5. Charges added to a conductor quickly spread over the surface of the object.
a. True
b. False

u = 0.077

Explanation:

Work done by friction is

Wf = ∆KE + ∆PE

-umgx = ∆KE,. ∆PE =0 (level ice surface)

-umgx = KEf - KEi = -(1/2)mv^2

Solving for u,

u = v^2/2gx

= (12 m/s)^2/2(9.8 m/s^2)(95 m)

= 0.077

Kinetic friction is the ratio of the friction force to the normal force experienced by a body in moving state.The coefficient of kinetic friction between the ice and skates is 0.077.

Given-

velocity of the ice skater is 12 m/ sec.

Work done by the friction is the sum of the change of the kinetic energy and the change in potential energy.

[tex]W_{f}=\bigtriangleup KE +\bigtriangleup PE[/tex]

The value for the potential energy will be equal to Zero in this case. Therefore the work done by the friction is,

[tex]W_{f}=\bigtriangleup KE +0[/tex]

Kinetic energy is directly proportional to the mass of the object and to the square of its velocity and work done can be given as,

[tex]W_{f} =u_{f} mgx[/tex]

Here,  [tex]u_{f}[/tex] is friction force, [tex]m[/tex] is mass, [tex]g[/tex] is gravity and x is the distance .

Equate the value of kinetic energy and work done of friction for further result, we get,

[tex]u_{f} mgx=\dfrac{1}{2} \times mv^2[/tex]

[tex]u_{f} =\dfrac{1}{2gx} \times v^2[/tex]

[tex]u_{f} =\dfrac{1}{9.8\times 95} \times 12^2[/tex]

[tex]u_{f} =0.077[/tex]

Hence, the coefficient of kinetic friction between the ice and skates is 0.077.

For more about the friction, follow the link below-

https://brainly.com/question/13357196

Answer:

[tex]12.35\:\mathrm{m/s}[/tex]

Explanation:

We can use the work-energy theorem to solve this problem. The work-energy theorem states that the work done on an object will be equal to that object's change in kinetic energy. Thus, we have the following equation:

[tex]W=\Delta KE,\\W=\frac{1}{2}mv^2-0,\\122=\frac{1}{2}\cdot 1.6\cdot v^2,\\v^2=152.5,\\v\approx \boxed{12.35\:\mathrm{m/s}}[/tex]

Answer:

the distance between the submarine and the ocean floor is 11,250 m

Explanation:

Given;

speed of the wave, v = 1500 m/s

time of motion of the wave, t = 15 s

The time taken to receive the echo is calculated as;

[tex]time \ of \ motion \ (t) = \frac{total \ distance }{speed \ of \ wave} = \frac{2d}{v} \\\\2d = vt\\\\d = \frac{vt}{2} \\\\d = \frac{1500 \times 15}{2} \\\\d = 11,250 \ m[/tex]

Therefore, the distance between the submarine and the ocean floor is 11,250 m

Answer:

v= IR then you can R= v/I 9v÷0.002 = 450 Ohms

Answer:

30 newtons

explanations

data given

mass=6kg

acceleration=5

f=m×a

6×5=30

Answer:

0.9378

Explanation:

Weight (W) of the rider = 100 kg;

since 1 kg = 9.8067 N

100 kg will be = 980.67 N

W = 980.67 N

At the slope of 12%, the angle θ is calculated as:

[tex]tan \ \theta = \dfrac{12}{100} \\ \\ tan \ \theta = 0.12 \\ \\ \theta = tan^{-1}(0.12) \\\\ \theta = 6.84^0[/tex]

The drag force D = Wsinθ

[tex]\dfrac{1}{2}C_v \rho AV^2 = W sin \theta[/tex]

where;

[tex]\rho = 1.23 \ kg/m^3[/tex]

A = 0.9 m²

V = 15 m/s

∴

Drag coefficient [tex]C_D = \dfrac{2 *W*sin \theta}{\rho *A *V^2}[/tex]

[tex]C_D =\dfrac{2 *980.67*sin 6.84}{1.23 *0.9 *15^2}[/tex]

[tex]C_D =0.9378[/tex]

Answer:

person that you know and like (not a member of your family), and who likes you

Answer:

the value of the final pressure is 0.168 atm

Explanation:

Given the data in the question;

Let p₁ be initial pressure, v₁ be initial volume.

After expansion, p₂ is final pressure and v₂ is final volume.

So using the following equations;

p₁v₁ = nRT

p₂v₂ = nRT

hence, p₁v₁ = p₂v₂

we find p₂

p₂ = p₁v₁ / v₂

given that; initial volume v₁ = 0.175 m³, Initial pressure p₁ = 0.350 atm,

final volume v₂ = 0.365 m³

we substitute

p₂ = ( 0.350 atm × 0.175 m³ ) / 0.365 m³

p₂ = 0.06125 atm-m³ / 0.365 m³

p₂ = 0.168 atm

Therefore, the value of the final pressure is 0.168 atm

If two objects are the SAME temperature and are physically touching,

then

. . .

. . .

. . .

The objects would be in thermodynamic equilibrium and as a result there would be no heat transfer.

Answer:

Resistance is as large as 2.8 ohm

Explanation:

Complete question

A 3.0 A current is maintained in a simple circuit that consists of a resistor between the terminals of an ideal battery. If the battery supplies energy at a rate of 25 W, how large is the resistance?

Solution -

The relation between Power and current is as follows  

P = I^2*R

R = P/I^2

Were P = Power

R = resistance and

I = current

Given-  

P = 25 W  

I = 3.0 A

Substituting the given values, in above equation, we get -  

R = 25/3.0^2

R = 2.8 ohm

a) v = 1.94 × 10^5 m/s

b) Ms = 2.09 × 10^24 kg

Explanation:

Given:

m = 0.257M (M = mass of earth = 5.972×10^24 kg)

= 1.535×10^24 kg

r = 7.18×10^9 m

T = 2.7 days × (24 hr/1 day) × (3600 s/1 hr)

= 2.3328×10^5 s

a) To find the orbital speed of the planet, we need to find the circumference of the planet's orbit first:

C = 2×(pi)×r

= 2(3.14)(7.18×10^9m)

= 4.51×10^10 m

The orbital speed v is then given by

v = C/T

= (4.51×10^10 m)/(2.33×10^5 s)

= 1.94 × 10^5 m/s

b) We know that centripetal force Fc is given by

Fc = mv^2/r

where v = orbital speed

r = average orbital radius

m = mass of planet

We also know that the gravitational force FG between the star K2-116 and the planet is given by

FG = GmMs/r^2

where m = mass of planet

Ms = mass of star K2-116

r. = average orbital radius

G = universal gravitational constant

= 6.67 × 10^-11 m^3/kg-s^2

Equating Fc and FG together, we get

Fc = FG

mv^2/r = GmMs/r^2

Note that m and one of the r's get cancelled out so we are left with

v^2 = GMs/r

Solving for the mass of the star Ms, we get

Ms = rv^2/G

=(7.18 × 10^9 m)(1.94 × 10^5 m/s)^2/(6.67 × 10^-11 m^3/kg-^2)

= 2.09 × 10^24 kg

Solution :

The conditions for the maximum in the Young's experiment is :

d sin θ = m λ,     where m = 0, 1, 2, 3, .....

The angle between the central maximum and the 1st order maximum can be determined by setting the m = 1. So,

d sin θ =  λ

[tex]$\theta = \sin^{-1}\left(\frac{\lambda}{d}\right)$[/tex]

Given : d = 100 λ

[tex]$\theta = \sin^{-1}\left(\frac{\lambda}{100 \lambda}\right)$[/tex]

[tex]$\theta = \sin^{-1}\left(\frac{1}{100}\right)$[/tex]

 [tex]$=0.573^\circ$[/tex]

  = 0.01 rad

Answer:

both the same

Explanation:

When a ball is thrown vertically upwards, it experiences that same acceleration due to gravity as an object thrown directly downwards.

This means that if we ignore the effects of air resistance, and the two balls have the same initial speed, they are expected both expected to hit the ground at the same speed as a result of the principle of conservation of energy.

Answer:

Both balls will hit the ground at the same speed because air resistance is ignored.

Explanation:

Answer:

(a) Negative Q

(b) Positive Q

Explanation:

Charge is the inherent property of matter due to the transference of electrons.

There are three methods of charging a body.

(i) Charging by friction: When two uncharged bodies rubbed together, then one body gets positive charged and the other is negatively charges it is due to the transference of electrons form one body to another.  

(ii) Conduction: when a charged body comes in contact with the another uncharged body, the uncharged body gets the same charge and the charge is distributed equally.

(iii) Induction: When a uncharged body keep near the charged body, the uncharged body gets the same amount of charge but opposite in sign.  

(a) When a small tack of charge Q is lowered into the hole, then due to the process of induction, the charge on the inner surface of the shell is - Q.

(b) Due to the process of conduction, the charge on the outer surface of the shell is Q.

The charge on the inner surface of the shell is negative whereas the charge on the outer surface of the shell is positive.

Due to the process of induction the inner surface of the shell creates negative charge because when a uncharged body bring near to the charged body, the uncharged body gets the same amount of charge but opposite in sign.

While on the other hand, there is no charge interaction with the outer surface so it remains positively charge so we can conclude that the charge on the inner surface of the shell is negative whereas the charge on the outer surface of the shell is positive.

Learn more about charge here: https://brainly.com/question/18102056

Answer:

false

Explanation:

Answer:

D

Explanation:

Because the y axis is meter. If it is straight line at time and meter graph then it velocity and speed is 0

Answer:

a) the wavelength that the light emits is 6.1875 × 10⁻⁷ m

b) the distance between the two n = 1 dark fringes is 5.5 × 10⁻³ m

Explanation:

Given the data in the question;

separation  between two slits  d = 0.36 mm = 0.00036 m

Separation between two adjacent fringes β = 5.5 mm = 0.0055 m

Distance of screen from slits D = 1.6 m

n = 2

a) the wavelength the light emits;

Using the formula;

β = (nD/d)λ

To find wavelength, we make λ the subject of formula;

βd = nDλ

λ = βd / nD

so we substitute

λ = ( 0.0055 m × 0.00036 m ) / ( 2 × 1.6 m )

λ = 0.00000198 / 3.2

λ = 6.1875 × 10⁻⁷ m

Therefore, the wavelength that the light emits is 6.1875 × 10⁻⁷ m

b) the distance between the two n = 1 dark fringes;

To find the distance between the two n = 1 dark fringes, we use the following formula;

y[tex]_m[/tex] = 2nλD / d

given that n = 1, we substitute

y[tex]_m[/tex] = ( 2 × 1 × ( 6.1875 × 10⁻⁷ m ) × 1.6 m ) / 0.00036 m

y[tex]_m[/tex] = 0.00000198 / 0.00036

y[tex]_m[/tex] = 0.0055 m

y[tex]_m[/tex] = 5.5 × 10⁻³ m

Therefore, the distance between the two n = 1 dark fringes is 5.5 × 10⁻³ m

Answer:

dino :)

Explanation:

Complete question:

For a galvanic cell to generate an electric current flowing from anode to cathode, what must be true?

(a) Electrons flow from the anode to the cathode

(b) Electrons flow from the more negatively charged electrode to the more positively charged electrode

(c) Electrons flow from higher potential energy to lower potential energy

(d) All of the above are true.

Answer:

(d) All of the above are true.

Explanation:

A galvanic or Voltaic cell is a primary type of electrochemical cell that is used to generate electrical energy from the chemical reactions that take place in it.

It consists of a positive electrode (cathode) and a negative electrode (anode) for the movement of charges.

(a) Electrons flow from the anode to the cathode. TRUE

Anode is the negative electrode and for electron current, electrons flow from negative electrode to positive electrode.

(b) Electrons flow from the more negatively charged electrode to the more positively charged electrode. TRUE

Based on electron current flow.

(c) Electrons flow from higher potential energy to lower potential energy. TRUE

The driving force of the electron flow is the potential difference. Electrons must flow from higher potential to lower potential.

All the options are correct, so we select option "D"

Hi there!

[tex]\large\boxed{1560 kgm/s}[/tex]

Recall that:

P = m · v, where:

P = momentum

m = mass (kg)

v = velocity (m/s)

Thus:

P = 156 · 10

P = 1560 kgm/s

Answer:  I think that its b, they would have difficulty sprinting in a race

Explanation:

Answer:

f = 1.255 cm

Explanation:

The Radius of Curvature:

The radius of that hollow sphere, whose part is the spherical mirror, is known as ‘The Radius of Curvature’ of  mirror.

Focal Length:

The distance between principal focus and pole is called ‘Focal Length’. It is denoted by ‘F’

The focal length is basically equal to the half of the radius of curvature of the mirror:

[tex]f = \frac{r}{2}[/tex]

where,

f = focal length = ?

r = radius of curvature = 2.51 cm

[tex]f = \frac{2.51\ cm}{2}[/tex]

f = 1.255 cm

Answer:

a)   a = 34.375 m / s²,  b)    v_f = 550 m / s

Explanation:

This problem is the launch of projectiles, they tell us to ignore the effect of the friction force.

a) Let's start with the final part of the movement, which is carried out from t= 16 s with constant speed

             v_f = [tex]\frac{x-x_1}{t}[/tex]

we substitute the values

             v_f = [tex]\frac{ 6600 -x_1}{4}[/tex]  

The initial part of the movement is carried out with acceleration

             v_f = v₀ + a t

             x₁ = x₀ + v₀ t + ½ a t²

the rocket starts from rest v₀ = 0 with an initial height x₀ = 0

             x₁ = ½ a t²

             v_f = a t

we substitute the values

              x₁ = 1/2  a 16²

              x₁ = 128 a

              v_f = 16 a

let's write our system of equations  

               v_f = [tex]\frac{6600 - x_1}{4}[/tex]

               x₁ = 128 a

               v_f = 16 a

we substitute in the first equation  

               16 a = [tex]\frac{6600 -128 a}{4}[/tex]

               16 4 a = 6600 - 128 a

                a (64 + 128) = 6600

                a = 6600/192

                 a = 34.375 m / s²

b) let's find the time to reach this height

                x = ½ to t²

                t² = 2y / a

                t² = 2 5100 / 34.375

                t² = 296.72

                t = 17.2 s

We can see that for this time the acceleration is zero, so the rocket is in the constant velocity part

               v_f = 16 a

               v_f = 16 34.375

               v_f = 550 m / s

Answer:

a)   w = 31.4 rad / s,  b)  a = 118.4 m / s²

Explanation:

a) let's reduce to the SI system

   w = 5 rev / s (2pi rad / 1 rev)

   w = 31.4 rad / s

b) the expression for the centripetal acceleration is

      a = v² / r

linear and angular variables are related

      v = w r

    we substitute

     a = w² r

     a = 31.4² 0.120

     a = 118.4 m / s²

Answer:

Option A (69.56 newtons) is the appropriate solution.

Explanation:

According to the question,

On the X-axis,

⇒ [tex]T_1Cos30^{\circ}-T_2Cos60^{\circ}=0[/tex]

or,

    [tex]T_1Cos 30^{\circ}=T_2Cos60^{\circ}[/tex]

On substituting the values, we get

      [tex]T_1\times \frac{\sqrt{3} }{2}=T_2\times \frac{1}{2}[/tex]

      [tex]T_1\times \sqrt{3} =T_2[/tex]....(equation 1)

On the Y-axis,

⇒ [tex]T_1Sin30^{\circ}+T_2Sin60^{\circ}=139.3 \ N[/tex]

                        [tex]\frac{T_1}{2} +\frac{\sqrt{3} }{2} =139.2 \ N[/tex]

                    [tex]T_1+\sqrt{3}T_2=139.2\times 2[/tex]

From equation 1, we get

           [tex]T_1+\sqrt{3}\times \sqrt{3}T_1 =278.4 \ N[/tex]

                        [tex]T_1+3T_1=278.4 \ N[/tex]

                                [tex]4T_1=278.4 \ N[/tex]

                                  [tex]T_1=\frac{278.4}{4}[/tex]

                                       [tex]=69.6 \ N[/tex]  

Answer:

69.58

Explanation:

c. a natural process

It is a natural process

Answer: The egg will get sucked into the bottle. To get the egg out of the bottle, turn the bottle upside down and blow into it, so that the egg acts as a one-way valve. The increased air pressure in the bottle will cause the egg to pop back out.

Explanation:

Quickly place the egg over the mouth of the bottle. The egg will get sucked into the bottle. To get the egg out of the bottle, turn the bottle upside down and blow into it, so that the egg acts as a one-way valve. The increased air pressure in the bottle will cause the egg to pop back out.

Answer:

yes

Explanation:

this means the answer is yes