563 nm.

1. The image distance, denoted as `i`, is determined by the lens formula: `1/f = 1/o + 1/i`, where `f` represents the focal length, `o` is the object distance, and `i` represents the image distance. Given `f = 28.3 cm` and `o = 1.78 m`, we need to convert the object distance from meters to centimeters: `o = 1.78 m = 178 cm`. Therefore, the image distance is calculated as follows:

i = (1/f - 1/o)^-1 = (1/28.3 - 1/178)^-1 = 24.53 cm.

The image height, denoted as `h'`, can be determined using the object height `h` and the magnification `m` relationship: `h' = m * h`. The magnification `m` is given by `m = -i/o`, where the negative sign indicates an inverted image. Thus,

m = -i/o = -(24.53 cm)/(178 cm) = -0.138.

The image height `h'` is obtained by multiplying `h` by `m`: `h' = m * h`, where `h = 2.08 m`. Therefore,

h' = (-0.138) * 2.08 = -0.287 m.

The negative sign signifies an inverted image. Hence, the height of the image is determined as `0.287 m`, and it is inverted.

2. Bright fringes are observed at angles `theta` satisfying the condition `d sin theta = m lambda`, where `d` represents the spacing between two slits, `m` is an integer indicating the fringe order, and `lambda` denotes the wavelength of light. In this case, given `d = 1/20 mm` and `m = 1`, the angle `theta` corresponding to the first bright fringe is given by `tan theta = x/L`, where `x` represents the separation between two fringes, and `L` is the distance from the grating to the screen. With `x = 27.2 mm` and `L = 2.41 m`, we can calculate:

tan theta = (27.2 mm)/(2.41 m) = 0.01126.

Therefore, `sin theta = tan theta = 0.01126`.

Consequently, the wavelength `lambda` is determined using the formula `lambda = d sin theta / m`, where `d = 1/20 x 10^-3 m`, `sin theta = 0.01126`, and `m = 1`:

lambda = (1/20 x 10^-3 m) x 0.01126 / 1 = 5.63 x 10^-7 m = 563 nm.

In summary:

1. The height of the image is 0.287 m.

2. The wavelength of the light is 563 nm.

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1) Solve the points A, B, C and D (20 points) step by step


A). Coulomb's law is used to find the force of attraction or repulsion between
two point charges, which are at a certain distance. could you describe
why two charges attract or repel each other, use examples for your argument.
B). If two point charges are at a distance of 2.5 m and between them is exerted
a force equal to the day of your birthday (Keep in mind that force is measured in
Newton). Determine the value of the other load if one of them has a value of
3.2x10-6C.
Note: For example, if your birthday is February 19, then the force you use in the
the exercise would be 19 Newton.
C). Find the distance between two equal charges, which are equal to the
last digit of your identity document multiplied by ten to the minus three,
knowing that there is a repulsive force of 5.8 N.
Note: For example, if the last digit of your identity document is 5, the charges
will have the following value Q= 5x10-3
D). If we have two charges q1 and q2 that are at a distance # meters (# equals
at your age). Find the force exerted between them knowing that:
q1 is equal to 5x10-4
q2 is equal to 3.3x10-3

Comment, in your own words, in no less than 5 lines, why the
Name Coulomb's law.
Note: Keep in mind to attach with your answers the evidence of the values ​​that you
use in each of the problems that require it.

Answers

A) Two charges attract or repel each other based on their electrical properties, B) Coulomb's constant k = 9x10^9 Nm^2/C^2, we can calculate the value of q2, c) Using the given values of the repulsive force F = 5.8 N, and the charge value Q = 5x10^(-3) C,D) Substituting the given values of q1 = 5x10^(-4) C, q2 = 3.3x10^(-3) C, and r = # meters (# equals your age), and using the value of the Coulomb's constant k = 9x10^9 Nm^2/C^2, we can calculate the force between the charges.

A) Two charges attract or repel each other based on their electrical properties. Like charges, such as two positive charges or two negative charges, repel each other because they have the same charge sign. This repulsion occurs because there is a force pushing them apart. On the other hand, unlike charges, such as a positive and a negative charge, attract each other because they have opposite charge signs. This attraction occurs because there is a force pulling them together. An example is the attraction between the positive and negative terminals of a battery, which allows for the flow of electric current.

B) Given that the force exerted between the two point charges is equal to the birthday value (19 Newton in this example), and one charge has a value of 3.2x10^(-6) C, we can use Coulomb's law to find the value of the other charge. Coulomb's law states that the force between two charges is given by the equation:

F = (k * q1 * q2) / r^2

Rearranging the equation, we can solve for q2:

q2 = (F * r^2) / (k * q1)

Substituting the given values, with F = 19 N, r = 2.5 m, q1 = 3.2x10^(-6) C, and using the value of the Coulomb's constant k = 9x10^9 Nm^2/C^2, we can calculate the value of q2.

C) To find the distance between two equal charges, given a repulsive force of 5.8 N, we can rearrange Coulomb's law equation to solve for distance (r):

r = sqrt((k * Q^2) / F)

Using the given values of the repulsive force F = 5.8 N, and the charge value Q = 5x10^(-3) C (assuming the last digit of the identity document is 5), and the Coulomb's constant k = 9x10^9 Nm^2/C^2, we can calculate the distance between the charges.

D) To find the force exerted between two charges q1 and q2 at a distance equal to the age in meters, we can use Coulomb's law:

F = (k * q1 * q2) / r^2

Substituting the given values of q1 = 5x10^(-4) C, q2 = 3.3x10^(-3) C, and r = # meters (# equals your age), and using the value of the Coulomb's constant k = 9x10^9 Nm^2/C^2, we can calculate the force between the charges.

The name Coulomb's law is given in honor of Charles-Augustin de Coulomb, a French physicist who formulated this law in the 18th century. Coulomb's law describes the fundamental relationship between electric charges and the force they exert on each other. It provides a mathematical framework for understanding and quantifying the forces between charged objects, making it a significant contribution to the field of electromagnetism.

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how to know which light is out on christmas lights

Answers

To determine which light is out on Christmas lights, one of the common ways is by gently wiggling each bulb to see which one is faulty. Alternatively, visually examining each bulb and utilizing a light tester can also help to determine the faulty bulb.

When Christmas lights go out, it is difficult to determine which lightbulb is causing the problem.

Here are a few methods for identifying which light is causing the issue:

Method 1:

One of the most common methods for determining which lightbulb is burnt out is the "half-string" approach. Gently wiggle each bulb and pay attention to which bulb causes the entire string to light up. The faulty lightbulb is the one that caused the string to go out.

Method 2:

Check each bulb visually. Make sure the lights are turned off and unplugged. Examine each bulb closely to see whether any are burnt out, broken, or loose. Replace any bulbs that are faulty.

Method 3:

Utilize a light tester. A light tester is an electronic device that can help you identify a burnt-out bulb. Connect the tool to one end of the Christmas lights' cord, turn on the tool, and use the tool to scan the cord. If the tool detects a faulty bulb, it will indicate it.

If there is no faulty bulb, the entire light string should light up.

In conclusion, one of the typical methods is to gently shake each bulb to see which one is broken in order to identify the out-of-service light on Christmas lights.

As an alternative, visually evaluating each bulb and using a light tester might also be helpful in identifying the problematic bulb.

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A self-driving car traveling along a straight section of road starts from rest, accelerating at 2.00 m/s
2
until it reaches a speed of 26.0 m/s
. Then

the vehicle travels for 81.0 s at constant speed until the brakes are applied, stopping the vehicle in a uniform manner in an additional 5.00.s. (a) How long is the self-driving car in motion (in s)? (b) What is the average velocity of the self-driving car for the motion described? (Enter the magnitude in m/s.) m/s

Answers

A. the total time the self-driving car is in motion is 99.0s. and B. the average velocity of the self-driving car for the motion described is 24.29 m/s

**(a) The self-driving car is in motion for a total of 112 seconds (s).**
To calculate the total time the self-driving car is in motion, we need to consider three segments: the acceleration phase, the constant speed phase, and the deceleration phase.
During the acceleration phase, the car starts from rest and reaches a speed of 26.0 m/s with an acceleration of 2.00 m/s². Using the equation of motion:
[tex]\(v = u + at\)[/tex]
where (v) is the final velocity, (u) is the initial velocity, (a) is the acceleration, and (t) is the time, we can rearrange the equation to solve for (t):
[tex]\(t = (v - u) / a\)[/tex]
[tex]\(t = (26.0 m/s - 0 m/s) / 2.00 m/s^2 = 13.0 s\)[/tex]
During the constant speed phase, the car travels for 81.0 seconds.
During the deceleration phase, the car stops in 5.00 seconds.
Therefore, the total time the self-driving car is in motion is [tex]\(13.0 s + 81.0 s + 5.00 s = 99.0 s\).[/tex]

**(b) The average velocity of the self-driving car for the motion described is 14.95 m/s (meters per second).**
Average velocity is calculated by dividing the total displacement by the total time taken. In this case, since the motion is along a straight line, the displacement is the same as the distance traveled.
The distance traveled during the acceleration phase can be calculated using the equation:
[tex]\(s = ut + \frac{1}{2}at^2\)[/tex]
where (s) is the distance, (u) is the initial velocity, (t) is the time, and (a) is the acceleration. Substituting the given values:
[tex]\(s = 0 + \frac{1}{2} \times 2.00 m/s^2 \times (13.0 s)^2 = 169 m\)[/tex]
During the constant speed phase, the distance traveled is:
[tex]\(s = (26.0 m/s) \times (81.0 s) = 2106 m\)[/tex]
During the deceleration phase, the distance traveled is:
[tex]\(s = (26.0 m/s) \times (5.00 s) = 130 m\)[/tex]
The total distance traveled is [tex]\(169 m + 2106 m + 130 m = 2405 m\)[/tex].

Average velocity is calculated as:

[tex]\(v_{\text{average}} = \frac{\text{total distance}}{\text{total time}}\)\\\(v_{\text{average}} = \frac{2405 m}{99.0 s} = 24.29 m/s\)[/tex]

Therefore, the average velocity of the self-driving car for the motion described is 24.29 m/s (rounded to two decimal places).

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A jetliner has a cruising air speed of 620mi/h relative to the air. Part A For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Beiative How long does it take this plane to fly round-trip trom San Francisco to Cricago, an east-west fight of 2000mi each way. velocity on the highway. Express your answer in hours. Part B How long does it take this plane to fly round tra from San Franciseo is Chicago an east-west fight of 2000 mi each waf. if the wind is blowing at 160mi/h from the west to the east? Express your answer in houns.

Answers

The time taken by the plane to fly round-trip from San Francisco to Chicago is 6.45 hours.the total time taken by the plane to fly round-trip is 6.91 hours.

Part AThe given the cruising air speed of the jetliner is 620 mi/h. Therefore, the time taken to fly 2000 miles in one direction is given by:Time = Distance/Speed

= 2000/620

= 3.225 hours. Therefore, the round-trip time taken is twice the time taken in one direction. So, the time taken by the plane to fly round-trip from San Francisco to Chicago is:Round-trip time = 2 × 3.225 = 6.45 hours. Hence, the required answer is 6.45 hours.

Part BThe cruising air speed of the jetliner is still 620 mi/h. However, now there is a wind blowing from west to east at 160 mi/h. This means that the plane will face a headwind while flying from west to east, and a tailwind while flying from east to west. The actual speed of the plane relative to the ground while flying in the two directions is given by:East to West Speed = 620 - 160

= 460 mi/h West to East Speed

= 620 + 160

= 780 mi/h. The time taken by the plane to fly 2000 miles in one direction is given by:Time taken to fly eastward = 2000/780

= 2.56 hours. Time taken to fly westward

= 2000/460

= 4.35 hours. Therefore, the total time taken by the plane to fly round-trip is:Total time taken = 2.56 + 4.35 = 6.91 hours. Hence, the required answer is 6.91 hours.

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H 5 A proton is moring in a ciccular ocbit of radius 29.3 cm with Q. spead of 3.7×10
6
m/s in a uniform magnetic field perpendicular to the velocity of the proton. Find the magnetic field (inT) in which the proton is moving mp=1.67×10^−27 kg, charge of proton =1.61×10^−19 C Abo calculate the time period (in s) of the proton.) * Colculate the magnetic field ond fime period to two decimal ploces.

Answers

The magnetic field and time period is 0.0662 T and 1.5 × 10^-7 s

Given data:

Radius of circular orbit, r = 29.3 cm = 0.293 m

Speed of proton, v = 3.7 × 10^6 m/s

Mass of proton, mp = 1.67 × 10^-27 kg

Charge on proton, q = 1.61 × 10^-19 C

Formula to calculate magnetic field is given by,

B = mv/qr

Here, m is mass of the proton, v is speed of the proton, q is charge on the proton and r is the radius of circular orbit.

Substituting the values, we get,

B = mv/qr

  = (1.67 × 10^-27 kg) × (3.7 × 10^6 m/s) / (1.61 × 10^-19 C) × (0.293 m)

  = 0.0662 T

The magnetic field in which the proton is moving is 0.0662 T.

Now, the time period of the proton is given by,

T = 2πr/v

Substituting the given values, we get,

T = 2πr/v= 2π (0.293 m) / (3.7 × 10^6 m/s)

  = 1.5 × 10^-7 s

The time period of the proton is 1.5 × 10^-7 s.

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Thermal neutrons are diffracted of (b.c.c) solid sample at diffraction angle 50.56 ^∘. If the diffraction take place of the plane (211) and the lattice constant of the unit cell is 2.868 A, what is the momentum of the incident neutron?

Answers

Thermal neutrons are diffracted of (b.c.c) solid sample at diffraction angle 50.56 ^∘. If the diffraction takes place of the plane (211) and the lattice constant of the unit cell is 2.868 A, what is the momentum of the incident neutron?

Thermal neutrons are slowed-down neutrons with energy lower than the energy of fast neutrons. They have a kinetic energy of about 0.025 eV (electron volts) or lower, which is roughly 2.2 kBT, where T is the temperature in kelvins. Thermal neutrons have a wavelength of about 0.2 nanometers, which is the same order of magnitude as the spacing between the atoms in a solid. As a result, when they collide with a solid, they can cause diffraction. When a neutron of momentum p and mass m is diffracted by a crystal, the diffraction angle θ is given by the Bragg equation:

2dsinθ = nλ,

where d is the spacing between the planes of atoms, λ is the wavelength of the neutron, n is an integer, and θ is the angle between the direction of the incident neutron and the plane of atoms.

In this problem, the diffraction angle is given as θ = 50.56°, and the spacing between the (211) plane of atoms is d = 2.868 Å.

The wavelength of the neutron is λ = h/p, where h is Planck's constant and p is the momentum of the neutron. Therefore, p = h/λ. Combining these equations gives:

2dsinθ = nλ

=> 2(2.868 Å)sin(50.56°) = n(h/p)

=> 2(2.868 × 10^−10 m)sin(50.56°) = n(6.63 × 10^−34 J s/p)

p=1.2 × 10^-24 kg m/s.

Approximately.

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when the central ray is angled, the structure situated _____ is projected the farthest.

Answers

When the central ray is angled, the structure situated **closest to the central ray** is projected the farthest.

A projectile is fired from the ground with an initial speed of 50.0 m/s at an angle of 23.0

to the horizontal, Assuming peceleration due to gravity to be 9.80 m/s
3
downwards, calculate: (a) The maximum height reached by the projectile (b) The time taken by the projectile to reach its maximum height (c) The total time taken by the projectile from the time it was fired, to the time it lands on the ground. (d) The horizontal distance travelled by the projectile.

Answers

The projectile fired with an initial speed of 50.0 m/s at an angle of 23.0 degrees to the horizontal reaches a maximum height of approximately 57.6 meters. It takes approximately 2.59 seconds to reach the maximum height, and the total time taken from firing to landing is approximately 5.18 seconds. The horizontal distance traveled by the projectile is approximately 296 meters.

(a) To calculate the maximum height reached by the projectile, we can use the following equation:

H = (V₀² * sin²(θ)) / (2 * g)

where V₀ is the initial speed (50.0 m/s), θ is the launch angle (23.0 degrees), and g is the acceleration due to gravity (9.8 m/s²). Plugging in the values:

H = (50.0² * sin²(23.0)) / (2 * 9.8)

H ≈ 57.6 meters

Therefore, the maximum height reached by the projectile is approximately 57.6 meters.

(b) The time taken by the projectile to reach its maximum height can be calculated using the following equation:

t = (V₀ * sin(θ)) / g

Plugging in the values:

t = (50.0 * sin(23.0)) / 9.8

t ≈ 2.59 seconds

Thus, the time taken by the projectile to reach its maximum height is approximately 2.59 seconds.

(c) The total time taken by the projectile from the time it was fired to the time it lands on the ground can be found by doubling the time taken to reach the maximum height:

Total time = 2 * t

Total time ≈ 2 * 2.59

Total time ≈ 5.18 seconds

Therefore, the total time taken by the projectile is approximately 5.18 seconds.

(d) The horizontal distance traveled by the projectile can be calculated using the formula:

R = V₀x * t

where V₀x is the initial horizontal velocity (V₀ * cos(θ)). Plugging in the values:

R = (50.0 * cos(23.0)) * 5.18

R ≈ 296 meters

Hence, the horizontal distance traveled by the projectile is approximately 296 meters.

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A ball is dropped from the top of a building and is
observed to hit the ground after 5.5 seconds. How far has it
traveled after 2.0s?

Answers

The ball has traveled 19.6 meters after 2.0 seconds. After 2.0 s, the distance traveled by the ball is 19.6 meters. We can use kinematic equations to solve this problem. The following equations are relevant in this scenario: vf = vi + atx

= vi(t) + 1/2(a)(t)^2

where vf = final velocity

= 0 m/s (since the ball has hit the ground, so its velocity is zero)

vi = initial velocity (unknown)

a = acceleration due to gravity = 5.5 s (time taken to hit the ground) In the vertical direction, we can write the second equation as follows: x = 150.125 m

This is the total distance traveled by the ball. However, we need to find the distance traveled by the ball in the first 2.0 s. We can use the same equation again, but with t = 2.0 s instead of 5.5 s:

x =  19.6 m Therefore, the ball has traveled 19.6 meters after 2.0 seconds.

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How many millicoulombs ( mC ) of charge is stored in a 121μF capacitor when 150 V is applied to it? Your answer should be a number with two decimal places, do not include the unit.

Answers

The amount of charge stored in the 121 μF capacitor when 150 V is applied to it is 18.15 mC.

To calculate the amount of charge stored in a capacitor, we can use the formula:

Q = C * V

where Q is the charge, C is the capacitance, and V is the applied voltage.

Given:

Capacitance (C) = 121 μF = 121 x 10⁻⁶ F

Applied voltage (V) = 150 V

Using the given values in the formula, we can calculate the charge (Q):

Q = C * V

= (121 x 10⁻⁶ F) * 150 V

Calculating the value:

Q = 121 x 10⁻⁶ F * 150 V

= 0.01815 C

To convert the charge to millicoulombs (mC), we multiply by 1000:

Q_mC = 0.01815 C * 1000

= 18.15 mC

Therefore, the amount of charge stored in the 121 μF capacitor when 150 V is applied to it is 18.15 mC.

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What is the slrengit of the electrie feed 3.0 cm from a amali plastic bead that has been charged to 7.4nC ? Express your answer to two significant figures and include the appropriate units. Part B What it me direction of the electric fiold 3.0 cm from a small plastic besd that has been charged to −7.AnC ?

Answers

To find the strength of the electric field at a distance of 3.0 cm from the charged plastic bead, we can use the equation:

Electric Field (E) = Charge (Q) / (4πε₀r²)

where E is the electric field, Q is the charge, ε₀ is the permittivity of free space (ε₀ ≈ 8.85 × 10^(-12) C²/(N·m²)), and r is the distance from the charge.

Given:

Charge (Q) = 7.4 nC = 7.4 × 10^(-9) C

Distance (r) = 3.0 cm = 3.0 × 10^(-2) m

Substituting these values into the equation, we get:

E = (7.4 × 10^(-9) C) / (4π(8.85 × 10^(-12) C²/(N·m²))(3.0 × 10^(-2) m)²)

Calculating this expression gives us:

E ≈ 7.89 × 10^7 N/C

Therefore, the strength of the electric field at a distance of 3.0 cm from the charged plastic bead is approximately 7.89 × 10^7 N/C.

Now, let's consider the direction of the electric field at the same distance but with a charge of -7.0 nC.

Given:

Charge (Q) = -7.0 nC = -7.0 × 10^(-9) C

Since the sign of the charge is negative, the direction of the electric field will be opposite to what it was in the previous case. Therefore, the direction of the electric field at 3.0 cm from the small plastic bead charged to -7.0 nC is in the opposite direction of the previous case.

Hence, the direction of the electric field is toward the charged plastic bead.

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On vacation, your 1250−kg car pulls a 570−kg trailer away from a stoplight with an acceleration of 1.80 m/s
2
. What is the net force exerted by the car on the trailer? You may want to review Assume that the positive x axis is directed toward the direction of motion. Part B What force does the trailer exert on the car? Assume that the positive x axis is directed toward the direction of motion. Assume that the positive x axis is directed toward the direction of motion. On vacation, your 1250−kg car pulls a 570-kg trailer away from a stoplight with an acceleration of 1.80 m/s
2
. You may want to review ( Part C What is the net force acting on the car? Assume that the positive x axis is directed toward the direction of motion.

Answers

On vacation, your 1250−kg car pulls a 570−kg trailer away from a stoplight with an acceleration of 1.80 m/s². The net force exerted by the car on the trailer and trailer on the car is 3276 N.

To find the net force exerted by the car on the trailer, we can use Newton's second law of motion, which states that the net force acting on an object is equal to its mass multiplied by its acceleration.

Given:

Mass of the car (m₁) = 1250 kg

Mass of the trailer (m₂) = 570 kg

Acceleration (a) = 1.80 m/s² (assuming it's in the positive x-axis direction)

Part A: Net force exerted by the car on the trailer

Using Newton's second law, we can calculate the net force (F_net) exerted by the car on the trailer:

F_net = (m₁ + m₂) * a

F_net = (1250 kg + 570 kg) * 1.80 m/s²

F_net = 1820 kg * 1.80 m/s²

F_net = 3276 N

Therefore, the net force exerted by the car on the trailer is 3276 N.

Part B: Force exerted by the trailer on the car

According to Newton's third law of motion, the force exerted by the trailer on the car is equal in magnitude but opposite in direction to the force exerted by the car on the trailer. Therefore, the force exerted by the trailer on the car is also 3276 N, but in the opposite direction (negative x-axis direction).

Part C: Net force acting on the car

Since there are no other external forces mentioned, the net force acting on the car will be equal in magnitude but opposite in direction to the force exerted by the trailer on the car. Thus, the net force acting on the car is -3276 N (negative x-axis direction).

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Write the azimuth for the displayed bearing. Remember to enter
only a number, no spaces, letters, or punctuation.

Answers

The azimuth is a numerical representation of a bearing's direction, determined by determining the angle between the reference direction and the bearing's direction. It should be entered without spaces, letters, or punctuation.

The azimuth for a displayed bearing is a numerical representation of the direction. To write the azimuth, you should enter only a number without any spaces, letters, or punctuation.

To find the azimuth, you need to determine the angle between the reference direction (usually north) and the direction of the bearing.

Here's a step-by-step example:

1. Identify the reference direction, which is typically north.
2. Determine the direction of the bearing from the reference direction.
3. Measure the angle between the reference direction and the bearing direction.
4. Express the angle in degrees, ranging from 0 to 360.

For instance, if the bearing is 30 degrees east of north, the azimuth would be 30. If the bearing is 45 degrees west of north, the azimuth would be 315 (since it is 360 - 45).

Remember, when writing the azimuth, only enter the numerical value without any spaces, letters, or punctuation.

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How lona (in s) would it take to reach the ground if it is thrown straight down with the same speed? (Enter a number.)

Answers

It would take approximately 0.612 seconds to reach the ground if the object is thrown straight down with the same speed.

If the object is thrown straight down with the same speed, the time it takes to reach the ground can be determined by considering the vertical motion. Since the object is in free fall, the only force acting on it is gravity.

The time it takes to reach the ground can be found using the equation:

y = y0 + v0t + (1/2)gt^2

Where:

y is the displacement (height) of the object

y0 is the initial height of the object (in this case, 2.8 m)

v0 is the initial velocity of the object (in this case, 30.0 m/s)

g is the acceleration due to gravity (approximately 9.8 m/s^2)

t is the time taken to reach the ground (what we want to find)

Since the object is thrown straight down, the initial velocity v0 is negative.

Plugging in the values, we have:

0 = 2.8 m + (-30.0 m/s)t + (1/2)(-9.8 m/s^2)t^2

Simplifying the equation:

4.9t^2 - 30t - 2.8 = 0

This is a quadratic equation, and we can solve it to find the time t. Using the quadratic formula:

t = (-b ± √(b^2 - 4ac)) / (2a)

For this equation, a = 4.9, b = -30, and c = -2.8.

Calculating the time using the quadratic formula:

t ≈ 0.612 s or t ≈ 4.848 s

Since we are interested in the time it takes to reach the ground, we consider the positive value:

t ≈ 0.612 s

Therefore, it would take approximately 0.612 seconds to reach the ground if the object is thrown straight down with the same speed.

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The electric field at a point in space is
E
=(200
i
^
+300
j
^

)N/C What is the x-component of the electric force on a proton at this point? What is the y-component of the electric force on a proton at this point? What is the x-component of the electric force on an electron at this point? What is the y-component of the electric force on a electron at this point? Express your answer with the appropriate units. What is the magnitude of the proton's acceleration? What is the magnitude of the electron's acceleration?

Answers

The x-component of the electric force on a proton is [tex]3.2\times10-17 N.[/tex]

The y-component of the electric force on a proton is[tex]-4.8\times10-17[/tex]N.

The y-component of the electric force on an electron is[tex]-4.8\times10-17[/tex]N.

The magnitude of the proton's acceleration is [tex]1.91\times10^{10} ms^{2} .[/tex]

The magnitude of the electron's acceleration is [tex]3.51\times10^{13} m/s^{2}[/tex].

The given electric field at a point in space is [tex]E=(200 ^i+300^j) N/C.[/tex]

We are supposed to find the x-component of the electric force on a proton at this point, y-component of the electric force on a proton at this point, x-component of the electric force on an electron at this point, y-component of the electric force on an electron at this point, the magnitude of the proton's acceleration, and the magnitude of the electron's acceleration.

X-component of the electric force on a proton:

[tex]F_{proton}=q_{proton} * E_xF_{proton }[/tex][tex]= proton's charge * x-component\ of electric field[/tex]

[tex]F_{proton} = 1.6\times10-19 C * 200 N/C = 3.2\times10-17 N.[/tex]

Hence the x-component of the electric force on a proton is [tex]3.2\times10-17 N.[/tex]

Y-component of the electric force on a proton:

[tex]F_{proton}=q_{proton} * E_yF_{proton }[/tex][tex]= proton's charge * x-component\ of electric field[/tex]

[tex]F_{proton} = 1.6\times10-19 C * 300 N/C = 4.8\times10-17 N.[/tex]

Hence the y-component of the electric force on a proton is[tex]4.8\times10-17 N.[/tex]

X-component of the electric force on an electron:

[tex]F_{electron} = q_{electron} * E_xF_{electron} = electron's charge * x-component of electric field[/tex]

[tex]F_{electron} = -1.6\times10-19 C * 200 N/C = -3.2\times10-17 N.[/tex]

Hence the x-component of the electric force on an electron is [tex]-3.2\times10-17 N.[/tex]

Y-component of the electric force on an electron:

[tex]F_{electron} = q_{electron} * E_yF_{electron} = electron's charge * x-component of electric field[/tex]

[tex]F_{electron} = -1.6\times10-19 C * 300 N/C = -4.8\times10-17 N.[/tex]

Hence the y-component of the electric force on an electron is [tex]-4.8\times10-17[/tex]N.

The magnitude of the proton's acceleration:

The force experienced by a proton, [tex]F = m_{proton} * a_{proton}a_{proton} = F_{proton} / m_{proton}a_{proton} =[/tex] [tex]3.2\times10-17 N / 1.67\times10-27 kg = 1.91\times10^{10}[/tex][tex]m/s^{2}[/tex]

Hence the magnitude of the proton's acceleration is [tex]1.91\times10^{10}[/tex][tex]m/s^{2}[/tex]

The magnitude of the electron's acceleration: The force experienced by an electron,

[tex]F = m_{proton} * a_{proton}a_{proton} = F_{proton} / m_{proton}a_{proton}[/tex][tex]= -3.2\times0-17 N / 9.11\times10−31 kg = -3.51\times10^{13}[/tex][tex]m/s^{2}[/tex]

Hence the magnitude of the electron's acceleration is [tex]3.51\times10^{13} m/s^{2}[/tex].

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A student in an undergraduate physics lab is studying Archimede's principle of bouyancy. The student is given a brass cylinder and, using a triple beam balance, finds the mass to be 1.25 kg. The density of this particular alloy of brass is 8.56 g/cm^3 . The student ties a massless string to one end of the cylinder and submerges it into a tank of water where there is an apparent reduction in the weight of the cylinder. With this information, calculate the volume, V, of the cylinder and the tension, T, in the string when it is submerged in the tank of water. The density of water is 1.00 g/cm3 , and the acceleration due to gravity is g=9.81 m/s ^2 V= cm^3 T= _____

Answers

The volume of the cylinder is 146 cm³ and the tension in the string when it is submerged in the tank of water is 10.83 N.

Given data:

Mass of the cylinder = 1.25 kg

Density of the alloy of brass = 8.56 g/cm³

Density of water = 1.00 g/cm³

The volume of the cylinder can be calculated using the formula:

Density = Mass / Volume

So, the formula for the volume of the cylinder is:

Volume = Mass / Density

             = 1.25 kg / 8.56 g/cm³

             = 1.460 × 10^5 mm³

             = 146 cm³

The apparent reduction in the weight of the cylinder is equal to the buoyant force on the cylinder. So, the weight of the cylinder is equal to its actual weight minus the buoyant force. The weight of the cylinder is given by:

Weight = Mass × Acceleration due to gravity

            = 1.25 kg × 9.81 m/s²

            = 12.2625 N

To find the buoyant force on the cylinder, we need to know the volume of the cylinder.

The buoyant force is equal to the weight of the water displaced by the cylinder.

Since the density of water is 1.00 g/cm³, the weight of the water displaced by the cylinder is equal to its volume times the density of water times the acceleration due to gravity.

Thus, the buoyant force is given by:

Buoyant force = Weight of displaced water

= Volume of cylinder × Density of water × Acceleration due to gravity

= 146 cm³ × 1.00 g/cm³ × 9.81 m/s²

= 1.43226 N

Thus, the tension in the string when it is submerged in the tank of water is equal to the weight of the cylinder minus the buoyant force:

Tension = Weight - Buoyant force

              = 12.2625 N - 1.43226 N

              = 10.83 N

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Suppose trucks in a convoy are spaced at intervals of 500 m, and the convoy is traveling past an observer at 75 km/hr. What is the frequency (in trucks/hr) that he observes?

Answers

The frequency (in trucks/hr) that the observer observes is 150 trucks/hr.

The speed of the convoy relative to the observer=75 km/hr=20.83 m/s.

Since the distance between the trucks is 500 m, the observer sees a truck passing every time it moves a distance of 500 m.

So, the time interval between each truck passing is the time taken to travel 500 m at a speed of 20.83 m/s.So, the time interval between each truck passing is 500/20.83 = 24.00 seconds.

frequency = (number of events) ÷ (time interval)Here, the event is a truck passing and the time interval is 24.00 seconds.

To convert km/hr to m/s, multiply the km/hr value by 0.277778The speed of the convoy = 75 km/hr x 0.277778 = 20.8333 m/s.

Distance between trucks = 500 m. Time interval between trucks = 500/20.8333 = 24.0012 seconds.

Frequency of trucks passing the observer = (number of events) ÷ (time interval) = (1) ÷ (24.0012 seconds) = 0.04167 trucks/second.

Now, let's convert this to trucks/hr:0.04167 trucks/second x 60 seconds/minute x 60 minutes/hour = 150 trucks/hour.

Therefore, the frequency (in trucks/hr) that the observer observes is 150 trucks/hr.

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N (b) Calculate the magnitude of the gravitational force exerted by the human on Mercury. N (c) For comparison, calculate the approximate magnitude of the gravitational force of this human on a similar human who is standing 2.5 meters away. N (d) What approximations or simplifying assumptions must you make in these calculations? (Note: Some of these choices are false because they are wrong physicsl) Treat the humans as though they were points or uniform-density spheres. Treat Mercury as though it were spherically symmetric. Use the same gravitational constant in (a) and (b) despite its dependence on the size of the masses. Ignore the effects of the Sun, which alters the gravitational force that one object exerts on another.

Answers

The magnitude of the gravitational force exerted by the human on Mercury is calculated using Newton's law of universal gravitation, treating them as point masses.

To calculate the magnitude of the gravitational force exerted by the human on Mercury, we need to use Newton's law of universal gravitation. The formula is given by:

F = [tex](G * m1 * m2) / r^2[/tex]

where F is the gravitational force, G is the gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between their centers of mass.

Assuming the mass of the human is m1 and the mass of Mercury is m2, we can calculate the gravitational force as:

F = [tex](G * m1 * m2) / r^2[/tex]

(c) To calculate the approximate magnitude of the gravitational force between two humans, we use the same formula as in (b) with the masses of the two humans (m1 and m2) and the distance between them (r). Let's assume both humans have the same mass.

F = [tex](G * m1 * m2) / r^2[/tex]

(d) The approximations or simplifying assumptions made in these calculations are:

Treating the humans and Mercury as though they were points or uniform-density spheres: This assumes that the mass is concentrated at a single point or evenly distributed within a spherical volume.

Treating Mercury as though it were spherically symmetric: This approximation assumes that Mercury's mass is uniformly distributed throughout the planet, resulting in a spherically symmetric gravitational field.

Using the same gravitational constant in (a) and (b) despite its dependence on the size of the masses:

The gravitational constant (G) is a universal constant that does not change based on the masses involved in the calculation.

Ignoring the effects of the Sun, which alters the gravitational force that one object exerts on another: This assumption neglects the influence of the Sun's gravitational pull on both the human and Mercury.

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A 5.20 kg steel ball is dropped onto a copper plate from a height of 10.0 m. If the ball leaves a dent 3.75 mm deep, what is the average force exerted by the plate on the ball during the impact? N

Answers

The average force exerted by the copper plate on the steel ball during the impact can be calculated using the principle of conservation of energy. The average force is found to be approximately 4052 N.

When the steel ball is dropped onto the copper plate, it falls from a height of 10.0 m and acquires gravitational potential energy. This potential energy is converted into kinetic energy as the ball falls. The initial kinetic energy of the ball is given by

KE = (1/2)mv², where

m is the mass of the ball and

v is its velocity just before impact.

The depth of the dent created on the plate is related to the work done by the average force exerted by the plate on the ball. The work done is given by the equation

W = Fd, where

F is the magnitude of the average force and

d is the depth of the dent.

The work done by the average force is equal to the change in kinetic energy of the ball.

By equating the work done to the change in kinetic energy, we have

(1/2)mv² = Fd.

Rearranging the equation, we can solve for the average force

F = (1/2)mv² / d.

Plugging in the values of m = 5.20 kg, v = √(2gh), where

h is the initial height of the ball, and

d = 3.75 mm,

The average force is found to be approximately 4052 N.

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If wind blows at 46 m/s over the roof of your housc, what is the pressure difference at the roof between the inside and outside air? Use an air density of 1,3 kg/m^3, A. 1.5×10 ^5Pa B. 7.5×10^5Pa C. 1.0×10^5Pa D. 1.4×10^3 Pa E. Orher 7. If the roof in problem 3 is 20 m×15 m what is the total force on the roof?

Answers

,The total force on the roof is 4,211,820 N.

Given:

Wind speed over the roof of the house, v1 = 46 m/s

Air density, ρ = 1.3 kg/m³

To find:

Pressure difference at the roof between the inside and outside air

Formula used:

Bernoulli's principle, which states that P1 + 0.5ρv1² = P2 + 0.5ρv2²

Where:

P1 = pressure at point 1

v1 = velocity at point 1

P2 = pressure at point 2

v2 = velocity at point 2

ρ = density of the fluid

Substituting the given values in the expression:

P1 + 0.5ρv1² = P2 + 0.5ρv2² ...(1)

Here, point 1 is outside the house and point 2 is inside the house.

Since the pressure inside the house is greater than the pressure outside the house, P2 > P1.

Given:

v1 = 46 m/s

v2 = 0 m/s (since the wind stops inside the house)

ρ = 1.3 kg/m³

Substituting these values in equation (1):

1.5P1 = P2

Since P2 > P1:

1.5P1 - P1 = 0.5P1

0.5P1 = 0.5 × 1.3 × 46²

P1 = 1403.94 Pa

Thus, the pressure difference at the roof between the inside and outside air is 1403.94 Pa. Hence, the correct option is D. 1.4×10³ Pa.

Now, let's find the total force on the roof using the formula:

F = P × A

Where:

F is the force

P is the pressure

A is the area of the roof

Substituting the given values in the above formula:

F = 1403.94 × 20 × 15

F = 4,211,820 N

Thus, the total force on the roof is 4,211,820 N.

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It a thin isotropic ply has a young's modulus of 60,000 N/mm and a Poisson's ratio of 5 determine the terms in the reduced stiffness and compliance matrices What is the significance of these two matrices in composite laminates?

Answers

In composite laminates, these matrices ([Q] and [S-bar]) are of significant importance as they determine the physical properties of the laminates, including stiffness, strength, and other mechanical properties.

Given:

Young's modulus (E) of isotropic ply = 60,000 N/mm

Poisson's ratio (μ) = 5

We can calculate the stiffness matrix [Q] for isotropic materials using the formula:

[Q] = E / (1 - μ^2) [1 μ 0] [μ 1 0] [0 0 (1-μ)/2]

Putting the given values of E and μ in the equation, we get:

[Q] = 60,000 / (1 - 5^2) [1 5 0] [5 1 0] [0 0 -2.5]

Simplifying further, we have:

[Q] = 150 [1 5 0] [5 1 0] [0 0 -1/4]

The reduced stiffness matrix [Q-bar] can be calculated as the inverse of [Q]:

[Q-bar] = [Q]^-1

Calculating the inverse, we have:

[Q-bar] = 1/150 [-1/25 5/125 0] [5/125 -1/25 0] [0 0 -4/15]

Similarly, the reduced compliance matrix [S-bar] can be obtained as the inverse of [Q-bar]:

[S-bar] = [Q-bar]^-1

Therefore, [S-bar] is given by:

[S-bar] = 150 [-1/25 5/125 0] [5/125 -1/25 0] [0 0 -4/15]

In composite laminates, these matrices ([Q] and [S-bar]) are of significant importance as they determine the physical properties of the laminates, including stiffness, strength, and other mechanical properties. The reduced stiffness and compliance matrices are used to calculate the effective properties of the composite laminates in different directions. They play a crucial role in the design and analysis of composite laminates to achieve desired mechanical properties.

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Explain two non-electrical (Mechanical) pressure measurements devices?

Answers

Pressure is a vital and common measurement parameter in physics and engineering. Numerous pressure measurement devices are available to measure pressure, which can be electrical or non-electrical.

The following are two non-electrical (Mechanical) pressure measurement devices:

Differential Pressure Gauge

Differential pressure gauges are mechanical pressure gauges that provide a direct pressure measurement of the difference between two pressures. This type of pressure gauge measures the difference in pressure between two points in a system, typically across a filter or other component.

A differential pressure gauge includes two pressure ports, one at each end of the gauge, and a diaphragm or bellows that flexes in response to a change in pressure. When the differential pressure changes, the gauge's diaphragm or bellows deflects, indicating the pressure difference. Bourdon Tube Gauge A Bourdon tube gauge is another mechanical pressure gauge that can be used to measure pressure.

A bourdon tube is a curved, hollow metal tube with an elliptical cross-section, usually in a C shape. When pressure is applied to the interior of the bourdon tube, the tube straightens or uncoils, causing the free end to move. The motion of the free end of the bourdon tube is transmitted through a mechanical linkage to the gauge needle, which indicates the pressure value. This is how the Bourdon tube gauge operates.

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Anthony carelessly rolls his toy car off a 72.0−cm-high table. The car strikes the floor at a horizontal distance of 96.0 cm from the edge of the table. (a) What was the velocity with which the car left the table? (Enter the magnitude.) m/s (b) What was the angle of the car's velocity with respect to the floor just before the impact?

Answers

Height of the table = 72 cm, Horizontal distance from the edge of the table = 96 cmWe need to find out the velocity with which the car left the table and the angle of the car's velocity with respect to the floor just before the impact.

a) Velocity with which the car left the table: Let's assume that the car is in free fall for time t and using the kinematic equation: Final velocity (v)² = Initial velocity (u)² + 2 x acceleration (a) x distance (d)Here, final velocity (v) = 0 (the car comes to rest on striking the floor), initial velocity (u) =? (to be calculated), acceleration (a) = g = 9.8 m/s² (acceleration due to gravity), distance (d) = height of the table = 0.72 m.

The equation will be:v² = u² + 2ad 0 = u² + 2 x 9.8 x 0.72 u² = 12.528u = √12.528 ≈ 3.54 m/sTherefore, the velocity with which the car left the table is 3.54 m/s (magnitude).

b) The angle of the car's velocity with respect to the floor just before the impact: From the above equation, we can also calculate the time (t) for which the car was in free fall by rearranging the equation:t = √(2d/a) = √(2 x 0.72/9.8) ≈ 0.38 sNow, we can calculate the horizontal distance (x) that the car traveled in 0.38 s by using the formula:x = vt = 3.54 x 0.38 ≈ 1.34 m.Therefore, the car traveled a horizontal distance of 1.34 m just before the impact.

Using the formula, tan θ = (vertical displacement/horizontal displacement)tan θ = 0.72/1.34 = 0.5373θ = tan⁻¹0.5373 ≈ 29.1°.Therefore, the angle of the car's velocity with respect to the floor just before the impact is 29.1°.

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Question 5 5 pts How long does it take to displace the following cement slurry (in hours)? . Cement pump rate is 67 SPM Pump output is 0.09 barrel/stroke • Mixing capacity is 67 sack per minute • Previous casing string is 9 5/8" x 8.921" Casing string is 7" x 6.18" in a hole of 8.5" Casing shoe TVD/MD is 7,888/8,021 ft Casing shoe track length is 80 ft . . . (round up to the nearest two decimal places)

Answers

To calculate the time it takes to displace the cement slurry, we need to consider the pumping rate, pump output, and the volume of the slurry. It takes approximately 0.13 hours (or 7.84 minutes) to displace the cement slurry.


First, let's calculate the volume of the cement slurry displaced in one stroke of the pump:
Volume per stroke = Pump output = 0.09 barrel/stroke.
Next, we need to calculate the volume of the casing string:
[tex]Volume of casing string = π/4 * (9.625^2 - 8.921^2) * 80 ft = 348.21 ft³.[/tex]
Now, let's calculate the total volume of the cement slurry:
[tex]Total volume of cement slurry = Volume of casing string + Volume of hole = 348.21 ft³ + (π/4 * (8.5^2 - 6.18^2) * (8021 - 7888) ft) = 348.21 ft³ + 432.25 ft³ = 780.46 ft³.[/tex]

To calculate the time it takes to displace the cement slurry, we need to divide the total volume by the mixing capacity:
[tex]Time = Total volume / Mixing capacity = 780.46 ft³ / (67 sack/min * 1.6 ft³/sack) = 7.84 min[/tex].
Finally, let's convert minutes to hours:
[tex]Time in hours = 7.84 min / 60 min/hour = 0.13 hours.[/tex]

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The plane was going at 150 mph towards 270 degrees. I get into a wind zone and its speed combines with the wind speed and now it's going 250 mph towards 323.13 degrees. Which flow slows down the speed of the wind?

Answer: ____mi/h to ____degrees

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When the plane is in the wind zone, the speed of the wind slows down the flow of the plane. As the wind is directly opposite to the plane, the wind resistance increases, and the speed of the plane decreases.

The original speed of the plane is 150 mph towards 270 degrees. After entering the wind zone, its speed combined with the wind speed, and it moved towards 323.
13 degrees at 250 mph. It is known that the wind resistance slows down the speed of the plane. So, the unknown answer would be less than 250 mph towards some unknown degree.

It cannot be determined without additional information. The question can't be fully answered, but the flow that slows down the speed of the plane is the wind.

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​​​​​​​ (a) What is the radius of the sphere? \( \mathrm{mm} \) (b) What is the thickness of the shell wall? \( \mathrm{mm} \)

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The radius of the sphere can be determined by dividing the diameter by 2, while the thickness of the shell wall can be found by subtracting the inner radius from the outer radius.

(a) To determine the radius of the sphere, we need more information. The radius is the distance from the center of the sphere to any point on its surface. If we have the diameter, we can divide it by 2 to find the radius. For example, if the diameter is 10 mm, then the radius would be 10 mm ÷ 2 = 5 mm.

(b) Similarly, to find the thickness of the shell wall, we need more details. The shell wall refers to the thickness of the hollow part of the sphere. If we know the outer radius and the inner radius of the shell, we can subtract the inner radius from the outer radius to find the thickness.

For example, if the outer radius is 8 mm and the inner radius is 6 mm, then the thickness of the shell wall would be 8 mm - 6 mm = 2 mm.

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A guitar string with a linear density of 3.0 g/m and a length of 0.80 m is oscillating in the first harmonic and second harmonic as the tension is gradually increased. When the tensionstudent submitted image, transcription available below passes through the value of 150 N, what is the rate df/dstudent submitted image, transcription available belowof the frequnecy change for: a) the first harmonic and b) the second harmonic?

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The rate of frequency change for the first harmonic and the second harmonic are 3.1 × 10^(-3) Hz/N and 4.4 × 10^(-3) Hz/N, respectively.

Given that the linear density (μ) of the string is 3.0 g/m,

                 the length (L) of the string is 0.80 m, the tension (T) passes through the value of 150 N.

We need to find the rate df/dT of frequency change for the first harmonic and the second harmonic.

The frequency (f) of a vibrating string is given as;

                                            f = (1/2L) √(T/μ)............(1)

The wavelength (λ) of the vibrating string is given as;

                                            λ = 2L/n............(2)

Where n is the mode of vibration or harmonic and takes the values n = 1, 2, 3, ....
Now, to find df/dT, we differentiate equation (1) with respect to T as follows;

                                        df/dT = (1/2L) (1/2) (T/μ)^(-1/2) (1/μ)........................(3)
Also, for the first harmonic, n = 1.

Substituting equation (2) into equation (1), we have;

                                       f1 = (1/2L) √(T/μ)........................................(4)
Taking the derivative of equation (4) with respect to T, we get;

                                       df1/dT = (1/2L) (1/2) (T/μ)^(-1/2) (1/μ)........................................(5)
Similarly, for the second harmonic, n = 2.

Substituting equation (2) into equation (1), we have;

                                        f2 = (1/2L) √(2T/μ)......................................(6)
Taking the derivative of equation (6) with respect to T, we get;

                                       df2/dT = (1/2L) (1/2) (2T/μ)^(-1/2) (2/μ)...................(7)
Substituting T = 150 N,

                    L = 0.80 m, and

                    μ = 3.0 g/m

                       = 0.0030 kg/m into equations (3), (5), and (7), we obtain;

                  df/dT = (1/2 × 0.80) (1/2) (150/0.0030)^(-1/2) (1/0.0030)

                  df1/dT = (1/2 × 0.80) (1/2) (150/0.0030)^(-1/2) (1/0.0030)

                  df2/dT = (1/2 × 0.80) (1/2) (2 × 150/0.0030)^(-1/2) (2/0.0030)
Simplifying the expressions, we get;

                  df/dT = 3.1 × 10^(-3) Hz/N

                   df1/dT = 3.1 × 10^(-3) Hz/N

                   df2/dT = 4.4 × 10^(-3) Hz/N

Therefore, the rate of frequency change for the first harmonic and the second harmonic are 3.1 × 10^(-3) Hz/N and 4.4 × 10^(-3) Hz/N, respectively.

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