If 5.4 J of work is needed to stretch a spring from 15 cm to 21 cm and another 9 J is needed to stretch it from 21 cm to 27 cm, what is the natural length (in cm) of the spring

Answer:

a. True

Explanation:

A semiconductor can be defined as a crystalline solid substance that has its conductivity lying between that of a metal and an insulator, due to the effects of temperature or an addition of an impurity. Semiconductors are classified into two main categories;

1. Extrinsic semiconductor.

2. Intrinsic semiconductor.

An intrinsic semiconductor is a crystalline solid substance that is in its purest form and having no impurities added to it. Examples of intrinsic semiconductor are Germanium and Silicon.

Basically, the number of free electrons in an intrinsic semiconductor is equal to the number of holes. Also, the number of holes and free electrons in an intrinsic semiconductor is directly proportional to the temperature; as the temperature increases, the number of holes and free electrons increases and vice-versa.

In an intrinsic semiconductor, each free electrons (valence electrons) produces a covalent bond.

Generally, a process referred to as doping can be used to increase the conductivity of an intrinsic semiconductor such as silicon or germanium, by adding small amounts of impurities found in group 3A or group 5A elements.

Answer:

Following are the solution to the given question:

Explanation:

Please find the complete question in the attached file.

The cost after 30 days is 60 dollars. As energy remains constant, the cost per hour over 30 days will be decreased.

[tex]\to \frac{\$60}{\frac{30 \ days}{24\ hours}} = \$0.08 / kwh.[/tex]

Thus, [tex]\frac{\$0.08}{\$0.12} = 0.694 \ kW \times 0.694 \ kW \times 1000 = 694 \ W.[/tex]

The electricity used is continuously 694W over 30 days.

If just resistor loads (no reagents) were assumed,

[tex]\to I = \frac{P}{V}= \frac{694\ W}{120\ V} = 5.78\ A[/tex]

Energy usage reduction percentage = [tex](\frac{60\ W}{694\ W} \times 100\%)[/tex]

This bulb accounts for [tex]8.64\%[/tex] of the energy used, hence it saves when you switch it off.

Answer:

R/2

Explanation:

The potential at a distance r is given by :

[tex]V=\dfrac{kq}{r}[/tex]

Where

k is electrostatic constant

q is the charge

The potential (relative to infinity) due to a point charge is V at a distance R from this charge. So,

[tex]\dfrac{V_1}{V_2}=\dfrac{r_2}{r_1}[/tex]

Put all the values,

[tex]\dfrac{V}{2V}=\dfrac{r_2}{R}\\\\\dfrac{1}{2}=\dfrac{r_2}{R}\\\\r_2=\dfrac{R}{2}[/tex]

Answer:

The distance of a mass 25g from the pivot​ is 18cm

Explanation:

Given

[tex]m_1 = 10g[/tex]

[tex]d_1 = 45cm[/tex]

[tex]m_2 = 25g[/tex]

Required

Distance of m2 from the pivot

To do this, we make use of:

[tex]m_1 * d_1 = m_2 * d_2[/tex] --- moments of the masses

So, we have:

[tex]10 * 45= 25* d_2[/tex]

[tex]450= 25* d_2[/tex]

Divide both sides by 25

[tex]18= d_2[/tex]

Hence:

[tex]d_2 = 18[/tex]

Answer:

The magnitude of the electric field is same while the direction at the left and at the right is opposite to each other.

Explanation:

The direction of the electric field due to the dipole on the axial line is same as  the direction of dipole moment.

The magnitude of the electric field due to an electric dipole on its axial line is

[tex]E=\frac{2kp}{r^3}[/tex]

where, k is the constant, p is the electric dipole moment and r is the distance from the center of dipole.

The magnitude of the electric field is same while the direction at the left and at the right is opposite to each other.  

Answer:

False

Explanation:

You have a circle so think back to circular motion. Theres 2 directions, centripetal and tangential. The problem tells you there's a constant tangential speed so tangential acceleration is 0. However there is a centripetal acceleration acting on the ball that holds it in its circular motion (i.e. tension, or gravity). Since centripetal is perpendicular to the tangential direction, acceleration and velocity are in different directions.

Answer:

[tex]K.E_{max}=0.8973J[/tex]

Explanation:

From the question we are told that:

Mass [tex]m=1.64kg[/tex]

Equation of Mass

[tex]X=0.33cos(3.17t)[/tex]...1

Generally equation for distance X is

[tex]X=Acos(\omega t)[/tex]...2

Therefore comparing equation

Angular Velocity [tex]\omega=3.17rad/s[/tex]

Amplitude A=0.33

Generally the equation for Max speed is mathematically given by

[tex]V_{max}=A\omega[/tex]

[tex]V_{max}=0.33*3.17[/tex]

[tex]V_{max}=1.0461m/s[/tex]

Therefore

[tex]K.E_{max}=0.5mv^2[/tex]

[tex]K.E_{max}=0.5*1.64*(1.0461)^2[/tex]

[tex]K.E_{max}=0.8973J[/tex]

Answer:

Option A. –5 m/s²

Explanation:

From the question given above, the following data were obtained:

Initial velocity (v₁) = 9 m/s

Initial time (t₁) = 0 s

Final velocity (v₂) = –6 m/s

Final time (t₂) = 3 s

Acceleration (a) =?

Next, we shall determine the change in the velocity and time. This can be obtained as follow:

For velocity:

Initial velocity (v₁) = 9 m/s

Final velocity (v₂) = –6 m/s

Change in velocity (Δv) =?

ΔV = v₂ – v₁

ΔV = –6 – 9

ΔV = –15 m/s

For time:

Initial time (t₁) = 0 s

Final time (t₂) = 3 s

Change in time (Δt) =?

Δt = t₂ – t₁

Δt = 3 – 0

Δt = 3 s

Finally, we shall determine the acceleration of the object. This can be obtained as follow:

Change in velocity (Δv) = –15 m/s

Change in time (Δt) = 3 s

Acceleration (a) =?

a = Δv / Δt

a = –15 / 3

a = –5 m/s²

Thus, the acceleration of the object is

–5 m/s².

Answer:

The number of revolutions is 44.6.

Explanation:

We can find the revolutions of the wheel with the following equation:

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

Where:

[tex]\omega_{0}[/tex]: is the initial angular velocity = 13 rad/s              

t: is the time = 8 s

α: is the angular acceleration

We can find the angular acceleration with the initial and final angular velocities:

[tex] \omega_{f} = \omega_{0} + \alpha t [/tex]

Where:

[tex] \omega_{f} [/tex]: is the final angular velocity = 57 rad/s

[tex] \alpha = \frac{\omega_{f} - \omega_{0}}{t} = \frac{57 rad/s - 13 rad/s}{8 s} = 5.5 rad/s^{2} [/tex]

Hence, the number of revolutions is:

[tex] \theta = \omega_{0}t + \frac{1}{2}\alpha t^{2} = 13 rad/s*8 s + \frac{1}{2}*5.5 rad/s^{2}*(8 s)^{2} = 280 rad*\frac{1 rev}{2\pi rad} = 44.6 rev [/tex]

Therefore, the number of revolutions is 44.6.

I hope it helps you!

Answer:

I DON'T UNDERSTAND

Explanation:

GUESS A MISUNDERSTANDING PLZ PUT A UNDERSTANDABLE QUESTION.

Answer:

mass

Explanation:

Answer:

45 × 8 units = A + B as formular

Answer:

P = 29.3 W

Explanation:

The magnetic field in a solenoid is

          B = μ₀  n i

          i = B /μ₀ n

where n is the density of turns

We can use a direct rule of proportions or rule of three to find the number of turns, 1 a turn has a diameter of 0.100 cm = 10⁻³ m, in the length of

L= 85.1 cm = 0.851 m how many turns there are

         #_threads = 0.851 / 10⁻³

         #_threads = 8.50 10³ turns

the density of turns is

          n = # _threads / L

          n = 8.51 103 / 0.851

          n = 104 turn / m

the current that must pass through the solenoid is

          i = 8.90 10-3 / 4pi 10-7 104

          i = 0.70823 A

now let's find the resistance of the copper wire

         R = ρ L / A

the resistivity of copper is ρ = 1.72 10⁻⁸ Ω m

wire area

         A = π r²

         A = π (5 10⁻⁴)

         A = 7,854 10⁻⁷ m²

let's find the length of wire to build the coil, the length of a turn is

         Lo = 2π r = ππ d

         Lo = π 0.100

         Lo = 0.314159 m / turn

With a direct proportion rule we find the length of the wire to construct the 8.5 103 turns

          L = Lo #_threads

          L = 0.314159 8.50 10³

          L = 2.67 10³ m

resistance is

         R = 1.72 10⁻⁸ 2.67 10₃ / 7.854 10⁻⁷

         R = 5,847 10¹

         R = 58.47 ohm

The power to be supplied to the coil is

          P = VI = R i²

          P = 58.47 0.70823²

          P = 29.3 W

Answer:

Look at work

Explanation:

a) I am not sure if you want tangential or centripetal but I will give both

Centripetal acceleration = r*α

Since ω is constant, α is 0 so centripetal acceleration is 0m/s^2

Tangential acceleration = ω^2*r

convert 200rev/min into rev/s

200/60= 10/3 rev/s

a= 100/9*1.2= 120/9= 40/3 m/s^2

b) Rotational Kinetic Energy = 1/2Iω^2

I= mr^2

Plug in givens

I= 43.2kgm^2

K= 1/2*43.2*100/9=2160/9=240J

Answer:

219 ft

Explanation:

Here we can define the value t = 0s as the moment when the car starts decelerating.

At this point, the acceleration of the car is given by the equation:

A(t) = -10 ft/s^2

Where the negative sign is because the car is decelerating.

To get the velocity equation of the car, we integrate over time, to get:

V(t) = (-10 ft/s^2)*t + V0

Where V0 is the initial velocity of the car, we know that this is 88 ft/s

Then the velocity equation is:

V(t) = (-10 ft/s^2)*t + 88ft/s

To get the position equation we need to integrate again, this time we get:

P(t) = (1/2)*(-10 ft/s^2)*t^2 + (88ft/s)*t + P0

Where P0 is the initial position of the car, we do not know this, but it does not matter for now.

We want to find the total distance that the car traveled in a 3 seconds interval.

This will be equal to the difference in the position at t = 3s and the position at t = 0s

distance = P(3s) - P(0s)

 = ( (1/2)*(-10 ft/s^2)*(3s)^2+ (88ft/s)*3s + P0) - ( (1/2)*(-10 ft/s^2)*(0s)^2 + (88ft/s)*0s + P0)

=  ( (1/2)*(-10 ft/s^2)*9s^2+ (88ft/s)*3s + P0) - ( P0)

=  (1/2)*(-10 ft/s^2)*9s^2+ (88ft/s)*3s = 219ft

The car advanced a distance of 219 ft in the 3 seconds interval.

Answer:

Initially let n cannonballs with a total mass of m be to the left of the center of mass at L /2 and the mass of the car at L/2

x1 =  [-m / (m + M)] * L / 2   is the original position of the CM

x2 = (m (x + L/2) + M x) / (m + M) * L/2 final position of CM with all cannon balls to the right

[-m x - m L / 2 + m x - M x] / (M + m) * L/2

= - ( m L / 2 + M x) / (m + M) * L/2 = Xcm

Check the math, but maximum distance occurs when the cannonballs of mass m move from -L/2 to L/2 and the car of mass M moves from zero to -x

Answer:

a)  y = 0.98 t², t=1s y= 0.98 m,  

b) he two blocks must move the same distance

c) v = 1.96 m / s,  d)  a = -1.96 m / s², e)  x = 0.98 m

Explanation:

For this exercise we can use Newton's second law

Big Block

Y axis

             N-W = 0

             N = M g

X axis

             T- fr = Ma

the friction force has the expression

             fr = μ N

             fr = μ Mg

small block

             w- T = m a

we write the system of equations

             T - fr = M a

             mg - T = m a

we add and resolved

             mg-  μ Mg = (M + m) a

             a = [tex]g \ \frac{m - \mu M}{m+M}[/tex]

             a = [tex]9.8 \ \frac{10- 0.2 \ 20}{ 10 \ +\ 20}[/tex]

             a = 9.8 (6/30)

             a = 1.96 m / s²

a) now we can use the kinematic relations

             y = v₀ t + ½ a t²

the blocks come out of rest so their initial velocity is zero

             y = ½ a t²

             y = ½ 1.96 t²

             y = 0.98 t²

for t = 1s y = 0.98 m

       t = 2s y = 1.96 m

b) Time is a scale that is the same for the entire system, the question should be oriented to how far the big block will move.

As the curda is in tension the two blocks must move the same distance

c) the velocity of the block M

           v = vo + a t

           v = 0 + 1.96 t

for t = 1 s v = 1.96 m / s

       t = 2 s v = 3.92 m / s

d) the deceleration if the chain is cut

when removing the chain the tension becomes zero

           -fr = M a

          - μ M g = M a

          a = - μ g

          a = - 0.2 9.8

          a = -1.96 m / s²

e) the distance to stop the block is

         v² = vo² - 2 a x

        0 = vo² - 2a x

        x = vo² / 2a

        x = 1.96² / 2 1.96

        x = 0.98 m

the time to travel this distance is

        v = vo - a t

        t = vo / a

        t = 1.96 /1.96

        t = 1 s

Answer:

In parts 1 and 3 the energy

In part 2  moment.  inelastic

conserved

Explanation:

In this exercise, we are asked to describe the conservation processes for each part of the exercise.

In parts 1 and 3 the energy is conserved since the bodies do not change

In part 2 the bodies change since they are united therefore the moment is conserved and part of the kinetic energy is converted into potential energy.

Energy

moment   .inelastic

conserved

The two balls swing up together just after the collision to their highest point. energy is conserved.

According to the law of conservation of momentum, the momentum of the body before the collision is always equal to the momentum of the body after the collision.

According to the law of conservation of momentum

Momentum before collision =Momentum after collision

When the first ball is released and just before it hits the stationary ball, The two balls collide, The two balls swing up together just after the collision to their highest point. energy is conserved.

The balls swing like pendulums. During the collision in part (2) energy is conserved as the collision is inelastic.

We are requested to describe the conservation methods for each element of the activity in this exercise.

Because the bodies do not change in sections 1 and 3, energy is conserved.

Because the bodies change in part 2 is joined, the moment is conserved and some of the kinetic energy is transformed into potential energy.

Hence the two balls swing up together just after the collision to their highest point. energy is conserved.

To learn more about the law of conservation of momentum refer to;

https://brainly.com/question/1113396

The acceleration will increase by 61.3%.

Explanation:

The centripetal acceleration [tex]a_c[/tex] is given by

[tex]a_c = \dfrac{v^2}{r}[/tex]

If the velocity of the object increases by 27.0%, then its new velocity v' becomes

[tex]v' = 1.270v[/tex]

The new centripetal acceleration [tex]a'_c[/tex] becomes

[tex]a'_c = \dfrac{(1.270v)^2}{r} = 1.613 \left(\dfrac{v^2}{r} \right)[/tex]

[tex]\:\:\:\:\:\:\:\:\:= 1.613a_c[/tex]

Answer:

 ΔL = 3.82 10⁻⁴ m

Explanation:

This is a thermal expansion exercise

          ΔL = α L₀ ΔT

          ΔT = T_f - T₀

where ΔL is the change in length and ΔT is the change in temperature

Let's reduce the length to SI units

          L₀ = 90.5 mm (1m / 1000 mm) = 0.0905 m

let's calculate

          ΔL = 25.10⁻⁶ 0.0905 (154.6 - (14.4))

          ΔL = 3.8236 10⁻⁴ m

using the criterion of three significant figures

          ΔL = 3.82 10⁻⁴ m

Answer:

a. 282.84 cm/s b. 100 cm/s c. 400 cm/s d. 200 cm/s

Explanation:

The speed of the wave v = √(T/μ) where T = tension and μ = mass per unit length = m/l where m = mass of string and l = length of string.

So, v = √(T/μ)

v = √(T/m/l)

v = √(Tl/m)

a. The string's tension is doubled?

If the tension is doubled, T' = 2T the new speed is

v' = √(T'l/m)

v' = √(2Tl/m)

v' = √2(√Tl/m)

v' = √2v

v' = √2 × 200 cm/s

v' = 282.84 cm/s

b. The string's mass is quadrupled (but its length is unchanged)?

If the mass is quadrupled, m' = 4m the new speed is

v' = √(Tl/m')

v' = √(Tl/4m)

v' = (1/√4)(√Tl/m)

v' = v/2

v' = 200/2 cm/s

v' = 100 cm/s

c. The string's length is quadrupled (but its mass is unchanged)?

If the length is quadrupled, l' = 4l the new speed is

v' = √(Tl'/m)

v' = √(T(4l)/m)

v' = √4)(√Tl/m)

v' = 2v

v' = 200 × 2 cm/s

v' = 400 cm/s

d. The string's mass and length are both quadrupled?

If the length is quadrupled, l' = 4l and mass quadrupled, m' = 4m, the new speed is

v' = √(Tl'/m')

v' = √(T(4l)/4m)

v' = √(Tl/m)

v' = v

v' = 200 cm/s

Solution :

It is given that a car ran off from a cliff and it crashes on canyon floor. Now the system of a car as well as the earth together have a  [tex]\text{ net momentum of zero}[/tex] when the car crashes on the canyon floor, thus reducing the momentum of the car to zero. The earth also stops its upward motion and it also reduces the momentum to zero.

Answer:

A) q = 1.67 × 10^(-7) C

B) q = 1.67 × 10^(-10) C

Explanation:

We are given;

Potential; V = 30 KV = 30000 V

Radius of sphere; r = diameter/2 = 0.1/2 = 0.05 m

A) To find the charge of the sphere, we will use the formula;

V = kq/r

Where;

q is the charge

k is electric force constant = 9 × 10^(9) N.m²/C²

Thus;

q = Vr/k

q = (30000 × 0.05)/(9 × 10^(9))

q = 1.67 × 10^(-7) C

B) Now, potential energy here is a formula; U = qV

However, for the drop of paint to move, the potential energy will be equal to the kinetic energy. Thus;

qV = ½mv²

q = mv²/2V

Where;

v is speed = 10 m/s

V = 30000 V

m = mass = 0.100 mg = 0.1 × 10^(-6) Thus;

q = (0.1 × 10^(-6) × 10²)/(2 × 30000)

q = 1.67 × 10^(-10) C

Answer:

Jupiter

Explanation:

Answer:

The answer is Jupiter.

Explanation:

Jupiter is an orange/yellow colored planet.

Answer:

The speed of the car, v = 21.69 m/s

Explanation:

The diameter is  = 48.01 m

Therefore, the radius of the loop R = 24.005 m

Weight at the top is n = mv^2/R - mg

Since the apparent weight is equal to the real weight.

So, mv^2/R - mg = mg

v = √(2Rg)

v = √[2(24.005 m)(9.8 m/s^2)]

The speed of the car, v = 21.69 m/s

Answer:

The speed is 15.34 m/s.

Explanation:

Diameter, d = 48.01 m

Radius, R = 24.005 m

Let the speed is v and the mass is m.

Here, the weight of the car is balanced by the centripetal force.

According to the question

[tex]m g = \frac{mv^2}{R}\\\\v =\sqrt{24.005\times9.8}\\\\v = 15.34 m/s[/tex]

Answer:

0.5 V

Explanation:

The electric potential distance between different locations in an electric field area is unaffected by the charge that is transferred between them. It is solely dependent on the distance. Thus, for two electrons pushed together at the same distance into the same field, the electric potential will remain at 1 V. However, the electric potential of one of the two electrons will be half the value of the electric potential for the two electrons.

Answer:

Greatest potential: moment before being dropped

Zero Kinetic: when it comes to rest

Greatest Kinetic: moment before first bounce

Explanation:

Answer:

Tempered glass

Explanation:

When warmed, an amorphous substance has a non-crystalline architecture that differentiates from its isochemical liquid, but this does not go through structural breakdown or the glass transition.

Answer:

A) nm, um, mm, cm

B) 1mm, 1cm, 1m, 1km

A) 3500g, B) 2500m, C) 7200 seconds