As an electron moves in the direction the electric field lines:________.
a. it is moving from low potential to high potential and gaining electric potential energy.
b. it is moving from low potential to high potential and losing electric potential energy.
c. it is moving from high potential to low potential and gaining electric potential energy.
d. it is moving from high potential to low potential and losing electric potential energy.
e. both its electric potential and electric potential energy remain constant.

Answer:

a) yes

b) no

c) yes

d)Cart 2 with mass [tex]\frac{M}{3}[/tex]   is expected to be more faster

e) u₂ = 48 m/s

Explanation:

a) the all out linear momentum of an arrangement of particles of Cart 1 not followed up on by external forces is constant.

b) the linear momentum of Cart 2 will be acted upon by external force by Cart 1 with mass M, thereby it's variable and the momentum is not conserved

c) yes, the momentum is conserved because no external force acted upon it and both Carts share the same velocity after the reaction

note: m₁u₁ + m₂u₂ = (m₁ + m₂)v

d) Cart 2 with mass [tex]\frac{M}{3}[/tex] will be faster than Cart 1 because Cart 2 is three times lighter than Cart 1.

e) Given

m₁=  M

u₁ = 16m/s

m₂ =[tex]\frac{M}{3}[/tex]

u₂ = ?

from law of conservation of momentum

m₁u₁= m₂u₂

M× 16 = [tex]\frac{M}{3}[/tex] × u₂(multiply both sides by 3)

therefore, u₂ = [tex]\frac{3(M .16)}{M}[/tex] ("." means multiplication)

∴u₂ = 3×16 = 48 m/s

Answer:

B

Explanation:

ANSEWER :B IN ROWS ONLY

Answer:

horizontal component = 10.54m/s

Explanation:

horizontal component = 13.2cos37°

horizontal component = 10.54m/s

Answer:

Since the angular acceleration of the objects will be proportional to the torque (due to gravity) acting on them and they will all experience the same torque their accelerations will be inversely proportional to their moments of inertia:

I disk = 1/2 M R^2

I sphere = 2/5 M R^2

I shell = 2/3 M R^2

Thus the sphere will experience the greatest angular acceleration and reach the bottom first, and then be followed by the disk and the shell.

By conservation of energy they will all have the same kinetic energy when they reach the bottom of the ramp.

(a) The ranking of the objects in order of how they will finish the race is

solid sphere > disk > hollow spherical shell

(b) The ranking of the objects in order of kinetic energy is

solid sphere > disk > hollow spherical shell

The moment of inertia of each object is calculated as follows;

The angular momentum of the objects is calculated as follows;

[tex]L =I \omega \\\\\omega = \frac{L}{I}[/tex]

The object with the least moment of inertia is will have the highest speed.

The ranking of the objects in order of how they will finish the race;

solid sphere > disk > hollow

The kinetic energy of the objects is calculated as follows;

[tex]K.E = \frac{1}{2} I \omega ^2[/tex]

The ranking of the objects in order of kinetic energy;

solid sphere > disk > hollow

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Complete question is;

(a) A light, rigid rod of length, l = 1.00 m joins two particles, with masses m = 4.00 kg and m, = 3.00 kg, at its ends. The combination rotates in the xy-plane about a pivot through the center of the rod (see figure below). Determine the angular momentum of the system about the origin when the speed of each particle is 6.4 m/s. (Enter the magnitude to at least two decimal places in kg. m/s.)

(b) What If? What would be the new angular momentum of the system (in kg. m/s) if each of the masses were instead a solid sphere 15.0 cm in diameter? (Round your answer to at least two decimal places.)

Answer:

A) L = 22.4 kg.m²/s and it's direction will be along the positive(+ve) z-axis

B) L = 3.36 kg.m²/s

Explanation:

The image is missing, so i have attached it.

Formula for the moment of Inertia would be; I = mr²

m1 = 4 kg

m2 = 3 kg

r = 1/2 = 0.5 m

So, sum of moment of inertia for the 2 masses would be;

I = (4 × 0.5²) + (3 × 0.5²)

I = 1.75 kg.m²

Now, angular velocity is given by the formula;

ω = v/r

We are given v = 6.4 m/s

So;

ω = 6.4/0.5

ω = 12.8 rad/s

Now, let's find angular momentum.

Angular momentum; L = Iω

L = 1.75 × 12.8

L = 22.4 kg.m²/s

Now, using the right hand rule, the direction will be along the positive(+ve) z-axis.

B) Now, the new diameter is 15 cm = 0.15 m

Thus,

radius;r = 0.15/2 = 0.075 m

Similar to a above;

I = (4 × 0.075²) + (3 × 0.075²)

I = 0.039375 kg.m²

ω = v/r

We are given v = 6.4 m/s

ω = 6.4/0.075

ω = 85.33 rad/s

Angular momentum; L = Iω

L = 0.039375 × 85.33

L = 3.36 kg.m²/s

The Complete question is;

(a) A light, rigid rod of length, l = 1.00 m joins two particles, with masses m = 4.00 kg and m, = 3.00 kg, at its ends. The combination rotates in the xy-plane about a pivot through the center of the rod (see figure below). Determine the angular momentum of the system about the origin when the speed of each particle is 6.4 m/s. (Enter the magnitude to at least two decimal places in kg. m/s.)

(b) What If? What would be the new angular momentum of the system (in kg. m/s) if each of the masses were instead a solid sphere 15.0 cm in diameter? (Round your answer to at least two decimal places.) Image check below.

A) L is = 22.4 kg.m²/s and it's direction will be along the positive(+ve) z-axis

B) L is = 3.36 kg.m²/s

We are applying the Formula for the moment of Inertia would be; I = mr²

Then, m1 = 4 kg

After that, m2 is = 3 kg

Now, r = 1/2 = 0.5 m

So, When the sum of the moment of inertia for the 2 masses would be;

Then, I = (4 × 0.5²) + (3 × 0.5²)

After that, I = 1.75 kg.m²

Now, when the angular velocity is given by the formula;

Then, ω = v/r

We are given v is = 6.4 m/s

Then, ω = 6.4/0.5

After that, ω = 12.8 rad/s

Now, let's find the angular momentum.

Then, the Angular momentum; L = Iω

L is = 1.75 × 12.8

Theregore, L = 22.4 kg.m²/s

Now, we are using the right-hand rule, the direction will be along the positive(+ve) z-axis.

B) Now, wehn the new diameter is 15 cm = 0.15 m

Thus, radius;r = 0.15/2 = 0.075 m

Then, Similar to a above;

After that, I = (4 × 0.075²) + (3 × 0.075²)

Now, I = 0.039375 kg.m²

Then, ω = v/r

We are given v = 6.4 m/s

ω is = 6.4/0.075

ω is = 85.33 rad/s

When the Angular momentum; L = Iω

Then, L = 0.039375 × 85.33

Therefore, L = 3.36 kg.m²/s

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Answer and Explanation:

There is no probability of obtaining such a circuit of closed track current carrying wire whose field of magnitude displays i.e.  [tex]B \alpha \frac{1}{r^2}[/tex]

The magnetic field is a volume of vectors

And [tex]\phi\ bds = 0[/tex]. This ensures isolated magnetic poles or magnetic charges would not exit

Therefore for a closed path,  we never received magnetic field that followed the [tex]B \alpha \frac{1}{r^2}[/tex] it is only for the simple current-carrying wire for both finite or infinite length.

Answer:

Velocity.

Explanation:

Projectile motion is characterized as the motion that an object undergoes when it is thrown into the air and it is only exposed to acceleration due to gravity.

As per the question, 'any change in the initial velocity of the projectile(object having gravity as the only force) would lead to a change in the range as well as the maximum height of the projectile.' To illustrate numerically:

Horizontal range: As per expression:

R= ([tex]u^{2}[/tex]*sin2θ)/g

the range depending on the square of the initial velocity.

Maximum height: As per expression:

H= ([tex]u^{2}[/tex] * [tex]sin^{2}[/tex]θ )/2g

the maximum distance also depends upon square of the initial velocity.

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Answer:

L = 130 decibels

Explanation:

The computation of the sound intensity level in decibels is shown below:

According to the question, data provided is as follows

I = sound intensity = 10 W/m^2

I0 = reference level = [tex]1 \times 10-12 W/m^2[/tex]

Now

Intensity level ( or Loudness)is

[tex]L = log10 \frac{I}{10}[/tex]

[tex]L = log10 \frac{10}{1\times 10^{-12}}[/tex]

[tex]L = log10 \times 1013[/tex]

[tex]= 13 \times 1 ( log10(10) = 1)[/tex]

Therefore  

L = 13 bel

And as we know that

1 bel = 10 decibels

So,

The  Sound intensity level is

L = 130 decibels

Answer:

x=22.57 m

Explanation:

Given that

35 m in W of S

angle = 40 degrees

25 m in east

From the diagram

The angle

[tex]\theta=90-40=50^o[/tex]

From the triangle OAB

[tex]cos40^o=\frac{35^2+25^2-x^2}{2\times 35\times 25}[/tex]

[tex]1340.57=35^2+25^2-x^2[/tex]

x=22.57 m

Therefore the answer of the above problem will be 22.57 m

Answer:

 θ  = 180

Explanation:

When an electric dipole is placed in an electric field, there is a torque due to the electric force

           τ = p x E

by rotating the dipole there is a change in potential energy

        ΔU = ∫ τ dθ

        ΔU = p E (cos θ₂ - cos θ₁)

when the dipole starts from an angle to the equilibrium position for θ = 0

          ΔU = pE (cos θ  - cos 0)

           cos θ  = 1 + DU / pE)

let's apply this expression to our case, the change in potential energy is ΔU = -0.2J

let's calculate

          cos θ  = 1 -0.2 / 0.1

          cos θ  = -1

           θ  = 180

Answer:

474.34 m

Explanation:

From the question,

E = kq/r²................. Equation 1

Where E = Electric Field, k = coulomb's constant, q = Charge, r = distance.

Make r the subject of the equation

r = √(kq/E)............ Equation 2

Given: q = 50 nC = 50×10⁻⁹ C, E = 0.002 N/C

Constant: k = 9×10⁹ Nm²/C².

Substitute these values into equation 2

r = √(50×10⁻⁹×9×10⁹/0.002)

r = √(450/0.002)

r = √(225000)

r = 474.34 m

Answer:

0.80 c

Explanation:

The computation of speed is shown below:-

Here, The speed of the captain ship A report for speed of the ship B which is

[tex]S = \frac{S_A + S_B}{1 + \frac{(S_AS_B)}{c^2} }[/tex]

where

[tex]S_A[/tex] indicates the speed of the ship A

[tex]S_B[/tex] indicates the speed of the ship B

and

C indicates the velocity of life

now we will Substitute 0.40c for A and 0.60 for B in the equation which is

[tex]S = \frac{0.40c + 0.60c}{1 + \frac{(0.40c)(0.60c)}{c^2} }[/tex]

after solving the above equation we will get

0.80 c

So, The correct answer is 0.80c

Answer:

Coulomb's law is:

[tex]F = \frac{1}{4*pi*e0} *(q1*q2)/r^2[/tex]

First, force has units of Newtons, the charges have units of Coulombs, and r, the distance, has units of meters, then, working only with the units we have:

N = (1/{e0})*C^2/m^2

then we have:

{e0} = C^2/(m^2*N)

And we know that N = kg*m/s^2

then the dimensions of e0 are:

{e0} = C^2*s^2/(m^3)

(current square per time square over cubed distance)

And knowing that a Faraday is:

F = C^2*S^2/m^2

The units of e0 are:

{e0} = F/m.

Answer:

The new angular speed is [tex]w = 6 \ rev/s[/tex]

Explanation:

From the  question we are told that

      The angular velocity of the spin is  [tex]w_o = 3 \ rev/s[/tex]

       The  original moment of inertia is  [tex]I_o[/tex]

        The new moment of inertia is  [tex]I =\frac{I_o}{2}[/tex]    

Generally angular momentum is mathematically represented as

      [tex]L = I * w[/tex]

Now according to the law of conservation of momentum, the initial momentum is equal to the final momentum hence the angular momentum is constant so

         [tex]I * w = constant[/tex]

=>       [tex]I_o * w _o = I * w[/tex]

where w is the new angular speed  

  So  

          [tex]I_o * 3 = \frac{I_o}{2} * w[/tex]

=>        [tex]w = \frac{3 * I_o}{\frac{I_o}{2} }[/tex]

=>         [tex]w = 6 \ rev/s[/tex]

Answer:

[tex]v_{2}=3.5 m/s[/tex]

Explanation:

Using the conservation of energy we have:

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

Let's solve it for v:

[tex]v=\sqrt{2gh}[/tex]

So the speed at the lowest point is [tex]v=7 m/s[/tex]

Now, using the conservation of momentum we have:

[tex]m_{1}v_{1}=m_{2}v_{2}[/tex]

[tex]v_{2}=\frac{1*7}{2}[/tex]

Therefore the speed of the block after the collision is [tex]v_{2}=3.5 m/s[/tex]

I hope it helps you!

Answer:

Acceleration of the asteroid relative to the spacecraft = 2ε[tex]E^{2}[/tex]A/m

Explanation:

The maximum electric field in the beam = 2E

cross-sectional area of beam = A

The intensity of an electromagnetic wave with electric field is

I = cε[tex]E_{0} ^{2}[/tex]/2

for [tex]E_{0}[/tex] = 2E

I = 2cε[tex]E^{2}[/tex]    ....equ 1

where

I is the intensity

c is the speed of light

ε is the permeability of free space

[tex]E_{0}[/tex]  is electric field

Radiation pressure of an electromagnetic wave on an absorbing surface is given as

P = I/c

substituting for I from above equ 1. we have

P = 2cε[tex]E^{2}[/tex]/c = 2ε[tex]E^{2}[/tex]    ....equ 2

Also, pressure P = F/A

therefore,

F = PA    ....equ 3

where

F is the force

P is pressure

A is cross-sectional area

substitute equ 2 into equ 3, we have

F = 2ε[tex]E^{2}[/tex]A

force on a body = mass x acceleration.

that is

F = ma

therefore,

a = F/m

acceleration of the asteroid will then be

a = 2ε[tex]E^{2}[/tex]A/m

where m is the mass of the asteroid.

Answer:

9.53 m

Explanation:

The computation of shortest organ pipe with both ends open that will resonate at this frequency is shown below:-

[tex]\lambda = \frac{velocity}{frequency}[/tex]

[tex]= \frac{343}{18}[/tex]

= 19.06 m

Now the

Shortest organ pipe  with both ends open is

=  [tex]\frac{\lambda}{2}[/tex]

[tex]= \frac{19.06}{2}[/tex]

= 9.53 m

Basically we applied the above formulas so that first we easily determined the shortest organ pipe for both ends at this frequency

Question:

A 10 gauge copper wire carries a current of 15 A. Assuming one free electron per copper atom, calculate the drift velocity of the electrons. (The cross-sectional area of a 10-gauge wire is 5.261 mm².)

Answer:

3.22 x 10⁻⁴ m/s

Explanation:

The drift velocity (v) of the electrons in a wire (copper wire in this case) carrying current (I) is given by;

v = [tex]\frac{I}{nqA}[/tex]

Where;

n = number of free electrons per cubic meter

q =  electron charge

A =  cross-sectional area of the wire

First let's calculate the number of free electrons per cubic meter (n)

Known constants:

density of copper, ρ = 8.95 x 10³kg/m³

molar mass of copper, M = 63.5 x 10⁻³kg/mol

Avogadro's number, Nₐ = 6.02 x 10²³ particles/mol

But;

The number of copper atoms, N, per cubic meter is given by;

N = (Nₐ x ρ / M)          -------------(ii)

Substitute the values of Nₐ, ρ and M into equation (ii) as follows;

N = (6.02 x 10²³ x 8.95 x 10³) / 63.5 x 10⁻³

N = 8.49 x 10²⁸ atom/m³

Since there is one free electron per copper atom, the number of free electrons per cubic meter is simply;

n = 8.49 x 10²⁸ electrons/m³

Now let's calculate the drift electron

Known values from question:

A = 5.261 mm² = 5.261 x 10⁻⁶m²

I = 23A

q = 1.6 x 10⁻¹⁹C

Substitute these values into equation (i) as follows;

v = [tex]\frac{I}{nqA}[/tex]

v = [tex]\frac{23}{8.49*10^{28} * 1.6 *10^{-19} * 5.261*10^{-6}}[/tex]

v = 3.22 x 10⁻⁴ m/s

Therefore, the drift electron is 3.22 x 10⁻⁴ m/s

Answer:

A: An upward force balances the downward force of gravity on the skydiver.

Explanation:

on edge! hope this helps!!~ (⌒▽⌒)☆

Answer:

Yes.

Explanation:

Law of relativity in relation to light states that the speed of light in a vacuum does not depend on all the motion of the observers and that all motion must be defined relative to a frame of reference and that space and time are relative, rather than absolute concepts. This was formulated by Albert Einstein in 1905.

Light travels more slowly in gas than in air because it interacts with atoms of glass that made it way through it and the refractive index of glass is more than air. This does contradict the theory of relativity as the speed of lights travel slower in glass because it's motion is slow and it is not relative.

Answer:

1ft per second

Explanation:

Physics is on my side!!!!!!!!!!

Answer: after 1.75 seconds

Explanation:

The only force acting on the ball is the gravitational force, so the acceleration will be:

a = -9.8 m/s^2

the velocity can be obtained by integrating over time:

v = -9.8m/s^2*t + v0

where v0 is the initial velocity; v0 = -7.95 m/s.

v = -9.8m/s^2*t - 7.95 m/s.

For the position we integrate again:

p = -4.9m/s^2*t^2 - 7.95 m/s*t + p0

where p0 is the initial position: p0 = 29m

p =  -4.9m/s^2*t^2 - 7.95 m/s*t + 29m

Now we want to find the time such that the position is equal to zero:

0 = -4.9m/s^2*t^2 - 7.95 m/s*t + 29m

Then we solve the Bhaskara's equation:

[tex]t = \frac{7.95 +- \sqrt{7.95^2 +4*4.9*29} }{-2*4.9} = \frac{7.95 +- 25.1}{9.8}[/tex]

Then the solutions are:

t = (7.95 + 25.1)/(-9.8) = -3.37s

t = (7.95 - 25.1)/(-9.8) = 1.75s

We need the positive time, then the correct answer is 1.75s

Answer:

a) Energy stored in the capacitor, [tex]E = 1.0125 *10^{-3} J[/tex]

b) Q = 45 µC

c) C' = 1.5 μF

d)  [tex]E = 6.75 *10^{-4} J[/tex]

Explanation:

Capacitance, C = 1 µF

Charge on the plates, Q = 45 µC

a) Energy stored in the capacitor is given by the formula:

[tex]E = \frac{Q^2}{2C} \\\\E = \frac{(45 * 10^{-6})^2}{2* 1* 10^{-6}}\\\\E = \frac{2025 * 10^{-6}}{2}\\\\E = 1012.5 *10^{-6}\\\\E = 1.0125 *10^{-3} J[/tex]

b) The charge on the plates of the capacitor will  not change

It will still remains, Q = 45 µC

c)  Electric field is non zero over (1-1/3) = 2/3 of d

From the relation V = Ed,

The voltage has changed by a factor of 2/3

Since the capacitance is given as C = Q/V  

The new capacitance with the conductor in place, C' = (3/2) C

C' = (3/2) * 1μF

C' = 1.5 μF

d) Energy stored in the capacitor with the conductor in place

[tex]E = \frac{Q^2}{2C} \\\\E = \frac{(45 * 10^{-6})^2}{2* 1.5* 10^{-6}}\\\\E = \frac{2025 * 10^{-6}}{3}\\\\E = 675 *10^{-6}\\\\E = 6.75 *10^{-4} J[/tex]

Answer:

a)  v_correcaminos = 22.95 m / s ,  b)  x = 512.4 m ,

c) v = (45.83 i ^ -109.56 j ^) m / s

Explanation:

We can solve this exercise using the kinematics equations

a) Let's find the time or the coyote takes to reach the cliff, let's start by finding the speed on the cliff

         v² = v₀² + 2 a x

they tell us that the coyote starts from rest v₀ = 0 and its acceleration is a=15 m / s²

         v = √ (2 15 70)

         v = 45.83 m / s

with this value calculate the time it takes to arrive

        v = v₀ + a t

        t = v / a

        t = 45.83 / 15

        t = 3.05 s

having the distance to the cliff and the time, we can find the constant speed of the roadrunner

         v_ roadrunner = x / t

         v_correcaminos = 70 / 3,05

         v_correcaminos = 22.95 m / s

b) if the coyote leaves the cliff with the horizontal velocity v₀ₓ = 45.83 m / s, they ask how far it reaches.

Let's start by looking for the time to reach the cliff floor

            y = y₀ + [tex]v_{oy}[/tex] t - ½ g t²

in this case y = 0 and the height of the cliff is y₀ = 100 m

          0 = 100 + 45.83 t - ½ 9.8 t²

          t² - 9,353 t - 20,408 = 0

we solve the quadratic equation

         t = [9,353 ±√ (9,353² + 4 20,408)] / 2

         t = [9,353 ± 13] / 2

         t₁ = 11.18 s

        t₂ = -1.8 s

Since time must be a positive quantity, the answer is t = 11.18 s

we calculate the horizontal distance traveled

        x = v₀ₓ t

        x = 45.83 11.18

        x = 512.4 m

c) speed when it hits the ground

         vₓ = v₀ₓ = 45.83 m / s

we look for vertical speed

         v_{y} = [tex]v_{oy}[/tex] - gt

         v_{y} = 0 - 9.8 11.18

         v_{y} = - 109.56 m / s

         v = (45.83 i ^ -109.56 j ^) m / s

Answer:

 θ₁ = 0.04º , θ₂ = 0.00118º

Explanation:

The equation that describes the diffraction pattern of a network is

             d sin θ = m λ

where the diffraction order is, in this case they indicate that the order

m = 1

           θ = sin⁻¹ (λ / d)

Trfuvsmod ls inrsd fr ll red s SI units

           d = 12000 line / inc (1 inc / 2.54cm) = 4724 line / cm

the distance between two lines we can look for it with a direct proportions rule

If there are 4724 lines in a centimeter, the distance for two hundred is

            d = 2 lines (1 cm / 4724 line) = 4.2337 10⁻⁴ cm

let's calculate the angles

λ = 300 10-9 m

            θ₁ = sin⁻¹ (300 10-9 / 4,2337 10-4)

            θ₁ = sin⁻¹ (7.08 10-4)

            θ₁ = 0.04º

λ = 5000

          θ₂ = sin-1 (500 10-9 / 4,2337 10-4)

          θ₂ = 0.00118º

Answer:

the time at which it passes through the equilibrum position is:

t = 0.1 second

Explanation:

given

w= 4pounds

k(spring constant) = 2lb/ft

g(gravitational constant) = 10m/s² = 32ft/s²

β(initial point above equilibrum) = 1

velocity = 14ft/s

attached is an image showing the calculations, because some of the parameters aren't convenient to type.

The time at which the mass passes will be "0.1 s".

According to the question,

Mass weighing, w = 4 pounds

Spring constant, k = 2 lb/ft

Gravitational constant, g = 10 m/s² = 32 ft/s²

Point above equilibrium, β = 1

Velocity = 14 ft/s

By using equation of motion,

→     x(t) =  (-1 + gt)

By substituting the values,

         0 =  (-1 + 10t)

-1 + 10t = 0

By adding "1" both sides, we get

-1 + 10t + 1 = 1

           10t = 1

               t = [tex]\frac{1}{10}[/tex]

                 = 0.1 s

Thus the above answer is right.

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Answer:

F = 17170.16 N = 17.17 KN

Explanation:

First we need to find the mass of helicopter by using its weight:

Weight = mg

2.75 x 10⁵ N = m(9.8 m/s²)

m = (2.75 x 10⁵ N)/(9.8 m/s²)

m = 28061.22 kg

Now, we find acceleration. We have position vector as:

r = (0.02 m/s³)(t³)i + (2.2 m/s)(t)j - (0.06 m/s²)(t²)k

taking its derivative twice, we can find acceleration:

a = (3)(2)(0.02 m/s³)(t)i + (0)j - (2)(1)(0.06 m/s²)k

a = (0.12 m/s³)(t)i - (0.12 m/s²)k

at, t = 5 sec

a = (0.12 m/s³)(5 s)i - (0.12 m/s²)k

a = (0.6 m/s²) i - (0.12 m/s²) k

Now, the magnitude of acceleration will be:

a = √[(0.6)² + (-0.12)²]

a = 0.61 m/s²

So, from Newton's Second Law, the net force on helicopter is given as:

F = ma

F = (28061.22 kg)(0.61 m/s²)

F = 17170.16 N = 17.17 KN

Answer:

τ = 0.26 N.m

Explanation:

First we find the moment of inertia of the wheel, by using the following formula:

I= mr²

where,

I = Moment of Inertia = ?

m = mass of wheel = 2.3 kg

r = radius of wheel = 50 cm/2 = 25 cm = 0.25 m

Therefore,

I = (2.3 kg)(0.25 m)²

I = 0.115 kg.m²

Now, we find the angular acceleration of the wheel:

α = (ωf - ωi)/t

where,

α = angular acceleration = ?

ωf = final angular velocity = (120 rpm)(2π rad/1 rev)(1 m/60 s) = 12.56 rad/s

ωi = Initial Angular Velocity = 0 rad/s

t = time = 5.5 s

Therefore,

α = (12.56 rad/s - 0 rad/s)/(5.5 s)

α = 2.28 rad/s²

Now, the torque is given as:

Torque = τ = Iα

τ = (0.115 kg.m²)(2.28 rad/s²)

τ = 0.26 N.m

Answer:

The angular momentum of the crate is [tex]M_{C} V_{1} d[/tex]

Explanation:

mass of the crate = [tex]M_{C}[/tex]

speed of forklift = [tex]V_{1}[/tex]

The distance between the center of the mass and the point A = d

Recall that the angular moment is the moment of the momentum.

[tex]L = P*d[/tex]    ..... equ 1

where L is the angular momentum,

P is the momentum of the system,

d is the perpendicular distance between the crate and the point on the axis about which the momentum acts. It is equal to d from the image

Also, we know that the momentum P is the product of mass and velocity

P = mv      ....equ 2

in this case, the mass = [tex]M_{C}[/tex]

the velocity = [tex]V_{1}[/tex]

therefore, the momentum P = [tex]M_{C}[/tex][tex]V_{1}[/tex]

we substitute equation 2 into equation 1 to give

[tex]L = M_{C} V_{1} d[/tex]

Answer:

The gray spheres is negatively charged while the red is positively charged

Explanation:

This is because theelectrophorus becomes less positive once it pulls some electrons away from the red sphere, but, the electrophorus is replaced on the slab and recharged by grounding it before it proceeds to charge the grey sphere, thereby giving it electrons and making it negatively charged

Answer:

The gray sphere has a positive charge and the red sphere has a positive charge.