You are at the controls of a particle accelerator, sending a beam of 2.10×107 m/s protons (mass m) at a gas target of an unknown element. Your detector tells you that some protons bounce straight back after a collision with one of the nuclei of the unknown element. All such protons rebound with a speed of 1.80×107 m/s. Assume that the initial speed of the target nucleus is negligible and the collision is elastic.
A) Find the mass of one nucleus of the unknown element.
B) What is the speed of the unknown nucleus immediately after such a collision?

Answer:

Let us consider the case of a bus turning around a corner with a constant velocity, as the bus approaches the corner, the velocity at say point A is Va, and is tangential to the curve with direction pointing away from the curve. Also, the velocity at another point say point B is Vb and is also tangential to the curve with direction pointing away from the curve. Although the velocity at point A and the velocity at point B have the same magnitude, their directions are different (velocity is a vector quantity), and hence we have a change in velocity. By definition, an acceleration occurs when we have a change in velocity, so the bus experiences an acceleration at the corner whose direction is away from the center of the corner.

The acceleration is not aligned with the direction of travel because the change in velocity is at a tangent (directed away) to the direction of travel of the bus.

Answer:

The heat rate is  

Explanation:

From the question we are told that

  The surface emissivity is  [tex]e=0.1[/tex]

   The length is  [tex]L = 0.1 \ m[/tex]

    The width is  [tex]W = 0.1 \ m[/tex]

     The surface temperature of one side is  [tex]T_1 = 500 \ K[/tex]

     The temperature of the quiescent air [tex]T_c = 300 \ K[/tex]

      The temperature of the surrounding is  [tex]T_s = 300 \ K[/tex]

The heat rate from the flat plate is mathematically represented as

         [tex]Q = \sigma A e (T_1^4 - T_a^4)[/tex]

Where [tex]\sigma[/tex] is the quiescent air Stefan-Boltzmann constant  and it value is

       [tex]\sigma = 5.67*10^{-8} m^{-2} \cdot K^{-4}[/tex]

     A is the area which is mathematically evaluated  as

           [tex]A = W * L[/tex]

substituting values

           [tex]A = 0.1 * 0.1[/tex]

           [tex]A = 0.01 \ m^2[/tex]

substituting values

          [tex]Q = 5.67 *10^{-8} * (0.01) *(500^4 -300^4)[/tex]

          [tex]Q =3.045 \ Watt[/tex]

Answer:

This statement is false , linear momentum is the product of mass and velocity

Answer:

Linear momentum is defined as the product of a system's mass multiplied by its velocity. In symbols, linear momentum is expressed as p = mv. ... Thus the greater an object's mass or the greater its velocity, the greater its momentum. Momentum p is a vector having the same direction as the velocity v.

Explanation:

hope this helps u stay safe

Answer:

  22.5°

Explanation:

Short answer: The trajectories will hit the same target when the projectile is launched at complementary angles. The second angle is 90° -67.5° = 22.5°.

__

Longer answer: The horizontal speed of the snowball launched at angle α with speed s is ...

  vh = s·cos(α)

Then the horizontal distance at time t is ...

  x = vh·t

and the time taken to get to some distance x is ...

  t = x/vh = x/(s·cos(α))

The equation for the vertical motion of the projectile is ...

  y = -4.9t² +s·sin(α)·t

Substituting the above expression for t, we have y in terms of x:

  y = -4.9x²/(s·cos(α))² +(s·sin(α)·x)/(s·cos(α))

Factoring gives ...

  y = (x/(cos(α))(-4.9x/(s²·cos(α)) +sin(α))

The height y will be zero at x=0 and at ...

  0 = -4.9x/(s²·cos(α)) +sin(α)

  x = s²·sin(α)·cos(α)/4.9 = (s²/9.8)sin(2α)

So, for some alternate angle β, we want ...

  sin(2α) = sin(2β)

We know this will be the case for ...

  2β = 180° -2α

  β = 90° -α

The second snowball should be thrown at an angle of 90°-67.5° = 22.5° to make it hit the same point as the first.

Given that,

Length of bar = 600 mm

Diameter of bar = 40 mm

Diameter of hole = 30 mm

Length of hole = 100 mm

Modulus of elasticity = 85 GN/m²

Load = 180 kN

We need to calculate the area of cross section without hole

Using formula of area

[tex]A=\dfrac{\pi\times d^2}{4}[/tex]

Put the value into the formula

[tex]A=\dfrac{\pi\times40^2}{4}[/tex]

[tex]A=1256.6\ mm^2[/tex]

We need to calculate the area of cross section with hole

Using formula of area

[tex]A=\pi\times\dfrac{(d_{b}^2-d_{h}^{2})}{4}[/tex]

Put the value into the formula

[tex]A=\pi\times\dfrac{(40^2-30^2)}{4}[/tex]

[tex]A=549.77\ mm^2[/tex]

We need to calculate the total contraction on the bar

Using formula of total contraction

Total contraction = contraction in bar without hole part + contraction in bar with hole part

[tex]Total\ contraction = \dfrac{F\times L_{1}}{A_{1}\times E}+\dfrac{F\times L_{2}}{A_{2}\times E}[/tex]

Where, F = load

L = length

A = area of cross section

E = modulus of elasticity

Put the value into the formula

[tex]Total\ contraction=\dfrac{180\times10^3}{85\times10^{3}}(\dfrac{500}{1256.6}+\dfrac{100}{549.77})[/tex]

[tex]Total\ contraction = 1.227\ mm^2[/tex]

Hence, The total contraction on the bar is 1.227 mm²

Answer:

false

Explanation:

Linear momentum is the product of an objects mass and velocity

p=m×v

Answer:

Linear momentum is defined as the product of a system's mass multiplied by its velocity. In symbols, linear momentum is expressed as p = mv. ... Thus the greater an object's mass or the greater its velocity, the greater its momentum. Momentum p is a vector having the same direction as the velocity v.

Explanation:

HOPE THIS HELPS U STAY SAFE

Complete Question

The complete question is shown on the first uploaded image

Answer:

The correct option is the second option

Explanation:

 Generally the electric force exerted by the charge Q  on the  charge  (q) is mathematically represented as

           [tex]F_Q = \frac{kqQ}{r^2}[/tex]

 Generally the electric force exerted by the charge 2Q  on the  charge  (q) is mathematically represented as

         [tex]F_{2Q} = \frac{kq2Q}{2r^2}[/tex]    

Now the net force exerted on q is

       [tex]F_{net} = \frac{kqQ}{r^2} - \frac{2k q Q}{4r^2}[/tex]

        [tex]F_{net} = \frac{4kqQ- 2kqQ}{4r^2}[/tex]

        [tex]F_{net} = \frac{kqQ}{2r^2}[/tex]

Looking at the resulting equation we see that [tex]F_{net} > 0[/tex]

This implies that the charge q would move to the right

Complete Question

The complete question is shown on the first uploaded image

Answer:

The  point charge is  [tex]Q_z = -0.0912 \ \mu C[/tex]

The inner shell is  [tex]Q_t = 0.4168 \ \mu C[/tex]

The outer shell is  [tex]Q_w = -0.6514 \ \mu C[/tex]

Explanation:

From the question we are told that

    The inner radius of thin first spherical conducting shell is  [tex]r_1[/tex]

    The outer radius of thin first spherical conducting shell is  [tex]r_2[/tex]

    The inner radius of second thin spherical conducting shell is [tex]R_1[/tex]

    The outer radius of second thin spherical conducting shell is [tex]R_2[/tex]

     The magnetic flux for different region is  [tex]\phi = -10.3 *10^3 N\cdot m^2 /C \ for \ r < r_1[/tex]

    The magnetic flux for first shell is [tex]\phi = 36 * 10^3 N \cdot m^2 /C \ for \ r_2 < r <R_1[/tex]

     The magnetic flux for second shell is [tex]\phi = -36 * 10^3 N \cdot m^2 /C \ for \ r <R_1[/tex]

The magnitude of the point charge is mathematically represented as

                [tex]Q_z = \ \phi_z * \epsilon _o[/tex]

               [tex]Q_z = -10.3*10^{3} * 8.85 *10^{-12}[/tex]

               [tex]Q_z = -9.115*10^{-8} \ C[/tex]

               [tex]Q_z = -0.0912 \ \mu C[/tex]

Considering the inner shell

        [tex]Q_a = \phi_a * \epsilon _o[/tex]

=>    [tex]Q_a = 36 .8 * 10^3 * 8.85*10^{-12}[/tex]

      [tex]Q_a = 32.56*10^{-8} \ C[/tex]

       [tex]Q_a =0.326} \ \mu C[/tex]

Charge on the inner shell is

       [tex]Q_t = Q_a - Q_z[/tex]

                    [tex]Q_t = 0.326} \ \mu - ( -0.0912 \ \mu)[/tex]

                      [tex]Q_t = 0.4168 \ \mu C[/tex]

Considering the outer  shell

     [tex]Q_y = \phi_y * \epsilon_o[/tex]

=>    [tex]Q_y = -36.8 *10^{3} * 8.85*10^{-12}[/tex]

        [tex]Q_y = -32.56*10^{-8} \ C[/tex]

        [tex]Q_y = - 0.326} \ \mu C[/tex]

Charge on the outer shell is

      [tex]Q_w = Q_y - Q_z[/tex]

      [tex]Q_w =- 0.326} \ \mu - ( -0.0912 \ \mu)[/tex]

       [tex]Q_w = -0.6514 \ \mu C[/tex]

Answer:

The correct answer will be:

(a) 2.2 rad/s

(b) 2.2 rad

Explanation:

The given values are:

Angular acceleration, [tex]\alpha = 0.35 \pi \ rad/s^2[/tex]

(a)...

At time t = 2.0 s,

The angular velocity will be:

⇒  [tex]\omega = \alpha t[/tex]

On putting the estimated values, we get

⇒     [tex]=0.35\pi \times 2.0[/tex]

∴ [[tex]\pi = 3.14[/tex]]

⇒     [tex]=0.35\times 3.14\times 2.0[/tex]

⇒     [tex]=2.2 \ rad/s[/tex]

(b)...

The angular displacement will be:

⇒  [tex]\theta = \frac{1}{2}\alpha t^2[/tex]

On putting the estimated values, we get

⇒     [tex]=\frac{1}{2}\times 0.35\pi\times (2.0)^2[/tex]

⇒     [tex]=\frac{1}{2}\times 0.35\times 3.14\times 4[/tex]

⇒     [tex]=1.099\times 2[/tex]

⇒     [tex]=2.2 \ rad[/tex]

Answer:

The nucleus of an atom contains the majority of the atom’s mass, and is composed of protons and neutrons, which are collectively referred to as nucleons. The much-lighter electrons orbit their atom’s nucleus. The Protons. Protons are positively charged particles found in an atom’s nucleus.

I hope u liked my answer. please mark me as branliest x

Answer:

A. The magnitude of the net force exerted on the disk

B. The distance between the center of the disk and where the net force is applied to the disk

Explanation:

To determine the change in angular momentum of the disk after a stipulated time, one must measure the above options.

The radius of the disk is fixed and does not vary with the experiment, and the mass of the disk is also constant and known.

One must first measure the magnitude of the net force exerted on the disk, and determine the torque as a result of this torque from the distance between the center of the disk and the point where the net force is applied.  The above statement also points out the necessity of measuring the distance between the center of the disk and the point where the net force is applied on the disk, as both the torque, and the moment of inertia is calculated from this point.

torque T = Force time distance of point of action of force from mid point of the disk

T = F X r

T x t = Δ(Iω)

Where t is the time,

and Δ(Iω) is change in angular momentum.

Answer: The force at t = 1s is 14 N

Explanation:

I guess that the position equation actually is: (by looking at the units of C and D)

S = A - B*t + C*t^2 - D*t^3

As you may know by Newton's third law, if we want to find the force, we first need to find the acceleration, and before that, the velocity.

the velocity can be found by integrating over time, the velocity is:

V = 0 - B + 2*C*T - 3*D*t^2

For the acceleration we integrate again:

A = 0 + 2*C - 6D*T

and we know that F = m*A

then:

Force(t) = 1kg*(2*10m/s^2 - 6*1m/s^3*t)

We want the force at t = 1s, so we replace t by 1s.

F(1s) = 1kg*(20m/s^2 - 6m/s^2) = 14 N

Answer:

 a

    The volume charge density is  [tex]\rho = nq[/tex]

b

    The surface charge density is  [tex]\sigma = n^{\frac{2}{3} } q[/tex]

Explanation:

From the question we are told that

    The radius is  R

     The length is L

       The velocity is  v

        The number of ions per unit volume is  n

         The charge is  q

          The thickness of the cylinder surface is  [tex]n^{\frac{1}{3} }[/tex]

The volume charge density is mathematically represented as

      [tex]\rho = nq[/tex]

The surface charge density is mathematically represented as

    [tex]\sigma = \rho n^{\frac{1}{3} }[/tex]

substituting for  [tex]\rho[/tex]

     [tex]\sigma = n * n^{\frac{1}{3} } q[/tex]

     [tex]\sigma = n^{\frac{2}{3} } q[/tex]

Answer:

Mass number - ⦁ The number of protons and neutrons in the nucleus of an atom.

Isotopes - ⦁ Atoms with the same number of protons, but different number of neutrons.

Nitrogen - ⦁ The name of the element with atomic number 7.

Atomic number - ⦁ The number of protons in the nucleus of an atom.

Explanation:

We have,

Mass of a particle is 0.6 kg

Speed at A is 2 m/s

Kinetic energy at B is 7.5 J

(a) The kinetic energy at A is given by :

[tex]K_A=\dfrac{1}{2}mv^2\\\\K_A=\dfrac{1}{2}\times 0.6\times 2^2\\\\K_A=1.2\ J[/tex]

(b) Kinetic energy at B is given by

[tex]K_B=\dfrac{1}{2}mV^2\\\\V=\sqrt{\dfrac{2K_B}{m}} \\\\V=\sqrt{\dfrac{2\times 7.5}{0.6}} \\\\V=5\ m/s[/tex]

(c) The work done on the particle as it moves form A to B is given by work energy theorem as :

[tex]W=\dfrac{1}{2}m(V^2-v^2)\\\\W=\dfrac{1}{2}\times 0.6\times (5^2-2^2)\\\\W=6.3\ J[/tex]

Answer:

[tex]y_n(x) =C_n \sin \sqrt{\frac{\rho}{T} } w_nx=C_n \sin \sqrt{\frac{\rho}{T} } \sqrt{\frac{T}{\rho} } \frac{n \pi}{L} x[/tex]

[tex]y_n(x) = C_n \sin \frac{n \pi x}{L}[/tex]

Explanation:

The given differential equation is

[tex]T\frac{d^2y}{dx^2} + \rho w ^2y=0[/tex] and y(0) = 0, y(L) =0

where T and ρ  are constants

The given rewrite as

[tex]\frac{d^2y}{dx^2} + \frac{\rho w^2}{T} y=0[/tex]

auxiliary equation is

[tex]m^2+ \frac{\rho w^2}{T} =0\\\\m= \pm\sqrt{\frac{\rho}{T} } wi[/tex]

Solution of this de is

[tex]y(x)=C_1 \cos \sqrt{\frac{\rho}{t} } wx + C_2 \sin \sqrt{\frac{\rho}{T} } wx[/tex]

y(0)=0 ⇒ C₁ = 0

[tex]y(x) = C_2 \sin \sqrt{\frac{\rho}{T} } wx[/tex]

y(L) = 0 ⇒

[tex]C_2 \sin \sqrt{\frac{\rho}{T} } wL=0[/tex]

we need non zero solution

⇒ C₂ ≠ 0 and

[tex]\sin \sqrt{\frac{\rho}{T} } wL=0[/tex]

[tex]\sin \sqrt{\frac{\rho}{T} } wL=0 \rightarrow \sqrt{\frac{\rho}{T} } wL=n \pi[/tex]

[tex]w_n = \sqrt{\frac{T}{\rho} } \frac{n \pi}{L}[/tex]

solution corresponding these [tex]w_n[/tex] values

[tex]y_n(x) =C_n \sin \sqrt{\frac{\rho}{T} } w_nx=C_n \sin \sqrt{\frac{\rho}{T} } \sqrt{\frac{T}{\rho} } \frac{n \pi}{L} x[/tex]

[tex]y_n(x) = C_n \sin \frac{n \pi x}{L}[/tex]

Answer:

0.2 W more power than nicad cell is delivered by alkaline cell

Explanation:

FOR NICKEL-CADMIUM CELL (nicads):

First we find the current supplied to radio by the cell. For this purpose, we use the formula:

I = E/(R+r)

where,

I = current supplied

E = emf of cell = 1.25 V

R = resistance of radio = 3.65 Ω

r = internal resistance of cell = 0.04 Ω

Therefore,

I = (1.25 V)/(3.65 Ω + 0.04 Ω)

I = 0.34 A

Now, we calculate the power delivered to radio by following formula:

P = VI

but, from Ohm's Law:   V = IR

Therefore,

P = I²R

where,

P = Power delivered = ?

I = current = 0.34 A

R = Resistance of radio = 3.65 Ω

Therefore,

P = (0.34 A)²(3.65 Ω)

P = 0.41 W

FOR ALKALINE CELL:

First we find the current supplied to radio by the cell. For this purpose, we use the formula:

I = E/(R+r)

where,

I = current supplied

E = emf of cell = 1.58 V

R = resistance of radio = 3.65 Ω

r = internal resistance of cell = 0.2 Ω

Therefore,

I = (1.58 V)/(3.65 Ω + 0.2 Ω)

I = 0.41 A

Now, we calculate the power delivered to radio by following formula:

P = VI

but, from Ohm's Law:   V = IR

Therefore,

P = I²R

where,

P = Power delivered = ?

I = current = 0.41 A

R = Resistance of radio = 3.65 Ω

Therefore,

P = (0.41 A)²(3.65 Ω)

P = 0.61 W

Now, fo the difference between delivered powers by both cells:

ΔP = (P)alkaline - (P)nicad

ΔP = 0.61 W - 0.41 W

ΔP = 0.2 W

Answer:

1:4

Explanation:

We have, two identical objects A and B fall from rest from different heights to the ground.

Object B takes twice as long as object A to reach the ground. It is required to find the ratio of the heights from which A and B fell. Let [tex]h_A\ \text{and}\ h_B[/tex] are the height for A and B respectively. So,

[tex]\dfrac{h_A}{h_B}=\dfrac{(1/2)gt_A^2}{(1/2)gt_B^2}\\\\\dfrac{h_A}{h_B}=\dfrac{t_A^2}{t_B^2}[/tex]

We have,

[tex]t_B=2t_A[/tex]

So,

[tex]\dfrac{h_A}{h_B}=\dfrac{t_A^2}{(2t_B)^2}\\\\\dfrac{h_A}{h_B}=\dfrac{1}{4}[/tex]

So, the ratio of the heights from which A and B fell is 1:4.

Answer:

v_r = 1.268 × 10⁸ mi/hr

Explanation:

We are given;

wavelength of the red light; λr = 693 nm = 693 × 10^(-9) m

wavelength of the yellow light; λy = 582 nm = 582 × 10^(-9) m

Frequency is given by the formula;

f = v/λ

Where v is speed of light = 3 x 10^(8) m

Frequency of red light; f_o = [3 x 10^(8)]/(693 × 10^(-9)) = 4.33 x 10¹⁴ Hz

Similarly, Frequency of yellow light;

f = [3 x 10⁸]/(582 × 10^(-9)) = 5.15 x 10¹⁴ Hz

To find the speed of the car, we will use the formula;

f = f_o[(c + v_r)/c)]

Where c is speed of light and v_r is speed of car.

Making v_r the subject;

cf/f_o = c + v_r

v_r = c(f/f_o - 1)

So, plugging in the relevant values, we have;

v_r = 3 × 10⁸[((5.15 x 10¹⁴)/(4.33 x 10¹⁴)) - 1]

v_r = 3 × 10⁸(0.189)

v_r = 5.67 x 10⁷ m/s

Converting to mi/hr, 1 m/s = 2.23694 mile/hr

So, v_r = 5.67 × 10⁷ × 2.23694

v_r = 1.268 × 10⁸ mi/hr

Answer:

a. Time = 16.11 s

b. Gauge Pressure = 1009400 Pa = 1 MPa  

c. Absolute Pressure = 1110725 Pa + 1.11 MPa

d. Force = 2.22 MN

Explanation:

a.

For the accelerated part of motion of submarine we can use equations of motion.

Using 1st equation of motion:

Vf = Vi + at₁

t₁ = (Vf - Vi)/a

where,

t₁ = time taken during accelerated motion = ?

Vf = final velocity = 4 m/s

Vi = Initial Velocity = 0 m/s   (Since, it starts from rest)

a = acceleration = 0.3 m/s²

Therefore,

t₁ = (4 m/s - 0 m/s)/(0.3 m/s²)

t₁ = 13.33 s

Now, using 2nd equation of motion:

d₁ = (Vi)(t₁) + (0.5)(a)(t₁)²

where,

d₁ = the depth covered during accelerated motion

Therefore,

d₁ = (0 m/s)(13.33 s) + (0.5)(0.3 m/s²)(13.33 s)²

d₁ = 88.89 m

Hence,

d₂ = d - d₁

where,

d₂ = depth covered during constant speed  motion

d = total depth = 100 m

Therefoe,

d₂ = 100 m - 88.89 m

d₂ = 11.11 m

So, for uniform motion:

s₂ = vt₂

where,

v = constant speed = 4 m/s

t₂ = time taken during constant speed  motion

11.11 m = (4 m/s)t₂

t₂ = 2.78 s

Therefore, total time taken by submarine to move down 100 m is:

t = t₁ + t₂

t = 13.33 s + 2.78 s

t = 16.11 s

b.

The gauge pressure on submarine can be calculated by the formula:

Pg = ρgh

where,

Pg = Gauge Pressure = ?

ρ = density of salt water = 1030 kg/m³

g = 9.8 m/s²

h = depth = 100 m

Therefore,

Pg = (1030 kg/m³)(9.8 m/s²)(100 m)

Pg = 1009400 Pa = 1 MPa

c.

The absolute pressure is given as:

P = Pg + Atmospheric Pressure

where,

P = Absolute Pressure = ?

Atmospheric Pressure = 101325 Pa

Therefore,

P = 1009400 Pa + 101325 Pa

P = 1110725 Pa + 1.11 MPa

d.

Since, the force to open the door must be equal to the force applied to the door by pressure externally.

Therefore, the  force required to open the door can be found out by the formula of pressure:

P = F/A

F = PA

where,

P = Absolute Pressure on Door = 1110725 Pa

A = Area of door = 2 m²

F = Force Required to Open the Door = ?

Therefore,

F = (1.11 MPa)(2 m²)

F = 2.22 MN

Answer in my own words:

Research reveals that most of the indoor microplastics contained in the air are created by synthetic plastic fibres, as well as by textiles used in furniture

Key words:

1: Research

2: air

3:  Plastic

Please mark brainliest

Hope this helps.

Answer:

The magnification of the image is  [tex]M = -1.5[/tex]

Explanation:

From the question we are told that

      The object distance is  [tex]u = 14 cm[/tex]

      The image distance is  [tex]v = 21 \ cm[/tex]

The magnification of the image is mathematically represented as

        [tex]M = - \frac{v}{u}[/tex]

substituting values

       [tex]M = - \frac{21}{14}[/tex]

        [tex]M = -1.5[/tex]

The negative value means that the image is real , inverted and enlarged

Explanation:

We have,

Electric field intensity is 700 N/Cj

It is required to find the magnitude and direction of the acceleration of this electron due to this field.

The electric force is balanced by the force due to its motion i.e.

qE =ma

a is acceleration of this electron

[tex]a=\dfrac{qE}{m}\\\\a=\dfrac{1.6\times 10^{-19}\times 700}{9.1\times 10^{-31}}\\\\a=1.23\times 10^{14}\ m/s^2[/tex]

So, the acceleration of the electron is [tex]1.23\times 10^{14}\ m/s^2[/tex] and it is moving in +y direction.

Answer:

A = Tan{-1} d/h

Explanation

Let the angle Theta be A

From identity the angle A has an opposite side d and an adjacent h.

From trigonometry ratio

Tan A = d/h

A = Tan{-1} d/h

Answer:

1800 J

Explanation:

Energy is conserved, so the maximum kinetic energy equals the change in gravitational energy.

Answer:

axial   V = 0

equatorial  V = k q 2a / (x² -a²),  V = k q 2x / (a² -x²)

Explanation:

A dipole is a system formed by two charges of equal magnitude, but different sign, separated by a distance 2a; let's look for the electrical potential in an axial line

       V = k (q / √(a² + y²) - q /√ (a² + y²))

        V = 0

the potential on the equator

we place the positive charge to the left and perform the calculation for a point outside the dipole

    V = k (q / (x-a) - q / (x + a))

    V = k q 2a / (x² -a²)

 we perform the calculation for a point between the dipo charges

     V = k (q / (a-x) - q / (a ​​+ x))

     V = k q 2x / (a² -x²)

Answer:

F = 187.5N.

Explanation:

So, from the question above we are given the following parameters or data or information which is going to assist us in solving the question/problem;

=> Mass= 200-kg , => radius = 1.50 m, => angular speed of 0.400 rev/s, and time = 2.0 seconds.

Step one: the first step is to calculate or determine the angular speed. Here, the angular speed is Calculated in rad/sec.

Angular speed, w = 0.400 × 2π

= 2.51 rad/s.

Step two: determine the value of a.

Using the formula below;

W = Wo + a × time,t.

2.51 = 0 + a(2.0).

a= 1.25 rad/s^2.

Step three: determine constant force from the Torque.

Torque = I × a.

F = 1/2 × (200 kg) × 1.50 × 1.25.

F = 187.5N

Answer:

a = -0.05 m/s² (negative sign shows deceleration)

Explanation:

In order, to find out the minimum average acceleration for a student starting at 5 m/s to slide to the end, we can use 3rd equation of motion. 3rd equation of motion is given as follows:

2as = Vf² - Vi²

where,

a = minimum acceleration required = ?

s = minimum distance covered = 250 m

Vf = Final Speed = 0 m/s (for minimum acceleration the student will barely cover 250 m and then stop)

Vi = Initial Velocity = 5 m/s

Therefore,

2a(250 m) = (0 m/s)² - (5 m/s)²

a = - (25 m²/s²)/(500 m)

a = -0.05 m/s² (negative sign shows deceleration)