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3 years ago

What affects the braking distance and thinking distance

2 answers:

8 0

Your speed is one of the only factors that has an effect on both your thinking distance and braking distance. Put simply, the faster you are going, the greater the distance travelled before you apply the brakes (thinking distance) and the vehicle comes to a complete stop (braking distance).

hodyreva [135]3 years ago

8 0

Several many things do.

Thinking distance: speed, driver's alertness or fatigue, light or dark outside, distractions inside the car, distractions outside the car, driver on the phone

Braking distance: speed, size and weight of the car or truck, condition of the tires, the road surface: wet or dry, concrete or tar, dirt or gravel

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A 60-m-long chain hangs vertically from a cylinder attached to a winch. Assume there is no friction in the system and that the

Mariulka [41]

Answer:

part (a). 176580 J

part (b). 197381 J

Explanation:

Given,

Density of the chain = $\rho\ =\ 10\ kg/m.$
Length of the chain = L = 60 m
Acceleration due to gravity = g = 9.81 $m/s^2$

part (a)

Let dy be the small element of the chain at a distance of 'y' from the ground.

mass of the small element of the chain = $\rho dy$

Work done due to the small element,

$dw\ =\ \rho g (60\ -\ y)dy\\$

Total work done to wind the entire chain = w

$w\ =\ \displaystyle\int_{0}^{L} \rho g(60\ -\ y)dy\\\Rightarrow w\ =\ \rho g\left |(60y\ -\ \dfrac{y^2}{2})\ \right |_{0}^{60}\\\Rightarrow w\ =\ 10\times 9.81\times (60\times 60\ -\ \dfrac{60^2}{2})\\\Rightarrow w\ =\ 176580\ J$

part (b)

mass of the block connected to the chain = m = 35 kg

Total work done to wind the chain = work done due to the chain + work done due to the mass

$\therefore W\ =\ w\ +\ mgL\\\Rightarrow W\ =\ 176580\ +\ 35\times 9.81\times 60\\\Rightarrow W\ =\ 176580\ +\ 20601\\\Rightarrow W\ =\ 197381\ J$

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2 years ago

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BaLLatris [955]

The part of the electromagnetic spectrum has a shorter wavelength than ultraviolet light is x-rays.

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3 years ago

Read 2 more answers

A 10-cm-long spring is attached to theceiling. When a 2.0 kg mass is hung from it,the spring stretches to a length of 15 cm.a.Wh

alekssr [168]

(a) 392 N/m

Hook's law states that:

$F=k\Delta x$ (1)

where

F is the force exerted on the spring

k is the spring constant

$\Delta x$ is the stretching/compression of the spring

In this problem:

- The force exerted on the spring is equal to the weight of the block attached to the spring:

$F=mg=(2.0 kg)(9.8 m/s^2)=19.6 N$

- The stretching of the spring is

$\Delta x=15 cm-10 cm=5 cm=0.05 m$

Solving eq.(1) for k, we find the spring constant:

$k=\frac{F}{\Delta x}=\frac{19.6 N}{0.05 m}=392 N/m$

(b) 17.5 cm

If a block of m = 3.0 kg is attached to the spring, the new force applied is

$F=mg=(3.0 kg)(9.8 m/s^2)=29.4 N$

And so, the stretch of the spring is

$\Delta x=\frac{F}{k}=\frac{29.4 N}{392 N/m}=0.075 m=7.5 cm$

And since the initial lenght of the spring is

$x_0 = 10 cm$

The final length will be

$x_f = x_0 +\Delta x=10 cm+7.5 cm=17.5 cm$

8 0

3 years ago

Read 2 more answers

Hydrogen has only one electron in its only (and outer) electron shell. If a hydrogen atom were to absorb a small amount of energ

Xelga [282]

Answer:

D. move up to another shell that would form

Explanation:

An atom has protons, neutrons and electrons. Protons and neutrons are present in the nucleus and electrons orbit the nucleus in fixed shells. An electron can jump to higher shell when it gains energy and lower one when it loses energy. Thus, when single electron in hydrogen atom is given a small amount of energy, it would jump to another higher shell.

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3 years ago

Evaluate x and y in the equation: E=Cm^xV^y , where E is kinetic energy , m is mass , V is velocity and C is a dimension less co

Korvikt [17]

Answer: Do I look like Einstein

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2 years ago