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3 years ago

11

Two cars are traveling along a straight line in the same direction, the lead car at 25 m/s and the other car at 35 m/s. At the m

oment the cars are 45 m apart, the lead driver applies the brakes, causing the car to have an acceleration of â2.0 m/sÂ².
a. How long does it take for the lead car to stop?
b. Assume that the driver of the chasing car applies the brakes at the same time as the driver of the lead car. What must the chasing carâs minimum negative acceleration be to avoid hitting the lead car?
c. How long does it take the chasing car to stop?

1 answer:

Phoenix [80]3 years ago

7 0

Answer:

a. $t_1=12.5\ s$

b. $a_2=-13.61\ m.s^{-2}$ must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping

c. $t_2=2.5714\ s$ is the time taken to stop after braking

Explanation:

Given:

speed of leading car, $u_1=25\ m.s^{-1}$

speed of lagging car, $u_{2}=35\ m.s^{-1}$

distance between the cars, $\Delta s=45\ m$

deceleration of the leading car after braking, $a_1=-2\ m.s^{-2}$

a.

Time taken by the car to stop:

$v_1=u_1+a_1.t_1$

where:

$v_1=0$ , final velocity after braking

$t_1=$ time taken

$0=25-2\times t_1$

$t_1=12.5\ s$

b.

using the eq. of motion for the given condition:

$v_2^2=u_2^2+2.a_2.\Delta s$

where:

$v_2=$ final velocity of the chasing car after braking = 0

$a_2=$ acceleration of the chasing car after braking

$0^2=35^2+2\times a_2\times 45$

$a_2=-13.61\ m.s^{-2}$ must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping

c.

time taken by the chasing car to stop:

$v_2=u_2+a_2.t_2$

$0=35-13.61\times t_2$

$t_2=2.5714\ s$ is the time taken to stop after braking

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A spotlight on a boat is y = 2.2 m above the water, and the light strikes the water at a point that is x = 8.5 m horizontally di

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Answer:

The answer to the question is

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Explanation:

To solve the question we note that Snell's law states that

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Answer:

They can be rank in the following way:

A radio signal from an AM radio station at 680 kHz on the dial

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The red light of a light-emitting diode, such as in a calculator

The yellow light from sodium vapor streetlights

The gamma rays produced by a radioactive nuclide used in medical

Explanation:

The electromagnetic spectrum is the distribution of radiation due to the different frequencies at which it radiates and its different intensities, that radiation is formed by electromagnetic waves, which are transverse waves formed by an electric field and a magnetic field perpendicular to it.

Radiation is distributed along that electromagnetic spectrum according to the wavelength or frequency.

Highest frequencies

X-rays

Ultraviolet rays

Visible region

Lower frequencies

Infrared

Microwave

Radio waves

Radio waves and the visible region (yellow light, red light) are part of the electromagnetic spectrum, any radiation of that electromagnetic spectrum has a speed of 3.00x10^{8}m/s in vacuum.

However, the following equation relates the velocity, the frequency, and the wavelength:

$c = \nu \cdot \lambda$ (1)

$\nu = \frac{c}{\lambda}$ (2)

It can be see in equation 2 that the frequency and the wavelength are inversely proportional (when the frequency increases the wavelength decreases).

Therefore, for what was already discussed, they can be rank in the next way:

A radio signal from an AM radio station at 680 kHz on the dial

Radiation from an FM radio station at 93.1 MHz on the dial

The red light of a light-emitting diode, such as in a calculator

The yellow light from sodium vapor streetlights

The gamma rays produced by a radioactive nuclide used in medical

Summary:

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$\lambda = \frac{c}{\nu}$

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$\lambda = \frac{c}{\nu}$

$\lambda = \frac{3x10^{8}m/s}{680000s^{-1}}$

$\lambda = 441.17m$

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Answer:

29.76245 rad/s², -117.80972 rad/s²

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3.95833

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$\omega_i$ = Initial angular velocity

$\alpha$ = Angular acceleration

$\theta$ = Angle of rotation

t = Time taken

Equation of rotational motion

$\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\frac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\frac{4.5\times 2\pi-0}{0.95}\\\Rightarrow \alpha=29.76245\ rad/s^2$

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Angular acceleration during spin down is -117.80972 rad/s²

Angular speed is given by

$\omega=2\pi 4.5=28.2743\ rad/s$

Maximum angular speed reached by the flywheel is 28.2743 rad/s

$\theta=\omega_it+\frac{1}{2}\alpha t^2\\\Rightarrow \theta=0\times t+\frac{1}{2}\times 29.76245\times 0.95^2\\\Rightarrow \theta=13.4303\ rad$

$\theta=\omega_it+\frac{1}{2}\alpha t^2\\\Rightarrow \theta=2\pi 4.5\times 0.24+\frac{1}{2}\times -117.80972\times 0.24^2\\\Rightarrow \theta=3.39292\ rad$

The ratio would be $\dfrac{13.4303}{3.39292}=3.95833$

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3 years ago