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Four students are assigned a project for which they must experimentally determine the coefficient of kinetic friction between a block of an unknown material and a table. Each student has their own idea of how to go about setting up the experiment. The group settles on two possible test to conduct: Student A's idea and Student B's idea. Student A believes the group can obtain the coefficient of kinetic friction by attaching a spring scale to the mass and pulling on the mass with the spring scale with a horizontally directed force. The reading on the spring scale at the moment the block begins to move is the coefficient of kinetic friction. Which of the following explains what is wrong with Student A's logic?

MC Options inserted in image below

Answer: i believe  its B Find the velocity of each block after they have moved apart sorry

Explanation: have a nice day buddy

Answer:

harmful effects

1. that will cause air pollution

2. that will destroy our earth

Answer:

useful effects of volcano are :-

harmful effects of volcano are:-

hope it is helpful to you ☺️

Answer:

#1 - True (touch) charging when electric conductors actually touch one another.

#2. True - electrical charges are conserved (not destroyed)

Answer:

[tex]291.67\ \text{m/s}[/tex]

Explanation:

[tex]m_1[/tex] = Mass of bullet = 4 g

[tex]m_2[/tex] = Mass of block = 1.3 kg

[tex]\mu[/tex] = Coefficient of friction = 0.17

[tex]s[/tex] = Displacement of block = 0.24 m

[tex]v_1[/tex] = Velocity of bullet

[tex]v[/tex] = Velocity of combined mass

[tex]g[/tex] = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]

The energy balance of the system is given by

[tex]\dfrac{1}{2}(m_1+m_2)v^2=\mu(m_1+m_2)gs\\\Rightarrow v=\sqrt{2\mu gs}[/tex]

As the momentum is conserved in the system we have

[tex]m_1v_1=(m_1+m_2)v\\\Rightarrow m_1v_1=(m_1+m_2)\sqrt{2\mu gs}\\\Rightarrow v_1=\dfrac{(m_1+m_2)\sqrt{2\mu gs}}{m_1}\\\Rightarrow v_1=\dfrac{(4\times 10^{-3}+1.3)\times \sqrt{2\times 0.17\times 9.81\times 0.24}}{4\times 10^{-3}}\\\Rightarrow v_1=291.67\ \text{m/s}[/tex]

The initial speed of the bullet is [tex]291.67\ \text{m/s}[/tex].

Answer:

speed: 35m/s

direction: left

Explanation:

Assuming the right side is the positive direction:

before explosion:

P = mv = 0

after explosion:

P' = 15P + 5P

(Set the velocity of the 15kg piece after explosion as v1' and the velocity of the 5kg piece after explosion as v2')

P' = 0.75mv1' + 0.25mv2'

P' = (15kg)v' + (5kg)(105m/s)

P' = 525kg/m/s + (15kg)v1'

P = P'

525kg/m/s + (15kg)v1' = 0

(15kg)v1' = -525kg/m/s

v1' = -35m/s

speed = |-35| = 35m/s

direction is to the left since the right side is the positive direction.

Answer:

a) v = 1.1 m/s

b) A = 0.315 m

c) v = 1.1 m/s  A= 0.15 m

Explanation:

a)

        [tex]v = \lambda * f (1)[/tex]

       [tex]f = \frac{1}{T} = \frac{1}{5.20s} = 0.19 Hz (3)[/tex]

      [tex]v = \lambda * f = 5.70 m * 0.19 Hz = 1.10 m/s (4)[/tex]

b)

      [tex]A = \frac{0.630m}{2} = 0.315 m (5)[/tex]

c)  

       v = 1.10 m/s (6)

       [tex]A = \frac{0.30m}{2} = 0.15 m (7)[/tex]

Answer:

The answer is "0.92 c"

Explanation:

[tex]v_1\ (earth) = 0.62 \ c \\\\v_2\ ( enterprise ) = -0.30[/tex]

so,

[tex]v_2 \ (earth) = 0.62 \ c - (-0.30 \ c) \\\\[/tex]

               [tex]= 0.62 \ c +0.30 \ c\\\\= 0.92 \ c[/tex]

Answer:

Blank 1: Gasses

Blank 2: More

Blank 3: Solids

Blank 4: Fluids

Blank 5: Liquid

Blank 6: Gas

Blank 7: Higher

Blank 8: Lower

Blank 9: Sun

Blank 10: Radiation

Blank 11: Conductors

P.S. order of answers does not matter between Blank 5 and 6.

Answer:

Approximately [tex]261\; \rm K[/tex], if this gas is an ideal gas, and that the quantity of this gas stayed constant during these changes.

Explanation:

Let [tex]P_1[/tex] and [tex]P_2[/tex] denote the pressure of this gas before and after the changes.

Let [tex]V_1[/tex] and [tex]V_2[/tex] denote the volume of this gas before and after the changes.

Let [tex]T_1[/tex] and [tex]T_2[/tex] denote the temperature (in degrees Kelvins) of this gas before and after the changes.

Let [tex]n_1[/tex] and [tex]n_2[/tex] denote the quantity (number of moles of gas particles) in this gas before and after the changes.

Assume that this gas is an ideal gas. By the ideal gas law, the ratios [tex]\displaystyle \frac{P_1 \cdot V_1}{n_1 \cdot T_1}[/tex] and [tex]\displaystyle \frac{P_2 \cdot V_2}{n_2 \cdot T_2}[/tex] should both be equal to the ideal gas constant, [tex]R[/tex].

In other words:

[tex]R = \displaystyle \frac{P_1 \cdot V_1}{n_1 \cdot T_1}[/tex].

[tex]R =\displaystyle \frac{P_2 \cdot V_2}{n_2 \cdot T_2}[/tex].

Combine the two equations (equate the right-hand side) to obtain:

[tex]\displaystyle \frac{P_1 \cdot V_1}{n_1 \cdot T_1} = \frac{P_2 \cdot V_2}{n_2 \cdot T_2}[/tex].

Rearrange this equation for an expression for [tex]T_2[/tex], the temperature of this gas after the changes:

[tex]\displaystyle T_2 = \frac{P_2}{P_1} \cdot \frac{V_2}{V_1} \cdot \frac{n_1}{n_2} \cdot T_1[/tex].

Assume that the container of this gas was sealed, such that the quantity of this gas stayed the same during these changes. Hence: [tex]n_2 = n_1[/tex], [tex](n_2 / n_1) = 1[/tex].

[tex]\begin{aligned} T_2 &= \frac{P_2}{P_1} \cdot \frac{V_2}{V_1} \cdot \frac{n_1}{n_2}\cdot T_1 \\[0.5em] &= \frac{1.67\; \rm atm}{1.12\; \rm atm} \times \frac{11.2\; \rm m^{3}}{15.7\; \rm m^{3}} \times 1 \times 245\; \rm K \\[0.5em] &\approx 261\; \rm K\end{aligned}[/tex].

Answer:

The equation of motion is [tex]x(t)=-[/tex][tex]\frac{1}{3} cos4\sqrt{6t}[/tex]

Explanation:

Lets calculate

The weight attached to the spring is 24 pounds

Acceleration due to gravity is [tex]32ft/s^2[/tex]

Assume x , is spring stretched length is ,4 inches

Converting the length inches into feet [tex]x=\frac{4}{12} =\frac{1}{3}feet[/tex]

The weight (W=mg) is balanced by restoring force ks at equilibrium position

mg=kx

[tex]W=kx[/tex] ⇒ [tex]k=\frac{W}{x}[/tex]

The spring constant , [tex]k=\frac{24}{1/3}[/tex]

                            = 72

If the mass is displaced from its equilibrium position by an amount x, then the differential equation is

    [tex]m\frac{d^2x}{dt} +kx=0[/tex]

    [tex]\frac{3}{4} \frac{d^2x}{dt} +72x=0[/tex]

  [tex]\frac{d^2x}{dt} +96x=0[/tex]

Auxiliary equation is, [tex]m^2+96=0[/tex]

                                 [tex]m=\sqrt{-96}[/tex]

                               =[tex]\frac{+}{} i4\sqrt{6}[/tex]

Thus , the solution is [tex]x(t)=c_1cos4\sqrt{6t}+c_2sin4\sqrt{6t}[/tex]

                                 [tex]x'(t)=-4\sqrt{6c_1} sin4\sqrt{6t}+c_2[/tex]  [tex]4\sqrt{6}[/tex] [tex]cos4\sqrt{6t}[/tex]

The mass is released from the rest x'(0) = 0

                    [tex]=-4\sqrt{6c_1} sin4\sqrt{6(0)}+c_2[/tex] [tex]4\sqrt{6}[/tex] [tex]cos4\sqrt{6(0)}[/tex] =0

                                                    [tex]c_2[/tex] [tex]4\sqrt{6} =0[/tex]

                                     [tex]c_2=0[/tex]

Therefore , [tex]x(t)=c_1[/tex] [tex]cos 4\sqrt{6t}[/tex]

Since , the mass is released from the rest from 4 inches

                    [tex]x(0)= -4[/tex] inches

[tex]c_1 cos 4\sqrt{6(0)} =-\frac{4}{12}[/tex] feet

   [tex]c_1=-\frac{1}{3}[/tex] feet

Therefore , the equation of motion is  [tex]-\frac{1}{3} cos4\sqrt{6t}[/tex]

Answer:

the total magnetic force on the 2 sides of the wire is 0.30 N

Option a) 0.30 N is the correct answer

Explanation:

Given the data in the question;

both the sides of the right angled triangle is the same and the magnetic is in the plane of the  triangle and is perpendicular to the hypotenuse.

hence, angle between the magnetic field and both the sides of the triangle is 45 degrees.

∅ = 45°

I = 2.0 A

L = 15 cm = 0.15 m

B = 0.7 T

Total magnetic force on the 2 sides of the wire will be;

F = 2BILsin∅

we substitute

F = 2 × 0.7 × 2.0 × 0.15 × sin(45°)

F = 0.42 × 0.70710678

F = 0.2970 ≈ 0.30 N

Therefore, the total magnetic force on the 2 sides of the wire is 0.30 N

Option a) 0.30 N is the correct answer

Answer:

r₂ = 4 r

Explanation:

For this exercise let's use Newton's second law with the magnetic force

          F = q v x B

bold letters indicate vectors, the magnitude of this expression is

          F = q v B sin θ  

in this case we assume that the angle is 90º between the speed and the magnetic field.

If we use the rule of the right hand with the positive charge, the thumb in the direction of the speed, the fingers extended in the direction of the magnetic field, the palm points in the direction of the force, which is towards the center of the circle, therefore the force is radial and the acceleration is centripetal

           a = v² / r

let's use Newton's second law

           F = ma

           q v B = m v² / r

           r = [tex]\frac{qB}{mv}[/tex]

Let's apply this expression to our case.

Proton 1

             r = \frac{qB_1}{mv_1}

Proton 2

             r₂ = [tex]\frac{q \ B_2}{m \ v_2}[/tex]

in the exercise indicate some relationships between the two protons

*    v₁ = 2 v₂

    v₂ = v₁ / 2

*   B₂ = 2B₁

we substitute

           r₂ = [tex]\frac{q \ 2B_1}{m \ \frac{v_1}{2} }[/tex]

           r₂ = 4 [tex]\frac{qB_1}{mv_1}[/tex]

           r₂ = 4 r

Wavelength = speed / frequency.

Wavelength = 3x10^8 m/s / 30 hz

Wavelength = 10 million meters

1/2 wavelength = 5 million meters

(that's about 3,100 miles)

I'm pretty sure the frequency is wrong in the question.

I think it's actually 30 kHz, not 30 Hz.

That makes the antenna about 3.1 miles long.

Answer:

como  ías

Explanation:

Answer:

B) a 1200 kg car driving 25 m/s

Explanation:

Momentum can be defined as the multiplication (product) of the mass possessed by an object and its velocity. Momentum is considered to be a vector quantity because it has both magnitude and direction.

Mathematically, momentum is given by the formula;

[tex] Momentum = mass * velocity [/tex]

This ultimately implies that, the mass of an object or body is directly proportional to its momentum. Thus, the higher the mass of an object or body, the greater would be its momentum and vice-versa.

By mere inspection of the data given, we can see that the object with the greatest amount of mass and velocity is the car weighing 1200 kilograms and moving at 25 meters per seconds.

Substituting into the formula, we have;

[tex] Momentum = 1200 * 25 [/tex]

Momentum = 30,000 Kgm/s

Answer:

Explanation:

Moment of inertia I = M k² , where M is mass and k is radius of gyration .

Putting the given values in the equation

5000 = 200 x k²

k² = 25

k = 5 cm .

Radius of gyration is 5 cm .

Answer:

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Explanation:

Using locally available resources for art help in the preservation of environment. A significant and practical aspects of art is material significance. The items used by artists while making an art piece affects both the form and the material. Every material delivers something special in the creative process.

According to Coulomb's law, the force between the given charges is 0.225N which is explained below.

Force on two identical charges q separate by a distance of r is given by:

F = kq²/r²

where k is Coulomb's constant

q is the charge

r is the separation between the charges

Given that q = 1×10⁻⁵C,

and r = 2m

So, the force between the given charges will be:

F = (9×10⁹)(1×10⁻⁵)²/2²

F = 0.225N is the required force.

Learn more about Coulomb's law:

https://brainly.com/question/506926?referrer=searchResults

Answer:The answer is because of the gravity and the mass formed in the sun, the magnetic field reacts to it and leaves a fault on earth. wind goes by the earth and procides to be ok

Explanation:that is it

Answer:

The second bulb will have thicker filament

Explanation:

Given;

First electric bulb: Power, P₁ = 100 W and Voltage, V₁ = 220 V

Second electric bulb: Power, P₂ = 200 W and Voltage, V₂ = 200 V

Resistivity of tungsten, ρ = 4.9 x 10⁻⁸ ohm. m

Resistance of the first bulb:

[tex]P = IV = \frac{V}{R} .V = \frac{V^2}{R} \\\\R = \frac{V^2}{P} \\\\R_1 = \frac{V_1^2}{P_1} = \frac{(220)^2}{100} = 484 \ ohms[/tex]

Resistance of the second bulb:

[tex]R_2 = \frac{V_2^2}{P_2} = \frac{(200)^2}{200} = 200 \ ohms[/tex]

Resistivity of the tungsten filament is given by the following equation;

[tex]\rho = \frac{RA}{L}[/tex]

where;

L is the length of the filament

R is resistance of each filament

A is area of each filament

[tex]A = \pi r^2[/tex]

where;

r is the thickness of each filament

[tex]\rho = \frac{R (\pi r^2)}{L} \\\\\frac{\rho L}{\pi} = Rr^2 \\\\Recall ,\ \frac{\rho L}{\pi} \ is \ constant \ for \ both \ filaments\\\\R_1r_1^2 = R_2r_2^2\\\\(\frac{r_1}{r_2} )^2 = \frac{R_2}{R_1} \\\\\frac{r_1}{r_2} = \sqrt{\frac{R_2}{R_1} } \\\\\frac{r_1}{r_2} = \sqrt{\frac{200}{484} } \\\\\frac{r_1}{r_2} = 0.64\\\\r_1 = 0.64 \ r_2\\\\r_2 = 1.56 \ r_1[/tex]

Therefore, the second bulb will have thicker filament

I believe the answer is a

Answer:

im sorry i would help but thats too much

Answer:

a.

Explanation:

there would be a new planet is our solar system which could cause different gravitation pull on all the planets also there could be possible be new life form or other valuable metals that haven't been discovered on this planet. hope this helps somewhat

Answer:

Explanation:

Time period of rotation

T = 2πR/ V where R is radius of orbit and V is orbital velocity

Orbital velocity V = √ ( GM/R ) , m is mass of the earth .

T = 2πR √R / GM

T² = 4π²R³ / GM

Putting the values

( 126 x 60 )² = 4 x 3.14² x R³ / 6.67 x 10⁻¹¹ x 5.97 x 10²⁴

57.15 x 10⁶ = 39.44 x R³ / 39.82 x 10¹³

R³ = 577 X 10¹⁸

R = 8.325 x 10⁶ m

= 8325 km

Radius of earth = 6400 km

height of satellite = 8325- 6400 = 1925 km .