Problem 2: A flat belt is used with a driving pulley (diameter 4 inches) and a driven pulley (diameter 18 inches) in a open configuration. The center distance between the pulleys is 48 inches. The friction coefficient between the belt and pulley is 0.6. Determine the following: a) If the belt is initially tensioned to 50 lbs, what is the force in the belt on the tight side just before slippage (neglect the centrifugal force of the belt). b) Find the maximum torque required at the driving pulley.

Answer:

a) 1504.8 J

b) 991.76 J

c) 0J

d) 0J

Explanation:

(a) The work done by the force P on the box is given by the following formula:

[tex]W_P=Px[/tex]

P: applied force = 171N

x: distance in which the for P is applied = 8.80m

you replace the values of P and x and obtain:

[tex]W_P=(171N)(8.80m)=1504.8J[/tex]

(b) The work don by the friction force is:

[tex]W_f=F_fx=\mu N x=\mu Mg x[/tex]

μ = coefficient of kinetic friction = 0.250

M: mass of the box = 46.0kg

g: gravitational constant = 9.8 m/s^2

[tex]W_f=(0.250)(46.0kg)(9.8m/s^2)(8.80m)=991.76J[/tex]

(c) The Normal force is

[tex]N=Mg=(46.0kg)(9,8m/s^2)=450.8N[/tex]

but this force does not do work on the box because the direction is perpendicular to the direction of the force P.

[tex]W_N=0J[/tex]

(d) the same as before:

[tex]W_g=0J[/tex]

Answer:

A.

Explanation:

Refraction light rays are bent as they pass into a different medium where its speed is different. As refracted light rays pass from a fast medium to a slow medium, the light ray bends toward the normal to the boundary between the two medium. Light refracts as it travels at an angle into a medium with a different refractive index.

Answer:

a) [tex]\vec{v}_{12}=2.86*10^{8} m/s[/tex]

b) [tex]p=2.81*10^{-16} kg*m/s[/tex]    

Explanation:

a) When we have two particles traveling in parallel directions, the formula for relative velocity is:

[tex]\vec{v}_{12}=\frac{\vec{v}_{1}-\vec{v}_{2}}{1-\frac{\vec{v}_{1}\vec{v}_{2}}{c^{2}}}[/tex]

Here we have that v(1) = -v(2), the speed of the of the second particle is the negative of the first one.

If we use these equivalence we have:

[tex]\vec{v}_{12}=\frac{2\vec{v}_{1}}{1+\frac{\vec{v}_{1}^{2}}{c^{2}}}[/tex]  

[tex]\vec{v}_{12}=\frac{2*2.19*10^{8}}{1+\frac{2.19*10^{16}}{3*10^{16}}}[/tex]  

[tex]\vec{v}_{12}=2.86*10^{8} m/s[/tex]          

And, [tex]\vec{v}_{21}=-2.86*10^{8} m/s[/tex]  

b) The relativistic momentum equation to one particle observed by the other particle, is:

[tex]p=\gamma mv[/tex]

Where gamma is:

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

Then gamma will be:

[tex]\gamma=\frac{1}{\sqrt{1-\frac{(2.86*10^{8})^{2}}{(3*10^{8})^{2}}}}[/tex]

[tex]\gamma=3.31[/tex]

Finally, the value of the momentum will be:

[tex]p=3.31*2.97*10^{-25}*2.86*10^{8}[/tex]    

[tex]p=2.81*10^{-16} kg*m/s[/tex]    

I hope it helps you!                        

Answer:

t = 2.94 x 10⁶ years

Explanation:

The equation used in the case of constant speed is:

s = vt

t = s/v

where,

s = distance = 12 light years

s = (12 light years)(9.461 x 10¹² km/light year) =  113.532 x 10¹² km

v = speed = 440 km/hr

t = time passed = ?

Therefore,

t = (113.532 x 10¹² km)/(440 km/hr)

t = 2.58 x 10¹¹ hr

Now, converting it to years:

t = (2.58 x 10¹¹ hr)(1 year/8766 hr)

t = 2.94 x 10⁶ years

Answer:171 N

Explanation:

Answer:

171 N.

Explanation:

Given Information:

Power = P = 100 Watts

Voltage = V = 220 Volts

Required Information:

a) Current = I = ?

b) Resistance = R = ?

Answer:

a) Current = I = 0.4545 A

b) Resistance = R = 484 Ω

Explanation:

According to the Ohm’s law, the power dissipated in the light bulb is given by

[tex]P = VI[/tex]

Where V is the voltage across the light bulb, I is the current flowing through the light bulb and P is the power dissipated in the light bulb.

Re-arranging the above equation for current I yields,

[tex]I = \frac{P}{V} \\\\I = \frac{100}{220} \\\\I = 0.4545 \: A \\\\[/tex]

Therefore, 0.4545 A current is flowing through the light bulb.

According to the Ohm’s law, the voltage across the light bulb is given by

[tex]V = IR[/tex]

Where V is the voltage across the light bulb, I is the current flowing through the light bulb and R is the resistance of the light bulb.

Re-arranging the above equation for resistance R yields,

[tex]R = \frac{V}{I} \\\\R = \frac{220}{0.4545} \\\\R = 484 \: \Omega[/tex]

Therefore, the resistance of the bulb is 484 Ω

Answer:

bulb will burn out!

Explanation:

Answer:

A.  falling water

Explanation:

I got it right on Edgenuity. Good luck on your quiz.

In hydroelectric plants, water falls on turbine and makes it rotate. In generator, this mechanical energy transforms into electrical energy.

Hydroelectric power is generated by turbines that turn the potential energy of falling or swiftly flowing water into mechanical energy, which is then used to power generators. The most popular renewable energy source in the early 21st century was hydroelectricity, which in 2019 accounted for more than 18% of the world's total power producing capacity.

Water is gathered or stored at a higher elevation during the production of hydroelectric power and then transported through substantial pipes or tunnels (penstocks) to a lower elevation; the difference between these two elevations is referred to as the head. The falling water turns turbines as it nears the bottom of the pipelines. In turn, the turbines power generators, which transfer the mechanical energy of the turbines into electricity.

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Answer:

Check the explanation

Explanation:

Kindly check the attached image below to see the step by step explanation to the question above.

Answer:

Explanation:

frequency n = 90 x 10⁶ Hz .

time period T = 1 / n

= 1 / 90 x 10⁶

= 1.11 x 10⁻⁸ S.

wavelength = velocity of light / frequency

= 3 x 10⁸ / 90 x 10⁶

= 3.33 m

maximum value of the magnetic field. ( B₀ )

E₀ / B₀ = c where E₀ and  B₀ are maximum electric and magnetic field .

E₀ / c=  B₀

200/ 3 x 10⁸

= 66.67 x 10⁻⁸ T .

expressions in SI units for the space and time variations of the electric field

[tex]E=E_{0y}sin(2\pi nt - \frac{2\pi x}{\lambda} )[/tex]

[tex]E=200sin(180\times 10^6\pi t - \frac{2\pi x}{\lambda} )[/tex] N/C

[tex]B=B_{0z}sin(2\pi nt - \frac{2\pi x}{\lambda} )[/tex]

[tex]B=66.67\times 10^{-8}sin(180\times 10^6\pi t - \frac{2\pi x}{\lambda} )[/tex] T

Answer:

coefficient of friction (μ) and normal force (N)

Answer: How hard the surfaces push together and the types of surfaces involved

Explanation:

Answer:A dielectric material is a material through which virtually no current flows, assuming normally operating voltage

Explanation:

Dielectric materials are ,which means that no current will flow through them, when a voltage is applied

Answer:

Total internal reflection (TIR) is the phenomenon that involves the reflection of all the incident light off the boundary. TIR only takes place when both of the following two conditions are met: the light is in the more dense medium and approaching the less dense medium.

Explanation: Hope i helped!!!

Answer:

The electric and magnetic fields gradually increase.

Explanation:

Time constant in RL circuits, denoted by τ, is equal to the value of L / R which is the value of the inductor over the resistor. It is used to calculate the point when the current will reach the maximum value in the steady state of the circuit. Because of this behavior of the circuit, the magnetic field and the electric field gradually increase to their maximum values.

I hope this answer helps.

Answer:

26.92 N

Explanation:

The normal reaction of the ball is due to two force component acting on it.

However ; the acceleration is in polar coordinate which is given by the relation:

[tex]a^ { ^ \to} = (r- r \omega^2) \hat {e_r} + ( r \theta + 2 r \omega ) \hat {e_ \theta}[/tex]

[tex]a_{\theta} = r \theta + 2 r \omega[/tex]

Given that :

ω = 5 rad/s

mass m = 6 kg

θ = 30 ◦

r = 0.9 m

speed v =  0.4 m/s

[tex]a_{\theta} = 0 + 2(-0.4)*5[/tex]

[tex]a_{\theta}= -4 \ m/s[/tex]

The normal force reaction (N) that the arm applies to the ball at this instant is :

N = mg cos θ  + [tex]ma_{\theta}[/tex]

N = (6 × 9.8× cos 30) + (6 ×(-4))

N = 26.92 N

Answer:

Explanation:

Let v₁ and v₂ be velocities at lowest and topmost position . Let r be the radius of the circle .

Let N₁ and N₂ be the normal reaction force .

At the top position

centripetal force = N₂ + mg ;  so

N₂ + mg  = m v₂² / r

At the bottom  position

centripetal force = N₁ - mg ;  so

N₁ - mg  = m v₁² / r

subtracting these two equations

N₁ - mg - N₂ - mg = m v₁² / r  - m v₂² / r

N₁ - N₂ - 2mg = 1/r (m v₁²   - m v₂²  )

N₁ - N₂ - 2mg = 1/r x mg x 2r  ( loss of potential energy = gain of kinetic energy )

N₁ - N₂ =  2mg +  2mg

= 4 mg .

Answer:

The acceleration is  [tex]a = 2.5 \ m/s^2[/tex]

Explanation:

From the question we are told that

   The mass of the metal block is [tex]m_b = 2 \ kg[/tex]

    The mass flow rate of the water is  [tex]\r m = 1\ kg/s[/tex]

    The speed of the water of the water release is [tex]v_w = 5 m/s[/tex]

Generally according to the law of conservation of linear momentum

    [tex]p_i = p_f[/tex]

Now  [tex]p_i[/tex] is the initial momentum of the system which mathematically represented as

       [tex]p_i = m_w * v_w + m_b * v_b[/tex]

Now [tex]m_w[/tex] is the mass of water at the point of contact with the block which can be deduced as [tex]m_w = 1 \ kg[/tex]

Now since at initial the block is at rest

       [tex]v_b = 0 \ m/s[/tex]

So

      [tex]p_i = 1 * 5[/tex]

     [tex]p_i = 5 \ kgm/ s[/tex]

And  [tex]p_f[/tex] is the final  momentum of the system which mathematically represented as

     [tex]p_f = m_w * v__{fw} } + m_b * v__{fb}}[/tex]

So   [tex]v__{fw} }[/tex] is the final velocity of water which is zero due to the fact that when the water hits the block it losses its momentum and eventually the velocity becomes zero

    So

          [tex]5 = 2 * v__{fb }[/tex]

Thus  [tex]v__{fb }} = \frac{5}{2}[/tex]

        [tex]v__{fb }} = 2.5 \ m/s[/tex]

Thus  

       [tex]p_f = 2.5 * 2[/tex]

      [tex]p_f = 5 \ kgm /s[/tex]

Now the average momentum change is  

        [tex]p_a = \frac{p_i +p_f}{2}[/tex]

       [tex]p_a = \frac{5+5}{2}[/tex]

        [tex]p_a =5 kgm/s[/tex]

Now the force acting on the block is  

      [tex]F = \frac{p_a }{t}[/tex]

and from the question the initial movement of the block took 1s as it is a mass of water moving at a rate of 1kg/s that caused the first movement of the block

  So

        [tex]F= \frac{5}{1}[/tex]

       [tex]F= 5 \ N[/tex]

Now the acceleration is  

       [tex]a = \frac{F}{m_b}[/tex]

=>     [tex]a = \frac{5}{2}[/tex]

        [tex]a = 2.5 \ m/s^2[/tex]

Answer:

18

Explanation:

got it right on edge

Answer:

The center of mass move with the velocity of -3.487 m/s.

Explanation:

Given values of block A.

Mass of block A, (M1) = 4 kg

Speed of block A, (V1) = 2 m/s

Given values of block B.

 Mass of block B, (M2) = 8.4 kg

Speed of block B, (V2) = -6.1 m/s

Below is the formula to find the velocity of center of mass.

[tex]Velocity = \frac{M1V1 + M2V2}{M1 + M2} \\[/tex]

[tex]= \frac{4 \times 2 + 8.4 \times (-6.1) }{4 + 8.4} \\[/tex]

[tex]= \frac{- 43.24}{12.4}\\[/tex]

[tex]= - 3.487 m/s[/tex]

Answer:

The force needed is the weight of the rock minus the buoyant force.

Explanation:

Answer:

v_max = (1/6)e^-1 a

Explanation:

You have the following equation for the instantaneous speed of a particle:

[tex]v(t)=ate^{-6t}[/tex]   (1)

To find the expression for the maximum speed in terms of the acceleration "a", you first derivative v(t) respect to time t:

[tex]\frac{dv(t)}{dt}=\frac{d}{dt}[ate^{-6t}]=a[(1)e^{-6t}+t(e^{-6t}(-6))][/tex]  (2)

where you have use the derivative of a product.

Next, you equal the expression (2) to zero in order to calculate t:

[tex]a[(1)e^{-6t}-6te^{-6t}]=0\\\\1-6t=0\\\\t=\frac{1}{6}[/tex]

For t = 1/6 you obtain the maximum speed.

Then, you replace that value of t in the expression (1):

[tex]v_{max}=a(\frac{1}{6})e^{-6(\frac{1}{6})}=\frac{e^{-1}}{6}a[/tex]

hence, the maximum speed is v_max = ((1/6)e^-1)a

Answer:

The thickness of the door is 0.4230 m

Explanation:

Given;

mass of bullet, m = 0.009 kg

initial velocity of the bullet, u = 803 m/s

final velocity of the bullet, v = 617 m/s

average resistive force of the door on the bullet, F = 5620 N

Apply Newton's second law of motion;

Force exerted by the door on the bullet = Force of the moving bullet

F = ma

where;

F is applied force

m is mass

a is acceleration

Also, Force exerted by the door on the bullet = Force of the moving bullet

[tex]F =ma, \ But \ a =\frac{dv}{dt} = \frac{u-v}{t} \\\\F = \frac{m(u-v)}{t}[/tex]

where;

v is the final velocity of the bullet

u is initial velocity of the bullet

t is time

We need to calculate the time spent by the bullet before it passes through the door.

[tex]t = \frac{m(u-v)}{F} \\\\t = \frac{0.009(803-617)}{5620} = 0.0002979 \ s[/tex]

Distance traveled by the bullet within this time period = thickness of the door

This distance is equivalent to the product of average velocity and time

[tex]S = (\frac{u+v}{2}) t[/tex]

where;

s is the distance traveled

[tex]S = (\frac{u+v}{2}) t\\\\S = (\frac{803+617}{2}) 0.0002979\\\\S = 0.4230 \ m[/tex]

Therefore, the thickness of the door is 0.4230 m

71% of the earth is water

Answer:

hope this helps

Explanation:

71% of earth is water

Answer:

The taken is  [tex]t_A = 19.0 \ s[/tex]

Explanation:

Frm the question we are told that

  The speed of car A is  [tex]v_A = 22 \ m/s[/tex]

   The speed of car B is  [tex]v_B = 29.0 \ m/s[/tex]

     The distance of car B  from A is  [tex]d = 300 \ m[/tex]

     The acceleration of car A is  [tex]a_A = 2.40 \ m/s^2[/tex]

For A to overtake B

    The distance traveled by car B  =  The distance traveled by car A - 300m

Now the this distance traveled by car B before it is overtaken by A is  

          [tex]d = v_B * t_A[/tex]

Where [tex]t_B[/tex] is the time taken by car B

Now this can also be represented as using equation of motion as

      [tex]d = v_A t_A + \frac{1}{2}a_A t_A^2 - 300[/tex]

Now substituting values

       [tex]d = 22t_A + \frac{1}{2} (2.40)^2 t_A^2 - 300[/tex]

Equating the both d

       [tex]v_B * t_A = 22t_A + \frac{1}{2} (2.40)^2 t_A^2 - 300[/tex]

substituting values

   [tex]29 * t_A = 22t_A + \frac{1}{2} (2.40)^2 t_A^2 - 300[/tex]

   [tex]7 t_A = \frac{1}{2} (2.40)^2 t_A^2 - 300[/tex]

  [tex]7 t_A =1.2 t_A^2 - 300[/tex]

   [tex]1.2 t_A^2 - 7 t_A - 300 = 0[/tex]

Solving this using quadratic formula we have that

     [tex]t_A = 19.0 \ s[/tex]

Answer: A.

tracking training through a leaming records store LRS.

Explanation:

An LRS uses xAPI to collect learner data, or experiences, from both online and offline sources. These experiences are reported back to the LRS in the form of xAPI statements, where they are stored. These statements can then be retrieved for reporting and interpretation of the learner data.

Answer: fewer health problems

Explanation:

The benefits of learned optimism that have been found in research are Fewer health problems, Making more money, and a lower divorce rate. The correct option is B.

Learned optimism has been associated with numerous benefits in research, including fewer health problems, making more money, and a lower divorce rate. Optimistic people tend to have better physical and mental health, which leads to fewer health problems. Additionally, optimistic people tend to be more successful in their careers and finances, which can lead to higher income and better financial stability. Finally, optimistic people tend to have better relationships, including lower divorce rates, as they are better able to handle conflicts and maintain positive attitudes toward their partners.

In summary, learned optimism has a range of benefits for individuals, including better physical and mental health, greater success in work and education, better relationships with others, and improved resilience. These benefits make learned optimism an important skill for individuals to develop in order to lead happier, healthier, and more successful lives.

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Answer:

The correct answer is c

Explanation:

When a body is in rotational equilibrium, the relationship must be met

     ∑ τ = 0

     Σ F d = 0

therefore the forces that are exerted on the body that are represented in the center of gravity are different from zero, therefore, to maintain balance the distance must be zero, therefore the center of gravity must be above the leg where the body is balancing

The correct answer is c

Answer is Weaker. If it is talking about the objects' gravitational forces.

Answer:

(b) 0.012 m/s

Explanation:

Given that:

mass (m₁) of the drop of water = 0.30 g = 3.0 × 10⁻⁴ kg

mass (M₂) of the fish = 65 g =  65 × 10⁻³ kg

speed (v₁) of the water = 2.5 m/s

speed (v₂) of the archerfish = ??

By conservation of momentum

m₁v₁ - M₂v₂ = 0

m₁v₁  = M₂v₂

v₂ = m₁v₁ / M₂

v₂ =  ( 3.0 × 10⁻⁴ × 2.5 ) /  65 × 10⁻³

v₂ = 0.0115 m/s

v₂ ≅ 0.012 m/s

Therefore, the speed of the archerfish immediately after it expels the drop of water 0.012 m/s.

Answer: [tex]0.13m/s^2[/tex]

Explanation:

[tex]Formula: a=\frac{V_2-V_1}{t}[/tex]

Where;

a = acceleration

V2 = final velocity

V1 = initial velocity

t = time

If John runs 1.0 m/s first, we assume this is V1. He accelerates to 1.6 m/s; this is V2.

[tex]a=\frac{1.6m/s-1.0m/s}{4.5s}[/tex]

[tex]a=\frac{0.6m/s}{4.5s}[/tex]

[tex]a=0.13m/s^2[/tex]

Answer:.13

Explanation: