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VMariaS [17]
3 years ago
15

An automobile with a standard differential turns sharply to the left. The left driving wheel turns on a 20-m radius. Distance be

tween right and left wheels is 1.5 m. What are the rotating speeds of each driving wheel as fractions of the drive shaft speed?
Physics
1 answer:
Inessa05 [86]3 years ago
5 0

Explanation:

The given data is as follows.

    Inner wheel Radius = 20 m,

   Distance between left and right wheel = 1.5m,

Let us assume speed of drive shaft is N rpm.

Formula to calculate angular velocity is as follows.

    Angular velocity of automobile = w = \frac{V}{R}

where,   V = linear velocity of automobile m/min,

              R = turning radius from automobile center in meter

In the given case, angular velocity remains same for inner and outer wheel but there is change in linear velocity of inner wheel and outer wheel.

Now, we assume that

         u = linear velocity of inner wheel

and,   u' = linear velocity of outer wheel.

Formula for angular velocity of inner wheel w = ,

Formula for angular velocity of outer wheel w =

Now, for inner wheels

                   w =

                      = \frac{u}{(R - d)}

                  u = V \times \frac{(R - d)}{R}

                    = V \times (1 - \frac{d}{R})

If radius of wheel is r it will cover  distance in one min.

Since, velocity of wheel is u it will cover distance u in unit time(min)

Thus,             u = 2\pi rn = V \times (1 - \frac{d}{R})

Now, rotation per minute of inner wheel is calculated as follows.

         n = \frac{V}{2 \pi r \times (1 - \frac{d}{R})}

            = \frac{V}{2 \pi r \times (1 - \frac{0.75}{20})} (since 2d = 1.5m given, d = 0.75m),

             = \frac{V}{r} \times 0.1532

So, rotation per minute of outer wheel; n' =  

                   = \frac{V}{2 \pi r \times (1 + \frac{0.75}{20})}

                   = \frac{V}{r} \times 0.1651

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raketka [301]

The impulse experienced is -18,000 kg m/s

Explanation:

The impulse exerted on an object is equal to the change in momentum of the object. Mathematically:

I=\Delta p = m(v-u)

where

m is the mass of the object

v is the final velocity of the object

u is the initial velocity

\Delta p is the change in momentum

I is the impulse

In the collision in this problem,

m = 1300 kg is the mass of the car

u = 11 m/s is the initial velocity

v = -2.5 m/s is the final velocity (negative, since it is in the opposite direction)

Substituting, we find

I=(1300)(-2.5-11)=-17,550 kg m/s

So the closest choice is

-18,000 kg m/s

Learn more about impulse and change in momentum:

brainly.com/question/9484203

#LearnwithBrainly

7 0
3 years ago
2. A person lifts 200kg seven times over the course of 11.8s. If they displaced the weight 2.2m up each time, how much power did
Aneli [31]

Answer:

<em>The person delivered a power of 2,558 Watt</em>

Explanation:

<u>Work and Power</u>

Mechanical work is the amount of energy transferred by a force. It's a scalar quantity, with SI units of joules.

Being  the force vector and  the displacement vector, the work is calculated as:

W=\vec F\cdot \vec s

If both the force and displacement are parallel, then we can use the equivalent scalar formula:

W=F.s

Power is the amount of energy transferred per unit of time. In the SI, the unit of power is the watt, equivalent to one joule per second.

The power can be calculated as:

\displaystyle P=\frac {W}{t}

Where W is the work and t is the time.

If the person lifts a mass of m=200 Kg, then exerts a force equal to its weight:

F = m.g = 200*9.8 = 1,960

F = 1,960 N

The work done when lifting the weight 7 times by a distance of s=2.2 m is:

W = 7*1,960*2.2=30,184

W = 30,184 J

Finally, the power delivered in t=11.8 seconds is:

\displaystyle P=\frac {30,184}{11.8}

P = 2,558 Watt

The person delivered a power of 2,558 Watt

4 0
3 years ago
Determine the magnitude and direction of the force on an electron traveling 3.58 E 6 m/s horizontally to the west in a verticall
olya-2409 [2.1K]

Answer:banana

Explanation:

Because the amount of potassium is that

4 0
3 years ago
A mass m = 75 kg slides on a frictionless track that has a drop, followed by a loop-the-loop with radius R = 19.2 m and finally
dalvyx [7]

Answer:

The velocity is 19.39 m/s

Solution:

As per the question:

Mass, m = 75 kg

Radius, R = 19.2 m

Now,

When the mass is at the top position in the loop, then the necessary centrifugal force is to keep the mass on the path is provided by the gravitational force acting downwards.

F_{C} = F_{G}

\frac{mv^{2}}{R} = mg

where

v = velocity

g = acceleration due to gravity

v = \sqrt{2gR} = \sqrt{2\times 9.8\times 19.2} = 19.39\ m/s

4 0
3 years ago
How much energy is needed to generate 0.71 x 10-16 kg of mass?
AleksAgata [21]

Answer:

6.39 J of energy is needed to generate 0.71 * 10⁻¹⁶ kg mass

Explanation:

According to the Equation: E = mc²

where the mass, m = 0.71 * 10⁻¹⁶ kg

the speed of light, c = 3 * 10⁸ m/s

The amount of energy needed to generate a mass of 0.71 * 10⁻¹⁶ kg is calculated as follows:

E = (0.71 * 10⁻¹⁶) (3 * 10⁸)²

E = 0.71 * 10⁻¹⁶ * 9 * 10¹⁶

E = 0.71 * 9

E = 6.39 J

6 0
3 years ago
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