how to find resultant force and magnitude of charges in right angled triangle​

To show that the adjacent planes is gin terms of the cube by d= a/(h2 +u2+l2)1/2, we have to fully analyse it. Therefore, let's go straight to the explanation.

The distance "d" between adjacent planes in a crystal lattice is given by the equation:

d = a / (h^2 + k^2 + l^2)^(1/2)

where "a" is the lattice parameter (length of one side of the unit cell) and (h,k,l) are the indices of the crystal plane. The indices specify the orientation of the plane in the crystal lattice and are related to the Miller indices of the plane.

The equation shows that the distance between the planes is inversely proportional to the square root of the sum of the squares of the indices.

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A substance with stronger molecular attraction will evaporate at a higher temperature because it requires more energy to be ADDED to overcome attraction between the molecules.

A phase change is when matter shifts from one state (solid, liquid, gas, or plasma) to another. These transitions take place when the system receives enough energy or loses enough energy, as well as when the pressure on the system is altered.

A phase change occurs when the kinetic energy decreases enough so that the attraction between molecules pulls them together.

A substance with weaker molecular attraction will freeze at a higher temperature because it requires more energy to be TAKEN OUT before the attraction pulls the molecules together.

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The runner who reaches their maximum speed more quickly has the larger maximum speed. In a 100-meter dash, reaching maximum speed more quickly generally indicates a higher level of acceleration, which is related to maximum speed.

The runner who reaches their maximum speed more slowly may have a longer time to build up speed, but once both runners have reached their maximum speed, they are both running at the same speed.

So, the runner who reached maximum speed more quickly will have had a higher maximum speed.

Speed is known as the rate of change of position of an object in any direction. Speed is calculated as the ratio of distance to the time in which the distance was covered.

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Answer:

Explanation:

First find the total resistance of the combination circuit:

R = V/I = 12 V / 2 A = 6 Ω

6Ω = R1 + R2 + 1/(1/R3 + 1/R4)

6Ω = 2Ω + 2Ω + 1/(1/4 + 1/R4)

2Ω = 1/(1/4Ω + 1/R4)

2Ω(1/4Ω + 1/R4) = 1

(1/4 + 1/R4) = 1/2

1/R4 = 1/4

R4 = 4Ω

Note:  R1 and R2 are in series, so you just add them together.  But R3 and R4 are in parallel, so Req = 1/R3 + 1/R4

The 2.5 kg block pulls on the 9 kg block with an 87.2 N force.

Newton's Third Law, often known as the law of action and reaction, can be used to determine the force that the 2.5 kg block exerts on the 9 kg block. The force applied to the 9 kg block is equivalent to, but applied in the opposite direction from, the force applied to the 2.5 kg block. Since they are in touch, the force of friction operating on the 9 kg block is equal to the force of friction acting on the 2.5 kg block.

Calculating the friction force is as simple as multiplying the normal force, or weight of the object, by the coefficient of friction:

Step 1: Determine the normal force or weight of the object. The weight of a 9 kg object can be calculated as 9 kg x 9.8 N/kg = 87.2 N.

Step 2: Determine the coefficient of friction. In this case, the coefficient of friction is 0.1.

Step 3: Calculate the friction force by multiplying the normal force by the coefficient of friction.

Friction force = Normal force x Coefficient of friction

= 87.2 N x 0.1 = 8.72 N.

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Answer:

need points srr

Explanation:

Answer:

2.8 A

Explanation:

since the resistances are in series, the net resistance will be,

R = R1 + R2 + R3 + .....

R = ( 5+10+12+16)Ω

R = 43Ω

From Ohm's law,

V = iR

120 = i * 43

i = 120/43 A

i = 2.8A

Using SOH,CAH,TOA, we can calculate the height of the building by taking the sine of the launch angle (60°) and the distance traveled (65 m). The height of the building is 56.9 m.

SOH,CAH,TOA stands for Sine = Opposite/Hypotenuse, Cosine = Adjacent/Hypotenuse and Tangent = Opposite/Adjacent. This is a set of trigonometric ratios that can be used to solve for angles and lengths of sides of a right triangle. It is used when the angle and length of two sides of a right triangle are known, but the length of the third side needs to be calculated.

Calculation using SOH, CAH, TOA

The distance from the ground to the top of the building is 65 m and the angle is 60 degrees. the height (h) of the building.has to be found

Using the trigonometric ratio of Sine,  calculate the height of the building:

Sin(60°) = Opposite/Hypotenuse

h/65 = sin(60°)

h = 65sin(60°)

h = 65(0.866)

h = 56.59 m

Therefore, the height of the building the projectile was launched to from the ground is 56.59 m.

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The vertical component of acceleration at time tm can be determined by taking the second derivative of the vertical position vector with respect to time.

Velocity is a vector quantity that measures the rate and direction of change in an object's position. It is typically expressed in units of meters per second (m/s). Velocity is equal to the distance traveled divided by the time it takes to travel that distance. Velocity is also related to acceleration, which is the rate of change of velocity.

At time tm, the vertical component of acceleration for the module would be the derivative of the vertical velocity vector with respect to time, which is the change in vertical velocity over time. This can be calculated by taking the second derivative of the vertical position vector with respect to time. Since the position vector is a function of time, the acceleration can be determined by taking the second derivative of the position vector with respect to time, which is defined as the rate of change of the velocity with respect to time. Therefore, the vertical component of acceleration at time tm can be determined by taking the second derivative of the vertical position vector with respect to time.

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The correct statement is (c): At the highest point in its motion, the ball's velocity is zero.

This is due to the fact that when anything is thrown straight up, it has a starting velocity and then accelerates as it descends towards the Earth. Its velocity is equal to zero as it reaches its highest point (due to the gravitational pull of the Earth). The ball's acceleration varies continuously as it travels because it accelerates towards the Earth when it is travelling upward and decelerates when it is travelling downward.

The ball is briefly at rest, and the only force acting on it is the gravitational pull of the Earth, therefore at its fastest point in motion, the acceleration is equal to zero.

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Heat and work are path functions because their values depend on the path taken from the initial to the final state of a system.

Heat is a form of energy that is transferred from one object to another due to a difference in temperature. Heat is a form of kinetic energy, the energy of motion, and can be transferred from one object to another by conduction, convection, and radiation.

Both heat and work can cross the boundary of a system, as energy transfer in the form of heat or work can occur when a system interacts with its surroundings.
Heat and work are associated with a process, which is a change in the state of a system from one thermodynamic state to another.
Heat and work are also associated with a thermodynamic state, as the amount of heat and work exchanged between a system and its surroundings is determined by the thermodynamic state of the system.
Finally, systems possess energy in the form of heat and work, which can be exchanged between a system and its environment.

Therefore, all the given option is correct
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In sequence from the position of where a stimulation is initially absorbed to whether a reply stimulus is conducted, dendrites, cell body and soma, axon hillock, axon, and synaptic terminal sections or areas of the neuron.

Dendrites are extensions made to receive messages from nearby cells. They appear to have a tree-like shape because they produce projections that are activated by numerous other neurons and carry the electromagnetic gradient to the neurone.

Dendrites receive impulses from numerous other neurons and pass them on to the nerve cell. If a neuron is sufficiently active, it will emit an electrical impulses, an electrical impulse that excites additional neurons. Huge networks of these neurons are set up, and they communicate among themselves in order to create ideas and actions.

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The limit integral of function x⁴ from c to d is [ d⁵-c⁵]/5.

In mathematics, an integral lends numerical values to functions to represent concepts like volume, area, and displacement that result from combining infinitesimally small amounts of data.

Integration is the action of locating integrals. . In addition to differentiation, integration is a fundamental, crucial calculus operation that helps to solve issues with the area of an arbitrary form.

The limit integral of function x⁴ from c to d is:

[tex]\int\limits^c_d {x^4} \, dx[/tex]

[tex]= [\frac{x^5}{5} ]_{x=c}^{x=d}[/tex]

= [ d⁵-c⁵]/5

Hence, the limit integral of function x⁴ from c to d is [ d⁵-c⁵]/5.

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After n half-lives, its radioactivity will be 100× [tex]2^{-n}[/tex] Bq.

A characteristic of some forms of matter known as radioactivity is the spontaneous emission of energy and subatomic particles. In essence, it is a characteristic of particular atomic nuclei.

The radioactivity of the material is 100 Bq, that means, the radioactivity of the material is 1000 disintegration per second.

the relation between radioactivity and half-life is : A = A₀ 2^-n

Hence, after n half-lives, its radioactivity will be = 100 × 2^-n Bq.

Therefore, after n half-lives, its radioactivity will be 100× [tex]2^{-n}[/tex] Bq.

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The gravitational potential energy of the car at the top of the ramp is 98,000 Joules. 

The gravitational potential energy (PE) of an object of mass m at ground level h is given by

PE = mgh

where g is the acceleration due to gravity, approximately 9.8 m/s^2 near the surface.

In this case, the car has a mass of 1000 kg and a height of 10 m above the ground. Therefore, the gravitational potential energy (PEg) is:

PEg = mgh = 1000kg * 9.8m/s^2 * 10m = 98,000J

Therefore, the gravitational potential energy of the car at the top of the ramp is 98,000 Joules. 

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Answer:

Explanation:

The only explanation I can think of is this:

Block 1 is moving to the left and strikes Block 2, which is at rest.  100% of the momentum of Block 1 transfers to Block 2.  So Block 2 now moves to the left at 4 m/s (you didn't specify a unit) and Block 1 comes to a complete stop and is now at rest.

The restoring force and acceleration are always in the same direction for a simple harmonic oscillator.

A simple harmonic oscillator is what?

A driven or dampened oscillator is known as a simple harmonic oscillator. It typically consists of a mass "m" that is pulled in the direction of the point x = 0 by a single force "F" that solely depends on the body's position "x" and a constant "k."

Consider a straightforward pendulum that displays SHM at low displacements. The location vector points upward while the acceleration and velocity vectors point downward during the downswing. The acceleration vector points downward while the location and velocity vectors point upward during an upswing. Therefore, unless they are both 0 at equilibrium, the acceleration always points in the opposite direction to the position vector. The acceleration and force of restoration are always in same direction .

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The transient heat conduction equation for a plane wall with constant thermal conductivity and no heat generation can be derived from the energy balance equation for a rectangular volume element.

Consider a plane wall with constant thermal conductivity, k, and no heat generation. Let's assume that the wall is at temperature T at time t and T + ΔT at time t + Δt. The energy balance equation for a rectangular volume element of the wall can be expressed as:

ΔQ/Δt = -kA(ΔT/Δx), where

Rearranging the equation:

ΔT/Δt = -(k/A)(ΔT/Δx)

The above equation represents the one-dimensional transient heat conduction equation for a plane wall with constant thermal conductivity and no heat generation.

This equation can be further simplified by using the thermal diffusivity, α, which is defined as:

α = k / (ρCp) where,

Substituting α into the equation:

ΔT/Δt = -α(ΔT/Δx^2)

This is the final form of the one-dimensional transient heat conduction equation for a plane wall with constant thermal conductivity and no heat generation.

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The average deceleration of the person while underwater can be estimated as 875 m/s².

To calculate the average deceleration, we need to find the change in velocity over the change in time. The change in velocity can be calculated as the initial velocity (which is assumed to be zero) minus the final velocity, which is the velocity just before the person stops moving downward. The change in time can be found by subtracting the starting height from the stopping height.

Starting height = 4.1 mm = 4.1 × 10⁻³ m

Stopping height = 2.3 mm = 2.3 × 10⁻³ m

Change in velocity = final velocity - initial velocity

Change in velocity = 0 m/s - (-√(2gh))

Change in velocity = √(2gh)

where g is the acceleration due to gravity (9.8 m/s²) and h is the height (4.1 mm)

Change in velocity = √(2 × 9.8 × 4.1 × 10⁻³)

Change in velocity = √(79.26 × 10⁻³) = 8.9 m/s

Change in time = final time - initial time

Change in time = 0 s - (stopping height - starting height)

Change in time = (starting height - stopping height)

Change in time = 4.1 × 10⁻³ - 2.3 × 10⁻³ = 1.8 × 10⁻³ s

Change in time = 1.8 × 10⁻³ s

Average deceleration = change in velocity / change in time

Average deceleration = √(2 × 9.8 × 4.1 × 10⁻³) / (h2 - h1)

Average deceleration = 8.9 / (1.8 × 10⁻³)

Average deceleration = 875 m/s²

Therefore, the average deceleration is estimated at 875 m/s².

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The surface velocity is approximately 2.07 m/s.

The average velocity of the liquid flow per unit depth,

Q/A = 1.25 [gal/min]/[1 ft x 0.5 in x (1/12) ft/in] = 10 [ft/min]

where Q is the flow rate and A is the cross-sectional area of the channel.

Use the laminar velocity profile to determine the velocity at the center of the channel (y=0),

u(y=0) = Uo(2(0) - 0^2)/(2h) = 0

Since there is no slip at the wall, the velocity at y=h/2,

u(y=h/2) = Uo(2(h/2) - (h/2)^2)/(h)

= Uo(2-h/2)

= 2Uo - Uh/2

Equating this to the average velocity,

2Uo - Uh/2 = 10 [ft/min]

Solving for Uo = (10 + Uh/2)/2

Uh = Ahu(y=h/2) = AhUo(2-h/2)/(h) = A*Uo(2-h/2)

Substituting A = 1 [ft^2] and h = 0.5 [in] = 0.042 [ft], we get:

Uh = Uo(2-0.042/2) = 0.979Uo

Uo = (10 + 0.979Uo/2)/2

Uo = 6.8 [ft/min] or 2.07 [m/s]

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The fraction of energy lost by the colliding body is 8/9.

In an elastic collision, the total kinetic energy of the system before the collision is equal to the total kinetic energy of the system after the collision.

Before the collision, the total kinetic energy of the system is given by:

[tex]K1 = 0.5 * m1 * u^2[/tex]

After the collision, the total kinetic energy of the system is given by:

[tex]K2 = 0.5 * m_{1} * v_{1} ^2 + 0.5 * m_{2} * v_{2}^2[/tex]

where m1 is the mass of body A, u is its initial velocity, v1 is its final velocity, m2 is the mass of body B, and v2 is its final velocity.

Conservation of momentum also applies in this case. The momentum of the system before the collision is equal to the momentum of the system after the collision.

Before the collision, the momentum of the system is given by:

[tex]p_{1} = m_{1} * u[/tex]

After the collision, the momentum of the system is given by:

[tex]p_{2} = m_{1} * v_{1} + m_{2} * v_{2}[/tex]

From the conservation of momentum, we have:

[tex]m_{1} * u = m_{1} * v_{} + m_{2} * v_{2}[/tex]

Solving for v1 and v2, we find:

[tex]v_{1} = u * (m_{1} - m_{2}) / (m_{1} + m_{2})[/tex]

[tex]v_{2} = 2 * u * m_{1} / (m_{1} + m_{2})[/tex]

Substituting these values into the equation for K2, we find:

[tex]K_{2} = 0.5 * m_{1} * u^2 * (m_{1} - m_{2})^2 / (m_{1} + m_{2})^2[/tex]

The fraction of energy lost by body A is given by:

[tex](K_{1} - K_{2}) / K_{1} = 1 - K_{2} / K_{1}[/tex]

= [tex]1 - 0.5 * m_{1} * u^2 * (m_{1} - m_{2})^2 / (m_{1} + m_{2})^2 / (0.5 * m_{1} * u^2)[/tex]

= [tex]1 - (m_{1} - m_{2})^2 / (m_{1} + m_{2})^2[/tex]

= 1 - (4 - 2)^2 / (4 + 2)^2

= 1 - (2)^2 / (6)^2

= 1 - 4/36 = 1 - 1/9 = 8/9

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The correct answer is (A) b < d < c < a.

we need to consider the placement of the resistors.

In circuits a, b and c, resistors are connected in series. The corresponding resistance of resistors combined in series is the sum of the particular resistances. So the equivalent resistance for these circuits is:

a) 3R + R + 3R = 7R

b) R + R + 3R + 3R = 8R

c) 3R + R + R + 3R = 8R

In circuit d, resistors are connected in parallel. The equivalent resistance of resistors connected in parallel is given by the formula

1/request = 1/R + 1/R + 1/R = 3/R

Therefore Req = R/3. Comparing the values ​​of the equivalent resistors reveals the following:

b < d < c < a

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The magnitude of the electric field, in newtons per coulomb, at the location of qa in the figure is 32737.87 N/C.

What is electric field?

An electric field is a physical field that surrounds electrically charged particles and acts as an attractor or repellent to all other charged particles in the vicinity. It can also refer to a system of charged particles' physical field.

When charge is present in any form, every point in space has an electric field associated with it. The value of E, often known as the electric field strength, electric field intensity, or just the electric field, expresses the strength and direction of the electric field.

[tex]$$The $\mathrm{x}$ component of $\vec{E}_q$ is given as:$$\begin{aligned}\vec{E}_{q, x} & =E_q \cos 45^{\circ} \hat{x} \\& =k \frac{\left|-1.1 \times 10^{-9}\right|}{1.25 \times 10^{-3}} \times \frac{1}{\sqrt{2}} \hat{x}\end{aligned}$$And its y component is given as:$$\begin{aligned}\vec{E}_{q, y} & =-E_q \sin 45^{\circ} \hat{y} \\& =-k \frac{\left|-1.1 \times 10^{-9}\right|}{1.25 \times 10^{-3}} \times \frac{1}{\sqrt{2}} \hat{y}\end{aligned}$$[/tex]

[tex]$$Therefore, the net electric field at the location of $q_a$ is given as:$$\begin{aligned}\vec{E} & =\vec{E}_b+\vec{E}_c+\vec{E}_{d, x}+\vec{E}_{d, y}+\vec{E}_{q, x}+\vec{E}_{q, y} \end{aligned}[/tex]

[tex]& =k \times 10^{-9} \times\left(\left(-\frac{5.9}{(0.05)^2}-\frac{5.9}{\sqrt{2} \times 5 \times 10^{-3}}+[/tex] [tex]\frac{1.1}{\sqrt{2} \times 1.25 \times 10^{-3}}\right) \hat{x}+\left(\frac{5.9}{(0.05)^2}+\frac{5.9}{\sqrt{2} \times 5 \times 10^{-3}}-\frac{1.1}{\sqrt{2} \times 1.25 \times 10^{-3}}\right) \hat{y}\right) \end{aligned}[/tex]

[tex]\begin{aligned}& =\left(9 \times 10^9\right) \times 10^{-9} \times(-2572.13 \hat{x}+2572.13 \hat{y}) \\& =23149.17 \times(-\hat{x}+\hat{y}) \mathrm{N} / \mathrm{C}\end{aligned}[/tex]

[tex]$And its magnitude is given by:$$\begin{aligned}|\vec{E}| & =23149.17 \times \sqrt{1^2+1^2} \\& =23149.17 \times \sqrt{2} \\& =32737.87 \mathrm{~N} / \mathrm{C}\end{aligned}$$[/tex]

Thus, the magnitude of the electric field, in newtons per coulomb, at the location of qa in the figure is 32737.87 N/C.

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Explanation:

the Newsweek shadow STD test like that card imperial how was

To keep the block at rest, an electric field with a magnitude equal to the gravitational force divided by the block's charge is required.

A physical quantity's magnitude can be used to gauge its size or strength, such as in the case of a vector. The term "magnitude" in the context of an electric field refers to the field's amplitude.

The "equilibrium electric field" is the electric field needed to hold an object still. A force will be applied to a charged object in an electric field, either in the direction of or away from the source of the field. The net force must be zero in order to maintain the item at rest, which necessitates counteracting the electric field's force with another force.

E = k * q / d2, where k is the Coulomb constant, q is the charge on the object, and d is the separation between the item and the field source, can be used to calculate the size of the electric field necessary for equilibrium. According to this equation, the electric field is inversely proportional to the square of the distance between the item and the field source and directly proportional to the charge on the object.

As a result, if an object is charged and placed in an electric field, the size of the field needed to keep the object at rest will depend on the object's charge and the distance between it and the field source.

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An expression for the magnitude of the electric field that enables the block to remain at rest is:

[tex]\mathrm{ E = \dfrac{m \times g\times sin(\theta) }{q} }[/tex]

A physical quantity's magnitude can be used tο gauge its size or strength, such as in the case of a vector

The "equilibrium electric field" is the electric field needed to hold an object still. A force will be applied to a charged object in an electric field, either in the directiοn of or away from the source of the field. The net force must be zero in order to maintain the item at rest, which necessitates cοunteracting the electric field's fοrce with another force.

According to this equation, the electric field is inversely proportional to the square οf the distance between the item and the field source and directly proportional to the charge on the οbject.

As with a typical inclined plane problem, we need tο find the compοnent of the gravitational fοrce parallel to the incline.

That is

[tex]\mathrm{ m \times g\times sin(\theta) }[/tex]

For the blοck to remain stationary, we set this equal to the coulοmb force

[tex]\mathrm{ qE = m \times g\times sin(\theta) }[/tex]

since E is parallel to the incline,

sοlving for E, we find the answer tο part a:

[tex]\mathrm{ E = \dfrac{m \times g\times sin(\theta) }{q} }[/tex]

Thus, an expression fοr the magnitude οf the electric field that enables the block tο remain at rest is:

[tex]\mathrm{ E = \dfrac{m \times g\times sin(\theta) }{q} }[/tex]

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The quantity of heat absorbed by 50 g of water that warms from 30°C to 90°C is 12552J.

The amount of heat absorbed or released by a substance can be calculated by using the following formula;

Q = mc∆T

Where;

According to this question, 50g of water warms from 30°C to 90°C. The quantity of heat absorbed is as follows:

Q = 50 × 4.184 × {90°C - 30°C}

Q = 209.2 × 60

Q = 12552J

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Answer:

Work = 1243.17 J

Explanation:

Work = Force × Distance × Cosine (Angle)

Work = 186 N x 6.5 m x Cosine (26°)

Work = 186 N x 6.5 m x 0.906

Work = 1243.17 J

The density of the metal is  determined as  6.1 g / cm³.

The density of the metal is defined as the ratio of the mass per unit volume of the metal.

Mathematically, the formula for density is given as;

ρ = m / V

where;

The density of the metal is  calculated as follows;

ρ = ( 0.47 kg x 1000 g/kg ) / ( 77 cm³ )

ρ = 6.1 g / cm³

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