A force of 30 N is required to hold a spring that has been stretched from its natural length of 20 cm to a length of 35 cm. How much work is done in stretching the spring from 35 cm to 40 cm

Answer:

[tex]v_i = 1.44\frac{m}{s}[/tex]

Explanation:

The computation of the initial velocity of the fly is shown below:-

But before that we need to do the following calculations

For 4.22 seconds

[tex]\bar v = \frac{-5.80 m}{4.22 s}[/tex]

[tex]= -1.37\frac{m}{s}[/tex]

For uniform acceleration

[tex]\bar v = \frac{v_i +v_f}{2}[/tex]

[tex]= v_i + v_f[/tex]

[tex]= -2.74\frac{m}{s}[/tex]

With initial and final velocities

[tex]= -1.33\frac{m}{s^2}[/tex]

[tex]= \frac{v_i +v_f}{4.22s}[/tex]

[tex]= -v_i + v_f[/tex]

[tex]= -5.61\frac{m}{s}[/tex]

So, the initial velocity is

[tex]v_i = 1.44\frac{m}{s}[/tex]

We simply applied the above steps to reach at the final solution i.e initial velocity

Answer:

number of electrons = 2.18*10^18 e

Explanation:

In order to calculate the number of electrons that move trough the second wire, you take into account one of the Kirchoff's laws. All the current that goes inside the junction, has to go out the junction.

Then, if you assume that the current of the wire 1 and 3 go inside the junction, then, all this current have to go out trough the second junction:

[tex]i_1+i_3=i_2[/tex]                 (1)

i1 = 0.40 A

i2 = 0.75 A

you solve the equation i3 from the equation (1):

[tex]i_3=i_2-i_1=0.75A-0.40A=0.35A[/tex]

Next, you take into account that 1A = 1C/s = 6.24*10^18

Then, you have:

[tex]0.35A=0.35\frac{C}{s}=0.35*\frac{6.24*10^{18}e}{s}=2.18*10^{18}\frac{e}{s}[/tex]

The number of electrons that trough the wire 3 is 2.18*10^18 e/s

Answer:

same direction = 80 (n)

opposite direction = 20 (n) going one direction

Explanation:

same direction means they are added to each other

and opposite means acting on eachother

Answer:

[tex]r_f=\frac{1}{6}r_i[/tex]

Explanation:

To find the new separation of the charges, you first take into account the formula for the electric force, when the force are separated a distance of ri.

You use the following expression:

[tex]F_i=k\frac{q_Aq_B}{r_i^2}[/tex]          (1)

k: Coulomb's constant

qA: charge of A particle

qB: charge of B particle

When the charges are separated to a new distance rf, the new force is 36Fi, if the charges have not changed, you have:

[tex]F_f=36F_i=k\frac{q_Aq_B}{r_f^2}[/tex]         (2)

To find the new separation you replace the expression for Fi of the equation (1) into the equation (2) and solve for rf in terms of ri:

[tex]36F_i=36k\frac{q_Aq_B}{r_i^2}=k\frac{q_Aq_B}{r_f^2}\\\\\frac{36}{r_i^2}=\frac{1}{r_f^2}\\\\r_f=\frac{1}{6}r_i[/tex]

The new separation of the charges is 1/6 times of the initial separation

Answer:

Explanation:

Given that

speed u=4*10^6 m/s

electric field E=4*10^3 N/c

distance b/w the plates d=2 cm

basing on the concept of the electrostatices

now we find the acceleration b/w the plates

acceleration a=qE/m=1.6*10^-19*4*10^3/9.1*10^-31=0.7*10^15 =7*10^14 m/s

now we find the horizantal distance travelled by electrons hit the plates

horizantal distance X=u[2y/a]^1/2

=4*10^6[2*2*10^-2/7*10^14]^1/2

=3*10^-2=3 cm

now we find the velocity f the electron strike the plate

v^2-(4*10^6)^2=2*7*10^14*2*10^-2

v^2=16*10^12+28*10^12

v^2=44*10^12

speed after hits =>V=6.6*10^6 m/s

Answer:

v(t) = s′(t) = −9sin(t)+9cos(t)

a(t) = v′(t) = −9cos(t) −9sin(t)

Explanation:

Given that

s = 9 cos(t) + 9 sin(t), t ≥ 0

Then acceleration and velocity is

v(t) = s′(t) = −9sin(t)+9cos(t)

a(t) = v′(t) = −9cos(t) −9sin(t)

Answer:

E = - dV/dx

Explanation:

Las superficies equipòtenciales son superficie donde el potencial eléctrico es constante por lo cual nos podemos desplazaren ella sin realizar nigun trabajo.

El campo electrico es el campo que existen algún punto en el espacio creado por alguna ddistribucion de carga.

De los antes expuesto las dos magnitudes están relacionadas

         E = - dV/dx

por lo cual el potenical es el gradiente del potencial eléctrico.

Como el campo eléctrico sobre un superficie equipotenciales constante, podemos colocar una punta de prueba con un potencial dado y seguir la linea que de una diferencia de potencial constar, lo cual permite visualizar las forma de cada linea equipotencial

Answer:

Radius r = 20.34 cm

The radius that can produces such a disk is 20.34 cm

Explanation:

Area of a circle;

A = πr^2

A = area

r = radius

Making r the subject of formula;

r = √(A/π) ........1

Given;

A = 1300 cm^2

Substituting into the equation 1;

r = √(1300/π)

r = 20.34214472564 cm

r = 20.34 cm

The radius that can produces such a disk is 20.34 cm

Answer:

A pie chart is a type of graph in which a circle is divided into sectors that each represents a proportion of the whole

Explanation:

pie charts are a useful way to organize data in order to see the size of components relative to the whole.

Answer:

[tex]\large \boxed{\text{3.3 cm}^{3}}[/tex]

Explanation:

Assume the stone consists of basalt, which has a density of 3.0 g/cm³.

[tex]\rho = \text{10 g}\times\dfrac{\text{1 cm}^{3}}{\text{3.0 g}} = \text{3.3 cm}^{3}\\\\\text{The volume of the stone is $\large \boxed{\textbf{3.3 cm}^{3}}$}[/tex]

Answer:

Only the horizontal component of a projectile’s momentum is conserved. Where as The vertical component of the momentum is not conserved, because the net vertical force Fy–net is not zero

Explanation:

Complete question:

A solid conducting sphere is placed in an external uniform electric field. With regard to the electric field on the sphere's interior, which statement is correct?

A. the interior field points in a direction parallel to the exterior field

B. There is no electric field on the interior of the conducting sphere.

C. The interior field points in a direction perpendicular to the exterior field.

D. the interior field points in a direction opposite to the exterior field.

Answer:

B. There is no electric field on the interior of the conducting sphere.

Explanation:

Conductors are said to have free charges that move around easily. When the conductor is now placed in a static electric field, the free charges react to attain electrostatic equilibrium (steady state).

Here, a solid conducting sphere is placed in an external uniform electric field. Until the lines of the electric field are perpendicular to the surface, the free charges will move around the spherical conductor, causing polarization. There would be no electric field in the interior of the spherical conductor because there would be movement of  free charges in the spherical conductor in response to any field until its neutralization.

Option B is correct.

There is no electric field on the interior of the conducting sphere.

Answer:

-1748*10^N/C

Explanation:

See attached file

Answer:

The atomic weight of the metal is 58.7 g/mol

Explanation:

Given;

density of the metal sample, ρ = 8.91 g/cm³

edge length of the face centered cubic cell, α = 352.4 pm = 352.4 x 10⁻¹⁰ cm

Volume of the unit cell of the metal;

V = α³

V = (352.4 x 10⁻¹⁰  cm)³

V = 4.376 x 10⁻²³ cm³

Mass of the metal in unit cell

mass = density x volume

mass = 8.91 g/cm³ x 4.376 x 10⁻²³ cm³

mass = 3.899 x 10⁻²² g

Atomic weight, based on 4 atoms per unit cell;

4 atoms = 3.899 x 10⁻²² g

6.022 x 10²³ atoms = ?

= (6.022 x 10²³atoms x 3.899 x 10⁻²² g) / (4 atoms)

= 58.699 g/mol

= 58.7 g/mol (this metal is Nickel)

Theerefore, the atomic weight of the metal is 58.7 g/mol

Answer:

a. 70 rad/s²

b. 5000 rev

Explanation:

As we know,

[tex]\omega = 20000\frac{rev}{min}\frac{2 \pi rad}{1 \ rev}\frac{1 \ min}{60 \ sec}[/tex]

then,

[tex]\omega=2100 \ rad/s[/tex]

a...

⇒  [tex]\bar{\alpha}=\frac{\omega-\omega_{0}}{\Delta t}[/tex]

On substituting the values, we get

⇒      [tex]=\frac{2100}{30}[/tex]

⇒      [tex]=70 \ rad/s^2[/tex]

b...

⇒  [tex]\theta=\theta_{0}=\omega_{0}t+\frac{1}{2}\alpha t^2[/tex]

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

       [tex]=\frac{1}{2}\times 70\times (30)^2[/tex]

       [tex]=31500 \ rad[/tex]

       [tex]=31500 \ rad\frac{1 \ rev}{2\pi rad}[/tex]

       [tex]=5000 \ rev[/tex]

(a) The average angular acceleration will be 70 rad/s².

(b) 20063.69 revolutions have the centrifuge rotor turned during its acceleration period.

Angular acceleration is defined as the pace of change of angular velocity with reference to time. It is denoted by α. Its unit is rad/s².

The given data in the problem is;

n is the revolution of centrifugal rotor =  20000 rpm

t is the time interval= the 30s

(α) is the Angular acceleration=?

n₁ is the revolution when the acceleration is constant =?

(a) The average angular acceleration will be  70 rad/s².

The value of the angular velocity is given by

[tex]\rm \omega_f = \frac{2\pi N}{60} } \\\\ \rm \omega_f = \frac{2 \times 3.14 \times 20000}{60} \\\\ \rm \omega_f= 2100\ rad/sec.[/tex]

The formula for angular acceleration is guven by;

[tex]\rm \alpha =\frac{ \omega_f-\omega_i}{dt} \\\\ \rm \alpha =\frac{ 2100-0}{3}\\\\ \rm \alpha =70\ rad/sec^2[/tex]

Hence the average angular acceleration will be 70 rad/s².

(b 5000 revolutions have the centrifuge rotor turned during its acceleration period.

[tex]\rm \theta= \theta_0+\frac{1}{2} \alpha t^2 \\\\ \rm \theta= \frac{1}{2} \times 70 \times (30)^2 \\\\ \rm \theta=31500\ rad[/tex]

As we know that the angular velocity is given by

[tex]\rm \omega = \frac{\theta}{t} \\\\ \rm \omega = \frac{31500}{30} \\\\ \rm \omega = 1050 \ rad/sec[/tex]

The relation of angular velocity and revolution will be

[tex]\rm n= \frac{ \omega \times 60}{2\pi} \\\\ \rm n= \frac{ 2100 \times 60}{2\times 3.14 } \\\\ \rm n = 20063.69 \ rev[/tex]

Hence 20063.69 revolutions have the centrifuge rotor turned during its acceleration period.

To learn more about angular acceleration refer to the link ;

https://brainly.com/question/408236

Answer:

[tex]I = 0.083 kg m^2[/tex]

Explanation:

Mass of the bucket, m = 23 kg

Radius of the pulley, r = 0.050 m

The bucket is released from rest, u = 0 m/s

The time taken to fall, t = 2 s

Speed, v = 8.0 m/s

Moment of Inertia of the pulley, I = ?

Using the equation of motion:

v = u + at

8 = 0 + 2a

a = 8/2

a = 4 m/s²

The relationship between the linear and angular accelerations is given by the equation:

[tex]a = \alpha r[/tex]

Angular acceleration, [tex]\alpha = a/r[/tex]

[tex]\alpha = 4/0.050\\\alpha = 80 rad/s^2[/tex]

Since the bucket is falling, it can be modeled by the equation:

mg - T = ma

T = mg - ma = m(g-a)

T = 23(9.8 - 4)

The tension, T = 133.4 N

The equation for the pulley can be modeled by:

[tex]T* r = I * \alpha\\133.4 * 0.050 = I * 80\\6.67 = 80 I\\I = 6.67/80\\I = 0.083 kg m^2[/tex]

Answer:

2.45 m/s

Explanation:

kinetic energy = 1/2 * m * v^2

then, 0.5 * 2500 * x^2 = 0.5 * 67 * 15^2

by solving for x, X = 2.45 m/s

Answer:

The potential at the center of the sphere is -924 V

Explanation:

Given;

radius of the sphere, R = 0.22 m

electric field at the surface of the sphere, E = 4200 N/C

Since the electric field is directed towards the center of the sphere, the charge is negative.

The Potential is the same at every point in the sphere, and it is given as;

[tex]V = \frac{1}{4 \pi \epsilon_o} \frac{q}{R}[/tex] -------equation (1)

The electric field on the sphere is also given as;

[tex]E = \frac{1}{4 \pi \epsilon _o} \frac{|q|}{R^2}[/tex]

[tex]|q |= 4 \pi \epsilon _o} R^2E[/tex]

Substitute in the value of q in equation (1)

[tex]V = \frac{1}{4 \pi \epsilon_o} \frac{-(4 \pi \epsilon _o R^2E)}{R} \ \ \ \ q \ is \ negative\ because \ E \ is\ directed \ toward \ the \ center\\\\V = -RE\\\\V = -(0.22* 4200)\\\\V = -924 \ V[/tex]

Therefore, the potential at the center of the sphere is -924 V

Answer:

Explanation:

given:

radius (r) = 1.88 km

velocity (v) = 136.3 km/hr

required:

banking angle ∡ ?

first:

convert 1.88 km to m = 1.88km * 1000m / 1km

r = 1880 m

convert velocity v = 136.3 km/hr to m/s = 136.3 km/hr * (1000 m/ 3600s)

v = 37.86 m/s

now.. calculate the angle

Ф = inv tan (v² / r * g)            we know that gravity = 9.8 m/s²

Ф = inv tan (37.86² / (1880 * 9.8))

Ф = 4.4°

Answer:

-i - 7j

Explanation:

The computation of components of the force is shown below:-

torque T = r cross F

T = (4.00 i + 5.00 j + 0 k) X (1.00 i + 7.00 j)

Now we will cross multiplying vectorilly

T = 4 × (iXi) + 28 × (jXi) + 0 + 5 × (iXj)35 × (jXj) + 0

T = 4 × 0 + 28 × (-k) + 5 × (k) + 35 × 0

T = 28 × (-k) + 5 × (k) = -23k

net torque |T| = 23 N - m

direction >> negative k

Or Simply we can do by the below method

r × f

28 - 5 = 23 K

-i - 7j

Answer:

the acceleration during the collision is: - 5  [tex]\frac{m}{s^2}[/tex]

Explanation:

Using the formula:

[tex]a=\frac{\Delta\,v}{\Delta\,t}[/tex]

we get:

[tex]a=\frac{-0.4-0.6}{0.2} \,\frac{m}{s^2} =\frac{-1}{0.2} \,\frac{m}{s^2} =-5\,\,\frac{m}{s^2}[/tex]

Answer:

The speed of the rod is 1.383 m/s

Explanation:

Given;

length of the conducting rod, L = 31 cm = 0.31 m

induced emf on the rod, emf = 0.75V

magnetic field around the rod, B = 1.75 T

Apply the following Faraday's equation for electromagnetic induction in a moving rod to determine the speed of the rod.

emef = BLv

where;

B is the magnetic field

L is length of the rod

v is the speed of the rod

v = emf / BL

v = (0.75) / (1.75 x 0.31)

v = 1.383 m/s

Therefore, the speed of the rod is 1.383 m/s

Answer:

(a)    f = 0.58Hz

(b)    vmax = 0.364m/s

(c)    amax = 1.32m/s^2

(d)    E = 0.1J

(e)    [tex]x(t)=0.1m\ cos(2\pi(0.58s^{-1})t)[/tex]

Explanation:

(a) The frequency of the oscillation, in a spring-mass system, is calulated by using the following formula:

[tex]f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}[/tex]          (1)

k: spring constant = 20.0N/m

m: mass = 1.5kg

you replace the values of m and k for getting f:

[tex]f=\frac{1}{2\pi}\sqrt{\frac{20.0N/m}{1.5kg}}=0.58s^{-1}=0.58Hz[/tex]

The frequency of the oscillation is 0.58Hz

(b) The maximum speed is given by:

[tex]v_{max}=\omega A=2\pi f A[/tex]     (2)

A: amplitude of the oscillations = 10.0cm = 0.10m

[tex]v_{max}=2\pi (0.58s^{-1})(0.10m)=0.364\frac{m}{s}[/tex]

The maximum speed of the mass is 0.364 m/s.

The maximum speed occurs when the mass passes trough the equilibrium point of the oscillation.

(c) The maximum acceleration is given by:

[tex]a_{max}=\omega^2A=(2\pi f)^2 A[/tex]

[tex]a_{max}=(2\pi (0.58s^{-1}))(0.10m)=1.32\frac{m}{s^2}[/tex]

The maximum acceleration is 1.32 m/s^2

The maximum acceleration occurs where the elastic force is a maximum, that is, where the mass is at the maximum distance from the equilibrium point, that is, the acceleration.

(d) The total energy of the system is:

[tex]E=\frac{1}{2}kA^2=\frac{1}{2}(20.0N/m)(0.10m)^2=0.1J[/tex]

The total energy is 0.1J

(e) The displacement as a function of time is:

[tex]x(t)=Acos(\omega t)=Acos(2\pi ft)\\\\x(t)=0.1m\ cos(2\pi(0.58s^{-1})t)[/tex]

Answer:

141178.8

Explanation:

use : density x velocity²

1.2 x 343² = 141178.8 pa

Answer:

STATIC,  STATIC

KINETIC friction is less than static friction

Explanation:

In this exercise you are asked to complete the sentences with the correct words.

STATIC friction prevents the relative movement of two surfaces in contact.

For moving surfaces the friction is STATIC is greater than the kinetic friction.

For the last two sentences I think they are misspelled, the correct thing is

KINETIC friction is less than static friction

Answer:

12.5 rad/s²

Explanation:

Angular Acceleration: This can be defined as the ratio of linear acceleration and radius. The S.I unit is rad/s²

From the question,

a = αr................... Equation 1

Where a = linear acceleration, α = angular acceleration, r = radius.

But,

a = (v-u)/t.............. Equation 2

Where v = final velocity, u = initial velocity, t = time.

Substitute equation 2 into equation 1

(v-u)/t = αr

make α the subject of the equation

α = (v-u)/tr................. Equation 3

Given: v = 30 m/s, u = 0 m/s, t = 6 s, r = 0.4 m

Substitute into equation 3

α = (30-0)/(0.4×6)

α = 30/2.4

α = 12.5 rad/s²

Answer:

[tex]\frac{50}{\pi }[/tex]Hz

Explanation:

In alternating current (AC) circuits, voltage (V) oscillates in a sine wave pattern and has a general equation as a function of time (t) as follows;

V(t) = V sin (ωt + Ф)            -----------------(i)

Where;

V = amplitude value of the voltage

ω = angular frequency = 2 π f        [f = cyclic frequency or simply, frequency]

Ф = phase difference between voltage and current.

Now,

From the question,

V(t) = 230 sin (100t)              ---------------(ii)

By comparing equations (i) and (ii) the following holds;

V = 230

ω = 100

Ф = 0

But;

ω = 2 π f = 100

2 π f = 100             [divide both sides by 2]

π f = 50

f = [tex]\frac{50}{\pi }[/tex]Hz

Therefore, the frequency of the voltage is [tex]\frac{50}{\pi }[/tex]Hz

Answer:

speed of puck acc. to the radar gun = 138 km/h

speed of player = 15 km/h

since the player is in motion when he shoots, the speed of the puck will be the sum of the speed of the player and the speed at which he shot. so,

speed of puck = speed of player + speed of puck acc. to player

138 = 15 + speed of puck acc. to player

speed of puck acc. to player = 138 -15

speed of puck acc. to player = 123 km/h

Brainly this answer if you think it deserves it

Answer:

208 V

Explanation:

resistor voltage = Vr = 124 V

capacitor voltage Vc = 167 V

the total voltage in the RC circuit is the resultant voltage of the resistor and the capacitor

total voltage i= [tex]\sqrt{Vr^{2} + Vc^{2} }[/tex]

==> [tex]\sqrt{124^{2} + 167^{2} } =[/tex] 208 V

Answer:

for the first interference m = 1   y = 2,839 10-3 m

for the second interference m = 2   y = 5,678 10-3 m

Explanation:

The double slit interference phenomenon, for constructive interference is described by the expression

                d sin θ = m λ

where d is the separation between the slits, λ the wavelength and m an integer that corresponds to the interference we see.

In these experiments in general the observation screen is L >> d, let's use trigonometry to find the angles

           tan θ = y / L

with the angle it is small,

          tan θ = sin θ / cos θ = sin θ

we substitute

         sin θ = y / L

         d y / L = m λ

the distance between the central maximum and an interference line is

        y = m λ L / d

let's reduce the magnitudes to the SI system

     λ = 546 nm = 546 10⁻⁹ m

     d = 0.25 mm = 0.25 10⁻³ m

let's substitute the values

      y = m 546 10⁻⁹ 1.3 / 0.25 10⁻³

      y =  m 2,839 10⁻³

the explicit value for a line depends on the value of the integer m, for example

for the first interference m = 1

the distance from the central maximum to the first line is y = 2,839 10-3 m

for the second interference m = 2

the distance from the central maximum to the second line is y = 5,678 10-3 m