Two small spheres with charges q1 = +9 nC and q2 = -3 nC are placed at marks x1 = -2 cm and x2 = 13 cm. Where along the x-axis should you place a third sphere with charge q3 so that the sum of the forces exerted by the first two spheres on the third one is zero?

The power dissipated in the power cord is approximately 65.5 kW, and in the heating element, it's around 1.49 kW.

The question pertains to the power dissipated in a circuit. We will use the formula Power (P) = Voltage (V) ^ 2 / Resistance (R) for each component of the circuit to find the desired values. Given a potential difference of 120V, we have:

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Answer:

The density of the liquid = 1278.95 kg/m³

Explanation:

Density: This can be defined as the ratio of the mass of  a body to its volume. The S.I unit of density is kg/m³.

From Archimedes principle,

R.d = Density of object/Density of liquid = Weight of object in air/Upthrust in liquid.

D₁/D₂ = W/U .......................... Equation 1

D₂ = D₁(U/W)........................ Equation 2

Where D₁ = Density of aluminum, D₂ = Density of liquid, W = weight of aluminum, U = upthrust.

m₁ = 3.8 kg, m₂ = 2.00 kg, g = 9.8 m/s²

W = m₁g = 3.8(9.8) = 37.24 N.

U = lost in weight of the aluminum = (m₁ - m₂)g = (3.8-2.0)9.8

U = 1.8(9.8) = 17.64 N.

Constant: D₁ = 2700 kg/m³

Substituting these values into equation 2

D₂ = 2700(17.64)/37.24

D₂ = 1278.95 kg/m³

Thus the density of the liquid = 1278.95 kg/m³

Answer:

The velocity of the water at the 8 cm section is 9 times the velocity of the water at 24 cm section.

Explanation:

For this question, we use the equation of continuity of an non-compressible  fluid in a pipe

[tex]A_{1} v_{1} =A_{2} v_{2}\\\\\pi r_{1}^{2}v^{2} = \pi r_{2}^{2}v^{2} \\\\\pi (24)^{2}v_{1} ^{2} = \pi (8)^{2}v_{2} ^{2}\\\\576v_{1} = 64v_{2}\\\\v_{2} = 9 v_{1}[/tex]

The velocity of the water at the 8 cm section is 9 times the velocity of the water at 24 cm section.

Answer:

(A)  111.77m

(B) 9.07m

(C)55.88m/s

(D)20.54m/s

Explanation:

step 1 " we have to calculate the time it took the projectile to get to its maximum height

(a)  t = usinθ/g

= 49sin 33/9.81

49×0.5446/9.81

=2.72s

the horizontal distance =  ucosθ×t  , because the projectile horizontal motion is unaffected by the force of graavity

= 49cos33 ×2.72

=111.77m

(B) with the same projectile fired the same way , the horizontal distance = 55.8m

55.8 = ucosθ×t

55.8 = 49cos33 ×t

t = 55.8/49cos33

t= 1.36s

height of the projectile = 1/2 ×g×t²

=1/2 ×9.81×1.36²

= 9.07m

(c) Velocity of the projectile when it hits the wall

V₀ = ucosθ×t

49cos 33 × 1.36

=55.88 m/s

(D) speed = distance / time

distance /2×t   ; total time of flight

= 55.88/ 2.72

=20.54m/s

Answer:

497.35 m.

Explanation:

Echo: This is the sound heard after the reflection of sound wave on a plane surface.

v = 2x/t.......................................... Equation 1

Where v = speed of sound in air, x = length of the lake, t = time taken to hear the echo.

Making x the subject of the equation,

x = vt/2.......................................... Equation 2

Given: v = 343 m/s, t = 2.9 s.

Substitute into equation 2

x = 343×2.9/2

x = 497.35 m

Thus the length of the lake  = 497.35 m.

Answer:

c. Satellite 1 has a shorter period and faster speed than satellite 2.

Explanation:

The period of an object in a circular orbit is given by:

[tex]T=\frac{2\pi r}{v}(1)[/tex]

Here r is the radius of the circular orbit and v is the speed of the object. So, we have [tex]r_1<r_2[/tex], according to (1) the period is proportional to the radius, thus [tex]T_1<T_2[/tex]. In other hand the period is inversely proportional to the speed, so [tex]v_1>v_2[/tex]

Answer:

it's b brainliest pls!

Explanation:

Answer:

a) v1 = 5.52m/s

b) v2 = -1.52m/s

c) v3 = 4.62m/s

d) vt = 3.85m/s

Explanation:

The velocity of the football wide receiver is his displacement per unit time.

Velocity v = (displacement d)/time t

v = d/t .....1

For each of the cases, equation 1 would be used to calculate the velocity.

a) v1 = d1/t1

d1= 16m

t1 = 2.9s

v1 = 16m/2.9s

v1 = 5.52m/s

b) v2 = d2/t2

d2 = -2.5m

t2 = 1.65s

v2 = -2.5/1.65

v2 = -1.52m/s

c) v3 = d3/t3

d3 = 24m

t3 = 5.2s

v3 = 24/5.2

v3 = 4.62m/s

d) vt = dt/tt

dt = 16m - 2.5m + 24m = 37.5m

tt = 2.9 + 1.65 + 5.2 = 9.75s

vt = 37.5/9.75

vt = 3.85m/s

Answer:

-6C

Explanation:

We are given that

Charge on an object,Q=-8C

Charge on one piece of object=[tex]q_1=-2C[/tex]

We have to find the exact charge of the other piece.

Let q be the charge of second piece of object.

By conservation of charge

Initial charge=Final charge

[tex]Q=q_1+q[/tex]

Substitute the values then we get

[tex]-8C=-2C+q[/tex]

[tex]q=-8C+2C[/tex]

[tex]q=-6C[/tex]

Hence, the exact charge of the other piece=-6C

2) The velocity of the wheel is increasing

3) The slope is the rate of change of velocity

4) The unit of the slope is metres per second squared.

5) Acceleration

6) The symbol used for acceleration is [tex]a[/tex]

7) The relationship between velocity and time is [tex]v=gsin \theta t[/tex]

Explanation:

2)

The graph in the problem represents the velocity of the wheel as a function of the time.

As we can see from the graph, the velocity is increasing as the time passes. This means that the wheel is accelerating (its velocity is changing constantly)

3)

The slope of the graph represents the rate of change of the velocity.

Mathematically, it can be written as:

[tex]m=\frac{\Delta v}{\Delta t}[/tex]

where

m is the slope

[tex]\Delta v[/tex] is the change in velocity

[tex]\Delta t[/tex] is the time interval considered

As we see from the graph, the slope of the line is positive and constant: this means that the velocity is increasing at a constant rate.

4)

The unit of the slope can be determined starting by the units of the two variables involved.

On the y-axis, we have the velocity, which is measured in metre per second ([tex]m/s[/tex])

On the x-axis, we have the time, which is measured in seconds ([tex]s[/tex])

The slope is the ratio between these two quantities:

[tex]m=\frac{\Delta v}{\Delta t}[/tex]

Therefore, the units of the slope are

[tex]m=\frac{[m/s]}{[s]}=[m/s^2][/tex]

So, metres per second squared.

5)

The rate of change of velocity is also known as acceleration.

In fact, acceleration is defined as the ratio between the change in velocity and the change in time:

[tex]a=\frac{\Delta v}{\Delta t}[/tex]

By comparing with the formula of the slope in part 3), we see that the two equations are identical, therefore the acceleration corresponds to the slope of the graph.

6)

The symbol used to represent the acceleration is [tex]a[/tex]:

[tex]a=\frac{\Delta v}{\Delta t}[/tex]

7)

In any uniformly accelerated motion, the relationship between velocity and time is given by the following suvat equation:

[tex]v=u+at[/tex]

where

u is the initial velocity

v is the final velocity

a is the acceleration

t is the time

In this problem, the wheel starts from rest, so

u = 0

Also, for an object rolling down a ramp, the acceleration is given by

[tex]a=g sin \theta[/tex]

where g is the acceleration of gravity and [tex]\theta[/tex] is the angle of the ramp. Substituting, we find the final expression of the velocity:

[tex]v=gsin \theta t[/tex]

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On a linear velocity vs. time graph, the velocity of the wheel is changing at a constant rate as time continues. The slope of the graph represents the acceleration. The equation describing the relationship between velocity and time is V = V0 + at.

In a velocity vs. time graph, if the graph is linear (i.e., a straight line), it indicates that the velocity is changing at a constant rate as time continues. The slope of the graph represents the acceleration of the wheel. The slope is measured in units of velocity divided by time, typically meters per second per second (m/s^2). The commonly used symbol to represent this rate is 'a', which stands for acceleration. The general equation describing the relationship between velocity and time for the wheel rolling down the incline from rest is given by V = V0 + at, where V is the velocity at a given time, V0 is the initial velocity (which is zero in this case), 'a' is the acceleration, and 't' is the time.

Answer:

IGNEOUS ROCKS

Explanation: Igneous rocks are those rocks that solidify from magma.

Igneous rock is divided into two ,they are:

1. Intrusive

Igneous rocks crystallized belowearth"s crust. Its cooling material is called lava.

2 Extrusive igneous rock is also known as known as volcanic rocks

Answer:

Please refer to the attachment below.

Explanation:

Please refer to the attachment below for explanation.

Answer:

a)  E = -4 10² N / C , b) x = 0.093 m, c)     a = 10.31 m / s², θ=-71.9⁰

Explanation:

For that exercise we use Newton's second Law, in the attached we can see a free body diagram of the ball

X axis

             [tex]F_{e}[/tex] - [tex]T_{x}[/tex] = m a

Axis y

            [tex]T_{y}[/tex] - W = 0

Initially the system is in equilibrium, so zero acceleration

            Fe = [tex]T_{x}[/tex]  

            T_{y} = W

Let us search with trigonometry the components of the tendency

            cos θ = T_{y} / T

            sin θ = [tex]T_{x}[/tex]  / T

           T_{y} = cos θ

           [tex]T_{x}[/tex]  = T sin θ

We replace

            q E = T sin θ

            mg = T cosθ

a) the electric force is

                [tex]F_{e}[/tex] = q E

                E = [tex]F_{e}[/tex] / q

                E = -0.032 / 80 10⁻⁶

                E = -4 10² N / C

b) the distance to this point can be found by dividing the two equations

                q E / mg = tan θ

                θ = tan⁻¹ qE / mg

Let's calculate

              θ = tan⁻¹ (80 10⁻⁶ 4 10² / 0.01 9.8)

              θ = tan⁻¹ 0.3265

               θ = 18 ⁰

               sin 18 = x/0.30

               x =0.30 sin 18

               x = 0.093 m

c) The rope is cut, two forces remain acting on the ball, on the x-axis the electric force and on the axis and the force gravitations

X axis

           [tex]F_{e}[/tex] = m aₓ

            aₓ = q E / m

           aₓ = 80 10⁻⁶ 4 10² / 0.01

           aₓ = 3.2 m / s²

Axis y

           W = m [tex]a_{y}[/tex]

           a_{y} = g

           a_{y} = 9.8 m/s²

The total acceleration is can be found using Pythagoras' theorem

             a = √ aₓ² + a_{y}²

             a = √ 3.2² + 9.8²

             a = 10.31 m / s²

The Angle meet him with trigonometry

               tan θ = a_{y} / aₓ

               θ = tan⁻¹ a_{y} / aₓ

               θ = tan⁻¹ (-9.8) / 3.2

               θ = -71.9⁰

Movement is two-dimensional type with acceleration in both axes

The magnitude of the electric field is 400,000 N/C directed to the right. The perpendicular distance from the wall to the ball is 0.0417 m. After the string is cut, the ball will accelerate at 10.29 m/s² and follow a parabolic trajectory.

To solve this problem, we will break it down into several steps.

1. Calculate the Electric Field

The electric force on the ball is given by:

Fe = qE

Where Fe is the electric force, q is the charge of the ball, and E is the electric field.

Given: Fe = 0.032 N and q = -80 μC = -80 × 10⁻⁶ C

Rearranging the formula to solve for E:

E = Fe / q = 0.032 N / (-80 × 10⁻⁶ C)

E = -400,000 N/C

So, the magnitude of the electric field is 400,000 N/C (since we take the absolute value) and it is directed to the right on the x-axis as the ball has a negative charge.

2. Determine the Perpendicular Distance from the Wall

The ball is in equilibrium under electric and gravitational fields. To find the distance from the wall, we need to consider the forces and the angle of the thread.

The electric field is horizontal (along x-axis) and the gravitational force is vertical (along y-axis). The thread is at an angle of 8° with the vertical.

tan(θ) = Fe / Fg

Where θ = 8°, Fe = 0.032 N, and Fg = mg = 0.01 kg × 9.8 m/s² = 0.098 N

tan(8°) = 0.032 / 0.098 ≈ 0.327

The perpendicular distance from the wall (d) is:

d = L sin(θ) = 0.30 m × sin(8°) ≈ 0.30 m × 0.139 ≈ 0.0417 m

3. Calculate the Resulting Acceleration When the String is Cut

When the string is cut, only the electric field will act on the ball horizontally and gravity will act vertically.

Horizontally (ax): ax = Fe / m = 0.032 N / 0.01 kg = 3.2 m/s²

Vertically (ay): ay = g = 9.8 m/s²

The magnitude of the resulting acceleration (a) is:

[tex]\[a = \sqrt{a_x^2 + a_y^2} = \sqrt{(3.2)^2 + (9.8)^2} \approx 10.29 \, \text{m/s}^2\][/tex]

4. Describe the Resulting Path of the Ball

The ball will follow a parabolic path due to the combined effects of the horizontal electric field and vertical gravitational acceleration, similar to projectile motion.

Answer:

0.64 m from the first charge

Explanation:

Force is given by

[tex]F_1=\dfrac{kq_1q_2}{r^2}\\\Rightarrow F_1=\dfrac{k2\times 10^{-9}\times 3.7\times 10^{-9}}{x^2}[/tex]

[tex]F_2=\dfrac{kq_2q_3}{r^2}\\\Rightarrow F_1=\dfrac{k4.5\times 10^{-9}\times 3.7\times 10^{-9}}{(1.6-x)^2}[/tex]

These forces are equal

[tex]\dfrac{k2\times 10^{-9}\times 3.7\times 10^{-9}}{x^2}=\dfrac{k4.5\times 10^{-9}\times 3.7\times 10^{-9}}{(1.6-x)^2}\\\Rightarrow \dfrac{2}{x^2}=\dfrac{4.5}{(1.6-x)^2}\\\Rightarrow \dfrac{2}{4.5}=\dfrac{x^2}{(1.6-x)^2}\\\Rightarrow \dfrac{4.5}{2}=\dfrac{(1.6-x)^2}{x^2}\\\Rightarrow \sqrt{\dfrac{4.5}{2}}=\dfrac{1.6-x}{x}\\\Rightarrow 1.5=\dfrac{1.6-x}{x}\\\Rightarrow 1.5x+x=1.6\\\Rightarrow x=\dfrac{1.6}{2.5}\\\Rightarrow x=0.64\ m[/tex]

The distance that charge should be placed is 0.64 m from the first charge

The charge +3.70 x [tex]10^{-9[/tex]C should be placed at approximately 0.96 meters from the origin for the net electric force to be zero. The solution involves balancing the Coulomb forces from two charges placed at different points. The calculations involve setting up the equation for the forces and solving it step-by-step.

Finding the Point Where the Net Electric Force is Zero

To determine the point where the charge +3.70 x [tex]10^{-9[/tex] C should be placed so that the net electric force on it is zero, we need to consider the forces exerted by both charges +2.00 x [tex]10^{-9[/tex] C (at the origin) and +4.50 x [tex]10^{-9[/tex] C (at x = 1.6 m).

Step-by-Step Explanation:

Let the position where the net force is zero be at distance x from the origin (charge +2.00 x [tex]10^{-9[/tex] C).

The distance from the charge +4.50 x [tex]10^{-9[/tex] C to this point will then be (1.6 - x) meters.

According to Coulomb's Law, the force due to a charge is given by F = k * |q1 * q2| / [tex]r^2[/tex], where k is the Coulomb constant (8.98755 x[tex]10^{9[/tex] N m²/C²).

For the net force on the charge +3.70 x [tex]10^{-9[/tex] C to be zero, the magnitudes of the forces due to the two other charges must be equal:

Set these forces equal to each other:

(8.98755 x[tex]10^{9[/tex] N m²/C²) * (2.00 x [tex]10^{-9[/tex]C) * (3.70 x [tex]10^{-9[/tex] C) / x² = (8.98755 x [tex]10^{9[/tex] N m²/C²) * (4.50 x [tex]10^{-9[/tex] C) * (3.70 x [tex]10^{-9[/tex] C) / (1.6 - x)²

Simplify and solve for x:

(2.00 x [tex]10^{-9[/tex]) / x² = (4.50 x [tex]10^{-9[/tex]) / (1.6 - x)²

2 / x² = 4.5 / (1.6 - x)²

x² / (1.6 - x)² = 2 / 4.5

x² / (1.6 - x)² = 0.444

To solve for x, take the square root of both sides:

Finally, solve the equation x = √0.444 * (1.6 - x):

Thus, the charge +3.70 x [tex]10^{-9[/tex] C should be placed at approximately 0.96 meters from the origin to have the net electric force on it be zero.

The final temperature of the coffee is approximately [tex]\( 57.21 \, ^\circ C \)[/tex] (rounded to 2 significant figures).

Given equations:

[tex]\[ Q_c = Q_a \][/tex]

[tex]\[ m_c \cdot c_w \cdot (T_{ic} - T_f) = m_a \cdot c_a \cdot (T_f - T_{ia}) \][/tex]

Substitute the given values:

[tex]\[ 0.5 \cdot 1000 \cdot (85 - T_f) = 0.2 \cdot 900 \cdot (T_f - (-20)) \][/tex]

Now, distribute and simplify:

[tex]\[ 42500 - 500 \cdot T_f = 180 \cdot T_f + 3600 \][/tex]

Combine like terms:

[tex]\[ 500 \cdot T_f + 180 \cdot T_f = 42500 - 3600 \][/tex]

[tex]\[ 680 \cdot T_f = 38900 \][/tex]

Now, solve for A [tex]\( T_f \):[/tex]

[tex]\[ T_f = \frac{38900}{680} \][/tex]

[tex]\[ T_f \approx 57.21 \, ^\circ C \][/tex]

So, the final temperature of the coffee is approximately [tex]\( 57.21 \, ^\circ C \)[/tex] (rounded to 2 significant figures).

Answer:

P = 17.28*10⁶ N

Explanation:

Given

L = 250 mm = 0.25 m

a = 0.54 m

b = 0.40 m

E = 95 GPa = 95*10⁹ Pa

σmax = 80 MPa = 80*10⁶ Pa

ΔL = 0.12%*L = 0.0012*0.25 m = 3*10⁻⁴ m

We get A as follows:

A = a*b = (0.54 m)*(0.40 m) = 0.216 m²

then, we apply the formula

ΔL = P*L/(A*E)  ⇒ P = ΔL*A*E/L

⇒  P = (3*10⁻⁴ m)*(0.216 m²)*(95*10⁹ Pa)/(0.25 m)

⇒  P = 24624000  N = 24.624*10⁶ N

Now we can use the equation

σ = P/A

⇒  σ = (24624000  N)/(0.216 m²) = 114000000 Pa = 114 MPa > 80 MPa

So σ > σmax  we use σmax

⇒  P = σmax*A = (80*10⁶ Pa)*(0.216 m²) = 17280000 N = 17.28*10⁶ N

Answer:

The resultant velocity of the boat is 3.6 mil/h

Explanation:

given information:

flowing rate to the east (positive x-axis), v₁ = 2 mil/h (0.2) = 2i

boat's speed to the north (positive y-axis), v₂ = 3 mil/h (3,0) = 3j

the resultant velocity of the boat:

[tex]v_{R}[/tex] = √3²+2²

    = √13

    = 3.6 mil/h

The resultant velocity of the boat is approximately 3.61 mph, making an angle of 56.31° north of east. This is calculated using the Pythagorean theorem for magnitude and the inverse tangent function for direction.

To find the resultant velocity of the boat when it heads north across a river at a rate of 3 miles per hour with an eastward current at 2 miles per hour, we can use vector addition. The velocity of the boat heading north (Vboat) is perpendicular to the velocity of the river's current (Vriver).

The northward velocity can be considered as the y-component (Vy = 3 mph), and the eastward current as the x-component (Vx = 2 mph). The resultant velocity (Vtot) is the vector sum of these two components and can be calculated using the Pythagorean theorem:

Vtot = √(Vx2 + Vy2)

Plugging in the values we get:

Vtot = √(22 + 32)

Vtot = √(4 + 9)

Vtot = √13

Vtot = 3.61 mph (approximately)

To find the direction of the resultant velocity, we can use the inverse tangent function (tan-1) to calculate the angle (θ) the resultant velocity vector makes with the positive x-axis (eastward direction).

θ = tan-1(Vy/Vx)

θ = tan-1(3/2)

θ = 56.31° (approximately)

Since the boat is heading north and the current flows east, the resultant velocity makes an angle of 56.31° north of east.

Answer:

After [tex]t=4.77\tau[/tex] voltage across the capacitor will be 0.85 % of the initial voltage across the capacitor

Explanation:

Let initially voltage across capacitor is [tex]v_0[/tex]

After discharging the voltage across the capacitor is .85% of its initial voltage

So final voltage [tex]v=0.0085v_0[/tex]

We know that voltage across capacitor is given by [tex]v=v_0e^{\frac{-t}{\tau }}[/tex]

So [tex]0.85v_0=v_0e^{\frac{-t}{\tau }}[/tex]

[tex]e^{\frac{-t}{\tau }}=0.0085[/tex]

[tex]{\frac{-t}{\tau }}=ln0.0085[/tex]

[tex]-t=-4.77\tau[/tex]

[tex]t=4.77\tau[/tex]

So after [tex]t=4.77\tau[/tex] voltage across the capacitor will be 0.85 % of the initial voltage across the capacitor

Time constant of capacitive circuit is equal to the ratio of resistance and capacitance

The number of time constants the voltage across a discharging cap has decayed to 0.85% of its' initial value is;

4.768τ

In RC time constants, the formula for voltage across capacitor when discharging is given by;

v = v₀e^(-t/τ)

where;

v₀ = initial voltage on the capacitor

v = the voltage after time t

t = time in seconds

τ = time constant

We are told that the capacitor decayed to 0.85% of its' initial vale. Thus;

v = 0.0085v₀

Thus;

0.0085v₀ = v₀e^(-t/τ)

v₀ will cancel out to give;

0.0085 = e^(-t/τ)

In 0.0085 = -t/τ

-4.768 = -t/τ

Since we want to find number of time constants, then let us make t the subject to get;

t = 4.768τ

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The magnitude of the vector sum of the electric forces exerted on particle 3 is 0.598 N, with the direction angle being 90 degrees, as both forces act along the positive y-axis due to symmetry.

To solve the student's question, we'll have to calculate the forces exerted on particle 3 by particle 1 and particle 2, respectively, and then find the vector sum of these forces. Since both particle 1 and particle 2 are on the x-axis at the same distance from the origin and carry the same charge, the forces they exert on particle 3 will be equal in magnitude and opposite in direction along the y-axis. Therefore, the resulting force on particle 3 will be twice the force exerted by one particle along the y-axis.

Using the formula:
F = k × |q₁ × q₂| / r²
For particle 1 or 2 exerting force on particle 3:
F = (8.9875 × 10⁹) × (3.0 × 10⁻⁹ C × 9.0 × 10⁻⁶ C) / (0.09 m)²
F = 8.9875 × 10⁹ × 2.7 × 10⁻¹⁴ C / 0.0081 m²
F = 0.299 N (Force exerted by one particle)
The total force on particle 3 will be twice this value: 0.598 N.

The direction angle of the vector sum will be 90 degrees since both forces act along the y-axis and hence the resultant force is directed straight up along the positive y-axis when measured counterclockwise from the positive x-axis.

Magnitude is √2 N, Direction is 45° counterclockwise from positive x-axis.

Coulomb's Law: formula for electric force between point charges.

[tex]\[ F = \frac{k \cdot |q_1 \cdot q_2|}{r^2} \][/tex]

[tex]\( q_1 = q_2 = 3.0 \, \text{nC} = 3.0 \times 10^{-9} \, \text{C} \),[/tex]

[tex]\( q_3 = 9.0 \, \text{μC} = 9.0 \times 10^{-6} \, \text{C} \),[/tex]

The distance between particle 3 and particles 1 and 2 is [tex]\( 90 \, \text{mm} = 0.09 \, \text{m} \)[/tex],

All charges are located on the x-axis.

The electric force exerted by particle 1 on particle 3:

[tex]\[ F_1 = \frac{k \cdot |q_1 \cdot q_3|}{r^2} \][/tex]

The electric force exerted by particle 2 on particle 3:

[tex]\[ F_2 = \frac{k \cdot |q_2 \cdot q_3|}{r^2} \][/tex]

Particles 1 and 2, equidistant from 3, have equal electric forces. The magnitude of each force:

[tex]\[ F_1 = \frac{(8.99 \times 10^9) \cdot |3.0 \times 10^{-9} \cdot 9.0 \times 10^{-6}|}{(0.09)^2} \][/tex]

[tex]\[ F_1 \approx \frac{8.09 \times 10^{-5}}{0.0081} \][/tex]

[tex]\[ F_1 \approx 1.00 \, \text{N} \][/tex]

Similarly, [tex]\( F_2 \approx 1.00 \, \text{N} \)[/tex].

Now, the total electric force [tex]\( F_{\text{total}} \)[/tex] exerted on particle 3 is the vector sum of [tex]\( F_1 \) and \( F_2 \)[/tex].

[tex]\[ F_{\text{total}} = \sqrt{F_1^2 + F_2^2} \][/tex]

[tex]\[ F_{\text{total}} = \sqrt{(1.00)^2 + (1.00)^2} \][/tex]

[tex]\[ F_{\text{total}} = \sqrt{2} \, \text{N} \][/tex]

Magnitude of total electric force on particle 3 is [tex]\( \sqrt{2} \, \text{N} \)[/tex].

[tex]\( \theta \)[/tex] is the angle from x-axis to [tex]\( F_{\text{total}} \)[/tex].

[tex]\[ \theta = \arctan\left(\frac{F_2}{F_1}\right) \][/tex]

[tex]\[ \theta = \arctan\left(\frac{1.00}{1.00}\right) \][/tex]

[tex]\[ \theta = \arctan(1) \][/tex]

[tex]\[ \theta \approx 45^\circ \][/tex]

Answer:

The answer to your question is v₁ = 5.74 m/s

Explanation:

Data

v₁ = ?

h₁ = 0 m

v₂ = 0.75 m/s

h₂ = 1.65 m

g = 9.81 m/s²

Formula

                mgh₁   +  1/2mv₁²   =   mgh₂  +  1/2mv₂²

mass is not consider (if we factor mass, it is cancelled)

                        gh₁ + 1/2v₁²  =   gh₂  +  1/2v₂²

Substitution

                        (9.81)(0) + 1/2v₁² = (9.81)(1.65) + 1/2(0.75)²

Simplification

                                 0    +  1/2v₁² = 16.19 + 0.28

Solve for v₁

                                            1/2v₁² = 16.47

                                                 v₁² = 2(16.47)

                                                 v₁² = 32.94

Result

                                                v₁ = 5.74 m/s

Minimum speed must 5.74 m/s in order to lift his center of mass 1.65 m and cross the bar with a speed of 0.75 m/s.

Given here,

v₁ - initial velocity = ?

h₁ - initial height = 0 m

v₂ - final velocity = 0.75 m/s

h₂ - final height = 1.65 m

g - gravitational acceleration = 9.81 m/s²

The speed can be calculated by using the formula,

[tex]\bold { mgh_1 + \dfrac 12 mv_1^2 = mgh_2 + \dfrac 12mv_2^2}[/tex]            

factor the mass,

 [tex]\bold { gh_1 + \dfrac 12 v_1^2 = gh_2 + \dfrac 12v_2^2}[/tex]

put the values in the formula, and solve it for V1

[tex]\bold { (9.81)(0) + \dfrac 12v_1^2 = (9.81)(1.65) + \dfrac 12(0.75)^2}\\\\\bold { \dfrac 12v_1^2 = 16.19 + 0.28}\\\\\bold { \dfrac 12v_1^2 = 16.47}\\\\ \bold {v_1^2 = 2(16.47)}\\\\\bold {v_1^2= 32.94}\\\\\bold { v_1 = 5.74\ m/s}[/tex]

Therefore,  minimum speed must 5.74 m/s in order to lift his center of mass 1.65 m and cross the bar with a speed of 0.75 m/s.

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Answer:

3.72 cm

Explanation:

the coordiante of the negative charge =( 0,0)

the  coordinate of the positive charge = (0,3.75)

the distance simply is  3.72-0 = 3.72 cm

Answer:

Distance = 0.738 m

Explanation:

Solution:  

First convert Km/h into m/s.  

75 km/h * 1000 m/km * 1 hr/3600 sec = 20.8333 m/s  

According to third equation of motion:

[tex]Vf^{2}[/tex] – [tex]Vi^{2}[/tex] = 2 * acceleration * distance

Vf= final velocity

Vi= initial velocity

putting values in third equation of motion....

[tex]0.2^{2}[/tex] – [tex]20.8333^{2}[/tex] = 2 * (-30 * 9.8) * distance

note:

negative sign is due to deceleration

Distance = 0.738 m

Given an initial speed of 75km/h and a maximum deceleration of 30 G's, the front end of the car must be designed to collapse over a distance of approximately 0.73m to prevent injuries that could be fatal in an accident.

The problem is asking for the distance over which the front end of the car must be designed to collapse if a crash brings the car to rest, given a certain initial speed and maximum deceleration. This is a problem of kinematics, which is a branch of physics. To solve it, we can use the equation d = (v_f^2 - v_i^2) / (2*a), where:

So, substituting these values into our equation, we find that the front end of the car must be designed to collapse over a distance of approximately 0.73 meters to prevent injuries that could be fatal.

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Answer:

The ratio is 0.53

Explanation:

There aren't external torques on the ice skater because we can assume ice surface frictionless and there're not external forces, that implies angular moment is conserved so it has the same value before and after him pulling his arms. Angular momentum is the product between angular velocity (ω) and moment of inertia (I).

The initial angular moment is:

[tex]L=I_{i}\omega_{i} [/tex] (1)

and final angular moment is:

[tex] L=I_{f}\omega_{f} [/tex] (2)

Because conservation of angular moment we can equate (1) and (2):

[tex]I_{i}\omega_{i}=I_{f}\omega_{f} [/tex]

rearranging the expression:

[tex]\frac{I_{f}}{I_{i}}=\frac{\omega_{i}}{\omega_{f}} [/tex]

So, the ratio of the skater's final moment of inertia to his initial moment of inertia is:

[tex] \frac{I_{f}}{I_{i}}=\frac{3.17\frac{rad}{s}}{5.96\frac{rad}{s}}[/tex]

[tex] \frac{I_{f}}{I_{i}}=0.53[/tex]

The required ratio of of the skater's final moment of inertia to his initial moment of inertia is 0.53 : 1.

Given data:

The initial angular speed of the ice skater is, [tex]\omega = 3.17 \;\rm rad/s[/tex].

The final angular speed of the ice skater is, [tex]\omega ' =5.96 \;\rm rad/s[/tex].

In this problem, we can apply the conservation of the angular momentum, which says that in the absence of external torque, the initial angular momentum is equal to the final angular momentum.

Initial angular momentum = final angular momentum

[tex]L = L'\\I \times \omega = I' \times \omega'[/tex]

Here,

I is the initial moment of inertia.

I' is the final moment of inertia.

Solving as,

[tex]\dfrac{I'}{I} = \dfrac{ \omega}{ \omega'} \\\\\dfrac{I'}{I} = \dfrac{3.17}{ 5.96}\\\\\dfrac{I'}{I} = 0.53[/tex]

Thus, the required ratio of of the skater's final moment of inertia to his initial moment of inertia is 0.53 : 1.

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Answer: 0.07meter

Pointing to the left.

Explanation:

The Electric potential V of the proton is 3.8v. while the Electric Field strength is 56V/m. We first calculate the distance r the proton would travel

E= V/r

r = V/E

r = 3.8V/(56V/m)

r =0.07m.

The proton which is positively charged will move in the direction of the Electric Field that is to the left.

When a proton enters a region with an electric field pointing to the left, it will experience a force in the opposite direction. The force on the proton can be calculated using the formula F = q*E. The distance the proton will get before coming to a stop can be calculated using the formula d = (1/2)at^2.

When a proton enters a region with an electric field pointing to the left, it will experience a force in the opposite direction. The magnitude of the force can be determined using the formula:

F = q*E

Where F is the force, q is the charge of the proton, and E is the magnitude of the electric field. In this case, the force on the proton is given by:

F = (1.6 x 10^-19 C)(56 N/C) = -8.96 x 10^-18 N

The negative sign indicates that the force is in the opposite direction of the electric field. Since the proton is moving to the right, it will slow down and eventually come to a stop. The distance the proton will get can be calculated using the formula:

d = (1/2)at^2

Where d is the distance, a is the acceleration, and t is the time. Since the proton starts from rest, its initial velocity is zero. The acceleration can be calculated using the formula:

a = F/m

Where a is the acceleration, F is the force, and m is the mass of the proton. The mass of a proton is approximately 1.67 x 10^-27 kg. Plugging in the values, we get:

a = (-8.96 x 10^-18 N) / (1.67 x 10^-27 kg) = -5.38 x 10^8 m/s^2

Substituting this value of acceleration and the given initial velocity into the distance formula, we can calculate the distance the proton will get:

d = (1/2)(-5.38 x 10^8 m/s^2)(t^2)

Where t is the time for which the proton travels. The time can be calculated using the formula:

t = v/a

Where t is the time, v is the initial velocity, and a is the acceleration. Plugging in the values, we get:

t = (3.8 m/s) / (-5.38 x 10^8 m/s^2) = -7.08 x 10^-9 s

Since the time cannot be negative, we take the magnitude of the time:

t = 7.08 x 10^-9 s

Substituting this value of time into the distance formula, we get:

d = (1/2)(-5.38 x 10^8 m/s^2)((7.08 x 10^-9 s)^2) = 1.05 x 10^-15 m

Therefore, the proton will get a distance of 1.05 x 10^-15 meters before coming to a stop.

Answer:

the electron jump goes from the negative terminal to the positive terminal

Explanation:

In a battery the energy is produced by a chemical reaction, which accumulates electrons in a terminal, called a negative terminal, but by convention the current goes from the positive to the negative terminal.

   We can see from this explanation that the negative wire is the one with an excess is electrons, so the electron jump goes from the negative terminal to the positive terminal

Answer:

2 and 3

Explanation:

The right answer is the option 2 and 3,

This is all about the electrons transfer from one material to the other material.

For example if the electrons in the valence shell of one material are loosely attached then the other material's atoms try to take those electron to complete their shells and that is how the charges transfers from one another.

And it could also be happen as in option 2.

Answer:

[tex]F_C=20800\ N[/tex]

Explanation:

Given:

During the motion of the car on a flat circular track there acts a force which pushes the car inwards to comply in the circular motion. This force is called centripetal force which is generated due to the frictional force between the road and the tyres of the car.

This centripetal force is given as:

[tex]F_C=m.\frac{v^2}{r}[/tex]

here: r = radius of the track = [tex]\frac{d}{2}[/tex]

[tex]F_C=1300\times \frac{40^2}{100}[/tex]

[tex]F_C=20800\ N[/tex] is the force which makes the car turn inwards going with the given uniform speed.

Final answer:

The net force on a 1300 kg car driving around a 200-m diameter track at 40 m/s is calculated using the centripetal force formula, resulting in a force of 20,800 N directed towards the center of the track.

Explanation:

The question asks to calculate the magnitude of the net force acting on a 1300 kg car driving around a 200-m diameter circular track at a speed of 40 m/s.

First, we convert the diameter to radius by dividing by 2, giving a radius (r) of 100 m. To find the net force (Fnet), we use the formula for centripetal force: Fnet = m * (v² / r), where m is mass, v is velocity, and r is radius.

Substituting the given values:

Mass (m) = 1300 kg

Velocity (v) = 40 m/s

Radius (r) = 100 m

This yields:

Fnet = 1300 kg * (40 m/s)² / 100 m

Calculating this, we get:

Fnet = 1300 kg * 1600 m²/s² / 100 m

Fnet = 1300 kg * 16 m/s²

The net force is 20,800 N directed towards the center of the circular track.

Answer:

The right option is C. 1/16 N

Explanation:

From coulomb's law,

F = kqq'/r²................ Equation 1

Where F = force between the charges, q = first charge, q' = second charge, r = distance of separation between the the charges, k = constant of proportionality.

qq' = Fr²/k............... Equation 2

Given: F = 1 N, r = 2 cm = 0.02 m

Substituting into equation 2, to get the value of the product of the charges in terms of k

qq' = 1×0.02²/k

qq' = 0.0004/k.................... Equation 3

When they are moved to a new separation of 8 cm,

Then r = 8 cm = 0.08 m

F = kqq'/0.08².................. Equation 4

Substituting the value of qq' in equation 3 into equation 4

F = k(0.0004k)/0.0064

F = 4/64

F = 1/6 N

Hence the electric force on each of the is 1/16 N

The right option is C. 1/16 N

Answer:

1.5 Hz

0.67 s

4 m

[tex]\pi[/tex]

2.82842 m

Explanation:

The equation is

[tex]x=4cos(3\pi t+\pi)[/tex]

It is of the form

[tex]x=Acos(2\pi ft+\phi)[/tex]

Comparing the equations we get

[tex]3\pi=2\pi f\\\Rightarrow f=\dfrac{3}{2}\\\Rightarrow f=1.5\ Hz[/tex]

Frequency is 1.5 Hz

Time period is given by

[tex]T=\dfrac{1}{f}\\\Rightarrow T=\dfrac{1}{1.5}\\\Rightarrow T=0.67\ s[/tex]

The period of the motion is 0.67 s

Amplitude

[tex]A=4\ m[/tex]

Amplitude is 4 m

Phase constant

[tex]\phi=\pi[/tex]

The phase constant is [tex]\pi[/tex]

At t = 0.25 s

[tex]x=4cos(3\pi t+\pi)\\\Rightarrow x=4cos(3\pi\times 0.25+\pi)\\\Rightarrow x=2.82842\ m[/tex]

The position of the particle is 2.82842 m

Final answer:

To determine the magnitude of the electric field through which an electron narrowly misses the upper plate, apply principles of kinematics and the formula for electric force. Calculate the time of flight using the horizontal motion, and use this to establish the vertical acceleration due to electric force. From the electron's charge and acceleration, derive the electric field magnitude.

Explanation:

To calculate the magnitude of the electric field through which an electron passes and just misses the upper plate upon exit, we need to apply the concepts of kinematics and electric forces. Since the electrical force is the only vertical force acting on the electron, it will cause a vertical acceleration according to Newton's second law (F = ma), where the force F is the electric force (F = qE) and the acceleration a is due to this force.

The vertical distance traveled by the electron just before it exits the field is equal to half the distance between the plates, which is 0.5 cm (or 0.005 m). The time it takes for the electron to travel this vertical distance can be calculated by using the initial horizontal speed and the horizontal distance (2 cm or 0.02 m) it travels. Using the formula t = d/v (where d is the horizontal distance and v is the horizontal velocity), we find t = 0.02 m / 5.35×10¶6 m/s = 3.74×10¶8 seconds.

The vertical acceleration a can be found by using the vertical distance (s), initial vertical speed (u which is 0 in this case), and time (t), using the equation s = ut + 0.5at². From this, we find a = 2s/t². Knowing the charge of an electron (e = 1.6×10¶19 C) and its mass (m = 9.11×10¶31 kg), the electric field E can then be calculated as E = F/q = ma/q. Plugging in the values, we get the magnitude of the electric field.

To calculate the angular velocity of the solid metal disk, we can use the equation for rotational motion: τ = Iα. We are given the torque applied to the disk (4.28 N) and the angle it is turned (3.32°). Using the formulas for moment of inertia and angular acceleration, we can calculate the final angular velocity.

To calculate the angular velocity of the solid metal disk, we can use the equation for rotational motion:

τ = Iα

Where τ is the torque applied to the disk, I is the moment of inertia of the disk, and α is the angular acceleration of the disk.

We are given the torque applied to the disk (4.28 N) and the angle it is turned (3.32°). We can calculate the moment of inertia of the disk using the formula:

I = (1/2)mr^2

Where m is the mass of the disk and r is the radius of the disk.

Using the given diameter (29.0 cm), we can calculate the radius (14.5 cm or 0.145 m).

Plugging in the values into the formulas, we can solve for the angular acceleration:

α = τ / I

Finally, we can use the equation for angular motion to find the angular velocity:

ω = ω0 + αt

Since the disk is released from rest, the initial angular velocity (ω0) is 0. Plugging in the values, we can solve for the final angular velocity.