A small, rigid object carries positive and negative 3.00 nC charges. It is oriented so that the positive charge has coordinates (−1.20 mm, 1.20 mm) and the negative charge is at the point (1.70 mm, −1.30 mm).

Required:
a. Find the electric dipole moment of the object.
b. The object is placed in an electric field E = (7.80 103 î − 4.90 103 ĵ). Find the torque acting on the object.
c. Find the potential energy of the object–field system when the object is in this orientation.
d. Assuming the orientation of the object can change, find the difference between the maximum and the minimum potential energies of the system,

Explanation:

A ball is thrown straight upward and falls back to Earth. It means that it is coming to the initial position. Displacement is given by the difference of final position and initial position. The displacement of the ball will be 0. As a result velocity will be 0.

Acceleration is equal to the rate of change of velocity. So, its acceleration is also equal to 0.

Hence, displacement, velocity and acceleration are zero.

Answer:

a) v₃ = 19.54 km, b)  70.2º north-west

Explanation:

This is a vector exercise, the best way to solve it is finding the components of each vector and doing the addition

vector 1 moves 26 km northeast

let's use trigonometry to find its components

         cos 45 = x₁ / V₁

         sin 45 = y₁ / V₁

         x₁ = v₁ cos 45

         y₁ = v₁ sin 45

         x₁ = 26 cos 45

         y₁ = 26 sin 45

         x₁ = 18.38 km

         y₁ = 18.38 km

Vector 2 moves 45 km north

        y₂ = 45 km

Unknown 3 vector

          x3 =?

          y3 =?

Vector Resulting 70 km north of the starting point

           R_y = 70 km

we make the sum on each axis

X axis

      Rₓ = x₁ + x₃

       x₃ = Rₓ -x₁

       x₃ = 0 - 18.38

       x₃ = -18.38 km

Y Axis

      R_y = y₁ + y₂ + y₃

       y₃ = R_y - y₁ -y₂

       y₃ = 70 -18.38 - 45

       y₃ = 6.62 km

the vector of the third leg of the journey is

         v₃ = (-18.38 i ^ +6.62 j^ ) km

let's use the Pythagorean theorem to find the length

         v₃ = √ (18.38² + 6.62²)

         v₃ = 19.54 km

to find the angle let's use trigonometry

           tan θ = y₃ / x₃

           θ = tan⁻¹ (y₃ / x₃)

           θ = tan⁻¹ (6.62 / (- 18.38))

           θ = -19.8º

with respect to the x axis, if we measure this angle from the positive side of the x axis it is

          θ’= 180 -19.8

          θ’= 160.19º

I mean the address is

          θ’’ = 90-19.8

          θ = 70.2º

70.2º north-west

Answer:

D

Explanation:

Answer:

It is D

Explanation: No cap

Answer:

   θ = 4.78º

with respect to the vertical or 4.78 to the east - north

Explanation:

This is a velocity compound exercise since it is a vector quantity.

The plane takes a direction, the air blows to the west and the result must be to the north, let's use the Pythagorean theorem to find the speed

                  v_fly² = v_nort² + v_air²

                  v_nort² = v_fly² + - v_air²

Let's use trigonometry to find the direction of the plane

        sin θ = v_air / v_fly

        θ = sin⁻¹ (v_air / v_fly)

let's calculate

        θ = sin⁻¹ (10/120)

         θ = 4.78º

with respect to the vertical or 4.78 to the north-east

Answer:

  r₂ = 1,586 m

Explanation:

For this problem we are going to solve it by parts, let's start by finding the sound intensity when we are 25 m

         β = 10 log (I / I₀)

where Io is the sensitivity threshold 10⁻¹² W / m²

          I₁ / I₀ = [tex]e^{\beta/10}[/tex]

          I₁ = I₀  e^{\beta/10}

let's calculate

          I₁ = 10⁻¹² e^{25/10}

          I₁ = 1.20 10⁻¹¹ W / m²

the other intensity in exercise is

          I₂ = 10⁻¹² e^{80/10}

          I₂ = 2.98 10⁻⁹ W / m²

now we use the definition of sound intensity

          I = P / A

where P is the emitted power that is a constant and A the area of ​​the sphere where the sound is distributed

         P = I A

the area a sphere is

         A = 4π r²

we can write this equation for two points of the found intensities

          I₁ A₁ = I₂ A₂

where index 1 corresponds to 25m and index 2 to the other distance

          I₁ 4π r₁² = I₂ 4π r₂²

          I₁ r₁² = I₂ r₂²

           r₂ = √ (I₁ / I₂) r₁

let's calculate

           r₂ = √ (1.20 10⁻¹¹ / 2.98 10⁻⁹) 25

           r₂ = √ (0.40268 10⁻²) 25

           r₂ = 1,586 m

Answer:

If the particle is an electron [tex]E_y = 3.311 * 10^3 N/C[/tex]

If the particle is a proton, [tex]E_y = 6.08 * 10^6 N/C[/tex]

Explanation:

Initial speed at the origin, [tex]u = 3 * 10^6 m/s[/tex]

[tex]\theta = 38^0[/tex] to +ve x-axis

The particle crosses the x-axis at , x = 1.5 cm = 0.015 m

The particle can either be an electron or a proton:

Mass of an electron, [tex]m_e = 9.1 * 10^{-31} kg[/tex]

Mass of a proton, [tex]m_p = 1.67 * 10^{-27} kg[/tex]

The electric field intensity along the positive y axis [tex]E_y[/tex], can be given by the formula:

[tex]E_y = \frac{2 m u^2 sin \theta cos \theta}{qx} \\[/tex]

If the particle is an electron:

[tex]E_y = \frac{2 m_e u^2 sin \theta cos \theta}{qx} \\[/tex]

[tex]E_y = \frac{2 * 9.1 * 10^{-31} * (3*10^6)^2 *(sin38)( cos38)}{1.6*10^{-19} * 0.015} \\[/tex]

[tex]E_y = 3311.13 N/C\\E_y = 3.311 * 10^3 N/C[/tex]

If the particle is a proton:

[tex]E_y = \frac{2 m_p u^2 sin \theta cos \theta}{qx} \\[/tex]

[tex]E_y = \frac{2 * 1.67 * 10^{-27} * (3*10^6)^2 *(sin38)( cos38)}{1.6*10^{-19} * 0.015} \\[/tex]

[tex]E_y = 6.08 * 10^6 N/C[/tex]

Answer:

mass 20 times of an amazing and all its motion

Answer:

a) WT = 137.5 J

b) v2 = 2.34 m/s

Explanation:

a) The total work done on the block is given by the following formula:

[tex]W_T=F_pd-F_fd=(F_p-F_f)d[/tex]          (1)

Fp: force parallel to the displacement of the block = 150N

Ff: friction force

d: distance = 5.0 m

Then, you first calculate the friction force by using the following relation:

[tex]F_f=\mu_k N=\mu_k Mg[/tex]        (2)

μk: coefficient of kinetic friction = 0.25

M: mass of the block = 50kg

g: gravitational constant = 9.8 m/s^2

Next, you replace the equation (2) into the equation (1) and solve for WT:

[tex]W_T=(F_p-\mu_kMg)d=(150N-(0.25)(50kg)(9.8m/s^2))(5.0m)\\\\W_T=137.5J[/tex]

The work done over the block is 137.5 J

b) If the block started from rest, you can use the following equation to calculate the final speed of the block:

[tex]W_T=\Delta K=\frac{1}{2}M(v_2^2-v_1^2)[/tex]     (3)

WT: total work = 137.5 J

v2: final speed = ?

v1: initial speed of the block = 0m/s

You solve the equation (3) for v2:

[tex]v_2=\sqrt{\frac{2W_T}{M}}=\sqrt{\frac{2(137.5J)}{50kg}}=2.34\frac{m}{s}[/tex]

The final speed of the block is 2.34 m/s

Answer:

blue  θ₂ = 22.26º

red    θ₂ = 22.79º

Explanation:

When a light beam passes from one material medium to another, it undergoes a deviation from the path, described by the law of refraction

         n₁ sin θ₁ = n₂ sin θ₂

where n₁ and n₂ are the incident and transmitted media refractive indices and θ are the angles in the media

let's apply this equation to each wavelength

λ = blue

in this case n₁ = 1, n₂ = 1,645

       sin θ₂ = n₁/ n₂ sin₂ θ₁

let's calculate

       sin θ₂ = 1 / 1,645 sint 38.55

       sin θ₂ = 0.37884

       θ₂ = sin⁻¹ 0.37884

       θ₂ = 22.26º

λ = red

n₂ = 1,609

         sin θ₂ = 1 / 1,609 sin 38.55

         sin θ₂ = 0.3873

         θ₂ = sim⁻¹ 0.3873

         θ₂ = 22.79º

the refracted rays are between these two angles

Answer:

A) M

Explanation:

The three blocks are set in series on a horizontal frictionless surface, whose mutual contact accelerates all system to the same value due to internal forces as response to external force exerted on the box of mass M (Newton's Third Law). Let be F the external force, and F' and F'' the internal forces between boxes of masses M and 2M, as well as between boxes of masses 2M and 3M. The equations of equilibrium of each box are described below:

Box with mass M

[tex]\Sigma F = F - F' = M\cdot a[/tex]

Box with mass 2M

[tex]\Sigma F = F' - F'' = 2\cdot M \cdot a[/tex]

Box with mass 3M

[tex]\Sigma F = F'' = 3\cdot M \cdot a[/tex]

On the third equation, acceleration can be modelled in terms of F'':

[tex]a = \frac{F''}{3\cdot M}[/tex]

An expression for F' can be deducted from the second equation by replacing F'' and clearing the respective variable.

[tex]F' = 2\cdot M \cdot a + F''[/tex]

[tex]F' = 2\cdot M \cdot \left(\frac{F''}{3\cdot M} \right) + F''[/tex]

[tex]F' = \frac{5}{3}\cdot F''[/tex]

Finally, F'' can be calculated in terms of the external force by replacing F' on the first equation:

[tex]F - \frac{5}{3}\cdot F'' = M \cdot \left(\frac{F''}{3\cdot M} \right)[/tex]

[tex]F = \frac{5}{3} \cdot F'' + \frac{1}{3}\cdot F''[/tex]

[tex]F = 2\cdot F''[/tex]

[tex]F'' = \frac{1}{2}\cdot F[/tex]

Afterwards, F' as function of the external force can be obtained by direct substitution:

[tex]F' = \frac{5}{6}\cdot F[/tex]

The net forces of each block are now calculated:

Box with mass M

[tex]M\cdot a = F - \frac{5}{6}\cdot F[/tex]

[tex]M\cdot a = \frac{1}{6}\cdot F[/tex]

Box with mass 2M

[tex]2\cdot M\cdot a = \frac{5}{6}\cdot F - \frac{1}{2}\cdot F[/tex]

[tex]2\cdot M \cdot a = \frac{1}{3}\cdot F[/tex]

Box with mass 3M

[tex]3\cdot M \cdot a = \frac{1}{2}\cdot F[/tex]

As a conclusion, the box with mass M experiments the smallest net force acting on it, which corresponds with answer A.

Complete question is;

After doing some exercises on the floor, you are lying on your back with one leg pointing straight up. If you allow your leg to fall freely until it hits the floor, what is the tangential speed of your foot just before it lands? Assume the leg can be treated as a uniform rod x = 0.98 m long that pivots freely about the hip.

Answer:

Tangential speed of foot just before it lands is; v = 5.37m/s

Explanation:

Let U (potential energy) be zero on the ground.

So, initially, U = mgh

where, h = 0.98/2 = 0.49m (midpoint of the leg)

Now just before the leg hits the floor it would have kinetic energy as;

K = ½Iω²

where ω = v/r and I = ⅓mr²

So, K = ½(⅓mr²)(v/r)²

K = (1/6) × (mr²)/(v²/r²)

K = (1/6) × mv²

From principle of conservation of energy, we have;

Potential energy = Kinetic energy

Thus;

mgh = (1/6) × mv²

m will cancel out to give;

gh = (1/6)v²

Making v the subject, we have;

v = √6gh

v = √(6 × 9.81 × 0.49)

v = √28.8414

v = 5.37m/s

Answer:

a

  [tex]KE = 7.17 *10^{7} \ J[/tex]

b

 [tex]t = 6411.09 \ s[/tex]

Explanation:

From the question we are told that

    The radius of the flywheel is  [tex]r = 1.50 \ m[/tex]

      The mass of the flywheel is [tex]m = 430 \ kg[/tex]

          The rotational speed of the flywheel is [tex]w = 5,200 \ rev/min = 5200 * \frac{2 \pi }{60} =544.61 \ rad/sec[/tex]

      The power supplied by the motor is  [tex]P = 15.0 hp = 15 * 746 = 11190 \ W[/tex]

     Generally the moment of inertia of the flywheel is  mathematically represented as

       [tex]I = \frac{1}{2} mr^2[/tex]

substituting values

       [tex]I = \frac{1}{2} ( 430)(1.50)^2[/tex]

       [tex]I = 483.75 \ kgm^2[/tex]

The kinetic energy that is been stored is  

       [tex]KE = \frac{1}{2} * I * w^2[/tex]

substituting values

        [tex]KE = \frac{1}{2} * 483.75 * (544.61)^2[/tex]

        [tex]KE = 7.17 *10^{7} \ J[/tex]

Generally power is mathematically represented as

          [tex]P = \frac{KE}{t}[/tex]

=>      [tex]t = \frac{KE}{P}[/tex]

substituting the value

        [tex]t = \frac{7.17 *10^{7}}{11190}[/tex]

        [tex]t = 6411.09 \ s[/tex]

Answer:

v₀ = 0.5058 m/s

Explanation:

From the question, for the block to hit the bottle, the elastic potential energy of the spring at the bottle (x = 0.08 m) should be equal to the sum of the elastic potential energy of the spring at x = 0.05 m and the kinetic energy of block at x = 0.05 m

Now, the potential energy of the block at x = 0.08 m is ½kx²

where;

k is the spring constant given by; k = ω²m

ω is the angular velocity of the oscillation

m is the mass of the block.

Thus, potential energy of the spring at the bottle(x = 0.08 m) is;

U = ½ω²m(0.08m)²

Also, potential energy of the spring at the bottle(x = 0.05 m) is;

U = ½ω²m(0.05m)²

and the kinetic energy of the block at x = 0.05 m is;

K = ½mv₀²

Thus;

½ω²m(0.08)² = ½ω²m(0.05)² + ½mv₀²

Inspecting this, ½m will cancel out to give;

ω²(0.08)² = ω²(0.05)² + v₀²

Making v₀ the subject, we have;

v₀ = ω√((0.08)² - (0.05)²)

So,

v₀ = 8.1√((0.08)² - (0.05)²)

v₀ = 0.5058 m/s

Answer: Part 1: Propellant Fraction (MR) = 8.76

Part 2: Propellant Fraction (MR) = 1.63

Explanation: The Ideal Rocket Equation is given by:

Δv = [tex]v_{ex}.ln(\frac{m_{f}}{m_{e}} )[/tex]

Where:

[tex]v_{ex}[/tex] is relationship between exhaust velocity and specific impulse

[tex]\frac{m_{f}}{m_{e}}[/tex] is the porpellant fraction, also written as MR.

The relationship [tex]v_{ex}[/tex] is: [tex]v_{ex} = g_{0}.Isp[/tex]

To determine the fraction:

Δv = [tex]v_{ex}.ln(\frac{m_{f}}{m_{e}} )[/tex]

[tex]ln(MR) = \frac{v}{v_{ex}}[/tex]

Knowing that change in velocity is Δv = 9.6km/s and [tex]g_{0}[/tex] = 9.81m/s²

Note: Velocity and gravity have different measures, so to cancel them out, transform km in m by multiplying velocity by 10³.

Part 1: Isp = 450s

[tex]ln(MR) = \frac{v}{v_{ex}}[/tex]

ln(MR) = [tex]\frac{9.6.10^{3}}{9.81.450}[/tex]

ln (MR) = 2.17

MR = [tex]e^{2.17}[/tex]

MR = 8.76

Part 2: Isp = 2000s

[tex]ln(MR) = \frac{v}{v_{ex}}[/tex]

ln (MR) = [tex]\frac{9.6.10^{3}}{9.81.2.10^{3}}[/tex]

ln (MR) = 0.49

MR = [tex]e^{0.49}[/tex]

MR = 1.63

Answer:

The beam used is a negatively charged electron beam with a velocity of

v = E / B

Explanation:

After reading this long statement we can extract the data to work on the problem.

* They indicate that when the beam passes through the plates it deviates towards the positive plate, so the beam must be negative electrons.

* Now indicates that the electric field and the magnetic field are contracted and that the beam passes without deviating, so the electric and magnetic forces must be balanced

           [tex]F_{e} = F_{m}[/tex]

           q E = qv B

           v = E / B

this configuration is called speed selector

They ask us what type of beam was used.

The beam used is a negatively charged electron beam with a velocity of v = E / B

Answer:

a) Jack does more work uphill

b) Numerically, we can see that Jill applied the most power downhill

Explanation:

Jack's mass = 75 kg

Jill's mass = [tex]1.5x = 75[/tex]

Jill's mass = [tex]x = \frac{75}{1.5}[/tex] = 50 kg

distance up hill = 15 m

a) work done by Jack uphill = mgh

where g = acceleration due to gravity= 9.81 m/s^2

work = 75 x 9.81 x 15 = 11036.25 J

similarly,

Jill's work uphill = 50 x 9.81 x 15 = 7357.5 J

this shows that Jack does more work climbing up the hill

b) assuming Jack's time downhill to be t,

then Jill's time = [tex]\frac{t}{3}[/tex]

we recall that power is the rate in which work id done, i.e

P = [tex]\frac{work}{time}[/tex]

For Jack, power = [tex]\frac{11036.25}{t}[/tex]

For Jill, power =  [tex]\frac{3*7357.5}{t}[/tex] =  [tex]\frac{22072.5}{t}[/tex]

Numerically, we can see that Jill applied the most power downhill

Answer:

A. Physics has changed the course of the world.

Explanation:

Answer:

9.6 Ns

Explanation:

Note: From newton's second law of motion,

Impulse = change in momentum

I = m(v-u).................. Equation 1

Where I = impulse, m = mass of the ball, v = final velocity, u = initial velocity.

Given: m = 2.4 kg, v = 2.5 m/s, u = -1.5 m/s (rebounds)

Substitute into equation 1

I = 2.4[2.5-(-1.5)]

I = 2.4(2.5+1.5)

I = 2.4(4)

I = 9.6 Ns

The magnitude of impulse will be "9.6 Ns".

According to the question,

Mass,

Final velocity,

Initial velocity,

By using Newton's 2nd law of motion, we get

→ Impulse, [tex]I = m(v-u)[/tex]

By substituting the values, we get

                     [tex]= 2.4[2.5-(1.5)][/tex]

                     [tex]= 2.4(2.5+1.5)[/tex]

                     [tex]= 2.4\times 4[/tex]

                     [tex]= 9.6 \ Ns[/tex]

Thus the above answer is right.    

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Answer:

μ_k = 0.1773

Explanation:

We are given;

Initial velocity;u = 20 m/s

Final velocity;v = 0 m/s (since it comes to rest)

Distance before coming to rest;s = 115 m

Let's find the acceleration using Newton's second law of motion;

v² = u² + 2as

Making a the subject, we have;

a = (v² - u²)/2s

Plugging relevant values;

a = (0² - 20²)/(2 × 115)

a = -400/230

a = -1.739 m/s²

From the question, the only force acting on the puck in the x direction is the force of friction. Since friction always opposes motion, we see that:

F_k = −ma - - - (1)

We also know that F_k is defined by;

F_k = μ_k•N

Where;

μ_k is coefficient of kinetic friction

N is normal force which is (mg)

Since gravity acts in the negative direction, the normal force will be positive.

Thus;

F_k = μ_k•mg - - - (2)

where g is acceleration due to gravity.

Thus,equating equation 1 and 2,we have;

−ma = μ_k•mg

m will cancel out to give;

-a = μ_k•g

μ_k = -a/g

g has a constant value of 9.81 m/s², so;

μ_k = - (-1.739/9.81)

μ_k = 0.1773

The coefficient of kinetic friction between the hockey puck and ice is equal to 0.178

Given the following data:

Scientific data:

To determine the coefficient of kinetic friction between the hockey puck and ice:

First of all, we would calculate the acceleration of the hockey puck by using the third equation of motion.

[tex]V^2 = U^2 + 2aS\\\\0^2 =20^2 + 2a(115)\\\\-400=230a\\\\a=\frac{-400}{230}[/tex]

Acceleration, a = -1.74 [tex]m/s^2[/tex]

Note: The negative signs indicates that the hockey puck is slowing down or decelerating.

From Newton's Second Law of Motion, we have:

[tex]\sum F_x = F_k + F_n =0\\\\F_k =- F_n\\\\\mu mg =-ma\\\\\mu = \frac{-a}{g}\\\\\mu = \frac{-(-1.74)}{9.8}\\\\\mu = \frac{1.74}{9.8}[/tex]

Coefficient of kinetic friction = 0.178

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Answer:

The velocity of the bird is [tex]v_f = 4.87 \ m/s[/tex]

Explanation:

From the question we are told that  

    The mass of the bird  is [tex]m_1 = 300 \ g = 0.3 \ kg[/tex]

    The initial speed of the bird is  [tex]u_1 = 6.2 \ m/s[/tex]

     The mass of the insect is [tex]m_2 = 10 \ g = 0.01 \ kg[/tex]

       The speed of the insect is [tex]u_ 2 =-35 \ m/s[/tex]

The negative sign is because it is moving in opposite direction  to the bird

According to the principle of linear momentum conservation

       [tex]m_1 u_1 + m_2 u_2 = (m_1 + m_2 )v_f[/tex]

substituting values

        [tex](0.3 * 6.2 ) + (0.01 * (-35)) = (0.3 + 0.01 )v_f[/tex]

    [tex]1.51 = 0.31 v_f[/tex]

     [tex]v_f = 4.87 \ m/s[/tex]

The Final velocity of Bird =  4.87 m/s

Mass of the bird = 300 g = 0.3 kg

Velocity of bird = 6.2 m/s

Momentum of Bird = Mass of bird [tex]\times[/tex] Velocity of Bird = 0.3 [tex]\times[/tex] 6.2 =  1.86 kgm/s

Mass of the insect = 10 g = 0.01 kg

Velocity of insect =   - 35 m/s

Momentum of the Insect = Mass of Insect [tex]\times[/tex] Velocity of Insect = - 0.35  kgm/s

According to the law of conservation of momentum We can write that

In the absence of external forces on the system , the momentum of system remains conserved in that particular direction.

The bird opens the mouth and enjoys the free lunch  hence

Let the final velocity of bird is [tex]v_f[/tex]

Initial momentum of the system = Final momentum of the system

1.86 -0.35 = [tex]v_f[/tex] ( 0.01 + 0.3 )

1.51 =  [tex]v_f[/tex] 0.31

[tex]v_f[/tex] = 4.87 m/s

The Final velocity of Bird =  4.87 m/s

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Answer:

q1 = 7.6uC , -2.3 uC

q2 = 7.6uC , -2.3 uC

( q1 , q2 ) = ( 7.6 uC , -2.3 uC ) OR ( -2.3 uC , 7.6 uC )

Explanation:

Solution:-

- We have two stationary identical conducting spheres with initial charges ( q1 and q2 ). Such that the force of attraction between them was F = 0.6286 N.

- To model the electrostatic force ( F ) between two stationary charged objects we can apply the Coulomb's Law, which states:

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

Where,

                     k: The coulomb's constant = 8.99*10^9

- Coulomb's law assume the objects as point charges with separation or ( r ) from center to center.  

- We can apply the assumption and approximate the spheres as point charges under the basis that charge is uniformly distributed over and inside the sphere.

- Therefore, the force of attraction between the spheres would be:

                             [tex]\frac{F}{k}*r^2 =| q_1|.|q_2| \\\\\frac{0.6286}{8.99*10^9}*(0.5)^2 = | q_1|.|q_2| \\\\ | q_1|.|q_2| = 1.74805 * 10^-^1^1[/tex] ... Eq 1

- Once, we connect the two spheres with a conducting wire the charges redistribute themselves until the charges on both sphere are equal ( q' ). This is the point when the re-distribution is complete ( current stops in the wire).

- We will apply the principle of conservation of charges. As charge is neither destroyed nor created. Therefore,

                             [tex]q' + q' = q_1 + q_2\\\\q' = \frac{q_1 + q_2}{2}[/tex]

- Once the conducting wire is connected. The spheres at the same distance of ( r = 0.5m) repel one another. We will again apply the Coulombs Law as follows for the force of repulsion (F = 0.2525 N ) as follows:

                          [tex]\frac{F}{k}*r^2 = (\frac{q_1 + q_2}{2})^2\\\\\sqrt{\frac{0.2525}{8.99*10^9}*0.5^2} = \frac{q_1 + q_2}{2}\\\\2.64985*10^-^6 = \frac{q_1 + q_2}{2}\\\\q_1 + q_2 = 5.29969*10^-^6[/tex]  .. Eq2

- We have two equations with two unknowns. We can solve them simultaneously to solve for initial charges ( q1 and q2 ) as follows:

                         [tex]-\frac{1.74805*10^-^1^1}{q_2} + q_2 = 5.29969*10^-^6 \\\\q^2_2 - (5.29969*10^-^6)q_2 - 1.74805*10^-^1^1 = 0\\\\q_2 = 0.0000075998, -0.000002300123[/tex]

                          [tex]q_1 = -\frac{1.74805*10^-^1^1}{-0.0000075998} = -2.3001uC\\\\q_1 = \frac{1.74805*10^-^1^1}{0.000002300123} = 7.59982uC\\[/tex]

-- As she lands on the air mattress, her momentum is (m v)

Momentum = (60 kg) (5 m/s down) = 300 kg-m/s down

-- As she leaves it after the bounce,

Momentum = (60 kg) (1 m/s up) = 60 kg-m/s up

-- The impulse (change in momentum) is

Change = (60 kg-m/s up) - (300 kg-m/s down)

Magnitude of the change = 360 km-m/s

The direction of the change is up /\ .

The direction of a body or object's movement is defined by its velocity.In its basic form, speed is a scalar quantity.In essence, velocity is a vector quantity.It is the speed at which distance changes.It is the displacement change rate.

Solve the problem ?

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Answer:

Explanation:

(a)

The true atmospheric pressure will has more value than the reading in the barometer. If Parm is the atmospheric

pressure in the tube then the resulting vapour pressure is

Patm - pgh = Prapor

The final reading ion the barometer is

pgh = Palm - Proper

Hence, the true atmospheric pressure is greater.

you can find the answer in this book

physics principles with Applications, Global Edition Problem 67P: Chapter: CH 13 Problem:67p

Answer:

The bottom two lines.

Explanation:

They need their own line of voltage quantity. A parallel circuit has the definition of 'two or more paths for current to flow through.' The voltage does stay the same in each line.

Complete question:

The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and the process is reversible and adiabatic. Use constant specific heat at 300 K to find the exit velocity.

Answer:

The exit velocity is 629.41 m/s

Explanation:

Given;

initial temperature, T₁ = 1200K

initial pressure, P₁ = 150 kPa

final pressure, P₂ = 80 kPa

specific heat at 300 K, Cp = 1004 J/kgK

k = 1.4

Calculate final temperature;

[tex]T_2 = T_1(\frac{P_2}{P_1})^{\frac{k-1 }{k}[/tex]

k = 1.4

[tex]T_2 = T_1(\frac{P_2}{P_1})^{\frac{k-1 }{k}}\\\\T_2 = 1200(\frac{80}{150})^{\frac{1.4-1 }{1.4}}\\\\T_2 = 1002.714K[/tex]

Work done is given as;

[tex]W = \frac{1}{2} *m*(v_i^2 - v_e^2)[/tex]

inlet velocity is negligible;

[tex]v_e = \sqrt{\frac{2W}{m} } = \sqrt{2*C_p(T_1-T_2)} \\\\v_e = \sqrt{2*1004(1200-1002.714)}\\\\v_e = \sqrt{396150.288} \\\\v_e = 629.41 \ m/s[/tex]

Therefore, the exit velocity is 629.41 m/s

Answer:

  c.  the net work a spring does on a body is zero when the body returns to its initial position

Explanation:

A force is conservative when the net work done over any path that returns to the initial position is zero. Choice C matches that definition.

An ideal spring of the kind used in physics problems has the characteristic that it applies the same force at the same distance always. So any work required to extend or compress the spring is reversed when the reverse motion takes place.

Answer:

About 6.26m/s

Explanation:

[tex]mgh=\dfrac{1}{2}mv^2[/tex]

Divide both sides by mass:

[tex]gh=\dfrac{1}{2}v^2[/tex]

Since the point of equality of kinetic and potential energy will be halfway down the cliff, height will be 4/2=2 meters.

[tex](9.8)(2)=\dfrac{1}{2}v^2 \\\\v^2=39.4 \\\\v\approx 6.26m/s[/tex]

Hope this helps!

The gravitational potential energy (relative to the base of the cliff) be equal to its kinetic energy for speed of rock of 8.85 m/s.

Given data:

The height of vertical cliff is, h = 4.0 m.

Since, we are asked for speed by giving the condition for gravitational potential energy (relative to the base of the cliff) be equal to its kinetic energy. Then we can apply the conservation of energy as,

Kinetic energy = Gravitational potential energy

[tex]\dfrac{1}{2}mv^{2}=mgh[/tex]

Here,

m is the mass of rock.

v is the speed of rock.

g is the gravitational acceleration.

Solving as,

[tex]v=\sqrt{2gh}\\\\v=\sqrt{2 \times 9.8 \times 4.0}\\\\v =8.85 \;\rm m/s[/tex]

Thus, we can conclude that the gravitational potential energy (relative to the base of the cliff) be equal to its kinetic energy for speed of rock of 8.85 m/s.

Learn more about the conservation of energy here:

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Answer:

t = 17.68s

Explanation:

In order to calculate the total elapsed time that skydiver takes to reache the ground, you first calculate the distance traveled by the skydiver in the first 5.00s. You use the following formula:

[tex]y=y_o-v_ot-\frac{1}{2}gt^2[/tex]            (1)

y: height for a time t

yo: initial height = 1000m

vo: initial velocity = 0m/s

g: gravitational acceleration = 9.8m/s^2

t: time = 5.00 s

You replace the values of the parameters to get the values of the new height of the skydiver:

[tex]y=1000m-\frac{1}{2}(9.8m/s^2)(5.00s)^2\\\\y=877.5m[/tex]

Next, you take this value of 877.5m as the initial height of the second part of the trajectory of the skydiver. Furthermore, use the value of 7.00m/s as the initial velocity.

You use the same equation (1) with the values of the initial velocity and new height. We are interested in the time for which the skydiver arrives to the ground, then y = 0

[tex]0=877.5-7.00t-4.9t^2[/tex]       (2)

The equation (2) is a quadratic equation, you solve it for t with the quadratic formula:

[tex]t_{1,2}=\frac{-(-7.00)\pm \sqrt{(-7.00)^2-4(-4.9)(877.5)}}{2(-4.9)}\\\\t_{1,2}=\frac{7.00\pm 131.33}{-9.8}\\\\t_1=12.68s\\\\t_2=-14.11s[/tex]

You use the positive value of t1 because it has physical meaning.

Finally, you sum the times of both parts of the trajectory:

total time = 5.00s + 12.68s = 17.68s

The total elapsed time taken by the skydiver to arrive to the ground from the airplane is 17.68s