Articulate succinctly (and use a sketch if helpful) why the sign of the wave-function matters when two or more atoms form bonds.

Answer:

\(\boxed {\boxed {\sf C. \ 5.0 \ m/s}}\)

Explanation:

Momentum is the product of mass and velocity. The formula is:

\(p=mv\)

The mass of the cart is 2.0 kilograms. The momentum is 10.0 kg m/s. The velocity is unknown.

\(m= 2.0 \ kg \\p= 10.0 \ kg \ m/s\)

Substitute the values into the formula.

\(10.0 \ kg \ m/s= (2.0 \ kg ) v\)

We want to solve for the velocity (v). Therefore we must isolate the variable. It is being multiplied and the inverse of multiplication is division. Divide both sides of the equation by 2.0 kg.

\(\frac { 10.0 \ kg \ m/s}{2.0 \ kg}=\frac{(2.0 \ kg )v}{2.0 \ kg}\)

The kilograms will cancel each other out.

\(\frac { 10.0 \ m/s}{2.0}=v\)

\(5.0 \ m/s=v\)

The velocity is 5 meters per second.

Answer:

The safe distance is 199 cm approximately 200 cm

Explanation:

Safe intensity = 1.00 W/m^2

wattage of radar leaked radar = 50.0 W

safe distance from the microwave will be = ?

We know that the intensity of a wave radiated uniformly in all direction is given as

\(I\) = \(\frac{W}{A}\)

where

W is the wattage of the leaked radar

A is the radial area, which is the area of a sphere that encapsulates the region through which this wave spreads uniformly.

From the equation above,

\(A\) = \(\frac{W}{I}\) = 50/1 = 50 m^2

But the area of this sphere \(A\) = \(4\pi r^{2}\)

where

r is the safe distance from the radar source

substituting for the area, we have

50 = 4 x 3.142 x \(r^{2}\)

50 = 12.568 \(r^{2}\)

\(r^{2}\) = 50/12.568 = 3.978

r = \(\sqrt{3.978}\) = 1.99 m = 199 cm ≅ 200 cm

Answer:

F-5-89

Explanation:

The magnitude of the induced electric field at a point near the center of the solenoid and 2 cm from the axis is approximately 0.000589 T*m²/s.

The magnitude of the induced electric field at a point near the center of the solenoid can be calculated using the formula for the magnetic field inside a solenoid. The induced electric field is directly proportional to the rate of change of the magnetic field with respect to time. First, let's calculate the magnetic field at the given point near the center of the solenoid. The formula for the magnetic field inside a solenoid is given by B = μ₀ * n * I, where B is the magnetic field, μ₀ is the permeability of free space (4π * 10⁻⁷ T*m/A), n is the number of turns per unit length (500 turns/m), and I is the current (60 A/s).

The magnetic field at the given point can be calculated as follows:
B = μ₀ * n * I = (4π * 10⁻⁷ T*m/A) * (500 turns/m) * (60 A/s) = 0.075 T
Next, we can calculate the induced electric field at the given point using the formula E = -dΦ/dt, where E is the induced electric field and Φ is the magnetic flux. The magnetic flux is given by Φ = B * A, where A is the area perpendicular to the magnetic field.

The area at the given point can be calculated as follows:
A = π * r² = π * (0.05 m)² = 0.00785 m²
Now, we can calculate the induced electric field:
E = -dΦ/dt = -(d(B * A)/dt) = -A * dB/dt = - (0.00785 m²) * (0.075 T/s) = -0.000589 T*m²/s

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Single-phase generators have much fewer than three phases because they create three separate waves and one of them is always at its peak ensuring a continuous heavy power supply.

Three-phase produces a superior control stream than single-phase since they make three isolated waves and one of them is continuously at its crest guaranteeing a ceaseless overwhelming control supply.

Indeed in case one conductor is out of arranging the other two will do the work so a three-phase will not close down totally at once but a single stage has the risk.

Difference between. single-phase generators and three-phase generators on different estimates and yields. As you might envision, a three-phase generator will ordinarily be much bigger than a single-phase generator.

This can be because the three-phase demonstration must account for the additional conductors. That additional estimate does cruel that you’ll devote more space to a three-phase generator.

Single-phase and three-phase control supply frameworks are the two most common shapes of control supply frameworks. Single-phase control is utilized in places where less control is required and to control unassuming loads.

When an enormous sum of control is required, the three stages are utilized in major undertakings, manufacturing plants, and fabricating units.

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Answer: Don't know what u mean

Explanation:

84 - 32 = 84 - (32 + 0) = 84 - 32 - 0

= 52

The equation that represents the relationship between the force, F, applied to the spring and its change in length, x, is F = kx, where k is the spring constant.

The relationship between the force applied to the spring and the extent to which it was compressed (stretched) is directly proportional. This means that as the force applied to the spring increases, the extent to which it is compressed (stretched) also increases.

The general equation describing the relationship between the applied force and the change in the length of the spring is F = kx, where F is the force applied, k is the spring constant, and x is the change in length.

For the second hoop, the same analysis can be conducted. The equation that represents the relationship between the force, F, applied to the spring and its change in length, x, is F = kx, where k is the spring constant. The relationship between the force applied to the spring and the extent to which it was compressed (stretched) is directly proportional. The general equation describing the relationship between the applied force and the change in the length of the spring is F = kx.

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Answer:

false gravity is not considered matter

Answer: +/-  

0 (18 - 18 )

Explanation:

plants

i hope its right and have a nice day

Answer:

number 1

Explanation:

Answer:

The force the archer need to pull in order to achieve the height is approximately 101.8 N

Explanation:

By energy conservation principle, puling an elastic bow with a force, for a given distance, performs work which is converted to the potential energy of the arrow at height

The given parameters are;

The mass of the arrow, m = 45.0 grams = 0.045 kg

The distance the elastic bow is pulled, d = 65.0 cm = 0.65 m

The height at which the arrow is reaches, h = 150.0 meters

Let 'F', represent the force the archer need to pull in order to achieve the height

Work done, W = Force × Distance moved in the direction of the force

Therefore;

The work done in pulling the arrow, W = F × d

By energy conservation, we have;

The work done in pulling the arrow, W = The potential energy gained by the arrow, P.E.

W = P.E.

The potential energy gained by the arrow, P.E. = m·g·h

Where;

m = The mass of the arrow

g = The acceleration due to gravity = 9.8 m/s²

h = The height the arrow reaches

∴ by plugging in the values, P.E. = 0.045 kg ×9.8 m/s² × 150 m = 66.15 J

W = F × d = F × 0.065 m

Also, W = P.E. = 66.15 J

∴ W = F × 0.065 m = 66.15 J

F × 0.065 m = 66.15 J

F = 66.15 J/(0.65 m) = 1323/13 N ≈ 101.8 N

The force the archer need to pull in order to achieve the height, F ≈ 101.8 N.

Answer: c a d b

Explanation:

Glaciologists use Glen–Nye Flow Law, to predict the movements of glaciers thus, The pattern they predict in the markers over time would be C D A B.

In some parts of the world, glaciers are an important natural resource because the nature and behaviour of glaciers are an impact the hydrologic, geologic, and ecological systems of their immediate location.

Due to this, Glaciologists monitor and try to predict the movement and morphology of glaciers.

One of the techniques used by "Glaciologists" in the monitoring and prediction of glaciers in the use of markers.

The movement of markers is measured relative to the edges of the valley down which the glacier flows.

Thus, The movement of the markers are then predicted using the Glen–Nye Flow Law.

The pattern they predict in the markers over time would be C D A B.

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The radial acceleration of the ball during its circular motion is approximately 59.4 m/s^2.

Centripetal acceleration, a = v^2 / r, where v is the speed of the ball and r is the radius of the circle.

The time it takes for the ball to reach the ground,

y = 1/2 g t^2

where y is the initial height of the ball (1.00 m), g is the acceleration due to gravity (9.81 m/s^2), and t is the time it takes for the ball to reach the ground.

t = sqrt(2y/g)

= sqrt(2 x 1.00 / 9.81)

≈ 0.45 s

Velocity, v = x/t

= 1.90 / 0.45

≈ 4.22 m/s

The radial acceleration of the ball during its circular motion can now be found using the equation,

a = v^2 / r

a = v^2 / r = (4.22)^2 / 0.300 ≈ 59.4 m/s^2

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Answer:

a) 102 meters

b) 14 meters

c) The direction is eastward.

Explanation:

Distance is said to be the length between two points. It is a scalar quantity.

a) The distance moved = 36 + 44 + 22

                                       = 102 meters

b) Displacement is the distance moved in a specified direction.

Representing the distance moved with specific direction as a directed number, 36 meters eastward = +36, 44 meters westward = -44, and 22 meters eastward = +22

/Displacement/ = +36 -44 +22

                         = +58-44

                         = 14

The magnitude of the displacement is 14 meters.

c) The direction is eastward.

So that the magnitude of displacement and his direction is 14 meters eastward.

Measurement of the distance traveled and the amount of time required to cover that distance are required in order to determine speed.

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The center of a 1.5 cm wide loop has a magnetic field of 3.0 mT. About 0.3 Amperes of current flow through the loop. A 3.0 mT magnetic field is present about 0.2 meters away from the wire.

To solve this problem, we can use the formula for the magnetic field produced by a current-carrying loop and the formula for the magnetic field produced by a current-carrying wire.

Answer of the following questions are below :  

(a) The magnetic field at the center of a current-carrying loop is given by the formula:

\(B = \frac{\mu_0 \cdot I \cdot N}{2 \cdot R}\)

Where B is the magnetic field, μ₀ is the permeability of free space (4π × 10⁻⁷ T·m/A), I is the current, N is the number of turns in the loop, and R is the radius of the loop.

We are given that the magnetic field is 3.0 mT (3.0 × 10⁻³ T) and the radius of the loop is 1.5 cm (0.015 m).

Substituting these values into the formula, we can solve for the current:

\(3.0 \times 10^{-3} \, \text{T} = \frac{4\pi \times 10^{-7} \, \text{T}\cdot\text{m/A} \cdot I}{2 \cdot 0.015 \, \text{m}}\)

Simplifying the equation, we find:

\(I = \frac{(3.0 \times 10^{-3} \, \text{T}) \cdot (2 \cdot 0.015 \, \text{m})}{4\pi \times 10^{-7} \, \text{T}\cdot\text{m/A}}\)

Calculating this expression, we get:

I ≈ 0.3 A

Therefore, the current in the loop is approximately 0.3 Amperes.

(b) The magnetic field produced by a long straight wire carrying current is given by the formula:

\(B = \frac{\mu_0 \cdot I}{2\pi \cdot r}\)

Where B is the magnetic field, μ₀ is the permeability of free space (4π × 10⁻⁷ T·m/A), I is the current, and r is the distance from the wire.

We are given that the magnetic field is 3.0 mT (3.0 × 10⁻³ T).

Substituting these values into the formula, we can solve for the distance from the wire:

\(3.0 \times 10^{-3} \, \text{T} = \frac{4\pi \times 10^{-7} \, \text{T} \cdot \text{m/A} \cdot I}{2\pi \cdot r}\)

Simplifying the equation, we find:

\(r = \frac{4\pi \times 10^{-7} \, \text{T} \cdot \text{m/A} \cdot I}{2\pi \cdot (3.0 \times 10^{-3} \, \text{T})}\)

Calculating this expression, we get:

r ≈ 0.2 m

Therefore, the magnetic field of 3.0 mT is found at a distance of approximately 0.2 meters from the wire.

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Aeronautical maps are usually meant to be used by pilots and air aviation professionals in other to navigate or traverse though the sky. With various elements such as vegetation, hills, valleys being depicted by color coded keys or legend. Hence, the absence of color on an aeronautical map make the representation of elements very difficult.

Visual map interpretation is usually aided by the use of legends. The legend hold the key to the elements which are represented on the map. Usually, a combination of colors and shapes makes up the legend and makes map interpretation easy.

Therefore, the absence of various color palletes for representation on a black and white aeronautical map will make it difficult to use.

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To solve this problem, we can use the formula for the magnetic field created by a current-carrying wire. Where B is the magnetic field, I is the current, r is the radius of the wire, and μ0 is the permeability of free space (a constant value)

B = μ0*I/(2*pi*r)

In this case, we are given the current (5.30 A), the magnetic field (8.80*10^5 T), and the angle swept out by the wire (0.100 radians). We want to find the radius of the arc.

We can start by rearranging the formula to solve for r:

r = μ0*I/(2*pi*B)

Substituting in the given values, we get:

r = (4*pi*10^-7)*(5.30)/(2*pi*8.80*10^5)

r = 0.00300 m

To convert this to centimeters, we multiply by 100:

r = 0.300 cm

Therefore, the radius of the arc is 0.300 cm.To find the radius of the arc, we can use the formula for the magnetic field at the center of a circular arc:

B = (μ₀ * I * θ) / (4 * π * r)

where B is the magnetic field, μ₀ is the permeability of free space (4π × 10⁻⁷ Tm/A), I is the current, θ is the angle in radians, and r is the radius.

Given:
B = 8.80 × 10⁵ T
I = 5.30 A
θ = 0.100 radians

We want to solve for r:

r = (μ₀ * I * θ) / (4 * π * B)

r = ((4π × 10⁻⁷ Tm/A) * (5.30 A) * (0.100)) / (4 * π * (8.80 × 10⁵ T))

r ≈ 1.885 × 10⁻⁶ m

Now, convert meters to centimeters:

r ≈ 1.885 × 10⁻⁶ m * (100 cm/1 m) = 1.885 × 10⁻⁴ cm

So, the radius of the arc is approximately 1.885 × 10⁻⁴ cm.

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Answer:

I think A. would be brighter

Explanation:

Circuit A has only one bulb so it gets 100% of the electricity

Circuit B has two bulbs which means that they only get 50% of the electricity because it has to be split equally.

Hope it helps!

A. The speed of the elevator after it has moved downward 1.00 m from the point where it first contacts a spring is 3.65m/s

B. The acceleration when the elevator is 1.00 {\rm m} below point where it first contacts a spring is 4m/s²

In calculating the speed of the elevator and acceleration, first we have to find the force of gravity F on the elevator, which is the force pulling the elevator in downward direction. Using the equation for force of gravity which is:

F = mg

Where:

Mass of the elevator; m= 2000kg

Acceleration due to gravity; g = 9.8m/s

2000kg × 9.8m/s²= 19600N

F = 19600

Force of opposing friction clamp of gravity = 17000N

Net force on the elevator = force of gravity - Force of opposing friction clamp

Net force on the elevator = 19600 - 17000

Net force on the elevator = 2600 N

We will also find the kinetic energy K.E; of the elevator at the point of contact with the spring using:

K.E = 1/2 mv²

Where

Mass of the elevator; m = 2000kg

Velocity of the elevator = 4.00m/s

K.E = (1/2)*2000kg*(4m/s)²

K.E = 16000J

The kinetic energy and energy gained will be absorbed by the spring across the next 2m

Therefore,

Energy; E = K.E + P.E

Where:

Kinetic energy K.E = 16000J

Potential Energy P.E = ?

P.E of spring = net force absorbed × distance at compression

Where:

Net force absorbed = 2600N

Distance at compression = 2.0m

P.E = 2600*2

P.E = 5200J

E = 16000J + 5200J

E = 21200J

Spring constant = k

To find k

Using:

E = (1/2)*k*(x)²

Where:

E = 21200J

k = ?

x = 2m

21200J = (1/2)*k*(2m)²

21200J*2 = (4m)k

K = 42400J/4m

K = 10600N/m

Therefore,

Acceleration at 1m compression = ?

Using:

F = K*X

Where

F is force provided by the spring = 10600N/m,

K = 10600 N/m

X = 1m

F = 10600N/m * 1m

F = 10600N (upward)

A. The speed of the elevator after it has moved downward 1.00 {\rm m} from the point where it first contacts a spring?

Using:

Original Kinetic energy + net force on the elevator = final kinetic energy + spring energy

16000N + 2600N = (1/2)mv² + (1/2)k x²

18600 = (1/2)(2000)(v²) + (1/2)(10600N)(1²)

18600 = 1000(v²) + 5300

18600 - 5300 = 1000(v²)

13300 = 1000(v²)

V² = 13.300

V =3.65m/s

B. The acceleration of the elevator is 1.00m below point where it first contacts a spring

Spring constant = net force on the elevator + resultant force

Where:

Spring constant = 10600N

Net force on the elevator = 2600N

Resultant force = ?

10600N = 2600N + resultant force

Resultant force = 10600N - 2600N

Resultant force = 8000N

Using the equation for Newton's 2nd law where F = ma,

a = F/m

Where:

Resultant force; F =8000N

Mass of the elevator; m =2000kg)

a = 8000 / 2000

a = 4m/s²

Here's the complete question:

In a "worst-case" design scenario, a 2000kg elevator with broken cables is falling at 4.00m/s when it first contacts a cushioning spring at thebottom of the shaft. The spring is supposed to stop the elevator,compressing 2.00m as it does so. During the motion a safety clampapplies a constant 17000N frictional force to the elevator.

1. What is the speed of the elevator after it has moved downward 1.00m from the point where it first contacts aspring?

2. When the elevator is 1.00m below point where it first contacts a spring, what is its acceleration?

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Part A: The x-component of vector B is 0.

Part B: The y-component of vector B is 0.

Given that vector A = 135i and A × B = 96k, we can determine the components of vector B as follows:

Part A:

Since A × B = 96k, and the cross product of two vectors is perpendicular to both vectors, the x-component of vector B would be zero. Therefore, the x-component of vector B is 0.

Part B:

To find the y-component of vector B, we can use the cross product formula. Since A × B = 96k, and the k-component of the cross product represents the y-component of the resultant vector, we have:

96 = Ay × 0 - Az × 0,

Ay = 0.

Therefore, the y-component of vector B is 0.

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Answer:

\(||\vec E_A||=1.11446 \times 10^6 \ \frac{N}{C}\), where the vector arrow will point from the charge towards point A.

Conceptual:

What is an electric field?

An electric field is a physical field produced by charged particles, these electric fields have the ability to exert forces on other charged particles. The following formula can be used to find the electric field (as a vector) at a point in space. "k_e" is Coulomb's constant and "\(\hat r\)" indicates the direction vector that point from the charge towards the field you are trying to calculate.

\(\boxed{\left\begin{array}{ccc} \text{\underline{Equation for Electric Field:}} \\\\\ \vec E=\frac{k_eq}{r^2} \hat r \\k_e=8.99 \times 10^9\frac{Nm^2}{C^2} \end{array}\right }\)

Explanation:

Given:

\(q=6 \times 10 ^{-6} \ C\\\\r=0.22 \ m\)

Find:

\(\vec E_A=?? \ \frac{N}{C}\)

\(\vec E_A=\frac{k_eq}{r^2} \hat r\\\\\Longrightarrow \vec E_A=\frac{(8.99 \times 10 ^9)(6 \times 10 ^{-6})}{(0.22)^2} \cdot\frac{ < 0,-0.22 > }{\sqrt{(0)^2+(-0.22)^2} } \\\\\Longrightarrow \vec E_A= 1.11446 \times 10^6 \cdot < 0,-1 > \\\\\Longrightarrow \vec E_A= < 0,-1.11446 \times 10^6 > \frac{N}{C} \\\\\Longrightarrow||\vec E_A||=\sqrt{(0)^2+(-1.11446 \times 10^6\))^2} \\\\\therefore \boxed{\boxed{||\vec E_A||=1.11446 \times 10^6 \ \frac{N}{C}}}\)

Thus, the electric field strength at point A is found. The vector arrow will point from the charge, q, towards point A.

Answer:

29.55 kg

Explanation:

force gravity =  G m1 m2 / (r^2)    now change the mass 1 and the r

                     =   G 2m1 m2 / ( 2r)^2

                      = 2/4  G m1 m2 / (r^2)

the weight will be 2/4   or  1/2 the original

The time period is 0.3 s and frequency is 3 Hz

Step 1 :

Given:

No. of cycles N = 24

Total time take t = 8 s

Step 1 : Calculating the frequency:

The frequency of a wave or oscillation is defined as the number of cycles or completed alternations per unit time.

f = N/t

f = 24/8

f = 3 Hz

Step 2: Calculating the time period:

Frequency and time period are inversely proportional.

The amount of time it takes for something to complete one oscillation is called its time period.

Time period = 1/Frequency

Time period = 1/3 = 0.3 s

So, the time period is 0.3 s and frequency is 3 Hz.

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A 10-v battery with just a 100- resistor placed across its terminals will conduct 0.1 A of current.

When a patterns is made to enable continuous movement of electric charge, a circuit is produced. A current seems to be the continuous passage of an electric charge thru the electrodes of a circuit; it is frequently referred to as a "flow," like a liquid might travel through with a straight tube.

Use Ohm's law

V = IR

where voltage (V), current (I), and resistance (R)

so,

I = V / R

I = 10 / 100

I = 0.1 A

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The complete question is-

How much current will flow through a 10-V battery with a 100Ω resistor connected across its terminals?

A. 0.1A

B. 1.0A

C. 0

D. 1000A

Answer:

Load = 80 Newton

Explanation:

Given the following data:

Effort = 20 N

Load arm = 15 cm

Effort arm = 60 cm

Conversion:

100 cm = 1 meters

15 cm = 15/100 = 0.15 meters

60 cm = 60/100 = 0.6 meters

To calculate the load, we would use the expression;

Effort * effort arm = load * load arm

Substituting into the expression, we have;

20 * 0.6 = load * 0.15

12 = load * 0.15

Load = 12/0.15

Load = 80 Newton

Answer:

50%.......

Explanation:

is the correct check it then send feedback

Explanation:

u=24

v=0

a=-0.29

s=

v^2=u^2+2as

s=(v^2-u^2)/2a

s=(0-576)/0.58

s=-575.42

Answer:

I think the answer might be the first one