Two identical metal balls of radii 2.50
cm are at a center to center distance of
1.00 m from each other. Each ball is
charged so that a point at the surface of
the first ball has an electric potential of
+1.20 x 103 V and a point at the surface
of the other ball has an electric
potential of -1.20 x 103 V. What is the total charge on each ball?

Answer:

d) The box’s initial potential energy is transformed into kinetic energy and

thermal energy.

Explanation:

Let us analyze the situation first. Initially we have a box at rest placed at some height on a ramp. So, the three energies associated with the box will be:

Kinetic Energy = 0 (due to zero velocity)

Thermal Energy = 0 (due to not temperature change)

Potential Energy = Greater than zero value (due to height)

Now, when the box is released to slide down the ramp with friction, the energies become:

Kinetic Energy = Increasing (due to increase in velocity)

Thermal Energy = Increasing (due to increase in temperature, because of friction)

Potential Energy = Decreasing (due to decrease in height)

So, from Law of Conservation of Energy, we can write:

Loss of Potential Energy = Gain in Kinetic Energy + Gain in Thermal Energy

So, the correct option is:

d) The box’s initial potential energy is transformed into kinetic energy and  thermal energy.

Answer:

9.96 s

Explanation:

Given in the y direction:

Δy = -115 m

v₀ = 65.0 m/s sin 35.0° = 37.28 m/s

a = -9.8 m/s²

Find: t

Δy = v₀ t + ½ at²

-115 = 37.28 t − 4.9 t²

4.9 t² − 37.28 t − 115 = 0

t = [ 37.28 ± √(37.28² − 4(4.9)(-115)) ] / 9.8

t = (37.28 ± 60.37) / 9.8

t = 9.96

2.0 kg into the bifactor of 1/4 so the mass struck would be 32

Answer:

The Ic will be zero.

Explanation:

Capacitors have a working principal as follows:

An AC voltage increases and decreases between certain maximum and minimum points periodically. So while the AC voltage is on the positive side, the capacitor charges up and when the AC voltage crosses to the negative side, the capacitor takes over and it's current starts increasing as the current coming from the AC source decreases.

So in this case, as the AC voltage crosses zero, the capacitor current was decreasing because the AC voltage was on the positive side and it was charging. The capacitor current will be zero as well and it will start to increase when AC voltage is on the negative.

I hope this answer helps.

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 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:

modem, digital music players, optical communication

Explanation:

Photos of devices are attached. An example of R2R logic circuit is also attached. DAC conversion is also done via pulse code modulation

Answer:

1111

1Explanation:

111111

1

Answer:

Explanation:

v(t) = 12 sin(913t + 71°) volts

a )  913° = (π / 180) x  913 radians

= 15.92 radians

a ) angular frequency  ω = 15.92 radians / s

b ) ω = 2πn

n = ω / 2π

= 15.92 / 2 x 3.14

= 2.53 Hz

c ) Period = 1 / n

= 1 / 2.53 = .4 s .

d )

Maximum voltage = 12 volt

e) Minimum volts = - 12 volts

g  ) rms volts = V / √2

= 12 / √2

= 8.48 V

h )

Average voltage = 0

j ) Average power

Vrms² / R

= 12 x 12 / 2 x 220

= .327 W.

k )

v(t) = 12 sin(913t + 71°)

v(t) = 12 sin(913x .003 + 71°)

=  12 sin(73.7°)

= 11.5 V .

Explanation:

We have,

Speed of plane a is 900 km/h

Plane b is moving at a rate of [tex]\dfrac{250\ km}{5\ h}=50\ km/h[/tex]

It is required to find which plane is faster. To find which plane is faster, we need to compare their speeds.

Speed of a plane a is 900 km/h and that of plane b is 50 km/h. So, we can say that plane a is moving faster.

Answer:

[tex]10.07m/s^2[/tex]

Explanation:

Information we have:

velocities:

initial velocity: [tex]v_{i}=0mph[/tex] (starts from rest)

final velocity: [tex]v_{f}=50mph[/tex]

time:  [tex]t=2.22s[/tex]

Since we need the answer in [tex]m/s^2[/tex], we nees to convert the speed to meters per second:

[tex]v_{f}=\frac{50miles}{1hour}(\frac{1hour}{3,600s} )(\frac{1609.34meters}{1mile} ) \\\\v_{f}=\frac{50*1609.34}{3600} m/s\\\\v_{f}=22.35m/s[/tex]

We find the acceleration with the following formula:

[tex]a=\frac{v_{f}-v_{i}}{t}[/tex]

substituting the known values:

[tex]a=\frac{22.35m/s-0m/s}{2.22s}\\ \\a=10.07m/s^2[/tex]

the acceleration is 10.07[tex]m/s^2[/tex]

Answer:

The acceleration of the Cheetah is found to be 10.07 m/s²

Explanation:

We have the following data in this question:

Vf = Final Velocity of the Cheetah = 50 miles/hr

Vf = (50 miles/hr)(1 hr/ 3600 s)(1609.34 m/1 mile)

Vf = 22.35 m/s

Vi = Initial Speed of Cheetah = 0 m/s (Since, Cheetah starts from rest)

t = time interval passed = 2.22 s

So, in order to find the acceleration, we simply use the formula for acceleration, that is:

Acceleration = a = (Vf - Vi)/t

a = (22.35 m/s - 0 m/s)/2.22 s

a = 10.07 m/s²

The acceleration of the Cheetah is found to be 10.07 m/s²

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:

24 rad/s

Explanation:

.................

Answer:

g = 15.5 m/s²

Explanation:

In order to find the acceleration due to gravity near the surface of this planet can be calculated by using 2nd equation of motion. The 2nd equation of motion is given as:

h = Vi t + (0.5)gt²

where,

h = height covered by the wrench = 5 m

Vi = Initial Velocity = 0 m/s

t = Time Taken to hit the ground = 0.804 s

g = acceleration due to gravity near the surface of the planet = ?

Therefore,

5 m = (0 m/s)(0.804 s) + (0.5)(g)(0.804 s)²

g = (5 m)/(0.3232 s²)

g = 15.5 m/s²

Answer:

14285.71N/m2

Explanation:

Young modulus known as modulus of elasticity is defined as;

E = σ/ε

Where E is young modulus

ε is strain defined as extention / length

σ is stress defined as force /area

Let's calculate σ;

Force = mass× gravity =200/1000. × 10 =0.2kg× 10 = 2N

The area impacted by the force is raduis surface of the wire and it's calculated thus;

πr^2 = 22/7 × (0.04)^2 ; 4cm in m = 0.04m=0.005m2

Hence σ = 2/0.005=400N/M2

Let's calculate ε;

ε= extension/ original length

extension = final length - original length

extension=25.7-25=0.7cm= 0.007m

ε= 0.7/25

Hence E = 400/ 0.7/25

E = 400 × 25/ 0.7= 14285.71N/m2

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) B = 0.015 T

b) ФB = 4.71*10⁻6 W

c) emf = 7.89*10^-4 V

d) B = 60.31 T

Explanation:

(a) To find the magnitude of the magnetic field inside the solenoid you use the following formula:

[tex]B=\frac{\mu_oNI}{L}[/tex]      (1)

B: magnitude of the magnetic field

μo: magnetic permeability of vacuum = 4π*10^-7 T/A

N: turns of the solenoid = 300

L: length = 2.50cm = 0.025 m

I: current = 12 A

You replace the values of the variables in the equation (1):

[tex]B=\frac{(4\pi *10^{-7}T/A)(300)(12A)}{0.3m}=0.015T[/tex]

(b) The magnetic flux is given by:

[tex]\Phi_B=BA[/tex]

A: area = π(0.01m)^2 = 3.1415*10^-4 m^2

[tex]\Phi_B=(0.015T)(3.1415*10^{-4}m^2)=4.71*10^{-6}W[/tex]

(c) The induced emf is given by the following formula:

[tex]emf=-\frac{\Delta \Phi_B}{\Delta t}=\frac{A(B_f-B_i)}{\Delta t}[/tex]      (2)

Bf and Bi are the final and initial magnetic field. You use the equation (1) into the equation (2):

[tex]emf=-A\mu_o\frac{N}{L}\frac{(I_2-I_1)}{\Delta t}\\\\emf=-\pi(0.01m)^2(4\pi*10^{-7}T/A)\frac{300}{0.3m}\frac{(10A-12A)}{0.001s}\\\\emf=7.89*10^{-4}V[/tex]

(d) With the ferromagnetic material inside the solenoid the magnetic field inside is modified as following:

[tex]B=\frac{\mu NI}{L}\\\\\mu=4000\mu_o=4000(4\pi*10^{-7}T/A)=5.026*10^{-3}T/A\\\\B=\frac{(5.026*10^{-3}T/A)(300)(12A)}{0.3m}=60.31T[/tex]

Following are the calculation to the given points:

Given:

[tex]r = 1.25\ cm \\\\l=30\ cm\\\\ N= 300 \\\\I=12\ A\\\\[/tex]

To find:

[tex]B=?\\\\\Phi_B=?\\\\[/tex]

Solution:

For point a:

Using formula:    

[tex]\to \mu_o= 4 \pi 10^{-7} \ \frac{T}{A}[/tex]

[tex]\to N= 300\\\\ \to L= 2.50\ cm = 0.025\ m\\\\ \to I= 12 \ A\\\\[/tex]

[tex]\to B = \frac{\mu_o \ N\ I}{L}\\[/tex]

        [tex]=\frac{4 \pi \times 10^{-7}\ \frac{T}{A} \times 300 \times 12\ A}{0.3\ m}\\\\=\frac{4 \times 3.14 \times 10^{-7}\ \frac{T}{A} \times 300 \times 12\ A}{0.3\ m}\\\\=\frac{48 \times 3.14\ T \times 300 }{0.3\ m \times 10^{7}}\\\\=\frac{ 150.72 \ T \times 3000 }{3\ m \times 10^{7}}\\\\=\frac{ 150.72 \ T \times 1000 }{10^{7}}\\\\= 0.015\ T\\\\[/tex]

For point b:

[tex]\to A : area = \pi (0.01m)^2 = 3.14 \times 10^{-4}\ m^2\\\\[/tex]

Using formula:

[tex]\to \Phi_B = BA\\\\[/tex]

          [tex]= (0.015\ T)(3.14 \times 10^{-4} \ m^2)\\\\ = 4.71 \times 10^{-6}\ W[/tex]

For point c:

 Using formula:

[tex]\to emf = -A \mu_o \frac{N}{L} \frac{(I_2-I_1)}{\Delta t}[/tex]

            [tex]= -\pi(0.01 \ m)^2(4\pi \times 10^{-7}\ \frac{T}{A})\frac{300}{0.3\ m} \frac{(10A-12A)}{0.001\ s} \\\\= -3.14 (0.01 \times 0.01 \ m)(4\times 3.14 \times 10^{-7}\ \frac{T}{A})\frac{300}{0.3} \frac{(-2A)}{0.001\ s} \\\\= -3.14 (1 \ m)(12.56\times 10^{-7}\ \frac{T}{A}) 10^6 \frac{(-2A)}{ 10^3\ s} \\\\= -3.14 (1 \ m)(12.56 \frac{T}{A}) \frac{(-2A)}{ 10^4\ s} \\\\= 7.89 \times 10^{-4}\ V\\\\[/tex]

For point d:  

[tex]\to B = \frac{\mu NI}{L}\\\\\to \mu = 4000 \mu_o = 4000(4\pi \times 10^{-7}\ \frac{T}{A}) = 5.026 \times 10^{-3} \frac{T}{A}\\\\ \to B = \frac{(5.026\times 10^{-3} \ \frac{T}{A})(300)(12\ A)}{0.3\ m} = 60.31\ T\\\\[/tex]

Learn more:

brainly.com/question/25562052

Answer:

The answer is 3,064x[tex]e^{-4}[/tex]

Explanation:

When the collision happens, the momentum of the first car is applied to the both of them.

So we can calculate the force that acts on both cars as:

Now we can find the force that acts on both of them by using the formula F = m.a which will give us the result as:

The friction force acts in the opposite direction and if they stop after moving 2.12 meters;

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

Answer:

Work done in moving an object in circular path is always 0 J

Explanation:

Work doen on any moving object, in general, is given by the following formula:

W = F d Cos θ

where,

W = Work Done

F = Force applied

d = Distance covered

θ = Angle between the direction of motion and the force applied

When the object is moving in a circular path the direction of its motion is always tangential to the circular path. While, the force applied on the object is the centripetal force. And centripetal force is always directed towards the center of the circular path. Therefore, the force and direction of motion are always perpendicular to each other. This means θ = 90°

Therefore,

W = F d Cos 90° = F d(0)

W = 0 J

Answer:

vx' = 8.7 m/s

vy' = - 2.9 m/s

Explanation:

The average acceleration during a time period is given by the formula:

a = (v' - v)/(t' - t)

where,

a = average acceleration

v' = Final Velocity

v = Initial Velocity

t' = Ending Time

t = Starting Time

Now, for x-component of velocity:

a = ax = 2 m/s²

v' = vx' = ?

v = vx = 0.5 m/s

t' = 4.1 s

t = 0 s

Therefore,

2 m/s² = (vx' - 0.5 m/s)/(4.1s - 0s)

(2 m/s²)(4.1 s) = vx' - 0.5 m/s

8.2 m/s + 0.5 m/s = vx'

vx' = 8.7 m/s

For y- component of velocity:

a = ay = -1 m/s²

v' = vy' = ?

v = vy = 1.2 m/s

t' = 4.1 s

t = 0 s

Therefore,

- 1 m/s² = (vy' - 1.2 m/s)/(4.1s - 0s)

(- 1 m/s²)(4.1 s) = vy' - 1.2 m/s

- 4.1 m/s + 1.2 m/s = vy'

vy' = - 2.9 m/s

Answer:

Palladium

Explanation:

Answer:

palladium-106

Explanation:

46 protons -46 electrons=no charge

46 electrons +60neutron = 106

Thus this is called palladium -106

Answer:

55.56kg

Explanation:

Given:

F= 52N

a=0.936m/s²

Applyinc Newton's second law, that states: force is equal to mass times acceleration.

F = ma

m=F/a =>52 / 0.936

m=55.56kg

Answer:

a) x = 0.098

b) x = 2.72 m

Explanation:

(a) To find the stretch of the fire net when the same person is lying in it, you can assume that the net is like a spring with constant spring k. It is necessary to find k.

When the person is falling down he acquires a kinetic energy K, this energy is equal to the elastic potential energy of the net when it is max stretched.

Then, you have:

[tex]K=U\\\\\frac{1}{2}mv^2=\frac{1}{2}kx^2[/tex]        (1)

m: mass of the person = 62kg

k: spring constant = ?

v: velocity of the person just when he touches the fire net = ?

x: elongation of the fire net = 1.4 m

Before the calculation of the spring constant, you calculate the final velocity of the person by using the following formula:

[tex]v^2=v_o^2+2gy[/tex]

vo: initial velocity = 0 m/s

g: gravitational acceleration = 9.8 m/s^2

y: height from the person jumps = 20.0m

[tex]v=\sqrt{2gy}=\sqrt{2(9.8m/s^2)(20.0m)}=14\frac{m}{s}[/tex]

With this value you can find the spring constant k from the equation (1):

[tex]mv^2=kx^2\\\\k=\frac{mv^2}{x^2}=\frac{(62kg)(14m/s)^2}{(1.4m)^2}=6200\frac{N}{m}[/tex]

When the person is lying on the fire net the weight of the person is equal to the elastic force of the fire net:

[tex]W=F_e\\\\mg=kx[/tex]

you solve the last expression for x:

[tex]x=\frac{mg}{k}=\frac{(62kg)(9.8m/s^2)}{6200N/m}=0.098m[/tex]

When the person is lying on the fire net the elongation of the fire net is 0.098m

b) To find how much would the net stretch, If the person jumps from 38 m, you first calculate the final velocity of the person again:

[tex]v=\sqrt{2gy}=\sqrt{2(9.8m/s^2)(38m)}=27.29\frac{m}{s}[/tex]

Next, you calculate x from the equation (1):

[tex]x=\sqrt{\frac{mv^2}{k}}=\sqrt{\frac{(62kg)(27.29m/s)^2}{6200N/m}}\\\\x=2.72m[/tex]

The net fire is stretched 2.72 m

A) If the person is lying on the net, the net would stretch by : 0.0679 m

B) If the person jumped from 38m, the net would stretch by : 1.61 m

Given data :

Mass of person (m) = 62 kg

Initial height ( H₁ ) = 20 m

Net stretch ( H[tex]_{f}[/tex] ) = -1.2 m

Initial potential energy before jump ( PE₁ ) = mgH₁ = 62 * 9.81 * 20 = 12164.4J

Potential energy after the jump ( PE₂ ) = mgH[tex]_{f}[/tex] = 62 * 9.81 * ( -1.2 ) = -729.86J

The potential spring constant ( SE ) = 1/2 kx²

                                                           = 1/2 * k * ( 1.2 )²

Applying the principle of energy conservation

PE₁ = PE₂ + SE

12164.4 = - 729.86 + 0.5 * k * 1.44

∴ k = ( 12164.4 + 729.86 ) / ( 0.5 * 1.44 )

     = 17908.69

1/2 kx² = mgx  ---- ( 1 )

Where x = stretch

equation ( 1 ) becomes equation ( 2 )

1/2 kx = mg ----- ( 2 )

x ( amount of stretch ) = ( 2mg ) / k

                                     = ( 2 * 62 * 9.81 ) / 17908.69

                                     = 0.0679 m

Hi = 38 m

MgHi = mgH[tex]_{f}[/tex] + 1/2 kx² ----- ( 3 )

where ; H[tex]_{f}[/tex] = -x

Back to equation ( 3 )

62 * 9.81 * 38 = 62 * 9.81 * (-x) + 1/2 ( 17908.69 x² )

23112.36 + 608.22x - [tex]\frac{17908.69 x^{2} }{2}[/tex] = 0

8954.345 x² - 608.22x - 23112.36 = 0 ----- ( 4 )

Resolving the quadratic equation ( 4 )

x = ± 1.60623 ≈ 1.61 m

Therefore we can conclude that the net would stretch 1.61 m if the person jumped from 38 m

Hence we can conclude that If the person is lying on the net, the net would stretch by : 0.0679 m, and If the person jumped from 38m the net would stretch by : 1.61 m

Learn more about simple spring : https://brainly.com/question/15727787

Answer:

a)  I = 1.5 A , b)  P = 18 W , c) A = 8 10⁻⁷ m², d)  σ = 0.625 10⁸ (Ω m)⁻¹,

e) C = 0.125 10⁻⁶ F

Explanation:

a) for this exercise let's use Ohm's Law

        V = I R

        I = V / R

        I = 12/8

        I = 1.5 A

b) the power is given by the expression

        P = V I

        P = V V / R = V² / R

        P = 12²/8

        P = 18 W

c) the resistance of the wire is given by

        R = ρ l / A

        A = ρ l / R

        A = 1.6 10⁻⁸ 100/8

        A = 8 10⁻⁷ m²

d) conductivity is the inverse of resistivity

        σ = 1 / ρ

        σ = 1 / 1,610⁻⁸

        σ = 0.625 10⁸ (Ω m)⁻¹

e) In an RC circuit the response time is

            τ = RC

            C = τ / R

            C = 1/8 10⁶

             C = 0.125 10⁻⁶ F

Answer:

2.4 × 10⁵ cal

Explanation:

Step 1: Given data

Step 2: Convert the mass to grams

We will use the relationship 1 kg = 1,000 g.

[tex]4kg \times \frac{1,000g}{1kg} = 4,000 g[/tex]

Step 3: Calculate the change in the temperature

[tex]\Delta T = T_f - T_i = 80 \° C - 20 \° C = 60 \° C[/tex]

Step 4: Calculate the heat required (Q)

We will use the following expression.

[tex]Q = c \times m \times \Delta T = \frac{1cal}{g. \° C} \times 4,000 g \times 60 \° C = 2.4 \times 10^{5} cal[/tex]

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:

I = 0.483 kgm^2

Explanation:

To know what is the moment of inertia I of the boxer's forearm you use the following formula:

[tex]\tau=I\alpha[/tex]  (1)

τ: torque exerted by the forearm

I: moment of inertia

α: angular acceleration = 125 rad/s^2

You calculate the torque by using the information about the force (1.95*10^3 N) and the lever arm (3.1 cm = 0.031m)

[tex]\tau=Fr=(1.95*10^3N)(0.031m)=60.45J[/tex]

Next, you replace this value of τ in the equation (1) and solve for I:

[tex]I=\frac{\tau}{\alpha}=\frac{60.45Kgm^2/s ^2}{125rad/s^2}=0.483 kgm^2[/tex]

hence, the moment of inertia of the forearm is 0.483 kgm^2

Answer:

Galvanómetro

Explanation:

Un galvanómetro es un dispositivo eléctrico utilizado para detectar la presencia de corriente y voltaje pequeños o para medir su magnitud. Los galvanómetros se utilizan principalmente en puentes y potenciómetros.

Answer:

The magnitude will be "2212 N".

Explanation:

The given values are:

Initial speed, u = 250 m/s

Final speed, v = 0

Mass, m = 16.0 = 0.016 kg

Distance, d = 22.6 cm

On converting cm into m, we get

d = 0.226 m

As we know,

⇒  [tex]v^2-u^2=-2ad[/tex]

⇒  [tex]a=\frac{v^2-u^2}{-2d}[/tex]

On putting the estimated values, we get

⇒    [tex]=\frac{(0)^2-(250)^2}{-2\times 0.226}[/tex]

⇒    [tex]=\frac{0-62500}{-0.452}[/tex]

⇒    [tex]=13,8276 \ m/s^2[/tex]

Now,

Force, [tex]F=ma[/tex]

On putting values, we get

⇒            [tex]=0.016\times 138276[/tex]

⇒            [tex]=2212 \ N[/tex] (magnitude)

The magnitude of the stopping force on the bullet is 2,212.4 N in opposite direction to the bullet.

The given parameters;

The acceleration of the bullet is calculated as follows;

v² = u² - 2ad

where;

0 = 250² - a(2 x 0.226)

0.452a = 62500

[tex]a = \frac{62500}{0.452} \\\\a = 138,274.34 \ m/s^2\\\\[/tex]

The magnitude of the stopping force on the bullet is calculated as follows;

F = ma

F = 0.016 x 138,274.34

F = 2,212.4 N

Thus, the magnitude of the stopping force on the bullet is 2,212.4 N in opposite direction to the bullet.

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