A 0.750kg block is attached to a spring with spring constant 13.5N/m . While the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of 32.0cm/s .

What are :

A) The amplitude of the subsequent oscillations?
Express your answer with the appropriate units.

B) The block's speed at the point where x= 0.750 A?
Express your answer with the appropriate units.

The speed of the wave in the string is 83.4 m/s

Explanation:

For a standing wave in a string, the speed of the wave is given by the equation:

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

where

L is the length of the string

T is the tension in the string

m is the mass of the string

In this problem, we have:

L = 0.72 m

m = 4.2 g = 0.0042 kg

T = 84.1 N

Solving the equation, we find the speed of the wave:

[tex]v=\frac{1}{2(0.72)}\sqrt{\frac{84.1}{0.0042/0.72}}=83.4 m/s[/tex]

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Answer:

Explanation:

For rate of flow of liquid in a pipe , the formula is

Q = π p r⁴ / 8 η l

where Q is volume of liquid flowing per unit time , p is pressure drop across pipe , r is radius of pipe , η is coefficient of viscosity and l is length of pipe

If η and l are constant for two pipes having equal rate of flow as in the given case

p₁ r₁⁴ = p₂ r₂⁴

p₁ = 1.789 p₂ ( given )

r₁ = D/2

1.789 p₂ x (D/2)⁴ = p₂ x r₂⁴

r₂ = [tex]\frac{D}{2}\times\sqrt[4 ]{1.789}[/tex]

2 x r₂ = D  x 1.1565

D₂ =  D  x 1.1565

To determine the diameter of the second pipe, apply Poiseuille's law to relate the known pressure drop ratio to the radii of the pipes, and then solve for the radius, r², of the second pipe.

The subject of this question pertains to the Physics of fluid mechanics, specifically dealing with the laminar flow of a fluid through pipes of different diameters and the associated pressure drops.

According to Poiseuille's law, which describes laminar flow in pipes, the pressure drop (ΔP) across a length of pipe is directly proportional to the length (L) and the viscosity (η) of the fluid, and inversely proportional to the fourth power of the radius (r) of the pipe.

Given that the pressure drop for the first pipe with diameter D is 1.789 times greater than the second pipe, we can establish the following relationship using Poiseuille's equation ΔP ∝ Lη/r^4 (ignoring constants for convenience since lengths are equal, and assuming the same fluid viscosity):

(ΔP1/ΔP2) = (r2⁴ / r1⁴)

1.789 = (r2⁴ / (D/2)⁴)

Therefore, solving for r2 (radius of the second pipe), we get:

r²= 0.5 * D * (1.789)^(1/4)

Let d2 be the diameter of the second pipe, then d2 = 2 * r2:

d2 = D * (1.789)^(1/4)

By calculating the fourth root of 1.789 and multiplying it by the diameter D of the first pipe, the diameter of the second pipe can be determined.

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Answer:

[tex]I=336.6kgm/s[/tex]

Explanation:

The equation for the linear impulse is as follows:

[tex]I=F\Delta t[/tex]

where [tex]I[/tex] is impulse, [tex]F[/tex] is the force, and [tex]\Delta t[/tex] is the change in time.

The force, according to Newton's second law:

[tex]F=ma[/tex]

and since [tex]a=\frac{v_{f}-v_{i}}{\Delta t}[/tex]

the force will be:

[tex]F=m(\frac{v_{f}-v_{i}}{\Delta t})[/tex]

replacing in the equation for impulse:

[tex]I=m(\frac{v_{f}-v_{i}}{\Delta t})(\Delta t)[/tex]

we see that [tex]\Delta t[/tex] is canceled, so

[tex]I=m(v_{f}-v_{i})[/tex]

And according to the problem [tex]v_{i}=0m/s[/tex], [tex]v_{f}=5.10m/s[/tex] and the mass of the passenger is [tex]m=66kg[/tex]. Thus:

[tex]I=(66kg)(5.10m/s-0m/s)[/tex]

[tex]I=(66kg)(5.10m/s)[/tex]

[tex]I=336.6kgm/s[/tex]

the magnitude of the linear impulse experienced the passenger is [tex]336.6kgm/s[/tex]

Answer:

31.42383 m/s

Explanation:

g = Acceleration due to gravity = 9.81 m/s²

[tex]\mu[/tex] = Coefficient of kinetic friction = 0.48

s = Displacement = 0.935 m

[tex]m_1[/tex] = Mass of bean bag = 0.354 kg

[tex]m_2[/tex] = Mass of empty crate = 3.77 kg

[tex]v_1[/tex] = Speed of the bean bag

[tex]v_2[/tex] = Speed of the crate

Acceleration

[tex]a=-\frac{f}{m}\\\Rightarrow a=-\frac{\mu mg}{m}\\\Rightarrow a=-\mu g[/tex]

[tex]a=--9.81\times 0.48=4.7088\ m/s^2[/tex]

From equation of motion

[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 4.7088\times 0.935+0^2}\\\Rightarrow v=2.96739\ m/s[/tex]

In this system the momentum is conserved

[tex]m_1v_1=(m_1+m_2)v_2\\\Rightarrow v_1=\frac{(m_1+m_2)v_2}{m_1}\\\Rightarrow u=\frac{(0.354+3.77)\times 2.69739}{0.354}\\\Rightarrow u=31.42383\ m/s[/tex]

The speed of the bean bag is 31.42383 m/s

The speed of the beanbag when it strikes the crate is 34.46 m/s.

In an inelastic collision, two bodies collide, stick together and move with the same velocity. The momentum is conserved while the kinetic energy is not.

As the beanbag sticks to the crate they move together with the same velocity until they stop after 0.935 m due to frictional force.

Let M be the combined mass of the crate and beanbag. the frictional force acting on the system is:

[tex]F=-\mu_k Mg\\\\Ma=-\mu_k Mg\\\\a=-\mu g\\\\a=-0.480\times9.8\;m/s^2\\\\a=4.7\;m/s^2[/tex]will be the acceleration of the system.

Let the initial velocity of the system be u and the final velocity is 0, then from the third equation of motion:

[tex]0=u^2-2as\\\\u=\sqrt[]{2as}\\\\u=\sqrt{2\times4.7\times0.935}\;m/s\\\\u=2.96\;m/s[/tex]

So the momentum after collision is

Mu = (0.354 + 3.77)×2.96 = 12.2 kgm/s

Which must be equal to the initial momentum that is contributed only by the beanbag, since the crate is on rest. Let the speed of the bean bag be v then initial momentum should be:

12.2 = 0.354 × v

v = 34.46 m/s

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To solve this problem it is necessary to apply the concept related to density and its definition with respect to mass and volume.

Density can be expressed as

[tex]\rho = \frac{m}{V}[/tex]

Where,

m = mass

V = Volume

The average density of fish air system is equal to the neutral bouyancy when the water displaced by the fish air system can be expressed as

[tex]\rho_{b} = \frac{m_a+m_f}{V_a+V_f}[/tex]

Where,

[tex]m_a[/tex]= Mass of air

[tex]m_f =[/tex] Mass of fish

[tex]V_a[/tex]= Volume of air

[tex]V_f[/tex]= Volume of fish

As we know that m = \rho V  we can replacing the equivalent value for the mass fro fish and air, then

[tex]\rho_{b} = \frac{m_a+m_f}{V_a+V_f}[/tex]

[tex]\rho_{b} = \frac{\rho_a V_a+\rho_f V_f}{V_a+V_f}[/tex]

[tex]\rho_{b}(V_a+V_f)=\rho_a V_a+\rho_f V_f[/tex]

[tex]\rho_{b} V_a+\rho_{b}V_f = \rho_a V_a+\rho_f V_f[/tex]

[tex]\rho_{b} V_a- \rho_a V_a=\rho_f V_f-\rho_{b}V_f[/tex]

[tex]V_a(\rho_{b}-\rho_a)=V_f(\rho_f-\rho_{b})[/tex]

[tex]\frac{V_a}{\rho_f}=\frac{(\rho_f-\rho_{b})}{(\rho_{b}-\rho_a)}[/tex]

Replacing with our values we have that

[tex]\frac{V_a}{\rho_f}=\frac{(1000-1080)}{(1.2-1000)}[/tex]

[tex]\frac{V_a}{\rho_f}=0.08[/tex]

[tex]\frac{V_a}{\rho_f}=8\%[/tex]

Therefore the percentage of Volume is increased by 8%

To achieve neutral buoyancy in fresh water, a fish with an initial density of 1080 kg/m^3 needs to increase its volume by 8%, utilizing its swim bladder.

The solution to this problem involves understanding Archimedes' Principle and the concept of density. In order to achieve neutral buoyancy, the fish needs to match its density to the density of freshwater, which is approximately 1000 kg/m³. It can do this by increasing its volume through inflating its swim bladder, thus reducing its overall density.

Since the average density of an object is mass divided by its volume, we can set up an equation involving the initial and final densities of the fish, with the initial density being 1080 kg/m³ and the final density being 1000 kg/m³. Let the initial volume of the fish be V1, the final volume be V2, and the mass of the fish (which doesn't change) be m. Therefore, m/V1 = 1080 and m/V2 = 1000. We can then solve these equations to find that V2 = 1.08V1.

In terms of percent increase, (V2 - V1) / V1 = 0.08 or 8%. Hence, the fish needs to increase its volume by 8% to achieve neutral buoyancy in freshwater.

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Answer:

Dare devil can cross 7 buses.

Explanation:

given,                                        

angle of inclination of ramp = 32°

speed of the motorcycle = 40 m/s

length of bus = 20 m                                    

how many buses daredevil can clear =?                                

to solve this we need to calculate the range of the daredevil

considering it as projectile

the range of motorcyclist

     [tex]R = \dfrac{V^2 sin (2\theta)}{g}[/tex]

     [tex]R = \dfrac{40^2 sin (2\times 32^0)}{9.8}[/tex]

     [tex]R =163.26 \times sin(2\times 32^0)[/tex]

     [tex]R =146.74\ m[/tex]                    

length of bus is given as 20 m            

Number of bus daredevil can cross            

     [tex]N = \dfrac{147.74}{20}[/tex]                

     [tex]N =7.34[/tex]                                

Dare devil can cross 7 buses.

Answer:

The number of buses are 7.

Explanation:

Given that,

Angle =32°

Speed = 40.0 m/s

Length of bus = 20.0

We need to calculate the range of bus

Using formula of range

[tex]R=\dfrac{v^2\sin2\theta}{g}[/tex]

Where, g = acceleration due to gravity

v = initial velocity

Put the value into the formula

[tex]R=\dfrac{(40.0)^2\times\sin(2\times32)}{9.8}[/tex]

[tex]R=150.20\ m[/tex]

We need to calculate the number of buses

Using formula of number of buses

[tex]N=\dfrac{R}{L}[/tex]

Where, R = range

L = length of bus

[tex]N=\dfrac{150.20}{20.0}[/tex]

[tex]N=7[/tex]

Hence, The number of buses are 7.

Answer:

6λ_film

Explanation:

Light of wavelength λ 600 nm

Assuming length of thin film of water to be 1200 nm

Total distance traveled = 2×1200 ×10^{-9} = [tex]\frac{2.4\times10^-6}{4\times10^{-7} }[/tex] m

Now, Wavelength of light in film = λ/n

n= refractive index

= [tex]\frac{600\times10^{-9}}{1.5}[/tex] = 4×10^{-7}

No. of wavelengths = distance traveled/Wavelength of light in film

= [tex]\frac{2.4\times10^-6}{4\times10^{-7} }[/tex] = 6

therefore,  the total (down-and-back) distance traveled by the light inside the film in terms of wavelength λfilm = 6λfilm

Final answer:

The total distance that the light travels inside the thin film is equal to the wavelength of light within the film (λfilm), as the light travels down and back up, covering a distance of 2t, which is equivalent to λfilm.

Explanation:

When light of wavelength 600 nm in vacuum (represented by λ) is incident on a thin film with an index of refraction of 1.5, we need to calculate the total distance traveled by the light within the film. The wavelength of light within the film (λfilm) is given by the formula λfilm = λ / n, where n is the index of refraction of the film. Given that light hits the film perpendicularly, it travels down to the bottom surface and back up, covering a distance of 2t, where t is the thickness of the film. To express the total distance in terms of λfilm, we use the relationship 2t = 2(λfilm / (2n)) = λfilm since n is already considered in the λfilm. Thus, the total distance traveled by the light inside the film is λfilm.

Answer:

f = 8 %

Explanation:

given,

density of body of fish = 1080 kg/m³

density of air = 1.2 Kg/m³

density of water = 1000 kg/m²

to protect the fish from sinking volume should increased by the factor f

density of fish + density of water x increase factor = volume changes in water                                                    

1080 +f x 1.2 =(1 + f ) x 1000                

1080 + f x 1.2 = 1000 + 1000 f      

998.8 f = 80                                  

f = 0.0800                            

f = 8 %                                        

the volume increase factor of fish will be equal to f = 8 %

Answer:

F₂ = 0 N

x = 0.5*d = 0.5 m

Explanation:

Given

m₁ = m₂ = m₃ = m = 1 Kg

d = distance between m₁ and m₃ = d₁₃ = 1 m

d₁₂ = 0.5*d = 0.5*(1 m) = 0.5 m = d₃₂

G = 6.673*10⁻¹¹ N*m²/ Kg²

a) Find the magnitude of the net gravitational force exerted by these objects on a kg object (F₃) placed midway between them.

we can apply

F₂ = F₁₂ + F₃₂

then we use the formula

F₁₂ = G*m₁*m₂ / d₁₂² = G*m² / (0.5*d)² = 4*G*m² / d²

⇒ F₁₂ = 4*G*m² / d² = 4*(6.673*10⁻¹¹ N*m²/ Kg²)*(1 Kg)² / (1 m)²

⇒ F₁₂ = 2.6692*10⁻¹⁰ N (←)

and

F₃₂ = G*m₃*m₂ / d₃₂² = G*m² / (0.5*d)² = 4*G*m² / d²

⇒ F₃₂ = 4*G*m² / d² = 4*(6.673*10⁻¹¹ N*m²/ Kg²)*(1 Kg)² / (1 m)²

⇒ F₃₂ = 2.6692*10⁻¹⁰ N (→)

we get

F₂ = F₁₂ + F₃₂ = (- 2.6692*10⁻¹⁰ N) + (2.6692*10⁻¹⁰ N) = 0 N

b) At what position other than an infinitely remote one can the kg object be placed so as to experience a net force of zero from the other two objects?

From a) It is known that x = 0.5*d = 0.5*(1 m) = 0.5 m

Nevertheless, we can find the distance as follows

If

F₂ = 0   ⇒   F₁₂ + F₃₂ = 0   ⇒   F₁₂ = - F₃₂

If

x = distance between m₁ and m₂

d - x = distance between m₂ and m₃

we get

F₁₂ = G*m² / x²

F₃₂ = G*m² / (d - x)²

If   F₁₂ = - F₃₂

⇒     G*m² / x² = - G*m² / (d - x)²

⇒     1 / x² = - 1 / (d - x)²  

⇒    (d - x)² = - x²

⇒  d²- 2*d*x + x² = - x²

⇒  2*x² - 2*d*x + d² = 0

⇒  2*x² - 2*1*x + 1² = 0

⇒  2*x² - 2*x + 1 = 0

Solving the equation we obtain

x = 0.5 m = 0.5*d

To solve this problem it is necessary to apply the concepts related to the Period of a body and the relationship between angular velocity and linear velocity.

The angular velocity as a function of the period is described as

[tex]\omega = \frac{2\pi}{T}[/tex]

Where,

[tex]\omega =[/tex]Angular velocity

T = Period

At the same time the relationship between Angular velocity and linear velocity is described by the equation.

[tex]v = \omega r[/tex]

Where,

r = Radius

Our values are given as,

[tex]T = 24 hours[/tex]

[tex]T = 24hours (\frac{3600s}{1 hour})[/tex]

[tex]T = 86400s[/tex]

We also know that the radius of the earth (r) is approximately

[tex]6.38*10^6m[/tex]

Usando la ecuación de la velocidad angular entonces tenemos que

[tex]\omega = \frac{2\pi}{T}[/tex]

[tex]\omega = \frac{2\pi}{86400}[/tex]

[tex]\omega = 7.272*10^{-5}rad/s[/tex]

Then the linear velocity would be,

[tex]v = \omega *r[/tex]

x[tex]v = \omega *r[/tex]

[tex]v= 463.96m/s[/tex]

The speed would Earth's inhabitants who live at the equator go flying off Earth's surface is  463.96

The measured length when the temperature rises to 29.4°C is 17.712 m.

To answer this question, we need to consider the thermal expansion of the steel tape and the aluminum support column. The coefficient of linear expansion for the steel tape is not given, so we will assume it to be the same as iron, which is approximately 12x10-6/°C.

We can use the formula:

ΔL = αL₀ΔT

where ΔL is the change in length, α is the coefficient of linear expansion, L₀ is the initial length, and ΔT is the change in temperature.

Given that the initial length (L₀) is 17.700 m and the initial temperature is 21.2°C, we can calculate the change in length (ΔL) when the temperature rises to 29.4°C.

ΔL = (12x10-6)(17.700)(29.4 - 21.2) = 0.01226352 m

Therefore, the measured length when the temperature rises to 29.4°C is 17.700 + 0.01226352 = 17.712 m (rounded to three decimal places).

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The change in length of the aluminum column when the temperature rises can be found using the equation ∆L = αL0∆T, where ∆L is the change in length, α is the coefficient of linear expansion, L0 is the initial length, and ∆T is the change in temperature. Plugging in the values provided, we find the change in length is 3.25848 x 10-4 m, rounded to three decimal places.

When a material undergoes thermal expansion, its length increases as the temperature rises. To calculate the change in length, we can use the equation ∆L = αL0∆T, where ∆L is the change in length, α is the coefficient of linear expansion, L0 is the initial length, and ∆T is the change in temperature. In this case, we need to find the change in length of the aluminum column when the temperature rises from 21.2°C to 29.4°C.

First, we need to determine the coefficient of linear expansion for aluminum. Using the information provided, we can find that the coefficient of linear expansion for aluminum is approximately 22·10-6 °C-1.

Next, we can plug in the values into the equation. Given L0 = 17.700 m, ∆T = (29.4 - 21.2)°C = 8.2°C, and α = 22·10-6 °C-1, we can calculate ∆L as follows:

∆L = (22·10-6 °C-1)(17.700 m)(8.2°C) = 3.25848 x 10-4 m

Rounded to three decimal places, the measured length of the aluminum support column when the temperature rises to 29.4°C is 0.000 m.

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To carry out this exercise, it is necessary to use the equations made to Centripetal Force and Gravitational Energy Conservation.

By definition we know that the Centripetal Force is estimated as

[tex]F_c = M\omega^2R[/tex]

Where,

M = mass

[tex]\omega =[/tex] Angular velocity

R = Radius

From the 'linear' point of view the centripetal force can also be defined as

[tex]F_c = \frac{GM^2}{R^2}[/tex]

PART A ) Equating both equations we have,

[tex]\frac{M}{\omega^2R}=\frac{GM^2}{R^2}[/tex]

Re-arrange to find \omega

[tex]\omega = \sqrt{\frac{Gm}{r^3}}[/tex]

Replacing with our values

[tex]\omega = \sqrt{\frac{(6.67*10^{-11})(4.6*10^{30})}{(1.9*10^{11})^3}}[/tex]

[tex]\omega = 2.115*!0^{-7}rad/s[/tex]

Therefore the angular speed is [tex]\omega = 2.115*!0^{-7}rad/s[/tex]

PART B) For energy conservation we have to

[tex]KE_{min} = PE_{cm}[/tex]

Where,

[tex]KE_{min} =[/tex] Minimus Kinetic Energy

[tex]PE_{cm} =[/tex] Gravitational potential energy at the center of mass

Then,

[tex]\frac{1}{2} mv^2_{min} = \frac{2GMm}{R}[/tex]

Re-arrange to find v,

[tex]v_min = \sqrt{\frac{4GM}{R}}[/tex]

[tex]v_min = \sqrt{\frac{4(6.67*10^{-11})(2.2*10^{30})}{(1.9*10^{11})}}[/tex]

[tex]v_min = 5.55*10^4m/s[/tex]

Therefore the minimum speed must it have at the center of mass if it is to escape to "infinity" from the two-star system is [tex]v_min = 5.55*10^4m/s[/tex]

Answer:

A) True B) false C) True D) False E) False

Explanation:

A) We know that the potential energy is transformed into kinetic energy since the ball is released.

[tex]EP1=m*g*h = 1 * 9.81 * 6 = 58.86J\\[/tex]

This energy will be the same at the moment that the ball is about to hit the floor. Therefore there was a transformation from potential to kinetic energy

EK1 = EP1

[tex]EK=\frac{1}{2}*m*v^{2}  \\v1=\sqrt{(58.86*2)/1} \\v1=10,84 m/s[/tex]

With the 2kg ball happens the same

[tex]EP2=m*g*h2=2*9.81*3=58.86J\\[/tex]

Velocity at the moment when is about to hit the floor (due to the energy transform) will be:

[tex]EK2=\frac{1}{2}*m*v^{2}  \\v1=\sqrt{(58.86*2)/2} \\v2=7.67 m/s[/tex]

A) Since both balls have the same potential energy, they will have the same kinetic energy

B) The velocities are differents as they were calculated before.

C) Therefore the 1 kg ball moves faster as it reaches the ground.

D) The kinetic energy is the same for both balls as it was calculated before.

E) The 2 kg ball will reach the ground first, since it is closer to the ground

With the velocities calculated in the previous steps we have now:

[tex]1 kg ball\\v=v0+g*t\\t1=10.85/9.81 = 1.1s\\\\2 kg ball\\t2=7.672/9.81 = 0.782s[/tex]

Answer:

28.3 ft/s

Explanation:

We are given that

Initial velocity of a car=[tex]u=0[/tex]

Acceleration of the car=[tex]a=3.22-0.004v^2[/tex]

We have to find the velocity [tex]v_B[/tex] at the bottom of the grade.

Distance covered by car=600 ft

We know that

Acceleration=a=[tex]\frac{dv}{dt}=\frac{dv}{ds}\times \frac{ds}{dt}=v\frac{dv}{ds}[/tex]

[tex]v=\frac{ds}{dt}[/tex]

[tex]ds=\frac{vdv}{a}[/tex]

Taking integration on both side and taking limit of s from 0 to 600 and limit of v from 0 to [tex]v_B[/tex]

[tex]\int_{0}^{600}ds=\int_{0}^{v_B}\frac{vdv}{3.22-0.004v^2}[/tex]

[tex][S]^{600}_{0}=-\frac{1}{0.008}[ln(3.22-0.004v^2)]^{v_B}_{0}[/tex]

Using substitution method  and [tex]\int f(x)dx=F(b)-F(a)[/tex]

[tex]600-0=-\frac{1}{0.008}[ln(3.22-0.004v^2_B)-ln(3.22)][/tex]

[tex]600=-\frac{1}{0.008}(ln(\frac{3.22-0.04v_B}{3.22}))[/tex]

[tex]ln(m)-ln(n)=ln(\frac{m}{n})[/tex]

[tex]600\times 0.008=-ln(\frac{3.22-0.04v_B}{3.22})[/tex]

[tex]-4.8=ln(\frac{3.22-0.04v_B}{3.22})[/tex]

[tex]\frac{3.22-0.04v_B}{3.22}=e^{-4.8}[/tex]

[tex]ln x=y\implies x=e^y[/tex]

[tex]3.22-0.004v^2_B=3.22e^{-4.8}[/tex]

[tex]3.22-0.004v^2_B=0.0265[/tex]

[tex]3.22-0.0265=0.004v^2_B[/tex]

[tex]0.004v^2_B=3.22-0.0265=3.1935[/tex]

[tex]v_B=\sqrt{\frac{3.1935}{0.004}}=28.3 ft/s[/tex]

Hence, the velocity [tex]v_B[/tex] at the bottom of the grade=28.3 ft/s

The velocity of the car at the bottom of the grade is 28.37 ft/s.

Acceleration is the rate of change of velocity with time.

The velocity of the car increases as the car moves downwards and it will be maximum when the car reaches equilibrium position. At equilibrium position (bottom of the grade) the acceleration will be zero but the velocity will be maximum.

a = 3.22 - 0.004v²

0 = 3.22 - 0.004v²

0.004v² = 3.22

v² = 3.22/0.004

v² = 805

v = √805

v = 28.37 ft/s

Thus, the velocity of the car at the bottom of the grade is 28.37 ft/s.

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To develop this problem it is necessary to apply the concepts related to Work and energy conservation.

By definition we know that the work done by a particle is subject to the force and distance traveled. That is to say,

[tex]W = F*d[/tex]

Where,

F= Force

d = Distance

On the other hand we know that the potential energy of a body is given based on height and weight, that is

[tex]PE = -mgh[/tex]

The total work done would be given by the conservation and sum of these energies, that is to say

[tex]W_{net} = W+PE[/tex]

PART A) Applying the work formula,

[tex]W = F*d\\W = 2750*1.25\\W = 3437.5J[/tex]

PART B) Applying the height equation and considering that there is an angle in the distance of 25 degrees and the component we are interested in is the vertical, then

[tex]PE = -mgh*sin25[/tex]

[tex]PE = -16*9.81*1.15sin25[/tex]

[tex]PE = -82.9J[/tex]

The net work would then be given by

[tex]W_{net} = W+PE[/tex]

[tex]W_{net} = 3437.5J-82.9J[/tex]

[tex]W_{net} = 3354.6J[/tex]

Therefore the net work done by these 2 forces on the cannonball while it is in the cannon barrel is 3355J

We must remember that a body in Earth orbit is under perpetual free fall, that is, that the body moves under a constant speed in constant around the earth.

The moment one of the charges is released, its displacement speed will be equal to that of the ship, so it will tend to continue rotating in the orbit in which the ship was traveling.

The correct answer is D. The payload will remain in the same orbit with the space shuttle.

Answer:

a = 3.52 m/s²

Explanation:

Newton's second law:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass s (kg)

a : acceleration  (m/s²)

Data

m= 27.4 kg : mas of the box

F= 170 N, at an angle of 25◦ above the horizontal :Force rope attached to the box

μk =  0.293 :Coefficient of friction between box and floor

g =   9.8 m/s² : acceleration due to gravity

We define the x-axis in the direction parallel to the movement of the  box and the y-axis in the direction perpendicular to it.

Forces acting on the box

W: Weight of the block : In vertical direction  ,downward

FN : Normal force : perpendicular to the floor  upward

f : Friction force: parallel to the floor  and opposite to the movement

F : force of the rope attached to the box , at an angle of 25◦ above the horizontal

Calculated of the W  ( weight of the box)

W= m*g

W=  27.4 kg* 9.8 m/s² = 268.52 N

x-y components  of the force of 170 N

Fx=170 N *cos 25° = 154.07 N

Fy=170 N *sin 25° =71.845 N

Calculated of the FN  ( Normal force)

We apply the formula (1)  

∑Fy = m*ay ay = 0  

FN + Fy - W = 0  

FN = W-  Fy

FN = 268.52 N -   71.845 N

FN =196. 675 N

Calculated of the f  (friction force)

f = μk*FN

f = 0.293*196. 675

f = 57.626 N

We apply the formula (1) to calculated acceleration of the box:

∑Fx = m*ax  ,  ax= a  : acceleration of the box

Fx-f = m*a

154.07-57.626 = (27.4)*a

96.45 =  (27.4)*a

a = (96.45)/ (27.4)

a = 3.52 m/s²

Final answer:

To calculate the box's acceleration, resolve the pulling force into components, find the normal and frictional forces, and apply Newton's second law with the net horizontal force and box's mass.

Explanation:

To find the acceleration of the box, we first need to resolve the applied force into horizontal and vertical components. The horizontal component of the force ( extit{F}_{horizontal}) is calculated by 170 N  imes  extit{cos}(25°), and the vertical component ( extit{F}_{vertical}) by 170 N  imes  extit{sin}(25°). The normal force ( extit{N}) is the sum of the vertical component and the weight of the box acting upwards, which is the box mass (27.4 kg) multiplied by the acceleration of gravity (9.8 m/s²), but since the box doesn't accelerate vertically, the normal force is equal to the weight minus  extit{F}_{vertical}.

Next, we calculate the frictional force ( extit{F}_{friction}) using the coefficient of friction (0.293) and the normal force. The net horizontal force ( extit{F}_{net}) is the difference between the horizontal component of the pull and the frictional force. Finally, we use Newton's second law ( extit{F} = m  imes  extit{a}) to find the acceleration by dividing the  extit{F}_{net} by the mass of the box (27.4 kg).

The pressure difference between the inside and outside of the window can be found using Bernoulli's principle. The force exerted on the window by air pressure can be calculated using the formula F = P * A.

A) The pressure difference between the inside and outside of the window can be found using Bernoulli's principle. According to Bernoulli's principle, the pressure difference is equal to the difference in kinetic energy per unit volume between the inside and outside air. This can be expressed as: ΔP = 1/2 ρv², where ΔP is the pressure difference, ρ is the density of air, and v is the speed of air. Substituting the given values, we get: ΔP = 1/2 * 1.14 kg/m³ * (150 m/s)² = 12,825 Pa.  

B) The force exerted on the window by air pressure can be calculated using the formula: F = P * A, where F is the force, P is the pressure difference, and A is the area of the window. Substituting the given values, we get: F = 12,825 Pa * (0.25 m * 0.45 m) = 1,826.25 N.

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Answer:

Work done, W = 84.57 Joules

Explanation:

It is given that,

Mass of the wooden block, m = 2 kg

Tension force acting on the string, F = 30 N

Angle made by the block with the horizontal, [tex]\theta=20^{\circ}[/tex]

Distance covered by the block, d = 3 m

Let W is the work done by the tension force. It can be calculated as :

[tex]W=F\ cos\theta\times d[/tex]

[tex]W=30\times cos(20)\times 3[/tex]

W = 84.57 Joules

So, the work done by the tension force is 84.57 Joules. Hence, this is the required solution.

The quantity of work done by this tension force is equal to 84.57 Joules.

Given the following data:

Mass of wooden block, m = 2 kg

Tension force, F = 30 N

Angle = 20°

Distance, d = 3 m.

In Science, work done is generally calculated by multiplying tension force and the vertical distance experienced by an object.

Mathematically, this is given by:

W = Fdcosθ

W = 30 × 3.0 × cos20

Work done, W = 84.57 Joules.

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Answer:

New time period, [tex]T_2=2.12\ s[/tex]

Explanation:

Given that,

Mass of the object 1, [tex]m_1=0.75\ kg[/tex]

Time period, [tex]T_1=1.5\ s[/tex]

If object 1 is replaced by object 2, [tex]m_2=1.5\ kg[/tex]

Let [tex]T_2[/tex] is the new period of oscillation.

The time period of oscillation of mass 1 is given by :

[tex]T_1=2\pi \sqrt{\dfrac{m_1}{k}}[/tex]

[tex]1.5=2\pi \sqrt{\dfrac{0.75}{k}}[/tex]............(1)

The time period of oscillation of mass 2 is given by :

[tex]T_2=2\pi \sqrt{\dfrac{m_2}{k}}[/tex]

[tex]T_2=2\pi \sqrt{\dfrac{1.5}{k}}[/tex]............(2)

From equation (1) and (2) we get :

[tex](\dfrac{T_1}{T_2})^2=\dfrac{m_1}{m_2}[/tex]

[tex](\dfrac{1.5}{T_2})^2=\dfrac{0.75}{1.5}[/tex]

[tex]\dfrac{1.5}{T_2}=0.707[/tex]

[tex]T_2=2.12\ s[/tex]

So, the new period of oscillation is 2.12 seconds. Hence, this is the required solution.

Final answer:

To find the new period of oscillation when the mass attached to a spring is increased, one can use the formula for the period T = 2π√(m/k). When the mass is doubled from 0.750 kg to 1.50 kg, the new period of oscillation is approximately 2.12 seconds.

Explanation:

The period of oscillation for a mass m suspended from a spring is given by the formula T = 2π√(m/k), where T is the period, m is the mass, and k is the spring constant. Using this relationship, we can compare the periods for different masses to find the new period when the mass changes.

When the mass was 0.750 kg, the period was 1.50 s. Now, if the mass is increased to 1.50 kg, we can expect the period to change. Noting that the spring constant k remains the same, we use the ratio of the periods T1/T2 = √(m1/m2). Solving for the new period T2 gives us T2 = T1 * √(m2/m1).

If we plug in the numbers, we get:

T1 = 1.50 s (given for 0.750 kg)

m1 = 0.750 kg

m2 = 1.50 kg

T2 = 1.50 s * √(1.50 kg / 0.750 kg) = 1.50 s * √2 ≈ 2.12 s

The new period of oscillation for the 1.50-kg object is therefore approximately 2.12 seconds.

Answer:

40.96%

Solution:

As per the question:

If the original radius be 'r' and the initial pressure difference be 'P':

After the drop, radius:

r' = 0.8r

After rise in pressure:

P' = 0.1 P

Now,

Rate of flow is given by:

R ∝ [tex]r^{4}P[/tex]

Thus

[tex]\frac{R'}{R} = (\frac{0.8r}{r})^{4}\frac{0.1P}{P}[/tex]

R' = 40.96%

Answer:

A.The dipole moment is a scalar quantity.

Explanation:

We know that

Dipole moment :It measure the polarity of chemical bond in a molecule .

The electric dipole moment is equal to product of any charge (positive or negative ) and the distance between the two charges.

Mathematical representation:

[tex]\mu=q\times d[/tex]

Where [tex]\mu[/tex] =Dipole moment

q=Charge on atom or particle

d=Distance between two charged particles

It helps to find out the molecule is polar or non- polar.

When the dipole moment is zero then the molecule is non-polar.

Dipole moment is  a vector quantity.

The direction of dipole moment from negative charge to positive.

When unit of charge is C and unit of distance is m.

Then, unit of dipole moment=C-m

Hence, option A is false.

Final answer:

The false statement about the electric dipole moment is that it is a scalar quantity; it is a vector quantity. So the correct option is A.

Explanation:

The statement concerning the electric dipole moment that is false is A) The dipole moment is a scalar quantity. This is incorrect because the electric dipole moment is a vector quantity, not a scalar. The correct statements about the dipole moment are that it does indeed have units of C·m (Statement B), it combines both the distance between the charges and the magnitude of the charges (Statement C), it is directed from the negative charge toward the positive charge (Statement D), and its direction specifies the orientation of the dipole (Statement E).

Answer:

The value of [tex]\dfrac{dU}{dM}[/tex] is [tex]6\times10^{-29}\ J/unit[/tex].

Explanation:

Given that,

Number [tex]N=10^{8}[/tex]

Energy difference = 6\times10^{-21}\ J[/tex]

Temperature T =300 K

We need to calculate the value of [tex]\dfrac{dU}{dM}[/tex]

We know that,

[tex]\dfrac{dU}{dM}=\dfrac{energy\ difference}{Change\ in\ number\ of\ system}[/tex]

[tex]\dfrac{dU}{dM}=\dfrac{6\times10^{-21}}{10^{8}}[/tex]

[tex]\dfrac{dU}{dM}=6\times10^{-29}\ J/unit[/tex]

Hence, The value of [tex]\dfrac{dU}{dM}[/tex] is [tex]6\times10^{-29}\ J/unit[/tex].

This problem can be resolved using the principle of conservation of angular momentum. Initially, the positive momentum of runner and negative of turntable cancel each other, making the total momentum zero. When the runner stops, the angular momentum of the turntable is expected to compensate for the total momentum of the system.

In Physics, this is a classic problem involving the conservation of angular momentum. Angular momentum is given as L = Iω for an object rotating about an axis, where I is the moment of inertia and ω is the angular velocity. The total initial angular momentum of the system is zero because the positive runner's angular momentum cancels out the negative turntable's angular momentum.

When the runner comes to rest, he no longer contributes an angular momentum. The turntable has to account for all of the angular momentum. We therefore use the conservation of angular momentum to solve for the new angular momentum of the turntable after the runner stops:

Equating moment of inertia of the runner (considering the runner as a point mass, moment of inertia, I = m*r^2) and moment of inertia of the turntable and using the equation L_initial = L_final, we can find the final angular velocity of the system:

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The final angular velocity of the system is 6.853 rad/s.

The initial angular momentum [tex]\(L_{i}\)[/tex] of the turntable is:

[tex]\[ L_{i} = I\omega_{i} \][/tex]

where I is the moment of inertia of the turntable.

The initial angular momentum [tex]\(L_{runner,i}\)[/tex] of the runner is:

[tex]\[ L_{runner,i} = mvr \][/tex]

since the angular momentum of a particle moving in a circle is the product of its mass, velocity, and the radius of the circle.

 The final angular momentum [tex]\(L_{f}\)[/tex] of the system is:

[tex]\[ L_{f} = I\omega_{f} + mvr_{f} \][/tex]

where [tex]\(r_{f}\)[/tex] is the final radius at which the runner is at rest relative to the turntable, which is the same as the initial radius r since the runner is still on the turntable.

By conservation of angular momentum:

[tex]\[ L_{i} + L_{runner,i} = L_{f} \] \[ I\omega_{i} + mvr = I\omega_{f} + mvr \][/tex]

Since the runner is at rest relative to the turntable in the final state, (vr = 0), and the equation simplifies to:

[tex]\[ I\omega_{i} + mvr = I\omega_{f} \][/tex]

Now we can solve for [tex]\(\omega_{f}\)[/tex]:

[tex]\[ \omega_{f} = \frac{I\omega_{i} + mvr}{I} \] \[ \omega_{f} = \omega_{i} + \frac{mvr}{I} \][/tex]

Given:

[tex]\(m = 56.0\) kg\\ \(v = 3.10\) m/s\\ \(r = 3.10\) m\\ \(I = 81.0\) kg\(\cdot\)m\(^2\)\\ \(\omega_{i} = 0.210\) rad/s[/tex]

Plugging in the values:

[tex]\[ \omega_{f} = 0.210 + \frac{56.0 \times 3.10 \times 3.10}{81.0} \] \[ \omega_{f} = 0.210 + \frac{56.0 \times 9.61}{81.0} \] \[ \omega_{f} = 0.210 + \frac{538.16}{81.0} \] \[ \omega_{f} = 0.210 + 6.643 \] \[ \omega_{f} = 6.853 \][/tex]

Answer:u=4.04 m/s

Explanation:

Given

Mass m=85 kg

mass of Raft M=130 kg

velocity of raft and man  v=1.6 m/s

Let initial speed of Tyrone is u

Conserving Momentum as there is no external Force

[tex]mu=(M+m)v[/tex]

[tex]85\times u=(85+130)\cdot 1.6[/tex]

[tex]u=\frac{215}{85}\cdot 1.6[/tex]

[tex]u=2.529\cdot 1.6=4.04 m/s[/tex]  

Final answer:

Tyrone's speed as he ran off the end of the pier was approximately 4.04 m/s.

Explanation:

To calculate Tyrone's speed as he ran off the end of the pier, we can use the principle of conservation of momentum. According to this principle, the initial momentum of the system (Tyrone and the raft) is equal to the final momentum. Tyrone's initial momentum is given by his mass (85 kg) multiplied by his speed (which we need to find). The mass of the raft is 130 kg and its initial velocity is 0 m/s since it was at rest. The final momentum of the system is given by the mass of Tyrone and the raft (215 kg) multiplied by their final velocity (1.6 m/s).

Using the equation for conservation of momentum, we have:

(85 kg)(v) = (215 kg)(1.6 m/s)

Solving for the velocity v:

v = (215 kg)(1.6 m/s) / 85 kg

v ≈ 4.04 m/s

Therefore, Tyrone's speed as he ran off the end of the pier was approximately 4.04 m/s.

Answer:

Ff = 6058.5N

Explanation:

The sum of forces is:

[tex]Ff = m*a_c[/tex]

[tex]Ff = m*V^2/R[/tex]

[tex]Ff = 1900*13^2/53[/tex]

[tex]Ff = 6058.5N[/tex]

To solve this problem it is necessary to apply the concepts based on Newton's second law and the Centripetal Force.

That is to say,

[tex]F_c = F_w[/tex]

Where,

[tex]F_c =[/tex]Centripetal Force

[tex]F_w =[/tex]Weight Force

Expanding the terms we have to,

[tex]mg = \frac{mv^2}{r}[/tex]

[tex]gr = v^2[/tex]

[tex]v = \sqrt{gr}[/tex]

Where,

r = Radius

g = Gravity

v = Velocity

Replacing with our values we have

[tex]v = \sqrt{(9.8)(11.8)}[/tex]

[tex]v = 10.75m/s[/tex]

Therefore the minimum speed must the car traverse the loop so that the rider does not fall out while upside down at the top is 10.75m/s

Answer:

I believe it's B

The amount of work done by friction in stopping a car is determined by the force of friction and the distance over which it acts. Car L will require the greatest force of friction to stop it, making the amount of work done by friction the greatest for Car L.

The amount of work done by friction in stopping a car is determined by the force of friction and the distance over which it acts. Since all three cars have identical coefficients of kinetic friction with the road surface, the force of friction will be the same for all three cars. However, the stopping distance for each car will vary based on its mass.

Car L has the greatest mass, so it will require the greatest force of friction to stop it. Therefore, the amount of work done by friction in stopping car L will be the greatest among the three cars.

Car M and Car N will require less force of friction compared to car L, so the amount of work done by friction in stopping them will be less than that of Car L, but equal to each other.

Answer:

Explanation:

Total internal reflection refers to a phenomenon that occurs when light travels from a denser medium to less dense medium, in  which the incidence ray is inclined at an angle greater than a certain critical angle to the normal ( angle at which the refracted ray is equal to 90°),  instead of refraction, the ray is reflected back into the material.

since it must start from a denser medium or a medium with higher refractive index, then it must start with plastics

to calculate critical angle,

we use Snell's law

n₁ sin C = n₂ sin r  since r is 90° where n₁ =1.54,    n₂  = 1.33

C = sin⁻¹ ( n₂ / n₁) = sin⁻¹ ( 1.33 / 1.54) = 59.7°

the critical angle = 59.7°

The light must start in the plastic for total internal reflection to occur. The critical angle, which is the angle of incidence where the refracted light is tangential to the boundary, is approximately 59.46 degrees.

For total internal reflection to occur, the light must go from a material with a higher refractive index to one with a lower refractive index. In this case, the light must start in the plastic.

The critical angle is the angle of incidence for which the angle of refraction equals 90 degrees. It can be calculated using Snell's Law (n1*sin(θ1) = n2*sin(θ2)), where n1 and n2 are the refractive indices and θ1 and θ2 are the angles of incidence and refraction, respectively. If we set θ2 to 90 degrees and rearrange the equation we get: θ1 (critical angle) = arcsin(n2/n1).

Plugging in the given index values, we get: θ1 = arcsin(1.33/1.54), which is approximately 59.46 degrees.

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Answer:

10.2 m .

Explanation:

Let object falls by angle θ .

At any moment after the fall , there are two forces acting on the sphere

1 ) mg cosθ which is a component of weight towards the centre 2 ) normal reaction of the surface R .

mgcosθ - R is net force acting, which provides centripetal force

mgcosθ - R = mv² / r

But v² = 2g r( 1-cosθ )   [ object falls by height ( r - r cosθ ).

mgcosθ - R = m / r x 2g r( 1-cosθ )

When the object is no longer in touch with sphere,

R = 0

mgcosθ  = m / r x 2g r( 1-cosθ )

3 gr cosθ = 2gr

cosθ = 2/3

height of fall

= r ( 1-cosθ )

r ( 1 - 2/3 )

1/3 r

1/3 x 15.3

5.1 m

Height from the ground

15.3 - 5.1

10.2 m .

The object sliding down the sphere loses contact at a point where its gravitational and centripetal forces become equal, corresponding to an angle θ where cosθ = 2/3. Given the sphere's radius of 15.3m, this corresponds to a height from the ground of 10.2 m.

To solve this problem, we need to apply the concept of conservative forces and energy conservation. Initially, the object is at rest at the top of the sphere, thus it possesses gravitational potential energy and no kinetic energy. As the object slides down, its potential energy converts into kinetic energy until a point where the object no longer remains in contact with the sphere.

This occurs when the gravitational force acting towards the center of the sphere becomes equal to the centripetal force required for the object in circular motion. Mathematically, this correlates to the equation: mgcosθ = mv²/r.

By simplifying this equation, considering all the potential energy converts into kinetic energy at the point of leaving contact (mgh = 1/2mv²), and knowing the radius of the sphere, we can deduce that the object loses contact with the sphere at a point where the angle θ (from the vertical axis) is cosθ = 2/3. The height (h) it loses contact can then be determined by the relationship h = r(1 - cosθ). Given the radius r = 15.3m, we find the height from the ground to be 10.2 m.

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