What is weightlessness ?

To test the performance of its tires, a car

velikii [3]

Answer:

The coefficient of static friction between the tires and the road is 1.987

Explanation:

Given:

Radius of the track, r = 516 m

Tangential Acceleration $a_r$ = 3.89 m/s^2

Speed,v = 32.8 m/s

To Find:

The coefficient of static friction between the tires and the road = ?

Solution:

The radial Acceleration is given by,

$a_{R = \frac{v^2}{r}$

$a_{R = \frac{(32.8)^2}{516}$

$a_{R = \frac{(1075.84)}{516}$

$a_{R = 2.085 m/s^2$

Now the total acceleration is

$\text{ total acceleration} = \sqrt{\text{(tangential acceleration)}^2 +{\text{(Radial acceleration)}^2$

=> $= \sqrt{ (a_r)^2+(a_R)^2}$

=> $\sqrt{ (3.89 )^2+( 2.085)^2}$

=> $\sqrt{ (15.1321)+(4.347)^2}$

=> $19.4791 m/s^2$

The frictional force on the car will be f = ma------------(1)

And the force due to gravity is W = mg--------------------(2)

Now the coefficient of static friction is

$\mu =\frac{f}{W}$

From (1) and (2)

$\mu =\frac{ma}{mg}$

$\mu =\frac{a}{g}$

Substituting the values, we get

$\mu =\frac{19.4791}{9.8}$

$\mu =1.987$

8 0

3 years ago

Two asteroids identical to those above collide at right angles and stick together; i.e, their initial velocities were perpendicu

11111nata11111 [884]

Answer:

velocity = 62.89 m/s in 58 degree measured from the x-axis

Explanation:

Relevant information:

Before the collision, asteroid A of mass 1,000 kg moved at 100 m/s, and asteroid B of mass 2,000 kg moved at 80 m/s.

Two asteroids moving with velocities collide at right angles and stick together. Asteroid A initially moving to right direction and asteroid B initially move in the upward direction.

Before collision Momentum of A = 1000 x 100 = $10^5$ kg - m/s in the right direction.

Before collision Momentum of B = 2000 x 80 = 1.6 x $10^5$ kg - m/s in upward direction.

Mass of System of after collision = 1000 + 2000 = 3000 kg

Now applying the Momentum Conservation, we get

Initial momentum in right direction = final momentum in right direction = $10^5$

And, Initial momentum in upward direction = Final momentum in upward direction = 1.6 x $10^5$

So, $V_x = \frac{10^5}{3000}$ = $\frac{100}{3}$ m/s

and $V_y=\frac{160}{3}$ m/s

Therefore, velocity is = $\sqrt{V_x^2 + V_y^2}$

= $\sqrt{(\frac{100}{3})^2 + (\frac{160}{3})^2}$

= 62.89 m/s

And direction is

tan θ = $\frac{V_y}{V_x}$ = 1.6

therefore, $\theta = \tan^{-1}1.6$

= $58 ^{\circ}$ from x-axis

4 0

3 years ago

What is the momentum of a 12 kg condor flying at 6 m/s?

rusak2 [61]

Answer:

72

Explanation:

Formula

$p=mv\\=12*6\\=72$

5 0

3 years ago

Compute the expected shell-model quadrupole moment of 209Bi () and compare with the experimental value, - 0.37 b

Over [174]

Answer:

0.22 b

Explanation:

Quadrupole moment of the nucleon is,

$Q=-\frac{2j-1}{2(j+1)}\frac{3}{5}R^{2}$

And also,

$R^{2}=R^{2} _{0}A^{\frac{2}{3} }$

And, $R _{0}=1.2\times 10^{-15}m$

Now,

$Q=-\frac{2j-1}{2(j+1)}\frac{3}{5}R^{2} _{0}A^{\frac{2}{3} }$

For Bismuth $j=\frac{9}{2}$ and A is 209.

$Q=-\frac{2\frac{9}{2} -1}{2(\frac{9}{2} +1)}\frac{3}{5}(1.2\times 10^{-15}) ^{2}(209)^{\frac{2}{3} }\\Q=0.628\times 35.28\times 10^{-30} \\Q=22.15\times 10^{-30} m^{2} \\Q=0.2215\times 10^{-28} m^{2} \\Q=0.22 barn$

Therefore, the expected value of quadrupole is 0.22 b which is quite related with experimental value which is 0.37 b

3 0

3 years ago

Calculate the mass of air in a room 5m × 4m ×2m given that the density of air is 1.3kgm-²

valentina_108 [34]

32.5 kg of air

Explanation:

To calculate the mass of the air, we use the density formula:

density = mass / volume

mass = density × volume

density of air = 1.3 kg/m³

volume = 5 × 3 × 2 = 25 m³

mass of the air = 1.3 kg/m³ × 25 m³

mass of the air = 32.5 kg

Learn more about:

density

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brainly.com/question/12982373

#learnwithBrainly

4 0

3 years ago

What is weightlessness ?​

What is weightlessness ?