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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:

[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:

a. 660 gallons

b.665 gallons

c. 1,800

Answer:

B motor

Explanation:

the exosphere is the least dense

Answer:

exosphere

Explanation: Just took the quiz and got an A, 100%

Answer:

please give me brainlist and follow

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.

Answer:

The border is 8km above sea level.

Explanation:

We know that:

Density = 1.25 kg/m^3

Pressure = 10^5 N/m^2

g = 10m/s^2

Now, suppose that we have a virtual rectangle, such that its bases have an area of 1m^2 and the rectangle has a height equal to H.

This virtual figure has a volume V = 1m^2*H, and it is filled with air (which we know that has a density 1.25 kg/m^3)

Then the total mass inside that volume is:

M = (1.25 kg/m^3)*V = (1.25 kg/m^3)*(1m^2*H)

The weight of this mass is:

W = g*M = (10m/s^2)*(1.25 kg/m^3)*(1m^2*H)

And if we divide the weight in a given surface, let's say 1 m^2, we get the pressure per square meter, which we know is equal to  10^5 N/m^2

then:

P = 10^5 N/m^2 = (10m/s^2)*(1.25 kg/m^3)*(1m^2*H)*(1/m^2)

Whit this equation we can find the value of H.

10^5 N/m^2 = (10m/s^2)*(1.25 kg/m^3)*(1m^2*H)*(1/m^2)

10^5 N =  (10m/s^2)*(1.25 kg/m^3)*(1m^2*H)

(10^5 N)/(10 m/s^2) = (1.25 kg/m^3)*(1m^2*H)

(10^4 kg) = (1.25 kg/m^3)*(1m^2*H)

(10^4 kg)/( 1.25 kg/m^3) = 1m^2*H

8,000 m^3 = 1m^2*H

(8,000 m^3)/(1m^2) =H

8,000 m = H

And we want this answer in km, knowing that 1,000m = 1km

8,000m = 8km = H

The border is 8km above sea level.

Height of boundaries is 8.2 km

Normal density = 1.25 kg/m³

1 atm = 101325 N/m²

Height of boundaries

Pressure = Height × Density × Gravitational acceleration

101325 = Height × 1.25 × 9.8

101325 = Height × 12.25

Height of boundaries = 101325 / 12.25

Height of boundaries = 8271.42 m

Height of boundaries = 8.2 km

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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:

t = 0.437 s

Explanation:

The speed of sound is a constant that is worth v = 343 m / s

           v = d / t

            t = d / v

the time it takes for the sound to reach Clark at d = 150 m is

           t = 150/343

           t = 0.437 s

This same sound takes much longer to reach you

          t₂ = 127 10³/343

          t₂ = 370 s

Answer: 0.516 ft/s

Explanation:

Given

Length of ladder L=20 ft

The speed at which the ladder moving away is v=2 ft/s

after 1 sec, the ladder is 5 ft away from the wall

So, the other end of the ladder is at

[tex]\Rightarrow y=\sqrt{20^2-5^2}=19.36\ ft[/tex]

Also, at any instant t

[tex]\Rightarrow l^2=x^2+y^2[/tex]

differentiate w.r.t.

[tex]\Rightarrow 0=2xv+2yv_y\\\\\Rightarrow v_y=-\dfrac{x}{y}\times v\\\\\Rightarrow v_y=-\dfrac{5}{19.36}\times 2=0.516\ ft/s[/tex]

Answer:

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Explanation:

isos

Answer:

C. Amount of oxygen

Explanation:

Options A and D are invalid as they aren't affecting factors.

Option B is false as the altitude increases as you go up a mountain.

Option C is true as the air pressure (atmospheric pressure) is inversely proportional to the height/altitude of the mountain.

Answer:

the answer is false

Explanation:

plastic doesnt conduct anything

Answer:

the length of the wire is 134.62 m.

Explanation:

Given;

resistivity of the copper wire, ρ = 2.6 x 10⁻⁸ Ωm

cross-sectional area of the wire, A  = 35 x 10⁻⁴ cm² = ( 35 x 10⁻⁴) x 10⁻⁴ m²

resistance of the wire, R = 10Ω

The length of the wire is calculated as follows;

[tex]R = \frac{\rho L}{A} \\\\L = \frac{RA}{\rho} \\\\L= \frac{10 \times (35\times 10^{-4}) \times 10^{-4}}{2.6 \times 10^{-8}} \\\\L = 134.62 \ m[/tex]

Therefore, the length of the wire is 134.62 m.

Answer:

the wave represents the second harmonic.

Explanation:

Given;

length of the cord, L = 64 cm

The first harmonic of a cord fixed at both ends is given as;

[tex]f_o = \frac{V}{2L}[/tex]

The wavelength of a standing wave with two antinodes is calculated as follows;

L = N---> A -----> N    +   N ----> A -----> N

Where;

N is node

A is antinode

L = N---> A -----> N    +   N ----> A -----> N =  λ/2  + λ/2

L = λ

The harmonic is calculated as;

[tex]f = \frac{V}{\lambda} \\\\f = \frac{V}{L} = 2(\frac{V}{2L} ) = 2(f_o) = 2^{nd} \ harmonic[/tex]

Therefore, the wave represents the second harmonic.

L = λ

Answer:

[tex]\lambda_s =6.43*10^-4m[/tex]

Explanation:

From the question we are told that:

Diffraction grating [tex]N=1250slits/cm[/tex]

Distance b/w Screen and grating length [tex]d_{sg}=17.5 cm[/tex]

Distance b/w neighboring maxima and Screen [tex]d_{ms}=8.44[/tex]

Generally the equation for grating space is mathematically given by

[tex]d(g)=\frac{1}{N}[/tex]

[tex]d(g)=\frac{100}{1250}[/tex]

[tex]d(g)=0.08[/tex]

Generally the equation for small angle approximation is mathematically given by

[tex]\triangle y=\frac{\lambda d}{L}[/tex]

Therefore for longest wavelength

[tex]\lambda _l=\frac{8.44*10^{-3}*(0.08)}{0.175m}[/tex]

[tex]\lambda _l=3.858*10^{-3}[/tex]

Therefore the third order maximum equation for the shorter wavelength as

[tex]\lambda_s =\frac{1}{6} \lambda_l[/tex]

[tex]\lambda_s =\frac{1}{6} (3.858*10^-^3)[/tex]

[tex]\lambda_s =6.43*10^-4m[/tex]

The wavelengths in the light is given as

[tex]\lambda_s =6.43*10^-4m[/tex]

Answer:

c. Double Replacement

Explanation:

As in Double Replacement reaction exchanges the cations (or the anions) of two ionic compounds.

Here, in BaCl2 , Ba has replaced with NO3 to form Ba(NO3)2

and in 2AgNo3 , Ag has replaced with Cl to form 2AgCl.

Answer:

Explanation:

mass per unit length ρ = .100 / 1.65 = .0606 . kg /m

length of wire L = 1.65 m

For fundamental frequency , the expression is as follows

n = [tex]\frac{1}{2L} \sqrt{\frac{T}{m} }[/tex]

L = 1.65 , T = 16 n and m = .0606

n = [tex]\frac{1}{2\times 1.65} \sqrt{\frac{16}{.0606} }[/tex]

= 4.9 /s .

This is fundamental frequency .

other mode of vibration ( first three ) will be as follows

4.9 x 2 = 9.8 /s ,

4.9 x 3 = 14.7 /s .

Answer:

The correct answer is -  91.4 nm

Explanation:

According to Bohr's model, the minimum wavelength to ionize Hydrogen atom from n= 1 state is expressed as:

(h×c)/λ=13.6eV

here,

h - Planck constant

c - the speed of light

λ - wavelength

Placing the value in the formula for the wavelength

(6.626×10^−34J.s × 3×10^8 m/s)/λ  =  13.6 ×1.6 × 10^−19 J

λ≈91.4nm

Thus, the correct answer would be = 91.4 nm

Answer:

a) The minimum thickness of the oil slick at the spot is 313 nm

b) the minimum thickness be now will be 125 nm

Explanation:  

Given the data in the question;

a) The index of refraction of the oil is 1.20. What is the minimum thickness of the oil slick at that spot?

t[tex]_{min[/tex] = λ/2n

given that; wavelength λ = 750 nm and  index of refraction of the oil n = 1.20

we substitute

t[tex]_{min[/tex] = 750 / 2(1.20)

t[tex]_{min[/tex] = 750 / 2.4

t[tex]_{min[/tex] = 312.5 ≈ 313 nm

Therefore, The minimum thickness of the oil slick at the spot is 313 nm

b)

Suppose the oil had an index of refraction of 1.50. What would the minimum thickness be now?

minimum thickness of the oil slick at the spot will be;

t[tex]_{min[/tex] = λ/4n

given that; wavelength λ = 750 nm and  index of refraction of the oil n = 1.50

we substitute

t[tex]_{min[/tex] = 750 / 4(1.50)

t[tex]_{min[/tex] = 750 / 6

t[tex]_{min[/tex] = 125 nm

Therefore, the minimum thickness be now will be 125 nm

Answer:

 P and S waves slow down when they reach this layer. The asthenosphere, also known as the magma chamber, is the uppermost component of the mantle. This layer is partially molten and is a ductile zone in a tectonically poor state.

It's almost hard and seismic waves move through the asthenosphere at a slow rate. The fragile lithosphere and the uppermost portion of the asthenosphere are assumed to be rigid.

seismic waves travel more quickly through denser materials and therefore generally travel more quickly with the depth it moves more slowly through a liquid than a solid. Molten areas within the Earth slow down P waves and stop S waves because their shearing motion cannot be transmitted through a liquid. Partially molten areas may slow down the P waves and attenuate or weaken S waves.

hope this helps...

S and P wave slow down and stop in  the uppermost part of the mantle. - For this, scientists have concluded that the uppermost part of the mantle is partially-molten.

A planetary body's mantle is a layer that is surrounded by the crust on top and the core underneath. The largest and most substantial layer of a planetary body, mantles are often comprised of rock or ice. Planetary bodies that have undergone density differentiation typically have mantles. Mantles are found on all terrestrial planets (including Earth), many asteroids, and a few planetary moons.

Between the crust and the outer core, there is a silicate rock layer known as the Earth's mantle. Despite being mostly solid, it behaves like a viscous fluid over geological time. Oceanic crust is created by the partial melting of the mantle at mid-ocean ridges, and continental crust is created by the partial melting of the mantle at subduction zones.

Learn more about mantle here:

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When light encounters transparent materials, almost all of it passes directly through them. Glass, for example, is transparent to all visible light. ... Most of the light is either reflected by the object or absorbed and converted to thermal energy. Materials such as wood, stone, and metals are opaque to visible light.

Answer:

6.12 m/s

Explanation:

Using the law of conservation of momentum

momentum before collision = momentum after collision

m₁v₁ + m₂v₂ = (m₁ + m₂)V    (since the train cars become attached to each other) where m₁ = mass of car 1 = 1,825 kg, m₂ = mass of car 2 = 2,645 kg, v₁ = initial velocity of car 1 = + 15.0 m/s (positive since it is moving in the positive x direction), v₂ = initial velocity of car 2 = 0 m/s (since it is initially stationary) and V = velocity of both cars after collision,

So, m₁v₁ + m₂v₂ = (m₁ + m₂)V  

m₁v₁ + m₂(0 m/s) = (m₁ + m₂)V  

m₁v₁ + 0 = (m₁ + m₂)V  

V = m₁v₁/(m₁ + m₂)

substituting the values of the other variables into the equation, we have

V = 1,825 kg × 15.0 m/s/(1,825 kg + 2,645 kg)

V = 27375 kgm/s/ 4470kg

V = 6.124 m/s

V ≅ 6.12 m/s

Answer:

[tex]63.38\ \text{mph}[/tex]

Explanation:

L = Lift force

[tex]\rho[/tex] = Density of air

A = Surface area

v = Velocity

[tex]v_1[/tex] = 45 mph

[tex]\rho_1=1.23\ \text{kg/m}^3[/tex]

[tex]\rho_2=0.62\ \text{kg/m}^3[/tex]

Coefficient of lift is given by

[tex]CL=\dfrac{2L}{\rho v^2A}\\\Rightarrow \rho=\dfrac{2L}{CL v^2A}[/tex]

So

[tex]\rho\propto \dfrac{1}{v^2}[/tex]

[tex]\dfrac{\rho_1}{\rho_2}=\dfrac{v_2^2}{v_1^2}\\\Rightarrow v_2=\sqrt{\dfrac{\rho_1}{\rho_2}}\times v_1\\\Rightarrow v_2=\sqrt{\dfrac{1.23}{0.62}}\times 45\\\Rightarrow v_2=63.38\ \text{mph}[/tex]

The velocity at the required altitude should be [tex]63.38\ \text{mph}[/tex] to maintain the same lift.

Answer:

[tex]65.6\ \text{J/m}^3[/tex]

[tex]0.11\ \text{J}[/tex]

Explanation:

B = Magnetic field = [tex]\mu_0 \dfrac{N}{l}I[/tex]

[tex]\mu_0[/tex] = Vacuum permeability = [tex]4\pi10^{-7}\ \text{H/m}[/tex]

N = Number of turns = 1270

[tex]l[/tex] = Length of solenoid = 93.9 cm = 0.939 m

[tex]I[/tex] = Current = 7.8 A

A = Area of solenoid = [tex]17.3\ \text{cm}^2[/tex]

Energy density of a solenoid is given by

[tex]u_m=\dfrac{B^2}{2\mu_0}\\\Rightarrow u_m=\dfrac{(\mu_0 \dfrac{N}{l}I)^2}{2\mu_0}\\\Rightarrow u_m=\dfrac{\mu_0N^2I^2}{2l^2}\\\Rightarrow u_m=\dfrac{4\pi\times 10^{-7}\times 1230^2\times 7.8^2}{2\times 0.939^2}\\\Rightarrow u_m=65.6\ \text{J/m}^3[/tex]

The energy density of the magnetic field inside the solenoid is [tex]65.6\ \text{J/m}^3[/tex]

Energy is given by

[tex]U_m=u_mAl\\\Rightarrow U_m=65.6\times 17.3\times 10^{-4}\times 0.939\\\Rightarrow U_m=0.11\ \text{J}[/tex]

The total energy in joules stored in the magnetic field is [tex]0.11\ \text{J}[/tex].