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givi [52]
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/s².
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?
Physics
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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nevsk [136]

Average speed is defined as total distance moved in total interval of time

so it is given as

v_{avg} = \frac{distance}{time}

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then we will have mathematical expression as follows

v = \frac{d}{t}

5 0
3 years ago
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
slava [35]

Answer:

The answer to the question is

The distance d, which locates the point where the light strikes the bottom is   29.345 m from the spotlight.

Explanation:

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

The product of the incident index and the sine of the angle of incident is equal to the product of the refractive index and the sine of the angle of refraction

n₁sinθ₁ = n₂sinθ₂

y = 2.2 m and strikes at x = 8.5 m, therefore tanθ₁ = 2.2/8.5 = 0.259 and

θ₁ =  14.511 °

n₁ = 1.0003 = refractive index of air

n₂ = 1.33 = refractive index of water

Therefore sinθ₂ =  \frac{n_1sin\theta_1}{n_2}  = \frac{1.003*0.251}{1.33} = 0.1885 and θ₂ = 10.86 °

Since the water depth is 4.0 m we have tanθ₂ = \frac{4}{x_2} or x₂ = \frac{4}{tan\theta_2 } =\frac{4}{tan(10.86)} = 20.845 m

d = x₂ + 8.5 = 20.845 m + 8.5 m = 29.345 m.

5 0
3 years ago
Which change will always result in an increase in gravitational force between two objects
grin007 [14]
Increasing the masses of the objects and decreasing the distance between the objects
7 0
3 years ago
Read 2 more answers
(a) the gamma rays produced by a radioactive nuclide used in medical imaging (b) radiation from an FM radio station at 93.1 MHz
lawyer [7]

Answer:

They can be rank in the following 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

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:

In the case of the radio waves can be used:

Case for \nu = 93.1 MHz:

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

\lambda = \frac{3x10^{8}m/s}{93100000s^{-1}}

\lambda = 3.22m

Case for \nu = 680 kHz:

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

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

\lambda = 441.17m

7 0
3 years ago
To start a lawn mower, you must pull on a rope wound around theperimeter of a flywheel. After you pull the rope for 0.95 s, thef
Molodets [167]

Answer:

29.76245 rad/s², -117.80972 rad/s²

28.2743 rad/s

3.95833

Explanation:

\omega_f = Final angular velocity

\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

Angular acceleration during speed up is 29.76245 rad/s²

\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\frac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\frac{0-4.5\times 2\pi}{0.24}\\\Rightarrow \alpha=-117.80972\ rad/s^2

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

3 0
3 years ago
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