Austin is interested in working for an intelligence branch of the government. A polygraph test is part of the interview
process. Though Austin intends to be honest, he is worried that the polygraph will say he is lying when he is not.
Austin's friend, Gabe, assures Austin that polygraph tests are infallible. Is Gabe right?

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: 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:

18.2 s

Explanation:

Given:

Δx = 738 m

v₀ = 0 m/s

a = 4.44 m/s²

Find: t

Δx = v₀ t + ½ at²

738 m = (0 m/s) t + ½ (4.44 m/s²) t²

t = 18.2 s

Answer:

41.667 per cent of the total kinetic energy is translational kinetic energy.

Explanation:

As the spherical shell is rolling without slipping at constant speed, the system can be considered as conservative due to the absence of non-conservative forces (i.e. drag, friction) and energy equation can be expressed only by the Principle of Energy Conservation, whose total energy is equal to the sum of rotational and translational kinetic energies. That is to say:

[tex]E = K_{t} + K_{r}[/tex]

Where:

[tex]E[/tex] - Total energy, measured in joules.

[tex]K_{r}[/tex] - Rotational kinetic energy, measured in joules.

[tex]K_{t}[/tex] - Translational kinetic energy, measured in joules.

The spherical shell can be considered as a rigid body, since there is no information of any deformation due to the motion. Then, rotational and translational components of kinetic energy are described by the following equations:

Rotational kinetic energy

[tex]K_{r} = \frac{1}{2}\cdot I_{g}\cdot \omega^{2}[/tex]

Translational kinetic energy

[tex]K_{t} = \frac{1}{2}\cdot m \cdot R^{2}\cdot \omega^{2}[/tex]

Where:

[tex]I_{g}[/tex] - Moment of inertia of the spherical shell with respect to its center of mass, measured in [tex]kg\cdot m^{2}[/tex].

[tex]\omega[/tex] - Angular speed of the spherical shell, measured in radians per second.

[tex]R[/tex] - Radius of the spherical shell, measured in meters.

After replacing each component and simplifying algebraically, the total energy of the spherical shell is equal to:

[tex]E = \frac{1}{2}\cdot (I_{g} + m\cdot R^{2})\cdot \omega^{2}[/tex]

In addition, the moment of inertia of a spherical shell is equal to:

[tex]I_{g} = \frac{2}{3}\cdot m\cdot R^{2}[/tex]

Then, total energy is reduced to this expression:

[tex]E = \frac{5}{6}\cdot m \cdot R^{2}\cdot \omega^{2}[/tex]

The fraction of the total kinetic energy that is translational in percentage is given by the following expression:

[tex]\%K_{t} = \frac{K_{t}}{E}\times 100\,\%[/tex]

[tex]\%K_{t} = \frac{\frac{1}{2}\cdot m \cdot R^{2}\cdot \omega^{2} }{\frac{5}{6}\cdot m \cdot R^{2}\cdot \omega^{2} } \times 100\,\%[/tex]

[tex]\%K_{t} = \frac{5}{12}\times 100\,\%[/tex]

[tex]\%K_{t} = 41.667\,\%[/tex]

41.667 per cent of the total kinetic energy is translational kinetic energy.

Answer:

Explanation:

Given data

length of cube= 0.2015 m

density = 19300 kg/m^3.

But the volume of cube is given as [tex]l*l*l= l^3[/tex]

[tex]volume -of- cube= 0.2015*0.2015*0.2015= 0.00818 m^3[/tex]

The density is expressed as = mass/volume

[tex]mass=19300*0.00818= 157.87m^3[/tex]

Answer:

Explanation:

Check the diagram attached below for the diagram.

Let the weight of the rock be W and the mass of the meter stick be M. Note that the mass of the meter stick will be placed at the middle of the meter stick i.e at the 50cm mark

Using the principle of moment to calculate the weight of the rock. It states that the sum of clockwise moments is equal to the sum of anti clockwise moment.

Moment = Force * perpendicular distance

The meterstick acts in the clockwise direction while the rock acys in the anti clockwise direction

Clockwise moment = 1kg * 25 = 25kg/cm

Anticlockwise moment = W * 25cm = 25W kg/cm

Equating both moments of forces

25W = 25

W = 25/23

W = 1 kg

The mass of the rock is also 1 kg

Answer:

The two masses are 3.39 Kg and 1.75 Kg

Explanation:

The gravitational force of attraction between two bodies is given by the formula;

F = Gm₁m₂/d²

where G is the gravitational force constant = 6.67 * 10⁻¹¹ Nm²Kg⁻²

m₁ = mass of first object; m₂ = mass of second object; d = distance of separation between the objects

Further calculations are provided in the attachment below

Answer:

It's more likely that the trailer is heavily loaded

Explanation:

Due to the fact that the frequency is proportional to the square root of the force constant and inversely proportional to the square root of the mass, it is very likely that the truck would be heavily loaded because the force constant would be the same whether the truck is empty or heavily loaded.

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:

The rubber band will be stretched 0.02 m.

The work done in stretching is 0.11 J.

Explanation:

Force 1 = 44 N

extension of rubber band = 0.080 m

Force 2 = 11 N

extension = ?

According to Hooke's Law, force applied is proportional to the extension provided elastic limit is not extended.

F = ke

where k = constant of elasticity

e = extension of the material

F = force applied.

For the first case,

44 = 0.080K

K = 44/0.080 = 550 N/m

For the second situation involving the same rubber band

Force = 11 N

e = 550 N/m

11 = 550e

extension e = 11/550 = 0.02 m

The work done to stretch the rubber band this far is equal to the potential energy stored within the rubber due to the stretch. This is in line with energy conservation.

potential energy stored = [tex]\frac{1}{2}ke^{2}[/tex]

==> [tex]\frac{1}{2}* 550* 0.02^{2}[/tex] = 0.11 J

Answer:

c.they have opposite charges.

Explanation:

because the protons have a positive charge and the neutrons have no charge.

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:

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.

Answer:

-    Force on A = 0.870N

-    charge of the object B = q = 2.1 μC

    charge of the object A = 2q = 4.2 μC

-    a = 0.966 m/s^2

Explanation:

- In order to determine the force on the object A, you take into account the third Newton law, which states that the force experienced by A has the same magnitude of the force experienced by B, but with an opposite direction.

Then, the force on A is 0.870N

- In order to calculate the charge of both objects, you use the following formula:

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

k: Coulomb's constant = 8.98*10^9 Nm^2/C^2

r: distance between the objects = 30.0cm = 0.30m

A has twice the charge of B. If the charge of B is qB=q, then the charge of A is qA=2qB = 2q.

You replace the expression for qA and qB into the equation (1), solve for q, and replace the values of the parameters.

[tex]F_e=k\frac{(2q)(q)}{r^2}=2k\frac{q^2}{r^2}\\\\q=\sqrt{\frac{r^2Fe}{2k}}\\\\q=\sqrt{\frac{(0.30m)^2(0.870N)}{2(8.98*10^9Nm^2/C^2)}}=2.1*10^{-6}C\\\\q=2.1\mu C[/tex]

Then, you have:

charge of the object B = q = 2.1 μC

charge of the object A = 2q = 4.2 μC

- In order to calculate the acceleration of A, you use the second Newton law with the electric force, as follow:

[tex]F_e=ma\\\\a=\frac{F_e}{m}[/tex]

m: mass of the object A = 900g = 0.900kg

[tex]a=\frac{0.870N}{0.900kg}=0.966\frac{m}{s^2}[/tex]

The acceleration of A is 0.966m/s^2

Answer:

95.5 m

Explanation:

The displacement is the position of the ending point relative to the starting point.

In this case, the magnitude of the displacement is the diameter of the circular track.

d = 300 m / π

d ≈ 95.5 m

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.

​

​

​

Answer:

I = 18 x 10⁻⁹ A = 18 nA

Explanation:

The current is defined as the flow of charge per unit time. Therefore,

I = q/t

where,

I = Average Current passing through nerve cell

q = Total flow of charges through nerve cell

t = time period of flow of charges

Here, in our case:

I = ?

q = (9 pC)(1 x 10⁻¹² C/1 pC) = 9 x 10⁻¹² C

t = (0.5 ms)(1 x 10⁻³ s/1 ms) = 5 x 10⁻⁴ s

Therefore,

I = (9 x 10⁻¹² C)/(5 x 10⁻⁴ s)

I = 18 x 10⁻⁹ A = 18 nA

Answer:

Crosstalk

Explanation:

The answer is Crosstalk as this phenomenon is most commonly associated with analog phone call.

Now, crosstalk is defined as a disturbance caused by the electric or magnetic fields of one telecommunication signal which affects a signal in an adjacent circuit. In a telephone circuit, crosstalk could result in hearing part of a voice conversation from another circuit. Hence, the phenomenon that causes crosstalk is called electromagnetic interference (EMI). This may occur in microcircuits within computers and audio equipments including within network circuits. This term is also usually applied to optical signals that interfere with each other.

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.

Find out more information about equilibrium here:

https://brainly.com/question/517289

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:

335.79cm/s

Explanation:

When a transverse wave of wavelength λ is produced during the vibration of a wire, the frequency(f), and the speed(v) of the wave pulses are related to the wavelength as follows;

v = fλ        ------------------(ii)

From the question;

f = 6.3Hz

λ = 53.3cm

Substitute these values into equation (i) as follows;

v = 6.3 x 53.3

v = 335.79cm/s

Therefore, the speed of the wave pulses along the wire is 335.79cm/s

Answer:

Option A is the only valid statement.

Explanation:

A)The electric field intensity is defined by the relationship:

E= -ΔV/Δr.

Now, according to the relationship above, the electric field would be the negative gradient of electric potential. Now, if the electric potential is constant throughout the given region of space, then the change in electric potential would be ΔV=0.

Thu,E= 0.

So the answer is that, E will be zero in this case.

So, the statement is valid.

B) Statement not valid because the field is the gradient of the potential. Hence, the field would be zero in any region where the potential is constant. However, constant does not necessarily mean a value of zero. With that being said, we can always change the definition of the potential function by adding a constant, to thus make it zero there. But then the potential will no longer be zero at infinity or in any different “flat” regions.

C) Statement not valid because, for the fact that electric field is zero at a particular point, it doesn't necessarily

imply that the electric potential is zero at that point. A good example would be the case of two identical charges which are separated by some distance. At the midpoint between the charges, the

electric field due to the charges would be zero. However, the electric potential due to the charges at that same point would not be zero. Thus, the potential will either have two positive contributions, if the charges are positive, or two negative contributions, if the charges are negative.

(D) Statement is not valid because, for the fact that electric potential is zero at a particular point, it does not necessarily imply that the electric field is zero at that point. A good example would be the case of a dipole, which

has two charges of the same magnitude, but opposite sign, and are separated by some distance. At

the midpoint between the charges, the electric potential due to the charges would be zero, but the electric field due to the charges at that same point would not be zero.

Answer:

horizontal component = 10.54m/s

Explanation:

horizontal component = 13.2cos37°

horizontal component = 10.54m/s

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:

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:

 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:

The waiting time is  [tex]t_w = 3 .47 \ s[/tex]

Explanation:

From the question we are told that

       The  height of the hot air balloon above the ground is  [tex]d = 50 \ m[/tex]

         The distance of the balloon from the target is  [tex]l = 100 \ m[/tex]

        The  velocity of the balloon is  [tex]v = 15 \ m/s[/tex]

Generally the time it will take to reach the ground  is

          [tex]t = \sqrt{2 * \frac{d}{g} }[/tex]

substituting values

         [tex]t = \sqrt{2 * \frac{50}{9.8} }[/tex]

        [tex]t = 3.2 \ s[/tex]

The distance that is covered at time with the given velocity is mathematically evaluated as  

            [tex]z = v * t[/tex]

substituting values

           [tex]z = 15 * 3.2[/tex]

           [tex]z = 48 \ m[/tex]

This implies that for the balloon moving at a velocity (v) to hit the target  it must be dropped at this distance (z)

Now the distance the balloonist has to wait before dropping in order to hit the target is  

        [tex]A = d - z[/tex]

substituting values

      [tex]A = 100 - 48[/tex]

      [tex]A = 52 \ m[/tex]

This implies that the time the balloonist has to wait is  

      [tex]t_w = \frac{A}{v}[/tex]

substituting values

      [tex]t_w = \frac{52}{15}[/tex]

      [tex]t_w = 3 .47 \ s[/tex]

Answer:

389 kg

Explanation:

The computation of mass is shown below:-

[tex]T = 2\pi \sqrt{\frac{m}{k} }[/tex]

Where K indicates spring constant

m indicates mass

For the new time period

[tex]T^' = 2\pi \sqrt{\frac{m'}{k} }[/tex]

Now, we will take 2 ratios of the time period

[tex]\frac{T}{T'} = \sqrt{\frac{m}{m'} }[/tex]

[tex]\frac{1.50}{2.00} = \sqrt{\frac{0.500}{m'} }[/tex]

[tex]0.5625 = \sqrt{\frac{0.500}{m'} }[/tex]

[tex]m' = \frac{0.500}{0.5625}[/tex]

= 0.889 kg

Since mass to be sum that is

= 0.889 - 0.500

0.389 kg

or

= 389 kg

Therefore for computing the mass we simply applied the above formula.

The mass added to the object to change the period to 2.00 s is 0.389 kg and this can be determined by using the formula of the time period.

Given :

A 0.500-kg mass suspended from a spring oscillates with a period of 1.50 s.

The formula of the time period is given by:

[tex]\rm T = 2\pi\sqrt{\dfrac{m}{K}}[/tex]   ---- (1)

where m is the mass and K is the spring constant.

The new time period is given by:

[tex]\rm T'=2\pi\sqrt{\dfrac{m'}{K}}[/tex]   ---- (2)

where m' is the total mass after the addition and K is the spring constant.

Now, divide equation (1) by equation (2).

[tex]\rm \dfrac{T}{T'}=\sqrt{\dfrac{m}{m'}}[/tex]

Now, substitute the known terms in the above expression.

[tex]\rm \dfrac{1.50}{2}=\sqrt{\dfrac{0.5}{m'}}[/tex]

Simplify the above expression in order to determine the value of m'.

[tex]\rm m'=\dfrac{0.5}{0.5625}[/tex]

m' = 0.889 Kg

Now, the mass added to the object to change the period to 2.00 s is given by:

m" = 0.889 - 0.500

m" = 0.389 Kg

For more information, refer to the link given below:

https://brainly.com/question/2144584

Answer:

Explanation:

a ) A good conducting material conducts heat easily from one point to another . So the metal spoon which is good conductor will become hot easily by conducting heat from hot water to tip of the spoon . So even after touching the far end of spoon , we can feel the heat of the hot water in the cup . For other material like wood or plastic we have to hold the part which is  very close to water to feel heat of water .

b ) The best conductor will be that which touches hot even when we touch its farthest  end . Hence metal spoon will be best conductor . Plastic pen will be worst conductor because even touching its part close to hot water , we do not feel much heat .

c ) metal spoon > glass rod  > wooden pencil > plastic pen .

Answer: The latitudinal value of tropic of cancer is 23.5° N on June 21, when the sun is directly up above the head at noon. The Equator is the circle at which sun is straight above the head.

Explanation: