If you slide down a rope, it’s possible to create enough thermal energy to burn your hands or your legs where they grip the rope. Suppose a 40 kg child slides down a rope at a playground, descending 2.0 m at a constant speed. How much thermal energy is created as she slides down the rope?

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:

the acceleration due to gravity g at the surface is proportional to the planet radius R (g ∝ R)

Explanation:

according to newton's law of universal gravitation ( we will neglect relativistic effects)

F= G*m*M/d² , G= constant , M= planet mass , m= mass of an object , d=distance between the object and the centre of mass of the planet

if we assume that the planet has a spherical shape,  the object mass at the surface is at a distance d=R (radius) from the centre of mass and the planet volume is V=4/3πR³ ,

since M= ρ* V = ρ* 4/3πR³ , ρ= density

F = G*m*M/R² = G*m*ρ* 4/3πR³/R²= G*ρ* 4/3πR

from Newton's second law

F= m*g = G*ρ*m* 4/3πR

thus

g = G*ρ* 4/3π*R = (4/3π*G*ρ)*R

g ∝ R

Answer:

Time period of the osculation will be 2.1371 sec

Explanation:

We have given mass m = (B+25)

And the spring is stretched by (8.5 A )

Here A = 13 and B = 427

So mass m = 427+25 = 452 gram = 0.452 kg

Spring stretched x= 8.5×13 = 110.5 cm

As there is additional streching of spring by 3 cm

So new x = 110.5+3 = 113.5 = 1.135 m

Now we know that force is given by F = mg

And we also know that F = Kx

So [tex]mg=Kx[/tex]

[tex]K=\frac{mg}{x}=\frac{0.452\times 9.8}{1.135}=3.90N/m[/tex]

Now we know that [tex]\omega =\sqrt{\frac{K}{m}}[/tex]

So [tex]\frac{2\pi }{T} =\sqrt{\frac{K}{m}}[/tex]

[tex]\frac{2\times 3.14 }{T} =\sqrt{\frac{3.90}{0.452}}[/tex]

[tex]T=2.1371sec[/tex]

The period of the resulting oscillation, named T, given a mass of (B + 25.0) g (with B = 427 g) and a spring displacement of (8.50 A) cm (with A = 13), when the object is pulled an additional 3.0 cm downward and released, is 2.345 seconds, when rounded to three significant figures.

The problem given can be solved using the concept of simple harmonic motion, which is encountered in Physics. The object attached to the spring can be visualized as a simple harmonic oscillator. The period of the resulting oscillation can be calculated using the formula T = 2π√(m/k), where T is the period, m is the mass and k is the spring constant.

First, we compute the mass in grams, which is given by (B + 25.0) g, where B = 427 g. Therefore, the mass is (427 g + 25.0 g) = 452 g or 0.452 kg (since 1 kg = 1000 g).

Next, the spring constant k is calculated using Hooke’s law (F=kx), where F is the force exerted by the mass on the spring (F=mg), m is the mass and g is the acceleration due to gravity, and x is the displacement of the spring. From the problem, m = 0.452 kg, g = 9.8 m/s², and x = (8.50 A) cm = (8.50 x 13) cm = 110.5 cm or 1.105 m (since 1 m = 100 cm). Substituting these values into Hooke’s law gives: k = mg/x = (0.452 kg x 9.8 m/s²) / 1.105 m = 4.00 N/m.

Finally, the period T of the resulting oscillation is found using the formula T = 2π√(m/k) = 2π√(0.452 kg/4.00 N/m) = 2.345 s, rounded to three significant figures.

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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:

The recoil velocity of the raft is 1.205 m/s.

Explanation:

given that,

Mass of the swimmer, [tex]m_1=55\ kg[/tex]

Mass of the raft, [tex]m_2=210\ kg[/tex]

Velocity of the swimmer, v = +4.6 m/s

It is mentioned that the swimmer then runs off the raft, the total linear momentum of the  swimmer/raft system is conserved. Let V is the recoil velocity of the raft.

[tex]m_1v+m_2V=0[/tex]

[tex]55\times 4.6+210V=0[/tex]

V = -1.205 m/s

So, the recoil velocity of the raft is 1.205 m/s. Hence, this is the required solution.

Answer:

The recoil velocity of the raft would be [tex]v_{r}\approx 1.2\frac{m}{s}[/tex] (pointing to the left if the swimmer runs to the right)

Explanation:

The problem states that the swimmer has a mass of m=55 kg, and the raft has a mass of M=210 kg. Then, it says that the swimmer runs off the raft with a (final) velocity of v=4.6 m/s relative to the shore.

To analyze it, we take a system of "two particles", wich means that we will consider the swimmer and the raft as a hole system, aisolated from the rest of the world.

Then, from the shore, we can put our reference system and take the initial moment when the swimmer and the raft are stationary. This means that the initial momentum is equal to zero:

[tex]p_{i}=0[/tex]

Besides, we can use magnitudes instead of vectors because the problem will develope in only one dimension after the initial stationary moment (x direction, positive to the side of the running swimmer, and negative to the side of the recoling raft), this means that we can write the final momentum as

[tex]p_{f}=mv-Mv_{r}=0[/tex]

The final momentum is equal to zero due to conservation of momentum (because there are no external forces in the problem, for the system "swimmer-raft"), so the momentum is constant.

Then, from that previous relation we can clear

[tex]v_{r}=\frac{m}{M}v=\frac{55}{210}*4.6\frac{m}{s}=\frac{253}{210}\frac{m}{s}\approx1.2\frac{m}{s}[/tex]

wich is the recoil velocity of the raft, and it is pointing to the left (we established this when we said that the raft was going to the negative side of the system of reference, and when we put a minus in the raft term inside the momentum equation).

Answer: 0.75 ft

Explanation:

This situation is due to Refraction, a phenomenon in which a wave (the light in this case) bends or changes its direction when passing through a medium with an index of refraction different from the other medium.  In this context, the index of refraction is a number that describes how fast light propagates through a medium or material.  

In addition, we have the following equation that states a relationship between the apparent depth [tex]{d}^{*}[/tex] and the actual depth [tex]d[/tex]:  

[tex]{d}^{*}=d\frac{{n}_{1}}{{n}_{2}}[/tex] (1)  

Where:

[tex]n_{1}=1[/tex] is the air's index of refraction  

[tex]n_{2}=\frac{4}{3}=1.33[/tex] water's index of refraction.

[tex]d=1 ft[/tex] is the actual depth of the quarters

Now. when [tex]n_{1}[/tex] is smaller than [tex]n_{2}[/tex] the apparent depth is smaller than the actual depth. And, when [tex]n_{1}[/tex] is greater than [tex]n_{2}[/tex] the apparent depth is greater than the actual depth.  

Let's prove it:

[tex]{d}^{*}=1 ft\frac{1}{1.33}[/tex] (2)  

Finally we find the aparent depth of the quarters, which is smaller than the actual depth:

[tex]{d}^{*}=0.75 ft[/tex]

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:

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: 624 Hz

Explanation:

If the ratio of the inductive reactance to the capacitive reactance, is 6.72, this means that it must be satified the following expression:

ωL / 1/ωC = 6.72

ω2 LC = 6.72 (1)

Now, at resonance, the inductive reactance and the capacitive reactance are equal each other in magnitude, as follows:

ωo L = 1/ωoC → ωo2 = 1/LC

So, as we know the resonance frequency, we can replace LC in (1) as follows:

ω2 / ωo2  = 6.72  

Converting the angular frequencies to frequencies, we have:

4π2 f2 / 4π2 fo2  = 6.72

Simplifying and solving for f, we have:

f = 240 Hz  . √6.72 = 624 Hz

As the circuit is inductive, f must be larger than the resonance frequency.

Answer:

f = 0.84 m

Explanation:

given,

near point of the distance from eye(di) = 0.60 m

focal length calculation = ?

distance he can read (d_o)= 0.350 m

now,

Using lens formula

[tex]\dfrac{1}{f} = \dfrac{1}{d_0} + \dfrac{1}{d_i}[/tex]

[tex]\dfrac{1}{f} = \dfrac{1}{0.35} + \dfrac{1}{-0.6}[/tex]

[tex]\dfrac{1}{f} = 1.19[/tex]

[tex] f = \dfrac{1}{1.19}[/tex]

f = 0.84 m  

The focal length of the lens f = 0.84 m

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:

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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a) The time taken for the wave to travel a distance of 8.40 m is 0.24 s

b) The time taken for a point on the string to travel a distance of 8.40 m is 6.46 s

c) The time in part a) does not change; the time in part b) is halved (3.23 s)

Explanation:

a)

In order to solve this part, we have to find the speed of the wave, which is given by the wave equation:

[tex]v=f \lambda[/tex]

where

v is the speed

f is the frequency

[tex]\lambda[/tex] is the wavelength

For the wave in this problem,

[tex]f = 62.0 Hz\\\lambda =0.560 m[/tex]

So its speed is

[tex]v=(62.0)(0.560)=34.7 m/s[/tex]

Now we want to know how long does it take for the wave to travel a distance of

d = 8.40 m

Since the wave has a constant velocity, we can use the equation for uniform motion:

[tex]t=\frac{d}{v}=\frac{8.40}{34.7}=0.24 s[/tex]

b)

In this case, we want to know how long does it take for a point on the string to travel a distance of

d = 8.40 m

A point on the string does not travel along the string, but instead oscillates up and down. The amplitude of the wave corresponds to the maximum displacement of a point on the string from the equilibrium position, and for this wave it is

A = 5.20 mm = 0.0052 m

The period of the wave is the time taken for a point on the string to complete one oscillation. It can be calculated as the reciprocal of the frequency:

[tex]T=\frac{1}{f}=\frac{1}{62.0}=0.016 s[/tex]

During one oscillation (so, in a time corresponding to one period), the distance covered by a point on the string is 4 times the amplitude:

[tex]d=4A=4(0.0052)=0.0208 m[/tex]

Since d is the distance covered by a point on the string in a time T, then we can find the time t needed to cover a distance of

d' = 8.40 m

By setting up the following proportion:

[tex]\frac{d}{T}=\frac{d'}{t}\\t=\frac{d'T}{d}=\frac{(8.40)(0.016)}{0.0208}=6.46 s[/tex]

c)

As we can see from the equation used in part a), the speed of the wave depends only on its frequency and wavelength, not on the amplitude: therefore, if the amplitude is changed, the speed of the wave is not affected, therefore the time calculated in part a) remains the same.

On the contrary, the answer to part b) is affected by the amplitude. In fact, if the amplitude is doubled:

A' = 2A = 2(0.0052) = 0.0104 m

Then the distance covered during one period is

[tex]d=4A' = 4(0.0104)=0.0416 m[/tex]

The time period does not change, so it is still

T = 0.016 s

Therefore, the time needed to cover a distance of

d' = 8.40 m

this time is

[tex]\frac{d}{T}=\frac{d'}{t}\\t=\frac{d'T}{d}=\frac{(8.40)(0.016)}{0.0416}=3.23 s[/tex]

So, the time taken has halved.

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The time taken for a wave to travel a distance of 8.40 m is approximately 0.120 seconds. The time taken for a point on the string to travel 8.40 m once the wave train has reached it is also approximately 0.120 seconds. Doubling the amplitude of the wave does not affect the time taken.

To determine how long it takes for the wave to travel a distance of 8.40 m, we can use the formula:

Time = Distance / Speed

The speed of a wave is equal to the product of its frequency and wavelength:

Speed = Frequency x Wavelength

Substituting the given values into the formulas, we find that it takes approximately 0.120 seconds for the wave to travel 8.40 m along the string.

To calculate the time it takes for a point on the string to travel a distance of 8.40 m once the wave train has reached the point and set it into motion, we need to divide the distance by the speed of the wave pulse:

Time = Distance / Speed

Since the speed of the wave pulse is the same as the speed of the wave, the time it takes for a point on the string to travel 8.40 m is also approximately 0.120 seconds.

If the amplitude of the wave is doubled, it does not affect the time it takes for the wave to travel a distance of 8.40 m or for a point on the string to travel 8.40 m. The only thing that changes is the maximum displacement of the particles in the wave, but it does not impact the time taken.

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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:

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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Answer:

a) [tex]a_c = 1.09m/s^2[/tex]

b) T = 720.85N

Explanation:

With a balance of energy from the lowest point to its maximum height:

[tex]m*g*L(1-cos\theta)-1/2*m*V_o^2=0[/tex]

Solving for [tex]V_o^2[/tex]:

[tex]V_o^2=2*g*L*(1-cos\theta)[/tex]

[tex]V_o^2=2.408[/tex]

Centripetal acceleration is:

[tex]a_c = V_o^2/L[/tex]

[tex]a_c = 2.408/2.21[/tex]

[tex]a_c = 1.09m/s^2[/tex]

To calculate the tension of the rope, we make a sum of forces:

[tex]T - m*g = m*a_c[/tex]

Solving for T:

[tex]T =m*(g+a_c)[/tex]

T = 720.85N

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:

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].

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:

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

Answer

given,

mass of cockroach = 0.117 Kg

radius = 17.3 cm

rotational inertia = 5.20 x 10⁻³ Kg.m²

speed of cockroach = 1.91 m/s

angular velocity of Susan (ω₀)= 2.87 rad/s

final speed of cockroach = 0 m/s

Initial angular velocity of Susan

L_s = I ω₀

L_s = 5.20 x 10⁻³ x 2.87

L_s=0.015 kg.m²/s

initial angular momentum of the cockroach

L_c = - mvr

L_c = - 0.117 x 1.91 x 0.173

L_c = - 0.0387 kg.m²/s

total angular momentum of Both

L = 0.015 - 0.0387

L = - 0.0237 kg.m²/s

after cockroach stop inertia becomes

I_f = I + mr^2

I_f = 5.20 x 10⁻³+ 0.117 x 0.173^2

I_f = 8.7  x 10⁻³ kg.m²/s

final angular momentum of the disk

L_f = I_f ω_f

L_f = 8.7  x 10⁻³ x ω_f

using conservation of momentum

L_i = L_f

-0.0237 =8.7  x 10⁻³ x ω_f

[tex]\omega_f = \dfrac{-0.0237}{8.7 \times 10^{-3}}[/tex]

[tex]\omega_f = -2.72\ rad/s[/tex]

angular speed of Susan is [tex]\omega_f = -2.72\ rad/s[/tex]

The value is negative because it is in the opposite direction of cockroach.

[tex]|\omega_f |= 2.72\ rad/s[/tex]

b) the mechanical energy is not conserved because cockroach stopped in between.  

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

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

Answer:2

Explanation:

Given

Velocity of bicycle is 5 m/s towards north

radius of rim [tex]r=20 cm[/tex]

mass of rim [tex]m=2 kg[/tex]

Angular momentum [tex]\vec{L}=I\cdot \vec{\omega }[/tex]

[tex]I=mr^2=2\times 0.2^2=0.08 kg-m^2[/tex]

[tex]\omega =\frac{v}{r}=\frac{5}{0.2}=25 rad/s[/tex]

[tex]L=0.08\times 25=2kg-m^2/s[/tex]  

direction of Angular momentum will be towards west

Answer:

a) 4 times larger. b) 16 times larger. c) 16 times smaller. d) Coulomb´s Law

Explanation:

Between any pair of charged particle, there exists a force, acting on the line that join the charges (assuming they can assimilated to point charges) directed from one to the other, which is directly proportional to the product of the charges, and inversely proportional to the square of the distance between them.

F= k q1q2 / (r12)2

a) If the distance is reduced to the half of the original distance of separation, and we introduce this value in the force equation, we get:

F(r/2) = k q1q2 / (r12/2)2 = k q1q2 /((r12)2/4) = 4 F(r)

b) By the same token, if r= r/4, we will have F(r/4) = 16 F(r)

c) If the distance increases 4 times, as the force is inversely proportional to the square of the distance, the force will be the original divided by 16, i.e., 16 times smaller.

The empirical law that allows to find out easily these values, is the Coulomb´s Law.

Explanation:

a) The spring constant is 772.1 N/m

b) The amplitude is 0.025 m

c) The period is 0.365 s, the frequency is 2.74 Hz

Explanation:

a)

At equilibrium, the weight of the fish hanging on the spring is equal to the restoring force of the spring. Therefore, we can write:

[tex]mg=kx[/tex]

where

(mg) is the weight of the fish

kx is the restoring force

m = 2.6 kg is the mass of the fish

[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity

k is the spring constant

x = 3.3 cm = 0.033 m is the stretching of the spring

Solving for k, we find:

[tex]k=\frac{mg}{x}=\frac{(2.6)(9.8)}{0.033}=772.1 N/m[/tex]

b)

Later, the fish is pulled down by 2.5 cm (0.025 m), and the system starts to oscillate.

The amplitude of a simple harmonic motion is equal to the maximum displacement of the system. In this case, when the fish is pulled down by 2.5 cm and then released, immediately after the releasing the fish moves upward, until it reaches the same displacement on the upper side (+2.5 cm).

This means that the amplitude of the motion corresponds also to the initial displacement of the fish when it is pulled down: therefore, the amplitude is

A = 2.5 cm = 0.025 m

c)

The period of oscillation of a mass-spring system is given by

[tex]T=2\pi \sqrt{\frac{m}{k}}[/tex]

where

m is the mass

k is the spring constant

Here we have

m = 2.6 kg

k = 772.1 N/m

So, the period is

[tex]T=2\pi \sqrt{\frac{2.6}{772.1}}=0.365 s[/tex]

And the frequency is given by the reciprocal of the period, therefore:

[tex]f=\frac{1}{T}=\frac{1}{0.365}=2.74 Hz[/tex]

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From the definition of equilibrium, at the moment in which both children are sitting facing each other at a certain distance the torque within the seesaw will be zero.

However, if one of the children approaches the pivot, the center of mass will move towards the end of the other child, which will immediately cause the child to rise.

Another way of observing this problem is considering the distance between the two, the distance is proportional to the Torque,

[tex]\tau_1 = \tau_2[/tex]

[tex]F*d_1 = F*d_2[/tex]

Therefore by decreasing the distance - when walking towards the pivot - the torque of the child sitting at the other end will be greater because it keeps its distance.

Being said higher torque will cause the approaching child to rise.

The correct answer is B.

The child's side will rise because of shifting of weight of the child from its place.

Two children are balanced on opposite sides of a seesaw. If one child leans inward toward the pivot point, her side will rise because of shifting of weight of the child from its place. Due to movement from its place, the weight of other child increases which leads to lowering of its side.

So we can conclude that the child's side will rise because of shifting of weight of the child from its place.

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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.

In order to develop this problem it is necessary to use the concepts related to the conservation of both potential cinematic as gravitational energy,

[tex]KE = \fract{1}{2}mv^2[/tex]

[tex]PE = GMm(\frac{1}{r_1}-(\frac{1}{r_2}))[/tex]

Where,

M = Mass of Earth

m = Mass of Object

v = Velocity

r = Radius

G = Gravitational universal constant

Our values are given as,

[tex]m = 910 Kg[/tex]

[tex]r_1 = 1200 + 6371 km = 7571km[/tex]

[tex]r_2 = 6371 km,[/tex]

Replacing we have,

[tex]\frac{1}{2} mv^2 =  -GMm(\frac{1}{r_1}-\frac{1}{r_2})[/tex]

[tex]v^2 =  -2GM(\frac{1}{r_1}-\frac{1}{r_2})[/tex]

[tex]v^2 = -2*(6.673 *10^-11)(5.98 *10^24) (\frac{1}{(7.571 *10^6)} -\frac{1}{(6.371 *10^6)})[/tex]

[tex]v = 4456 m/s[/tex]

Therefore the speed of the object when striking the surface of earth is 4456 m/s

The speed of a 910-kg object striking the Earth after falling from 1200 km above the North Pole can be found using conservation of mechanical energy and the formula for gravitational potential energy.

To determine the speed at which a 910-kg object strikes the surface of the Earth when released from rest at an altitude of 1200 km above the North Pole, we can use the conservation of mechanical energy. The total mechanical energy (kinetic plus potential) at the initial and final points must be equal since no external work is done on the system (we ignore atmospheric friction).

Initially, the object has potential energy due to its altitude above Earth and no kinetic energy since it's at rest. When it strikes the Earth, it has kinetic energy and minimal potential energy. Setting the initial and final total energies equal gives:

Ui + Ki = Uf + Kf,

where U is potential energy and K is kinetic energy. The potential energy can be calculated using the formula U = -GMEarthm/r, where m is the object's mass, r is the distance from the object to the center of Earth, and G is the gravitational constant. The kinetic energy is given by K = (1/2)mv2, where v is the velocity of the object.

The initial distance ri to the center of Earth is 6357 km + 1200 km (release altitude), and when the object strikes Earth, rf = 6357 km (polar radius of Earth). We know Ki = 0 and Uf is small enough to be negligible at the Earth's surface.

By plugging in the values, solving the equation for v, and taking the square root, we find the speed of the object when it strikes the ground.

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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