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ArbitrLikvidat [17]

3 years ago

6

A policeman investigating an accident measures the skid marks left by a car on the horizontal road. He determines that the dista

nce between the point that the driver slammed on the brakes (thereby locking the wheels) and the point where the car came to a stop was 28.0 m. From a reference manual he determines that the coefficient of kinetic friction between the tires and the road under the prevailing conditions was 0.300. How fast was the car going when the driver applied the brakes

1 answer:

weeeeeb [17]3 years ago

8 0

Answer:

The car was 12.8m/s fast when the driver applied the brakes.

Explanation:

The equations of motion of the car in the horizontal and vertical axes are:

$x: f_k=ma\\\\y: N-mg=0$

Since the kinetic friction is defined as $f_k=\mu_kN$ and $N=mg$ we have:

$\mu_kmg=ma\\\\a=\mu_kg$

Next, from the kinematics equation of speed in terms of distance, we have:

$v^2=v_0^2-2ax\\\\v^2=v_0^2-2\mu_kgx$

Since the car came to a stop, the final velocity $v$ is zero, and we get:

$0=v_0^2-2\mu_kgx\\\\v_0=\sqrt{2\mu_kgx}$

Finally, plugging in the known values, we obtain:

$v_0=\sqrt{2(0.300)(9.81m/s^2)(28.0m)}\\\\v_0=12.8m/s$

It means that the car was 12.8m/s fast when the driver applied the brakes.

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One of the harmonics on a string 1.30m long has a frequency of 15.60 Hz. The next higher harmonic has a frequency of 23.40 Hz. F

Alja [10]

Answer:

$\large \boxed{\text{(a) 7.800 Hz; (b) 20.3 m/s; 40.6 m/s; 60.8 m/s}}$

Explanation:

a) Fundamental frequency

A harmonic is an integral multiple of the fundamental frequency.

$\dfrac{\text{23.40 Hz}}{\text{15.60 Hz}} = \dfrac{1.500}{1} \approx \dfrac{3}{2}$

$f = \dfrac{\text{24.30 Hz}}{3} = \textbf{7.800 Hz}$

b) Wave speed

(i) Calculate the wavelength

In a fundamental vibration, the length of the string is half the wavelength.

$\begin{array}{rcl}L & = & \dfrac{\lambda}{2}\\\\\text{1.30 m} & = & \dfrac{\lambda}{2}\\\\\lambda & = & \text{2.60 m}\\\end{array}$

(b) Calculate the speed s

$\begin{array}{rcl}v_{1}& = & f_{1}\lambda\\& = & \text{7.800 s}^{-1} \times \text{2.60 m}\\& = & \textbf{20.3 m/s}\\\end{array}$

$\begin{array}{rcl}v_{2}& = & f_{2}\lambda\\& = & \text{15.60 s}^{-1} \times \text{2.60 m}\\& = & \textbf{40.6 m/s}\\\end{array}$

$\begin{array}{rcl}v_{3}& = & f_{3}\lambda\\& = & \text{23.40 s}^{-1} \times \text{2.60 m}\\& = & \textbf{60.8 m/s}\\\end{array}$

4 0

3 years ago

While standing on a balcony a child drops a penny. The penny lands on the ground floor 1.5 s later. How fast was the penny trave

sveticcg [70]

Answer:

14.7 m/s.

Explanation:

From the question given above, the following data were obtained:

Time (t) = 1.5 s

Acceleration due to gravity (g) = 9.8 m/s².

Height = 11.025 m

Final velocity (v) = 0 m/s

Initial velocity (u) =?

We, can obtain the initial velocity of the penny as follow:

H = ½(v + u) t

11.025 = ½ (0 + u) × 1.5

11.025 = ½ × u × 1.5

11.025 = u × 0.75

Divide both side by 0.75

u = 11.025/0.75

u = 14.7 m/s

Therefore, the penny was travelling at 14.7 m/s before hitting the ground.

8 0

3 years ago

Calculate the maximum capillary rise/fall of mercury in a 0.5 mm radius glass capillary. Assume that the surface tension for mer

tekilochka [14]

Answer: 0.01 m

Explanation: The formulae for capillarity rise or fall is given below as

h = (2T×cosθ)/rpg

Where θ = angle mercury made with glass = 50°

T = surface tension = 0.51 N/m

g = acceleration due gravity = 9.8 m/s²

r = radius of tube = 0.5mm = 0.0005m

p = density of mercury.

h = height of rise or fall

From the question, specific gravity of density = 13.3

Where specific gravity = density of mercury/ density of water, where density of water = 1000 kg/m³

Hence density of mercury = 13.3×1000 = 13,300 kg/m³.

By substituting parameters, we have that

h = 2×0.51×cos 50/0.0005×9.8×13,300

h = 0.6556/65.17

h = 0.01 m

8 0

3 years ago

Explain why secondary succession usually occurs more rapidly than primary succession.

kiruha [24]

Answer:

Explanation:

As it mostly occurs on a surface where an ecosystem has previously existed. Primary succession only occurs on a surface where no ecosystem existed ever before!

6 0

3 years ago

The time period T of a simple pendulum is given by the relation

Vanyuwa [196]

Answer:

$T^2 \propto L$

Explanation:

The period of a simple pendulum is given by:

$T=2\pi \sqrt{\frac{L}{g}}$

where

L is the length of the pendulum

g is the acceleration of gravity

From this equation we can write

$T\propto \sqrt{L}\\T\propto \frac{1}{\sqrt{g}}$

Taking the square of this equation, we get:

$T^2 = (2\pi)^2 \frac{L}{g}$

So we see that $T^2$ is proportional to L and inversely proportional to g. So, we can write:

$T^2 \propto L\\T^2 \propto \frac{1}{g}$

So the only correct option is

$T^2 \propto L$

5 0

3 years ago

Read 2 more answers