momentum is closely related to

Answer:

Explanation:

The equilibrium position will be in between q₁ and q₂ .

Let this position of third charge be x distance from q₁

q₁ will pull the third charge towards it with force F₁

F₁ =9 x 10⁹x 5 x 10⁻⁹x15 x 10⁻⁹ / x²

= 675 x 10⁻⁹ / x²

q₂ will pull the third charge towards it with force F₂

F₂ =9 x 10⁹x 2 x 10⁻⁹x15 x 10⁻⁹ /( .40-x )²

= 270 x 10⁻⁹ / ( .40-x )²

For equilibrium

675 x 10⁻⁹ / x² = 270 x 10⁻⁹ / ( .40-x )²

5  / x² = 2 / ( .40-x )²

( .40-x )² / x² = 2/5 = .4

.4 - x / x = .632

.4 - x = .632x

.4 = 1.632 x

x = .245 .

24.5 cm

so third charge must be placed at 24.5 cm away from q₁ charge.

Answer:

accelerated

Explanation:

The motion of the body where the acceleration is constant is known as uniformly accelerated motion. The value of the acceleration does not change with the function of time.

Newton's second law of motion

Answer:

F = 800N

the magnitude of the average force exerted on the wall by the ball is 800N

Explanation:

Applying the impulse-momentum equation;

Impulse = change in momentum

Ft = m∆v

F = (m∆v)/t

Where;

F = force

t = time

m = mass

∆v = v2 - v1 = change in velocity

Given;

m = 0.80 kg

t = 0.050 s

The ball strikes the wall horizontally with a speed of 25 m/s, and it bounces back with this same speed.

v2 = 25 m/s

v1 = -25 m/s

∆v = v2 - v1 = 25 - (-25) m/s = 25 +25 = 50 m/s

Substituting the values;

F = (m∆v)/t

F = (0.80×50)/0.05

F = 800N

the magnitude of the average force exerted on the wall by the ball is 800N

Answer:

Explanation:

Given that:

length of side , a = 0.5 m

charge , q = 6.65 mC

length of diagonal , d = 0.5 * sqrt(2)

d = 0.707 m

F is the force due to adjacent particle ,

F1 is the force due to diagonal particle

Now , for the net charge on a particle

Fnet = 2 * F * cos(45) + F1

Fnet = 2*cos(45) * k * q^2/a^2 + k * q^2/d^2

Fnet = 9*10^9 * 0.00665^2 * (2* cos(45)/.5^2 + 1/.707^2)

Fnet = 3.05 *10^6 N

the magnitude of net force acting on each particle is 3.05 *10^6 N

part B)

for the direction of particle

d) along the line between the charge and the center of the square outward of the center

Answer:

[tex]\Delta H_{tot} = 2258.025\,kJ[/tex]

Explanation:

The amount of heat released from water is equal to the sum of latent and sensible heats. Let suppose that water is initially at a temperature of [tex]25^{\circ}C[/tex]. Then:

[tex]\Delta H_{tot} = \Delta H_{s, w} + \Delta H_{f,w} + \Delta H_{s,i}[/tex]

[tex]\Delta H_{tot} = n\cdot (c_{w}\cdot \Delta T_{w} + L_{f} + c_{i}\cdot \Delta T_{i})[/tex]

Finally, the amount of heat released from water is now computed by replacing variables:

[tex]\Delta H_{tot} = (1\,mol)\cdot \left[\left(75.3\,\frac{kJ}{mol\cdot K} \right)\cdot (25^{\circ}C-0^{\circ}C)+ 6.025\,\frac{kJ}{mol} + \left(37.7\,\frac{kJ}{mol\cdot K} \right)\cdot (0 + 10^{\circ}C)\right][/tex][tex]\Delta H_{tot} = 2258.025\,kJ[/tex]

Answer:

   W₃ = 3310.49 J ,  W3 = 3310.49 J

Explanation:

We can solve this exercise in parts, the first with acceleration, the second with constant speed and the third with deceleration. Therefore it is work we calculate it in these three sections

We start with the part with acceleration, the distance traveled is y = 5.90 m and the final speed is v = 2.30 m / s. Let's calculate the acceleration with kinematics

       v2 = v₀² + 2 a₁ y

as they rest part of the rest the ricial speed is zero

        v² = 2 a₁ y

        a₁ = v² / 2y

        a₁ = 2.3² / (2 5.90)

        a₁ = 0.448 m / s²

with this acceleration we can calculate the applied force, using Newton's second law

         F -W = m a₁

         F = m a₁ + m g

         F = m (a₁ + g)

         F = 69 (0.448 + 9.8)

         F = 707.1 N

Work is defined by

         W₁ = F.y = F and cos tea

As the force lifts the man, this and the displacement are parallel, therefore the angle is zero

          W₁ = 707.1 5.9

           W₁ = 4171.89 J  W3 = 3310.49 J

Let's calculate for the second part

the speed is constant, therefore they relate it to zero

           F - W = 0

           F = W

           F = m g

           F = 60 9.8

           F = 588 A

the job is

           W² = 588 5.9

            W2 = 3469.2 J

finally the third part

in this case the initial speed is 2.3 m / s and the final speed is zero

           v² = v₀² + 2 a₂ y

            0 = vo2₀² + 2 a₂ y

            a₂ = -v₀² / 2 y

            a₂ = - 2.3²/2 5.9

            a2 = - 0.448 m / s²

we calculate the force

            F - W = m a₂

            F = m (g + a₂)

            F = 60 (9.8 - 0.448)

            F = 561.1 N

we calculate the work

            W3 = F and

            W3 = 561.1 5.9

          W3 = 3310.49 J

total work

          W_total = W1 + W2 + W3

          W_total = 4171.89 +3469.2 + 3310.49

           w_total = 10951.58 J

Answer:

Approximately [tex]25\; \rm cm[/tex], assuming that the permeability between the two wires is constant, and that the two wires are of infinite lengths.

Explanation:

Note that [tex]y_1 < y_2[/tex], which means that the first wire (with current [tex]I_1[/tex]) is underneath the second wire (with current [tex]I_2[/tex].) Let [tex]y[/tex] denote the [tex]y[/tex]-coordinate (in [tex]\rm cm[/tex]) of a point of interest. The two wires partition this region into three parts:

Apply the right-hand rule to find the direction of the magnetic field in each of the three regions. Assume that [tex]I_1[/tex] points to the left while [tex]I_2[/tex] points to the right. Let [tex]B_1[/tex] and [tex]B_2[/tex] denote the magnetic field due to [tex]I_1[/tex] and [tex]I_2[/tex], respectively.

The (net) magnetic field on this plane would be zero only in regions where [tex]B_1[/tex] and[tex]B_2[/tex] points in opposite directions. That rules out the region between the two wires.

At a distance of [tex]R[/tex] away from a wire with current [tex]I[/tex] and infinite length, the formula for the magnitude of the magnetic field [tex]B[/tex] due to that wire is:

[tex]\displaystyle B = \frac{\mu\, I}{2\pi\, R}[/tex],

where [tex]\mu[/tex] is the permeability of the space between the wire and the point of interest (should be constant.) The exact value of [tex]\mu[/tex] does not affect the answer to this question, as long as it is constant throughout this region.

Note that the value of [tex]R[/tex] in this formula is supposed to be positive. Let [tex]R_1[/tex] and [tex]R_2[/tex] denote the distance between the point of interest and the two wires, respectively.

Make sure that given the corresponding range of [tex]y[/tex], these distances are all positive.

For the net magnetic field to be zero at a certain [tex]y[/tex]-value, the strength of the two magnetic fields at that point should match. That is:

[tex]B_1 = B_2[/tex].

[tex]\displaystyle \frac{\mu\, I_1}{4\pi\, R_1} = \frac{\mu\, I_2}{4\pi\, R_2}[/tex].

Simplify this equation:

[tex]\displaystyle \frac{I_1}{R_1} = \frac{I_2}{R_2}[/tex].

These two equations give the same result: [tex]y = 25[/tex]. However, based on the respective assumptions on the value of [tex]y[/tex], this value corresponds to the region above the second wire.

Answer:

I think the distance should be given in the question because the formula for getting the

average speed= total distance/total time taken

Answer:

735.75 N

Explanation:

We are given;

Area of small lift; A1 = 5 cm²

Area of large lift; A2 = 300 cm²

Mass required to lift large lift;m = 4500 kg

Since Force = mg, then, Force required to lift large lift;F2 = 4500 × 9.81 = 44145 N

By Pascal's law,

F1/A1 = F2/A2

We want to find minimum force that must be applied to the small piston;F1, so let's make F1 the subject.

Thus;

F1 = (F2 × A1)/A2

Plugging in the relevant values, we have;

F1 = (44145 × 5)/300

F1 = 735.75 N

b) 16 cm

Magnification, m = v/u

3 = v/u

⇒ v = 3u

Lens formula : 1/v – 1/u = 1/f

1/3u = 1/u = 1/12

-2/3u = 1/12

⇒ u = -8 cm

V = 3 × (-8) = -24

Distance between object and image = u – v = -8 – (-24) = -8 + 24 = 16 cm

Answer:

Ask a pro,bro

Explanation:

I just ryhmed

Answer:

V₀ₓ = 10.94 m/s

V₀y = 18.87 m/s

Explanation:

To find the launch velocity, we use 1st equation of motion.

Vf = Vi + at

where,

Vf = Final Velocity of Ball = Launch Speed = V₀ = ?

Vi = Initial Velocity = 0 m/s (Since ball was initially at rest)

a = acceleration = 376 m/s²

t = time = 0.058 s

Therefore,

V₀ = 0 m/s + (376 m/s²)(0.058 s)

V₀ = 21.81 m/s

Now, for x-component:

V₀ₓ = V₀ Cos θ

where,

V₀ₓ = x-component of launch velocity = ?

θ = Angle with horizontal = 59.9⁰

V₀ₓ = (21.81 m/s)(Cos 59.9°)

V₀ₓ = 10.94 m/s

for y-component:

V₀ₓ = V₀ Sin θ

where,

V₀y = y-component of launch velocity = ?

θ = Angle with horizontal = 59.9⁰

V₀y = (21.81 m/s)(Sin 59.9°)

V₀y = 18.87 m/s

Complete Question

The complete question iws shown on the first uploaded image  

Answer:

a

    [tex]y \to z \to x[/tex]

b

  [tex]x \to z \to y[/tex]

Explanation:

Now looking at the diagram let take that the magnetic field is moving in the x-axis

 Now the magnetic force is mathematically represented as

             [tex]F = I L[/tex] x B

Note (The x is showing cross product )

Note the force(y-axis) is perpendicular to the field direction (x-axis)

Now when the loop is swinging forward

 The motion of the loop is  from   y to z to to x to y

Now since the force is perpendicular to the motion(velocity) of the loop

Hence the force would be from z to y and back to z  

and from lenze law the induce current opposes the force so the direction will be from y to z to x

Now when the loop is swinging backward

   The motion of the induced current will now be   x to z to y

Answer:

Explanation:

When sheet A is removed

[tex]I_B=32\cos^230=24W/m^2\\\\I_C=24 \cos^260=6W/m^2\\\\I_D=6\cos^230=4.5W/m^2[/tex]

When sheet B is removed

[tex]I_A=32\cos^20=32W/m^2\\\\I_C=24 \cos^290=0W/m^2\\\\I_D=0\cos^230=0W/m^2[/tex]

When sheet C is removed

[tex]I_A=32\cos^20=32W/m^2\\\\I_D=32 \cos^230=24W/m^2\\\\I_B=24\cos^290=0W/m^2[/tex]

When sheet D is removed

[tex]I_A=32\cos^20=32W/m^2\\\\I_B=32\cos^230=24W/m^2\\\\I_C=24\cos^260=6W/m^2[/tex]

Before the engines fail, the rocket's altitude at time t is given by

[tex]y_1(t)=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2[/tex]

and its velocity is

[tex]v_1(t)=80.6\dfrac{\rm m}{\rm s}+\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t[/tex]

The rocket then reaches an altitude of 1150 m at time t such that

[tex]1150\,\mathrm m=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2[/tex]

Solve for t to find this time to be

[tex]t=11.2\,\mathrm s[/tex]

At this time, the rocket attains a velocity of

[tex]v_1(11.2\,\mathrm s)=124\dfrac{\rm m}{\rm s}[/tex]

When it's in freefall, the rocket's altitude is given by

[tex]y_2(t)=1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2[/tex]

where [tex]g=9.80\frac{\rm m}{\mathrm s^2}[/tex] is the acceleration due to gravity, and its velocity is

[tex]v_2(t)=124\dfrac{\rm m}{\rm s}-gt[/tex]

(a) After the first 11.2 s of flight, the rocket is in the air for as long as it takes for [tex]y_2(t)[/tex] to reach 0:

[tex]1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies t=32.6\,\mathrm s[/tex]

So the rocket is in motion for a total of 11.2 s + 32.6 s = 43.4 s.

(b) Recall that

[tex]{v_f}^2-{v_i}^2=2a\Delta y[/tex]

where [tex]v_f[/tex] and [tex]v_i[/tex] denote final and initial velocities, respecitively, [tex]a[/tex] denotes acceleration, and [tex]\Delta y[/tex] the difference in altitudes over some time interval. At its maximum height, the rocket has zero velocity. After the engines fail, the rocket will keep moving upward for a little while before it starts to fall to the ground, which means [tex]y_2[/tex] will contain the information we need to find the maximum height.

[tex]-\left(124\dfrac{\rm m}{\rm s}\right)^2=-2g(y_{\rm max}-1150\,\mathrm m)[/tex]

Solve for [tex]y_{\rm max}[/tex] and we find that the rocket reaches a maximum altitude of about 1930 m.

(c) In part (a), we found the time it takes for the rocket to hit the ground (relative to [tex]y_2(t)[/tex]) to be about 32.6 s. Plug this into [tex]v_2(t)[/tex] to find the velocity before it crashes:

[tex]v_2(32.6\,\mathrm s)=-196\frac{\rm m}{\rm s}[/tex]

That is, the rocket has a velocity of 196 m/s in the downward direction as it hits the ground.

Answer:

c

Explanation:

Answer:

(i) v = 44 m/s

(ii) a = 72 m/s^2

Explanation:

You have the following equation for the potion of a car:

[tex]x=20t+6t^4[/tex]

(i) The instantaneous velocity is the derivative of x in time:

[tex]\frac{dx}{dt}=20+(6)(4)t^3=20+24t^3[/tex]

for t = 1 is:

[tex]v(t=1)=\frac{dx}{dt}=20+24(1)^3=44m/s[/tex]

(ii) The instantaneous acceleration is the derivative of the velocity:

[tex]\frac{dv}{dt}=24(3)t^2=72t^2[/tex]

for t = 1

[tex]a(t=1)=\frac{dv}{dt}=72(1)^2=72m/s^2[/tex]

Answer:

E = 9.62*10^-8 J

Explanation:

The energy stored in a capacitor is given by the following formula:

[tex]E=\frac{1}{2}CV^2[/tex]     (1)

E: energy stored

C: capacitance

V: potential difference of the capacitor = 4 V

The capacitance for a concentric cylindrical capacitor is:

[tex]C=\frac{2\pi \epsilon_o L}{ln(\frac{r_2}{r_1})}[/tex]     (2)

L: length of the capacitor = 150m

r2: radius of the outer cylinder  = 8mm = 8*10^-3m

r1: radius of the inner cylinder = 4mm = 4*10^-3m

εo: dielectric permittivity of vacuum = 8.85*10^-12C^2/Nm^2

You replace the expression (2) into the equation (1) and replace the values of all parameters:

[tex]E=\frac{1}{2}(\frac{2\pi \epsilon_o L}{ln(\frac{r_2}{r_1})})V^2\\\\E=\frac{\pi \epsilon_o L}{ln(\frac{r_2}{r_1})}V^2\\\\E=\frac{\pi (8.85*10^{-12}C^2/Nm^2)(150m)}{ln(\frac{8*10^{-3}m}{4*10^{-3}m})}(4V)^2\\\\E=9.62*10^{-8}J[/tex]

The energy stored in the cylindrical capacitor is 9.62*10-8 J

Answer:

the answer is D.

Explanation: Mark me brainlest

Answer:

a) 20.29N

b) 19.43N

c) 15N

Explanation:

To find the magnitude of the resultant vectors you first calculate the components of the vector for the angle in between them, next, you sum the x and y component, and finally, you calculate the magnitude.

In all these calculations you can asume that one of the vectors coincides with the x-axis.

a)

[tex]F_R=(9cos(30\°)+12)\hat{i}+(9sin(30\°))\hat{j}\\\\F_R=(19.79N)\hat{i}+(4.5N)\hat{j}\\\\|F_R|=\sqrt{(19.79N)^2+(4.5N)^2}=20.29N[/tex]

b)

[tex]F_R=(9cos(45\°)+12)\hat{i}+(9sin(45\°))\hat{j}\\\\F_R=(18.36N)\hat{i}+(6.36N)\hat{j}\\\\|F_R|=\sqrt{(18.36N)^2+(6.36N)^2}=19.43N[/tex]

c)

[tex]F_R=(9cos(90\°)+12)\hat{i}+(9sin(90\°))\hat{j}\\\\F_R=(12N)\hat{i}+(9N)\hat{j}\\\\|F_R|=\sqrt{(12N)^2+(9N)^2}=15N[/tex]

Answer:

vₓ = xg/2y

Explanation:

In this question, let us  find the time it takes for the ball on the right that has zero initial velocity to reach the ground.

By newton equation of motion we know that

y = v₀ t - ½ g t²

t = 2y / g

This is the time it takes for the ball on the right to reach the ground; at this time the ball on the left travels a distance

vₓ = x/t

vₓ = xg/2y

vₓ = xg/2y

Where we assume that x and y are known.

Answer:

The answer is 3.0

Explanation:

Answer:

The height of the image will be "1.16 mm".

Explanation:

The given values are:

Object distance, u = 25 cm

Focal distance, f = 1.8 cm

On applying the lens formula, we get

⇒  [tex]\frac{1}{v} -\frac{1}{u} =\frac{1}{f}[/tex]

On putting estimate values, we get

⇒  [tex]\frac{1}{v} -\frac{1}{(-25)} =\frac{1}{1.8}[/tex]

⇒  [tex]\frac{1}{v} =\frac{1}{1.8} -\frac{1}{25}[/tex]

⇒  [tex]v=1.94 \ cm[/tex]

As a result, the image would be established mostly on right side and would be true even though v is positive.

By magnification,

[tex]m=\frac{v}{u}[/tex] and [tex]m=\frac{h_{1}}{h_{0}}[/tex]

⇒  [tex]\frac{v}{u} =\frac{h_{1}}{h_{0}}[/tex]

⇒  [tex]\frac{1.94}{25}=\frac{{h_{1}}}{15}[/tex]

⇒  [tex]{h_{1}}=1.16 \ mm[/tex]

An object with momentum must also have impulse.

An object with momentum must also have: c. impulse.

Momentum can be defined as the product of the mass possessed by an object and its velocity. Also, it is a vector quantity and as such has both magnitude and direction.

Mathematically, momentum is giving by the formula;

[tex]Momentum = Mass \times Velocity[/tex]

In Physics, the impulse experienced by an object is always equal to the change in momentum of the object. This is due to the force acting on an object.

[tex]Impulse = Change\;in\;momentum\\\\Force \times time = m \Delta V[/tex]

In conclusion, an object with momentum must also have impulse.

Read more: https://brainly.com/question/15517471

Answer:

8924.6

Explanation:

We are given that

Viscosity of air,[tex]\eta=1.81\times 10^{-5}kg/m-s[/tex]

Density of air,[tex]\rho=1.21kg/m^3[/tex]

Diameter,d=0.15 m

v=0.89m/s

We have to find the Reynold's number.

Reynold's number,R=[tex]\frac{\rho vd}{\eta}[/tex]

Substitute the values then we get

[tex]R=\frac{1.21\times 0.89\times 0.15}{1.81\times 10^{-5}}[/tex]

R=[tex]8924.6[/tex]

Hence, the value of Reynold's number=8924.6

Answer:

c 250 kg-m/s

Explanation:

happy to help!!

Answer: 250 kg

Explanation:

Answer:

The shortest distance in which you can stop the automobile by locking the brakes is 53.64 m

Explanation:

Given;

coefficient of kinetic friction, μ = 0.84

speed of the automobile, u = 29.0 m/s

To determine the  the shortest distance in which you can stop an automobile by locking the brakes, we apply the following equation;

v² = u² + 2ax

where;

v is the final velocity

u is the initial velocity

a is the acceleration

x is the shortest distance

First we determine a;

From Newton's second law of motion

∑F = ma

F is the kinetic friction that opposes the motion of the car

-Fk = ma

but, -Fk = -μN

-μN = ma

-μmg = ma

-μg = a

- 0.8 x 9.8 = a

-7.84 m/s² = a

Now, substitute in the value of a in the equation above

v² = u² + 2ax

when the automobile stops, the final velocity, v = 0

0 = 29² + 2(-7.84)x

0 = 841 - 15.68x

15.68x = 841

x = 841 / 15.68

x = 53.64 m

Thus, the shortest distance in which you can stop the automobile by locking the brakes is 53.64 m

Answer: The value of the dielectric constant k = 1.8

Explanation:

If C= ε A/d and

Electrostatic energy W = 1/2CV^2

Substitutes C in the first formula into the energy formula.

W = 1/2 ε A/d × V^2

Let us remember that electric field strength E is the ratio of potential V to the distance d. Where V = Ed

Substitute V = Ed into the energy W.

W = 1/2 × ε A/d ×( Ed )^2

W = 1/2 × ε A/d × E^2 × d^2

d will cancel one of the ds

W = 1/2 × ε Ad × E^2

W/Ad = 1/2 × ε × E^2

W/V = 1/2 × ε E^2

Where Ad = volume V

E = dielectric strength

εo = permittivity of free space = 8.84 x 10^-12 F/m

W/V = 2800 J/m^3

Let first calculate the dielectric strength

2800 = 1/2 × 8.84×10^-12 × E^2

5600 = 8.84×10^-12E^2

E^2 = 5600/8.84×10^-12

E = sqrt( 6.3 × 10^14)

E = 25 × 10^7

75% of E = 18.9 × 10^6Jm

The permittivity of the material will be achieved by using the same formula

2800 = 1/2 × ε E^2

2800 = 0.5 × ε × (18.9×10^6)^2

2800 = ε × 1.78 × 10^14

ε = 2800/1.78×10^14

ε = 1.57 × 10^-11

Dielectric constant k = relative permittivity

Relative permittivity is the ratio of the permittivity of the material to the permittivity of the vacuum in a free space. That is

k = 1.57×10^-11/8.84×10^-12

k = 1.776

k = 1.8 approximately

Therefore, the value of the dielectric constant k is 1.8

Answer:

a) a = 2.35 m/s^2

Explanation:

(a) In order to calculate the magnitude of the acceleration of the ball, you use the following formula, for the position of the ball:

[tex]x=v_ot+\frac{1}{2}at^2[/tex]     (1)

x: position of the ball after t seconds = 87 m

t: time  = 8.6 s

a: acceleration of the ball = ?

vo: initial velocity of the ball = 0 m/s

You solve the equation (1) for a:

[tex]x=0+\frac{1}{2}at^2\\\\a=\frac{2x}{t^2}[/tex]

You replace the values of the parameters in the previous equation:

[tex]a=\frac{2(87m)}{(8.6s)^2}=2.35\frac{m}{s^2}[/tex]

The acceleration of the ball is 2.35 m/s^2

Hope it helps

Good luck on your assignment

we have seven groups in the periodic table