Which of the following devices can be used to measure force?

A. Bathroom scale
(b) spring scale
c. Force sensor
d. All of the above.

Answer:

(a) g = 8.82158145[tex]m/s^2[/tex].

(b) 7699.990192m/s.

(c)5484.3301s = 1.5234 hours.(extremely fast).

Explanation:

(a) Strength of gravitational field 'g' by definition is

[tex]g = \frac{M_{(earth)} }{r^2} G[/tex] , here G is Gravitational Constant, and r is distance from center of earth, all the values will remain same except r which will be radius of earth + altitude at which ISS is in orbit.

r = 6721,000 meters, putting this value in above equation gives g = 8.82158145[tex]m/s^2[/tex].

(b) We have to essentially calculate centripetal acceleration that equals new 'g'.

[tex]a_{centripetal}=\frac{V^2}{r} =g[/tex] here g is known, r is known and v is unknown.

plugging in r and g in above and solving for unknown gives V = 7699.990192m/s.

(c)  S = vT,  here T is time period or time required to complete one full revolution.

S =  earth's circumfrence , V is calculated in (B) T is unknown.

solving for unknown gives T = 5484.3301s = 1.5234hours.

The wave equation is missing and it is y(x,t) = (8.50 mm)cos(172 rad/m x − 4830 rad/s t)

Answer:

A) 0.0534 seconds

B) 0.67N

C) 41

D) (8.50 mm)cos(172 rad/m x + 4830 rad/s t)

Explanation:

we are given weight of string = 0.0125N

Thus, since weight = mg

Then, mass of string = 0.0125/9.8

Mass of string = 1.275 x 10⁻³ kg

Length of string; L= 1.5 m .

mass per unit length; μ = (1.275 x 10⁻³)/1.5

μ = 0.85 x 10⁻³ kg/m

We are given the wave equation: y(x,t) = (8.50 mm)cos(172 rad/m x − 4830 rad/s t)

Now if we compare it to the general equation of motion of standing wave on a string which is:

y(x,t) = Acos(Kx − ω t)

We can deduce that

angular velocity;ω = 4830 rad/s

Wave number;k = 172 rad/m

A) Velocity is given by the formula;

V = ω/k

Thus, V = 4830/172 m/s

V = 28.08 m /s

Thus time taken to go up the string = 1.5/28.08 = 0.0534 seconds

B) We know that in strings,

V² = F/μ

Where μ is mass per unit length and V is velocity.

Thus, F = V²*μ =28.08² x 0.85 x 10⁻³

F = 0.67N

C) Formula for wave length is given as; wave length;λ = 2π /k

λ = 2 x π/ 172

λ = 0.0365 m

Thus, number of wave lengths over whole length of string

= 1.5/0.0365 = 41

D) The equation for waves traveling down the string

= (8.50 mm)cos(172 rad/m x + 4830 rad/s t)

The required time is: 0.395 s. Weight and wavelengths are 0.013 and 1.5. The equation for waves traveling down the string is: y(x, t) = (8.50 mm)cos(172 rad/m)x+4830 rad/s)t).

a) To find the time it takes for a pulse to travel the full length of the string, we need to determine the velocity of the wave. The velocity of a wave on a string is given by the equation [tex]v = \sqrt{(T/\mu)[/tex], where T is the tension and μ is the linear mass density of the string.  

Given that the weight is 0.0125 N and the length is 1.50 m.

Here, mass = w/g = 0.0125/9.8 = 1.275 x 10⁻³ kg

we can calculate μ as follows: μ = m/L = (1.275 x 10⁻³)/(1.50 m) = 0.00085 kg/m.

Substituting the values into the equation:

we get v = [tex]\sqrt{W/0.00085}[/tex]. As it is given that the tension of the string is constant and equal to W.

Thus, v = [tex]\sqrt{14.7}[/tex] = 3.8 m/s

The time it takes for a pulse to travel the full length of the string can be calculated using the formula time = distance/velocity = 1.50/3.8. So, time = 0.395 s.

b) In case of string, V² = F/μ

μ x V² = F

This implies:

F = 0.00085 x [tex](3.8)^2[/tex] = 0.0123

C)  We know, f (frequency) = 1/t = 1/0.395 = 2.53

Also, v = λf. This implies:

λ = v/f = 3.8/2.53 = 1.5

D) Finally, to find the equation for waves traveling down the string, we need to adjust the sign inside the cosine function to account for the wave traveling in the opposite direction. It is given that:

y(x,t) = (8.50 mm)cos(172 rad/m x − 4830 rad/s t)

So, the equation becomes y(x, t) = (8.50 mm)cos(172 rad/m)x+4830 rad/s)t).

Complete Question:

Question

A 1.50-m string of weight 0.0125 N is tied to the ceil-ing at its upper end, and the lower end supports a weight W. Ignore the very small variation in tension along the length of the string that is produced by the weight of the string. When you pluck the string slightly, the waves traveling up the string obey the equation. Assume that the tension of the string is constant and equal to W.

y(x, t) = (8.50 mm)cos(172 rad/m)x+4830 rad/s)t).

(a) How much time does it take a pulse to travel the full length of the string?

(b) What is the weight W?

(c) How many wavelengths are on the string at any instant of time?

(d) What is the equation for waves traveling ?down? the string?

49.5 miles per hour will be the new speed.

This problem can be resolved using one-dimensional kinematics: v² = v₀² - 2 a d

Let's use this equation to find the beginning speed of the vehicle. In the first example, the distance is half (d´ = ½ d), the vehicle's acceleration is the same, and we find the fine velocity zero 0 = v₀² - 2 a d a = v₀² / 2d.

v₀´² = 2 a d = 2 (v₀² / 2d) 0 = v₀´² - 2 a d´ d´ v₀´ = v₀ √ 1/2

Let's compute: v₀ = 49.5 miles per hour; v₀´ = 70 ra 0.5.

Consequently, 49.5 miles per hour will be the new speed.

Answer:

option D.

Explanation:

Given,

distance, d = 1 x 10⁵ km

speed , v = 0.217 c

time of dilation , T₀= ?

Using the formula of time dilation

[tex]T=\frac{T_{0}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}[/tex]

[tex]T_{0}=T \sqrt{1-\frac{v^{2}}{c^{2}}} [/tex]

[tex]=\left(\frac{d}{v}\right) \sqrt{1-\frac{v^{2}}{c^{2}}} [/tex]

[tex]=\left(\frac{1.00 \times 10^{8}}{0.217 \times 2.99 \times 10^{8}}\right) \sqrt{1-\frac{(0.217 c)^{2}}{c^{2}}} [/tex]

[tex]=1.54 \times 0.976 [/tex]

[tex]=1.50\ s[/tex]

Time he should be allowed between passing the buoy and ejecting is equal to 1.50 s.

The correct answer is option D.

Answer:

a) h=3.16 m, b)  v_{cm }^ = 6.43 m / s

Explanation:

a) For this exercise we can use the conservation of mechanical energy

Starting point. Highest on the hill

           Em₀ = U = mg h

final point. Lowest point

           [tex]Em_{f}[/tex] = K

Scientific energy has two parts, one of translation of center of mass (center of the sphere) and one of stationery, the sphere

           K = ½ m [tex]v_{cm }^{2}[/tex] + ½ [tex]I_{cm}[/tex] w²

angular and linear speed are related

           v = w r

           w = v / r

            K = ½ m v_{cm }^{2} + ½ I_{cm} v_{cm }^{2} / r²

            Em_{f} = ½ v_{cm }^{2} (m + I_{cm} / r2)

as there are no friction losses, mechanical energy is conserved

             Em₀ = Em_{f}

             mg h = ½ v_{cm }^{2} (m + I_{cm} / r²)         (1)

             h = ½ v_{cm }^{2} / g (1 + I_{cm} / mr²)

for the moment of inertia of a basketball we can approximate it to a spherical shell

             I_{cm} = ⅔ m r²

we substitute

            h = ½ v_{cm }^{2} / g (1 + ⅔ mr² / mr²)

            h = ½ v_{cm }^{2}/g    5/3

             h = 5/6 v_{cm }^{2} / g

let's calculate

           h = 5/6 6.1 2 / 9.8

           h = 3.16 m

b) this part of the exercise we solve the speed of equation 1

          v_{cm }^{2} = 2m gh / (1 + I_{cm} / r²)

in this case the object is a frozen juice container, which we can simulate a solid cylinder with moment of inertia

              I_{cm} = ½ m r²

we substitute

             v_{cm } = √ [2gh / (1 + ½)]

             v_{cm } = √(4/3 gh)

let's calculate

             v_{cm } = √ (4/3 9.8 3.16)

             v_{cm }^ = 6.43 m / s

Answer: 1.3 ×10^-31

Explanation:

the required probability is P = e^(-2αL)

Firstly, evaluate (-2αL)

α= 1/hc √2mc^2 (U - E)

h= modified planck's constant

where,

(-2αL)

= -(2L)/(h/2π ) ×√2mc^2 (U - E)

= -(2L) / (hc^2/π )×√2mc^2 (U - E)

(hc^2/2pi) = 197*eV.nm (standard constant)

2*L = 8 nm

mc^2 = 0.511×10^6 eV

Where m = mass electron

C= speed of light

(-2αL) = [-8nm/(197 eV.nm)] × (1.022× 10^6 eV*×3 eV)^0.5

(-2αL) = -71.1

Probability = e^(-2αL) = e^-71.1 = 1.3 ×10^-31

Therefore, the probability that a ondu tion ele tron in one wire arriving at the gap will pass through the gap into the other wire 1.3 ×10^-31.

Answer:

 k = 11,564 N / m,   w = 6.06 rad / s

Explanation:

In this exercise we have a horizontal bar and a vertical spring not stretched, the bar is released, which due to the force of gravity begins to descend, in the position of Tea = 46º it is in equilibrium;

 let's apply the equilibrium condition at this point

Axis y

          W_{y} - Fr = 0

          Fr = k y

let's use trigonometry for the weight, we assume that the angle is measured with respect to the horizontal

             sin 46 = [tex]W_{y}[/tex] / W

             W_{y} = W sin 46

 we substitute

           mg sin 46 = k y

           k = mg / y sin 46

If the length of the bar is L

          sin 46 = y / L

           y = L sin46

we substitute

           k = mg / L sin 46 sin 46

           k = mg / L

for an explicit calculation the length of the bar must be known, for example L = 1 m

           k = 1.18 9.8 / 1

           k = 11,564 N / m

With this value we look for the angular velocity for the point tea = 30º

let's use the conservation of mechanical energy

starting point, higher

          Em₀ = U = mgy

end point. Point at 30º

         [tex]Em_{f}[/tex] = K -Ke = ½ I w² - ½ k y²

          em₀ = Em_{f}

          mgy = ½ I w² - ½ k y²

          w = √ (mgy + ½ ky²) 2 / I

the height by 30º

           sin 30 = y / L

           y = L sin 30

           y = 0.5 m

the moment of inertia of a bar that rotates at one end is

          I = ⅓ mL 2

          I = ½ 1.18 12

          I = 0.3933 kg m²

let's calculate

          w = Ra (1.18 9.8 0.5 + ½ 11,564 0.5 2) 2 / 0.3933)

          w = 6.06 rad / s

Answer:

E₄ = - 0.85 eV

E₂ = - 3.4 eV

Ephoton = 2.55 eV

Explanation:

The sum of Kinetic Energy (K) and Potential Energy (U) of the Helium atom is equal to the total energy of Helium atom in the specified state N. From Bohr's atomic model, the energy of a hydrogen atom in state N is given as:

En = K + U = (-1/n²)(13.6 eV)

a)

Here,

n = 4

Therefore,

E₄ = (-1/4²)(13.6 eV)

E₄ = - 0.85 eV

b)

Here,

n = 2

Therefore,

E₂ = (-1/2²)(13.6 eV)

E₂ = - 3.4 eV

c)

The energy of photon emitted in the transition from level 4 to level 2 will be equal to the difference in the energy of both levels:

Ephoton = ΔE =  E₄ - E₂

Ephoton = - 0.85 eV - (- 3.4 eV)

Ephoton = 2.55 eV

The electrons energy will be:

As we know. the formula:

→ [tex]E_n = K+U[/tex]

        [tex]= (-\frac{1}{n^2} ) (13.6 \ eV)[/tex]

(a)

Given:

then,

→ [tex]E_4 = (-\frac{1}{4^2} )(13.6 \ eV)[/tex]

        [tex]= -0.85 \ eV[/tex]

(b)

Given:

then,

→ [tex]E_2 = (-\frac{1}{2^2} )(13.6 \ eV)[/tex]

        [tex]= -3.4 \ eV[/tex]

(c)

The energy of photon emitted will be:

→ [tex]E_{photon} = \Delta E[/tex]

               [tex]= E_4 -E_2[/tex]

               [tex]=-0.85 \ eV-(-3.4 \ eV)[/tex]

               [tex]= 2.55 \ eV[/tex]  

Thus the above answers are appropriate.

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The object strikes the surface of the Earth with a speed of approximately 11.2 km/s.

Ignoring atmospheric friction, the object's total mechanical energy (the sum of its gravitational potential energy and kinetic energy) is conserved throughout its fall. We can use this principle to calculate the object's speed at the Earth's surface.

Here's how:

Calculate the gravitational potential energy (PE) at the initial altitude:

PE_i = -G * M_earth * m / (R_earth + h)

where:

G is the gravitational constant (6.67 × 10^-11 N m²/kg²)

M_earth is the mass of the Earth (5.97 × 10^24 kg)

m is the mass of the object (910 kg)

R_earth is the Earth's polar radius (6357 km)

h is the initial altitude (1200 km)

Set the final PE (PE_f) to 0:

Since all the PE will be converted to kinetic energy when the object reaches the surface, PE_f = 0.

Apply the conservation of mechanical energy:

E_i = PE_i + KE_i = PE_f + KE_f = KE_f (as PE_i = PE_f = 0)

where:

KE_i is the initial kinetic energy (0 J, as the object is at rest)

KE_f is the final kinetic energy

Solve for the final velocity (v_f):

KE_f = 1/2 * m * v_f^2

v_f = sqrt(2 * E_i / m)

Plug in the values and calculate v_f:

v_f = sqrt(2 * (-G * M_earth * m / (R_earth + h)) / m)

v_f ≈ 11,180 m/s

Convert m/s to km/s:

v_f ≈ 11.2 km/s

Answer:

m = 41.39 m/s

Explanation:

For the source alone , we can write next equation.  

V_r = m/2πr

Stagnation point will occur where r=b, thus;  

V = m/2πb

Now we can find source strength, and therefore we have;  

m = 2*π*b*V

 b = 0.16/2*π

b = 0.0254

m = 2*π*0.32*20.6

m = 41.39 m/s

Final answer:

To simulate flow around a half-body with a thickness of 0.16 m in a 20.6 m/s air stream, the required source strength is 6.592 m^3/s, calculated using the formula Q = 2 * Vθ * d. = 6.592 m3/s

Explanation:

Calculation of Source Strength for Flow Simulation

To simulate flow around a streamlined half-body in a uniform air stream, a combination of a uniform flow and a source flow is used. In the given scenario, the thickness of the half-body is 0.16 m, and it is placed in an air stream moving at 20.6 m/s. The strength of the source, often represented by the symbol Q (and measured in m3/s), is the volume of fluid introduced into the flow per unit of time. For a two-dimensional flow over a half-body, the source strength required to simulate the flow pattern is proportional to the product of the freestream velocity and the thickness of the half-body.

To calculate the required source strength mathematically, you would use the formula:

Q = 2 * V∞ * d

Where V∞ is the freestream velocity, and d is the maximum thickness of the half-body. Substituting the given values:

Q = 2 * 20.6 m/s * 0.16 m

Q = 6.592 m^3/s

Thus, the source strength required to simulate flow around the half-body in uniform flow would be 6.592 m3/s. It is essential to ensure that the flow remains laminar and non-turbulent to maintain the accuracy of this simulation. Factors such as the Reynolds number would also be considered in a more complex analysis to determine the flow regime.

Answer:

3.78V

Explanation:

The resistance in the circuit is

The potential difference across the 4.00 V battery, Vbc, in a circuit where it is in parallel with a series combination of 10.30 V and 8.00 V batteries, is equal to the total emf of the two batteries in series minus the potential drop due to their current through the 0.50 Ω internal resistance.

To determine the potential difference Vbc across the terminals of the 4.00 V battery, we first need to understand the circuit's behavior. The circuit now consists of two batteries connected in series - one with 10.30 V (internal resistance 0.50 Ω) and another of 8.00 V - and the 4.00 V battery is in parallel with this series combination.

Because the batteries in series share the same current, we can calculate the total current (I) using Ohm's law and considering the total emf and resistance:

I= (10.30 V + 8.00 V)/( R + 0.50 Ω)

Note that R includes any circuit resistance (which hasn't been mentioned) and the 4.00 V battery's own internal resistance. Therefore, Vbc, the potential difference across the 4.00 V battery will be equal to the total potential provided by the series batteries minus the potential drop due to their current through the 0.50 Ω internal resistance.

Vbc = (10.30 V + 8.00 V) - I * 0.50 Ω

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Answer:

(D) 3

Explanation:

The angular momentum is given by:

[tex]\vec{L}=\vec{r}\ X \ \vec{p}[/tex]

Thus, the magnitude of the angular momenta of both solar systems are given by:

[tex]L_1=Rm_1v_1=Rm_1(\omega R)=R^2m_1(\frac{2\pi}{T_1})=2\pi R^2\frac{m_1}{T_1}\\\\L_2=Rm_2v_2=2\pi R^2\frac{m_2}{T_2}[/tex]

where we have taken that both systems has the same radius.

By taking into account that T1=3T2, we have

[tex]L_1=2\pi R^2\frac{m_1}{3T_2}=\frac{1}{3}2\pi R^2\frac{1}{T_2}m_1=\frac{1}{3}\frac{L_2}{m_2}m_1[/tex]

but L1=L2=L:

[tex]L=\frac{1}{3}L\frac{m_1}{m_2}\\\\\frac{m_1}{m_2}=3[/tex]

Hence, the answer is (D) 3

HOPE THIS HELPS!!

Answer:

The ocean absorbs a significant portion of carbon dioxide (CO2) emissions from human activities, equivalent to about one-third of the total emissions for the past 200 years from fossil fuel combustion, cement production and land-use change (Sabine et al., 2004). Uptake of CO2 by the ocean benefits society by moderating the rate of climate change but also causes unprecedented changes to ocean chemistry, decreasing the pH of the water and leading to a suite of chemical changes collectively known as ocean acidification. Like climate change, ocean acidification is a growing global problem that will intensify with continued CO2 emissions and has the potential to change marine ecosystems and affect benefits to society.

The average pH of ocean surface waters has decreased by about 0.1 unit—from about 8.2 to 8.1—since the beginning of the industrial revolution, with model projections showing an additional 0.2-0.3 drop by the end of the century, even under optimistic scenarios (Caldeira and Wickett, 2005).1 Perhaps more important is that the rate of this change exceeds any known change in ocean chemistry for at least 800,000 years (Ridgewell and Zeebe, 2005). The major changes in ocean chemistry caused by increasing atmospheric CO2 are well understood and can be precisely calculated, despite some uncertainty resulting from biological feedback processes. However, the direct biological effects of ocean acidification are less certain

image

1 “Acidification” does not mean that the ocean has a pH below neutrality. The average pH of the ocean is still basic (8.1), but because the pH is decreasing, it is described as undergoing acidification.

Page 2

Suggested Citation:"Summary." National Research Council. 2010. Ocean Acidification: A National Strategy to Meet the Challenges of a Changing Ocean. Washington, DC: The National Academies Press. doi: 10.17226/12904. ×

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and will vary among organisms, with some coping well and others not at all. The long-term consequences of ocean acidification for marine biota are unknown, but changes in many ecosystems and the services they provide to society appear likely based on current understanding (Raven et al., 2005).

In response to these concerns, Congress requested that the National Research Council conduct a study on ocean acidification in the Magnuson-Stevens Fishery Conservation and Management Reauthorization Act of 2006. The Committee on the Development of an Integrated Science Strategy for Ocean Acidification Monitoring, Research, and Impacts Assessment is charged with reviewing the current state of knowledge and identifying key gaps in information to help federal agencies develop a program to improve understanding and address the consequences of ocean acidification (see Box S.1 for full statement of task). Shortly after the study was underway, Congress passed another law—the Federal Ocean Acidification Research and Monitoring (FOARAM) Act of 2009—which calls for, among other things, the establishment of a federal ocean acidification program; this report is directed to the ongoing strategic planning process for such a program.

Although ocean acidification research is in its infancy, there is already growing evidence of changes in ocean chemistry and ensuing biological impacts. Time-series measurements and other field data have documented the decrease in ocean pH and other related changes in seawater chemistry (Dore et al., 2009). The absorption of anthropogenic CO2 by the oceans increases the concentration of hydrogen ions in seawater (quanti-

Explanation:

Answer:

a) [tex]v = 2.886\,\frac{m}{s}[/tex], b) [tex]\mu_{k} = 0.014[/tex]

Explanation:

a) The final speed is determined by the Principle of Momentum Conservation:

[tex](62.2\,kg)\cdot (3.80\,\frac{m}{s} ) = (81.9\,kg)\cdot v[/tex]

[tex]v = 2.886\,\frac{m}{s}[/tex]

b) The deceleration experimented by the system person-sled is:

[tex]a = \frac{\left(0\,\frac{m}{s} \right)^{2}-\left(2.886\,\frac{m}{s} \right)^{2}}{2\cdot (30\,m)}[/tex]

[tex]a = -0.139\,\frac{m}{s^{2}}[/tex]

By using the Newton's Laws, the only force acting on the motion of the system is the friction between snow and sled. The kinetic coefficient of friction is:

[tex]-\mu_{k}\cdot m\cdot g = m\cdot a[/tex]

[tex]\mu_{k} = -\frac{a}{g}[/tex]

[tex]\mu_{k} = -\frac{\left(-0.139\,\frac{m}{s^{2}} \right)}{9.807\,\frac{m}{s^{2}} }[/tex]

[tex]\mu_{k} = 0.014[/tex]

Explanation:

Given that,

A uniform magnetic field passes through a horizontal circular wire loop at an angle 15.1∘ from the vertical, [tex]\theta=15.1^{\circ}[/tex]

The magnitude of the magnetic field B changes in time according to the equation :

[tex]B(t)=3.75+2.75 t-7.05 t^2[/tex]

Radius of the loop, r = 0.21 m

We need to find the magnitude of the induced emf in the loop when t=5.63 s. The induced emf is given by :

[tex]\epsilon=\dfrac{-d\phi}{dt}\\\\\epsilon=\dfrac{-d(BA\cos \theta)}{dt}[/tex]

B is magnetic field

A is area of cross section

[tex]\epsilon=A\dfrac{-dB}{dt}\\\\\epsilon=\pi r^2\dfrac{-d(3.75+2.75 t-7.05 t^2)}{dt}\times \cos\theta\\\\\epsilon=\pi r^2\times(2.75-14.1t)\times \cos\theta[/tex]

At t = 5.63 seconds,

[tex]\epsilon=-\pi (0.21)^2\times(2.75-14.1(5.63))\times \cos(15.1)\\\\\epsilon=10.25V[/tex]

So, the magnitude of induced emf in the loop when t=5.63 s is 10.25 V.

The EMF generated at time t = 5.63 is 10.18V.

Given a horizontal circular wire loop with a radius, r = 0.21m.

A time-dependent magnetic field B(t) = 3.75 + 2.75t -7.05t².

At an angle of θ = 15.1° to the area of the loop.

The magnetic flux passing through the loop is given by:

Ф = B(t)Acosθ

where A = πr² is the are of the loop.

Since the magnetic field is time-dependent, the magnetic flux through the loop changes with time, therefore an EMF is generated in the loop, given by:

[tex]E=-\frac{d\phi}{dt}\\\\E =-\frac{dB(t)}{dt}Acos\theta\\\\E=-\pi r^2cos\theta\frac{d}{dt}[ 3.75 + 2.75t -7.05t^2] \\\\E=\pi r^2cos\theta[14.10t-2.75]\\\\[/tex]

At time t = 5.63s

[tex]E=3.14\times(021)^2\times cos15.1\times[14.1\times5.63-2.75][/tex]

E = 10.18V

Learn more about magnetic flux:

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Answer:

Assuming the mass of the garbage truck is [tex]1.6*10^{4}[/tex] kg, the speed of the garbage truck is [tex]approx. = 27.8503\frac{m}{s}[/tex].

Explanation:

This is conservation of momentum where both objects stick at the end so they have the same final velocity so our equation is:

[tex]m_{1}v_{1initial} + m_{2}v_{2initial} = v_{both final}(m_{1} + m_{2})[/tex]

To solve for the final velocity, just divide by the sum of both masses:

[tex]\frac{ m_{1}v_{1initial} + m_{2}v_{2initial}}{(m_{1} + m_{2})} = v_{both final}[/tex]

So, plug in the known values (remember initial velocity for the trash can is 0):

[tex]\frac{ 1.6*10^{4}*28 + 86*0}{1.6+10^{4} + 86} = v_{both final}[/tex]

Answer:

If you need to measure much longer lengths - for example the length of a football pitch - then you could use a trundle wheel. You use it by pushing the wheel along the ground. It clicks every time it measures one metre.

You could employ a trundle wheel if you need to estimate considerably greater distances, like the size of a football field. By moving the wheels along the surface, you can use it. It makes a click sound each time it measures a meter.

Football, commonly known as association soccer, is a sport in which two groups of 11 players attempt to advance the ball into the goal of the other team by using any component of their bodies other than their hands and arms.

Only the goalie is allowed to handle the ball, and only within the towns near the goal that is designated as the penalty area. The team with the most goals scored wins.

According to the number of players and spectators, football is the most watched sport in the world.

The sport may be played practically everywhere, from official football fields (pitches) to gyms, streets, school playground, parks, or beaches, thanks to its basic rules and necessary equipment. The Federation Internationale Football Association is the governing organization of football.

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Answer:

d. a2 = 9a1

Explanation:

We can apply the following equation of motion to calculate the angular acceleration:

[tex]\omega^2 - \omega_0^2 = 2\alpha\theta[/tex]

Since both wheel starts from rest, their [tex]\omega_0 = 0 rad/s[/tex]

[tex]\omega^2 = 2\alpha\theta[/tex]

We can take the equation for the 1st wheel, divided by the equation by the 2nd wheel:

[tex]\frac{\omega_1^2}{\omega_2^2} = \frac{2\alpha_1\theta_1}{2\alpha_2\theta_2}[/tex]

As they were rotating through the same angular displacement [tex]\theta_1 = \theta_2[/tex], these 2 cancel out

[tex]\left(\frac{\omega_1}{\omega_2}\right)^2 = \frac{\alpha_1}{\alpha_2}[/tex]

[tex]\left(\frac{1}{3}\right)^2 = \frac{\alpha_1}{\alpha_2}[/tex]

[tex]\frac{1}{9} = \frac{\alpha_1}{\alpha_2}[/tex]

[tex]\alpha_2 = 9\alpha_1[/tex]

So d is the correct answer

The angular acceleration of the first wheel is four times higher than that of the second. Option D is correct.

It can be defined as the rate of change in the angular velocity of an object or body. It can be calculated by the equation of motion:

[tex]\omega ^2 - \omega _0^2 = 2\alpha \theta[/tex]

Since initial angular rotation is zero for both the wheels,

[tex]\omega ^2 = 2\alpha \theta[/tex]

Compare the angular acceleration of both wheels,

[tex]\dfrac {\omega_1^2}{\omega_2^2} = \dfrac {2\alpha_1 \theta}{2\alpha_2\theta }[/tex]

Put the values,

[tex]\begin{aligned} (\dfrac 13)^2&= \dfrac {\alpha_1 }{\alpha_2 }\\\\ \dfrac 19 &= \dfrac {\alpha_1 }{\alpha_2 }\\\\\alpha_2 &= 9\alpha_1 \end {aligned}[/tex]

Therefore, The angular acceleration of the first wheel is four times higher than that of the second. Option D is correct.

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Answer:   In order to walk on a floor (or any other surface), your foot must push backward on the floor (action force), so that the floor pushes you forward (reaction force).

Explanation:

Answer:

21.5°

Explanation:

Given,

Refractive index of water, n₁ = 1.33

Refractive index of polystyrene, n₂ = 1.49

Angle of reflection = ?

Angle of refraction = 19.1°

Using Snell's law

n₁ sin θ₁ = n₂ sin θ₂

1.33 x sin θ₁ = 1.49 x sin 19.1°

sin θ₁ = 0.366

θ₁  = 21.5°

According to law of reflection angle of incidence is equal to angle of reflection.

Angle of reflection =  21.5°

Final answer:

The angle of reflection of the reflected ray in polystyrene that is submerged in water will be equal to the angle of refraction, so it is 19.1°.

Explanation:

The question is about the reflection and refraction of light as it strikes a flat surface of a polystyrene block that is in water. According to the Law of Reflection, the angle of reflection is equal to the angle of incidence.

Since we know the angle of refraction is 19.1°, and by Snell's Law (n₁ × sin(i) = n₂ × sin(r)), we can infer that the angle of incidence is also 19.1° because the boundary is between two different mediums (water and polystyrene), resulting in the light being split into reflected and refracted rays.

Therefore, the angle of reflection of the reflected ray is also 19.1°, identical to the angle of refraction given in the question, because the angle of incidence equals the angle of reflection.

Answer:

189.47nm

Explanation:

We can solve this problem by taking into account the time that light takes in crossing the distance between the laser and the photocell, and the time in crossing the slab.

By using the values of c and 16.5ns we can calculate the value of d

[tex]d=(3*10^{8}m/s)(16.5*10^{-9}s)=4.95m[/tex]

We have to compute the time that light takes in crossing d-0.820m:

[tex]4.95m-0.820m=4.13m\\\\t=\frac{4.13m}{3*10^8m/s}=6.22*10^{-8}s=13.7ns[/tex]

Now, we can calculate the speed of the light in the slab by using the time difference between 21.5 ns and 13.7ns:

[tex]\Delta t=21.5ns-13.7ns=7.8ns\\\\v_l=\frac{0.82m}{7.8ns}=1.05*10^8m/s[/tex]

Then, the index of refraction will be:

[tex]n=\frac{c}{v_l}=2.85[/tex]

Finally, we have that:

[tex]\frac{\lambda_2}{\lambda_1}=\frac{n_1}{n_2}\\\\\lambda_2=\lambda_1 \frac{n_1}{n_2}=(540nm)\frac{1.00}{2.85}=189.47nm[/tex]

hope this helps!!

Answer:

ΔE =  20 eV

Explanation:

In a Helium atom we have two electrons in the s layer, so they can accommodate one with the spin up and the other with the spin down, give us a total spin of zero (S = 0) this state is singlet, in general this very stable states,

  When you transition to the 1s state to complete the two electrons allowed by layers

    ΔE = -5 - (-25) = 20 eV

this is the energy of the transition,

It should be mentioned that there can also be transitions with the two spins of the same orientation, but in this case the energy is a little different due to the electron-electron repulsion, this state is called ortho helium S = 1

Answer:

5.9*10^{-4}m

Explanation:

to find the uncertainty of the displacement it is necessary to compute the uncertainty for the angular frequency:

[tex]\omega=\frac{2\pi}{T}=\frac{2\pi}{0.40s}=15.707rad/s\\\\\frac{d \omega}{\omega}=\frac{dT}{T}\\\\d\omega=\omega \frac{dT}{T}=(15.707rad/s)\frac{0.020s}{0.40s}=0.785rad/s[/tex]

then, you can calculate the uncertainty in angular displacement:

[tex]\theta=\omega t\\\\\frac{d\theta}{\theta}=\sqrt{(\frac{d\omega}{\omega})^2+(\frac{dt}{t})^2}\\\\d\theta=\theta\sqrt{(\frac{d\omega}{\omega})^2+(\frac{dt}{t})^2}=0.0422[/tex]

finally, by using:

[tex]y=Acos(\omega t)\\\\dy=dAcos(\omega t)d(\omega t)=(dA)cos(\theta)d\theta=(0.002m)cos(0.785)(0.0422)\\\\dy=5.9*10^{-4}m[/tex]

The uncertainty of her displacement in mm is  : 0.59 mm

First step : determine the uncertaintiy of the angular frequency

w = [tex]\frac{2\pi }{T}[/tex]  = [tex]\frac{2\pi }{0.40} = 15.707 rad/s[/tex]

[tex]\frac{dw}{w} = \frac{dT}{T}[/tex]

therefore :

dw = 0.785 rad/s

Next step : determine the uncertainty of the angular displacement

θ = wt

dθ / θ = [tex]\sqrt{(\frac{dw}{w} )^2 + (\frac{dt}{t} )^2}[/tex]

therefore :

dθ = 0.0422

Final step : determine the uncertainty of displacement

y = Acos(wt)

dy = dAcos(wt)d(t) = (dA)cosθdθ

                              = ( 0.002m )cos (0.785)(0.0422)

                              = 5.9 * 10⁻⁴ m = 0.59 mm

Hence we can conclude that the  uncertainty of her displacement in mm is  : 0.59 mm

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Answer:

A) The thickness of the glass is 868 nm

B) The wavelength is 3472 nm

Explanation:

We are given;

Refractive index = 1.30

Wavelength = 496 nm

Next wavelength = 386 nm

Now, we need to calculate the thickness of the glass

Using formula for constructive interference which is given as;

2nt = (m + ½)λ

Where;

m is the order of the interference

λ is wavelength

t is thickness

Now, for the first wavelength, we have;

2nt = (m + ½)496 - - - - eq1

for the second wavelength, we have;

2nt = (m + 1 + ½)386

2nt = (m + 3/2)386 - - - - eq2

Thus, combining eq1 and eq2, we have;

(m + ½)496 = (m + 3/2)386

496m + 248 = 386m + 579

496m - 386m = 579 - 248

110m = 331

m = 331/110

m = 3

Put 3 for m in eq 1;

2nt = (3 + ½)496

2nt = 1736

t = 1736/(2 x 1)

t = 868 nm

B) now we need to calculate the longest wavelength.

From earlier, we saw that ;

2nt = (m + ½)λ

Making wavelength λ the subject, we have;

λ = 2nt/(m + ½)

The longest wavelength will be at m = 0

Thus,

λ = (2 x 1 x 868)/(0 + ½)

λ = 3472 nm

Amplitude is a measurement of the magnitude of displacement (or maximum disturbance) of a medium from its resting state, as diagramed in the peak deviation example below (it can also be a measurement of an electrical signal's increased or decreased strength above or below a nominal state).

Answer:

a) [tex]\mu_{k} = 0.704[/tex], b) [tex]R = 0.312\,m[/tex]

Explanation:

a) The minimum coeffcient of friction is computed by the following expression derived from the Principle of Energy Conservation:

[tex]\frac{1}{2}\cdot k \cdot x^{2} = \mu_{k}\cdot m\cdot g \cdot \Delta s[/tex]

[tex]\mu_{k} = \frac{k\cdot x^{2}}{2\cdot m\cdot g \cdot \Delta s}[/tex]

[tex]\mu_{k} = \frac{\left(8050\,\frac{N}{m} \right)\cdot (0.06\,m)^{2}}{2\cdot (0.3\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot (7\,m)}[/tex]

[tex]\mu_{k} = 0.704[/tex]

b) The speed of the block is determined by using the Principle of Energy Conservation:

[tex]\frac{1}{2}\cdot k \cdot x^{2} = \frac{1}{2}\cdot m \cdot v^{2}[/tex]

[tex]v = x\cdot \sqrt{\frac{k}{m} }[/tex]

[tex]v = (0.06\,m)\cdot \sqrt{\frac{8050\,\frac{N}{m} }{0.3\,kg} }[/tex]

[tex]v \approx 9.829\,\frac{m}{s}[/tex]

The radius of the circular loop is:

[tex]\Sigma F_{r} = -90\,N -(0.3\,kg)\cdot (9.807\,\frac{m}{s^{2}} ) = -(0.3\,kg)\cdot \frac{v^{2}}{R}[/tex]

[tex]\frac{\left(9.829\,\frac{m}{s}\right)^{2}}{R} = 309.807\,\frac{m}{s^{2}}[/tex]

[tex]R = 0.312\,m[/tex]

Answer:

f =3.4*10^3 hz

Since f is in the range of (20 Hz - 20,000 Hz] which is the range of audible frequencies, the frequency is audible.

Explanation:

The relation that describes the pressure amplitude for a sound wave is  

P_MAX = B*k*A                                              (1)  

Where the bulk modulus of the air is B = 1.30 x 10^5 Pa and the displacement  amplitude of the waves produced by the machine is 1.00 μmμm.  

Using (1) we can calculate k then we can use k to determine the wavelength  A of the wave, and remember that λ = 2π/k.  

So, substitute into (1) with 10 Pa for P_max, (1.30 x 10^5 Pa) for B and  

1 x 10^-6 m for A  

10 Pa = (1.30 x 10^5 Pa) x k x (1 x 10^-6 m)  

k = 62.5 m^-1

We can use the following relation to calculate the wavelength  

λ = 2π/k.  

λ = 0.100 m

Finally, the relation between the wavelength and the frequency of a sound  

wave is given by the following equation  

f = v/ λ

 =344/0.100 m

f =3.4*10^3 hz

Since f is in the range of (20 Hz - 20,000 Hz] which is the range of audible frequencies, the frequency is audible.

Answer:

a) [tex]F = 660.576\,N[/tex], b) [tex]a_{c} = 6.953\,\frac{m}{s^{2}}[/tex], c) [tex]v \approx 7255.423\,\frac{m}{s}[/tex], [tex]\omega = 9.583\times 10^{-4}\,\frac{rad}{s}[/tex], d) [tex]T \approx 1.821\,h[/tex]

Explanation:

a) The gravitational force exerted by the Earth on the satellite is:

[tex]F = G\cdot \frac{m\cdot M}{r^{2}}[/tex]

[tex]F = \left(6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} \right)\cdot \frac{(95\,kg)\cdot (5.972\times 10^{24}\,kg)}{(7.571\times 10^{6}\,m)^{2}}[/tex]

[tex]F = 660.576\,N[/tex]

b) The centripetal acceleration of the satellite is:

[tex]a_{c} = \frac{660.576\,N}{95\,kg}[/tex]

[tex]a_{c} = 6.953\,\frac{m}{s^{2}}[/tex]

c) The speed of the satellite is:

[tex]v = \sqrt{a_{c}\cdot R}[/tex]

[tex]v = \sqrt{\left(6.953\,\frac{m}{s^{2}} \right)\cdot (7.571\times 10^{6}\,m)}[/tex]

[tex]v \approx 7255.423\,\frac{m}{s}[/tex]

Likewise, the angular speed is:

[tex]\omega = \frac{7255.423\,\frac{m}{s} }{7.571\times 10^{6}\,m}[/tex]

[tex]\omega = 9.583\times 10^{-4}\,\frac{rad}{s}[/tex]

d) The period of the satellite's rotation around the Earth is:

[tex]T = \frac{2\pi}{\left(9.583\times 10^{-4}\,\frac{rad}{s} \right)} \cdot \left(\frac{1\,hour}{3600\,s} \right)[/tex]

[tex]T \approx 1.821\,h[/tex]

Answer:

The final angular velocity is [tex]w_f = \frac{MR^2}{MR^2+ mr^2} w[/tex]

Explanation:

From the question we are told that

     The mass of the first disk is  m

      The radius of the first  disk is  r

      The mass of  second disk is  M

      The radius of second disk is R

       The speed of rotation is w

       The moment of inertia of second disk is  [tex]I = \frac{1}{2} MR^2[/tex]

Since the first disk is at rest initially

        The initial angular momentum would be due to the second disk  and this is mathematically represented as

       [tex]L_i = Iw = \frac{1}{2} MR^2 w[/tex]

Now when the first disk is then dropped the angular momentum of the whole system now becomes

       [tex]L_f = (I_1 + I_2 ) w_f= ( \frac{1}{2} MR^2 + \frac{1}{2} m^2 r^2) w_f[/tex]

This above is because the formula for moment of inertia is the same for every disk

       According to the law  conservation of  angular momentum

                [tex]L_f = L_i[/tex]

    [tex]( \frac{1}{2} MR^2 + \frac{1}{2} m^2 r^2) w_f = \frac{1}{2} MR^2 w[/tex]

=>              [tex]w_f = \frac{\frac{1}{2} MR^2 w }{( \frac{1}{2} MR^2 + \frac{1}{2} m^2 r^2)}[/tex]

                  [tex]w_f = \frac{MR^2}{MR^2+ mr^2} w[/tex]

Answer: 4 kg

Explanation:

Given

Mass of the first shell, m1 = 1 kg

Diameter of the first shell, d1 = 2 m

Radius of the first shell, r1 = 1 m

Diameter of the second shell, d2 = 1 m

Radius of the second shell, r2 = 1/2 m

The moment of inertia of a spherical shell is given by the relation

I = mr²

This means that if two sphere's have the same moment of ineria:

I1 would be equal to I2. And thus

m1.r1² = m2.r2²

If we solve for the second mass m2

m2 = m1.r1²/r2²

m2 = m1 (r1 / r2)² and we substitute the values

m2 = 1 * (1 / 0.5)²

m2 = 2²

m2 = 4 kg

The needed mass of the second shell for their shells to have the same moment of inertia is 4 kg