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sweet [91]
3 years ago
11

If the coefficient of kinetic friction between tires and dry pavement is 0.98, what is the shortest distance in which you can st

op an automobile by locking the brakes when traveling at 34.7 m/s
Physics
1 answer:
Ilia_Sergeevich [38]3 years ago
7 0

Answer:

The shortest distance is 62.7 m

Explanation:

Given;

coefficient of kinetic friction, μk = 0.98

initial velocity, u = 34.7 m/s

Frictional force on the tire;

Fk = -μkN

where;

N is normal reaction = mg

ma = -μkN

ma = -μkmg

a = -μkg

a = - 0.98 x 9.8 = -9.604 m/s²

The shortest distance in which you can stop an automobile by locking the brakes:

Apply equation of motion;

v² = u² + 2ax

where;

v is the final velocity

u is the initial velocity

a is the acceleration of the  automobile

0 = 34.7² + 2(-9.604)x

0 = 1204.09 - 19.208x

19.208x = 1204.09

x = 1204.09/19.208

x = 62.7 m

Therefore, the shortest distance in which you can stop an automobile by locking the brakes when traveling at 34.7 m/s is 62.7 m

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Air at the poles tends to flow close to the surface toward the equator. What can you conclude about the characteristics of this
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Answer:

That the polar air has has more pressure than the air at the equator.

Explanation:

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3 years ago
The moon has a diameter of 3.48 x 106 m and is a distance of 3.85 x 108 m from the earth. The sun has a diameter of 1.39 x 109 m
Mrrafil [7]

Answer:

0.00903 rad

0.00926 rad

6.268\times 10^{-6}

Explanation:

s = Diameter of the object

r = Distance between the Earth and the object

Angle subtended is given by

\theta=\frac{s}{r}

For the Moon

\theta_m=\dfrac{3.48\times 10^6}{3.85\times 10^8}\\\Rightarrow \theta_m=0.00903\ rad

The angle subtended by the Moon is 0.00903 rad

For the Sun

\theta_s=\dfrac{1.39\times 10^9}{1.5\times 10^{11}}\\\Rightarrow \theta_s=0.00926\ rad

The angle subtended by the Sun is 0.00926 rad

Area ratio is given by

\frac{A_m}{A_s}=\dfrac{\pi r_m^2}{\pi r_s^2}\\\Rightarrow \frac{A_m}{A_s}=\dfrac{d_m^2}{d_s^2}\\\Rightarrow \frac{A_m}{A_s}=\dfrac{(3.48\times 10^{6})^2}{(1.39\times 10^9)^2}\\\Rightarrow \frac{A_m}{A_s}=6.268\times 10^{-6}

The area ratio is 6.268\times 10^{-6}

3 0
3 years ago
(a) Calculate the absolute pressure at the bottom of a freshwater lake at a point whose depth is 30.0 m. Assume the density of t
Law Incorporation [45]

Answer:

(a) The absolute pressure at the bottom of the freshwater lake is 395.3 kPa

(b) The force exerted by the water on the window is 36101.5 N

Explanation:

(a)

The absolute pressure is given by the formula

P = P_{o} + \rho gh

Where P is the absolute pressure

P_{o} is the atmospheric pressure

\rho is the density

g is the acceleration due to gravity (Take g = 9.8 m/s^{2} )

h is the height

From the question

h = 30.0 m

\rho = 1.00 × 10³ kg/m³ = 1000 kg/m³

P_{o} = 101.3 kPa = 101300 Pa

Using the formula

P = P_{o} + \rho gh

P = 101300 + (1000×9.8×30.0)

P = 101300 + 294000

P =395300 Pa

P = 395.3 kPa

Hence, the absolute pressure at the bottom of the freshwater lake is 395.3 kPa

(b)

For the force exerted

From

P = F/A

Where P is the pressure

F is the force

and A is the area

Then, F = P × A

Here, The area will be area of the window of the underwater vehicle.

Diameter of the circular window = 34.1 cm = 0.341 m

From Area = πD²/4

Then, A = π×(0.341)²/4 = 0.0913269 m²

Now,

From F = P × A

F = 395300 × 0.0913269

F = 36101.5 N

Hence, the force exerted by the water on the window is 36101.5 N

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Anyone know a GOOD show on Netflix please kid shows pleaseeeee
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