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Alona [7]
3 years ago
8

The tires on your truck have 0.35 m radius. In a straight line, you drive 2600 m. What is the angular displacement of the tire,

during this trip? Would the angular displacement be more or less if you put on big mud tires with a radius of .60m Mud tires have (more or less) angular displacemen
Physics
1 answer:
Travka [436]3 years ago
8 0
1) In a circular motion, the angular displacement \theta is given by
\theta =  \frac{S}{r}
where S is the arc length and r is the radius. The problem says that the truck drove for 2600 m, so this corresponds to the total arc length covered by the tire: S=2600 m. Using the information about the radius, r=0.35 m, we find the total angular displacement:
\theta =  \frac{2600 m}{0.35 m} =7428 rad

2) If we put larger tires, with radius r=0.60 m, the angular displacement will be smaller. We can see this by using the same formula. In fact, this time we have:
\theta =  \frac{2600 m}{0.60 m}=4333 rad
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Model rocket engines are sized by thrust, thrust duration, and total impulse, among other characteristics. A size C5 model rocke
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Answer:

v_{f} = 115.95 m / s

Explanation:

This is an exercise of a variable mass system, let's form a system formed by the masses of the rocket, the mass of the engines and the masses of the injected gases, in this case the system has a constant mass and can be solved using the conservation the amount of movement. Which can be described by the expressions

        Thrust = v_{e}  \frac{dM}{dt}

        v_{f}-v₀ = v_{e} ln ( \frac{M_{o} }{M_{f}} )

where v_{e} is the velocity of the gases relative to the rocket

let's apply these expressions to our case

the initial mass is the mass of the engines plus the mass of the fuel plus the kill of the rocket, let's work the system in SI units

       M₀ = 25.5 +12.7 + 54.5 = 92.7 g = 0.0927 kg

     

The final mass is the mass of the engines + the mass of the rocket

      M_{f} = 25.5 +54.5 = 80 g = 0.080 kg

thrust and duration of ignition are given

       thrust = 5.26 N

       t = 1.90 s

Let's start by calculating the velocity of the gases relative to the rocket, where we assume that the rate of consumption is linear

          thrust = v_{e} \frac{M_{f} - M_{o}  }{t_{f} - t_{o}  }

          v_{e} = thrust  \frac{\Delta t}{\Delta M}

          v_{e} = 5.26 \frac{1.90}{0.080 -0.0927}

          v_{e} = - 786.93 m / s

the negative sign indicates that the direction of the gases is opposite to the direction of the rocket

now we look for the final speed of the rocket, which as part of rest its initial speed is zero

            v_{f}-0 = v_{e} ln ( \frac{M_{o} }{M_{f} } )

we calculate

            v_{f} = 786.93 ln (0.0927 / 0.080)

            v_{f} = 115.95 m / s

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Read 2 more answers
A ray of light incident in water strikes the surface separating water from air making an angle of 10 ° with the normal to the su
labwork [276]

Answer:

a

 \theta _2  = 13^o

b

 \theta _1  =32.94^o

c

 \theta_c  =  53.05^o    

Explanation:

From the question we are told that

    The angle of incidence is  \theta_1 =  10^o

    The refractive index of water is  n_1 = 1.3

  Generally Snell's law is mathematically represented as

          n_1 sin(\theta_1) =  n_2 sin(\theta_ 2)

Here n_2 is the refractive index of air with value  n_2 =  1

         \theta_2  is the angle of refraction

So  

        \theta _2  =  sin^{-1}[\frac{n_1 * sin(\theta _1)}{n_2} ]

=>     \theta _2  =  sin^{-1}[\frac{1.3 * sin(10)}{1} ]

=>     \theta _2  = 13^o

Given that the angle should not be greater than \theta _2 =45^o  then the angle of incidence will be

       \theta _1  =  sin^{-1}[\frac{n_2 * sin(\theta _2)}{n_1} ]

=>     \theta _1  =  sin^{-1}[\frac{1 * sin(45)}{1.3} ]

=>     \theta _1  =32.94^o

Generally for critical angle is mathematically represented as

        \theta_c  =  sin^{-1}[\frac{n_2}{n_1} ]

=>     \theta_c  =  sin^{-1}[\frac{1}{1.3} ]  

=>     \theta_c  =  53.05^o            

4 0
3 years ago
How to Answer questions 12 and 13
zubka84 [21]

12) The weight of the plane is 7.35\cdot 10^5 N

13) The lift provided by the wing is 7.35\cdot 10^5 N

Explanation:

12)

The weight of an object is equal to the force of gravity acting on the object. Near the Earth's surface, it can be calculated as

W=mg

where

W is the weight

m is the mass of the object

g is the acceleration of gravity

In this problem, we know that the mass of the plane is

m = 75 tonnes

Since 1 ton = 1000 kg, the mass in kg is

m = 75,000 kg

Also, the acceleration of gravity is

g=9.8 m/s^2

Therefore, the weight of the plane is:

W=(75,000)(9.8)=7.35\cdot 10^5 N

13)

We can solve this question by applying Newton's second law along the vertical direction of motion. In fact, the net force acting along the vertical direction must be equal to the product between the mass of the plane and the vertical acceleration:

\sum F_y = ma_y

where

\sum F_y is the net force in the y-direction

m is the mass of the plane

a_y is the acceleration in the y-direction

There are two forces acting in the vertical direction:

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  • The weight W, acting  downward

So the vertical net force is

\sum F_y = L-W

And the equation becomes

L-W = ma_y

Also, we know that the plane is travelling in level flight, which means that the vertical acceleration is zero:

a_y = 0

Therefore, we get

L-W=0

And so the lift is

L=W=7.35\cdot 10^5 N

Learn more about  forces:

brainly.com/question/8459017

brainly.com/question/11292757

brainly.com/question/12978926

#LearnwithBrainly

4 0
3 years ago
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