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3 years ago

PLEASE HELPPPP! a safe driving skill for a driver and a Motorcyclist

1 answer:

8 0

An answer that fits a “driver” (assuming this means of a driver other than a
motorcyclist) and a motorcyclist is: safe following distance.
Mathematically you should allow one car length for every 10mph of movement to stop safely.
For example if a vehicle you are following is traveling at 50mph and you are following behind them, you should be a minimum of 5 car lengths away from their rear bumper. If going 70mph, it’s 7 car lengths, and so on.

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A dc shunt motor rated at 240 V has a field winding resistance of 120 Ω and an armature resistance of 0.12 Ω. The motor is suppl

patriot [66]

Answer:

a) see attached pic

b) No-load speed: 1368 rpm

c) IL at 1% below no-load speed = 22.4 A

Explanation:

(See attached pic)

b) The No- Load Speed:

Since, Ea = KФn [ Ea=Internal Generated Voltage, k=Machine constant,Ф=flux, n=speed]

As the Vt (terminal voltage) and field resistance (Rf) is constant, If (field current is constant. Assuming negligible armature reaction, the following relationship can be written,

Ea2 / Ea1 = K Ф n2 / K Ф n1

Since K and flux is contant,

Ea2 = (n2 / n1) x Ea1

Or, n2 = (Ea2 / Ea1) x n1 -------- (1)

Since, Ea = Vt - IaRa

But at no load, Ia = 0, therefore,

Ea = Vt

At load condition, IL = 29 A

Therefore,

Ia = IL - Vt/Rf

Ia = 29 - 240/120

Ia = 27 A

Therefore, Ea at this load will be,

Ea = Vt - IaRa

Ea = 240 - (27)(0.12)

Ea = 236.76 V

The No-Load speed of motor can be calculated using equation (1),

n2 = (Ea2 /Ea1 ) x n1

1350 = (236.76/240) x n1

n1 = 1368 rpm = No-load Speed

Now, if the speed is 1% below the no-load speed, which is,

n1 = no load speed = 1368 rpm

n2 = 1% less speed = 1354 rpm [ 1368 - (0.01)(1368) ]

Using equation (1),

n2 = (Ea2/Ea1) x n1

1354 = (Ea2/240) x 1368

Ea2 = 237.5 V

Since, Ea = Vt - IaRa

Ia = (Vt - Ea)/ Ra

Ia = (240-237.5)/(0.12)

Ia = 20.4 A

Now,

IL = Ia + If

IL = 20.4 + 240/120

When, Speed = 1368 rpm, IL = 4.75 A

When , Speed = 1354 rpm, IL = 22.4 A

When, Speed = 1350 rpm., IL = 29 A

<em>(See attached pic) </em>

7 0

4 years ago

two cars travel on a straight road from the point. A to point B both cars accelerate to their maximum speed and then continue at

Lady bird [3.3K]

Answer:

Part A

The acceleration of car 2, is larger than the acceleration of car 1

Part B

The average speed of car 2 is larger than the average speed of car 1

Explanation:

The question relates to kinematic motion

Using a similar question for the parameters, we have;

The time it takes car 1 to accelerate to v₁ = 20 m/s is t₁ = 30 s

The time it takes car 1 to reach point B, t₁₂ = 35 s

The time it takes car 2 to accelerate to v₂ = 20 m/s is t₂ = 20 s

The time it takes car 2 to reach point B is t₂₂ = 30 s

Part A

From the kinematic equation of motion, v = u + at, we have;

Acceleration, a = (v - u)/t

Where;

a = The acceleration

v = The final velocity

u = The initial velocity = 0 m/s for both cars as they start from rest

t = The time taken

The acceleration of car 1, a₁ = (v₁ - u₁)/t₁

∴ a₁ = (20 m/s - 0 m/s)/(30 s) = 2/3 m/s²

The acceleration of car 2, a₂ = (v₂ - u₂)/t₂

∴ a₂ = (20 m/s - 0 m/s)/(20 s) = 1 m/s²

The acceleration of car 2, a₂ = 1 m/s² is larger than the acceleration of car 1, a₁ = 2/3 m/s²

Part B

Average speed = Total distance/(Total time taken to reach the distance)

Let 'B' represent the distance to point in meters, we have;

The average speed of car 1 = B/(The time it takes car 1 to reach point B)

∴ The average speed of car 1 = B/35 m/s

The average speed of car 2 = B/(The time it takes car 2 to reach point B)

∴ The average speed of car 2 = B/30 m/s

Noting that the larger velocity is given by the smaller divisor to 'B', we have;

The average speed of car 2 = B/30 m/s is larger than the average speed of car 1 = B/35 m/s

6 0

3 years ago

The small washer is sliding down the cord OA. When it is at the midpoint, its speed is 28 m/s and its acceleration is 7 m/s 2 .

Neporo4naja [7]

Answer:

Velocity components

$V_r = -16.28 m/s$

$V_z = -22.8 m/s$

$V_q = 0 m/s$

For Acceleration components;

$a_r = -4.07m/s^2$

$a_z = -5.70m/s^2$

$a_q = 0m/s^2$

Explanation:

We are given:

$Speed v_o = 28 m/s$

$Acceleration a_o= 7 m/s^2$

We first need to find the radial position r of washer in x-y plane.

Therefore

$r = \sqrt{300^2 + 400^2}$

r = 500 mm

To find length along direction OA we have:

$L = \sqrt{500^2 + 700^2}L = 860 mm$

Therefore, the radial and vertical components of velocity will be given as:

$V_r = V_o*cos(Q)$

$V_z = V_o*sin(Q)$

Where Q is the angle between OA and vector r.

Therefore,

$V_r = 28 * \frac{r}{L} = > 28 * \frac{500}{860}$

$V_r = -16.28 m/s$

• $V_z = 28 * \frac{700}{860} = -22.8$

• $V_q = 0 m/s$

The radial and vertical components of acceleration will be:

$a_r = a_o*cos(Q)$

$a_z = a_o*sin(Q)$

Therefore we have:

• $a_r = 7* \frac{500}{860} = -4.07m/s^2$

• $a_z = 7 * \frac{700}{860} = -5.70 m/s^2$

• $a_q = 0 m/s^2$

Note : image is missing, so I attached it

3 0

4 years ago

A life cycle assessment (LCA) determines the environmental impact at all stages of a product's life cycle, including production,

lisabon 2012 [21]

Confusing very very confusing

4 0

3 years ago

Yield and tensile strengths and modulus of elasticity . with increasing temperature. (increase/decrease/independent)

meriva

Answer:

Yield strength, tensile strength decreases with increasing temperature and modulus of elasticity decreases with increasing in temperature.

Explanation:

The modulus of elasticity of a material is theoretically a function of the shape of curve plotted between the potential energy stored in the material as it is loaded versus the inter atomic distance in the material. The temperature distrots the molecular structure of the metal and hence it has an effect on the modulus of elasticity of a material.

Mathematically we can write,

$E(t)=E_o[1-a\frac{T}{T_m}]$

where,

E(t) is the modulus of elasticity at any temperature 'T'

$E_o$ is the modulus of elasticity at absolute zero.

$T_{m}$ is the mean melting point of the material

Hence we can see that with increasing temperature modulus of elasticity decreases.

In the case of yield strength and the tensile strength as we know that heating causes softening of a material thus we can physically conclude that in general the strength of the material decreases at elevated temperatures.

6 0

3 years ago