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3 years ago

If a bus travels 6 km in 10 minutes, what distance does it travel in 1 second

Physics

2 answers:

loris [4]3 years ago

4 0

Answer:

Your answer is 10 m/s...

Masteriza [31]3 years ago

4 0

Answer:

I think it is 1 foot???

Explanation:

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Par 1/2

BaLLatris [955]

part 1

mass = ρ x V

mass = 1739 kg/m³ x 3.8 km³ = 6608.2 kg

PE (potential energy)= mgh

PE = 6608.2 kg x 9.81 x 403

PE = 2.61 x 10⁷ J

part 2

megaton of TNT (Mt) =4.2 x 10¹⁵ J

convert PE to Mt:

2.61 x 10⁷ J : 4.2 x 10¹⁵ J = 6.21 x 10⁻⁹ Mt

4 0

3 years ago

Using Rayleigh's criterion, calculate the diameter of an earth-based telescope that gives this resolution with 700 nm light.

pogonyaev

Complete Question

Due to blurring caused by atmospheric distortion, the best resolution that can be obtained by a normal, earth-based, visible-light telescope is about 0.3 arcsecond (there are 60 arcminutes in a degree and 60 arcseconds in an arcminute).Using Rayleigh's criterion, calculate the diameter of an earth-based telescope that gives this resolution with 700 nm light

Answer:

The diameter is $D = 0.59 \ m$

Explanation:

From the question we are told that

The best resolution is $\theta = 0.3 \ arcsecond$

The wavelength is $\lambda = 700 \ nm = 700 *10^{-9 } \ m$

Generally the

1 arcminute = > 60 arcseconds

=> x arcminute => 0.3 arcsecond

$x = \frac{0.3}{60 }$

=> $x = 0.005 \ arcminutes$

Now

60 arcminutes => 1 degree

0.005 arcminutes = > z degrees

=> $z = \frac{0.005}{60 }$

=> $z = 8.333 *10^{-5} \ degree$

Converting to radian

$\theta = z = 8.333 *10^{-5} * 0.01745 = 1.454 *10^{-6} \ radian$

Generally the resolution is mathematically represented as

$\theta = \frac{1.22 * \lambda }{ D}$

=> $D = \frac{1.22 * \lambda }{\theta }$

=> $D = \frac{1.22 * 700 *10^{-9} }{ 1.454 *10^{-6} }$

=> $D = 0.59 \ m$

4 0

3 years ago

Which property is closely related to its temperature

Lilit [14]

Answer:

Mass – The single most important property that determines other properties of the star. Luminosity – The total amount of energy (light) that a star emits into space. Temperature – surface temperature, closely related to the luminosity and color of the star

Explanation:

8 0

3 years ago

Read 2 more answers

A hollow cylinder that is rolling without slipping is given a velocity of 5.0 m/s and rolls up an incline to a vertical height o

inysia [295]

Answer:

The hollow cylinder rolled up the inclined plane by 1.91 m

Explanation:

From the principle of conservation of mechanical energy, total kinetic energy = total potential energy

$M.E_T = \frac{1}{2}mv^2 + \frac{1}{2} I \omega^2 + mgh$

The total energy at the bottom of the inclined plane = total energy at the top of the inclined plane.

$\frac{1}{2}mv_i^2 + \frac{1}{2} I \omega_i^2 + mg(0) = \frac{1}{2}mv_f^2 + \frac{1}{2} I \omega_f^2 + mgh$

moment of inertia, I, of a hollow cylinder = ¹/₂mr²

substitute for I in the equation above;

$\frac{1}{2}mv_i^2 + \frac{1}{2} (\frac{1}{2}mr^2 \omega_i^2) = \frac{1}{2}mv_f^2 + \frac{1}{2} (\frac{1}{2}mr^2 \omega_f^2) + mgh\\\\ but \ v = r \omega\\\\\frac{1}{2}mv_i^2 + \frac{1}{2} (\frac{1}{2}m v_i^2 ) = \frac{1}{2}mv_f^2 + \frac{1}{2} (\frac{1}{2}m v_f^2) + mgh\\\\\frac{1}{2}mv_i^2 +\frac{1}{4}mv_i^2 = \frac{1}{2}mv_f^2 +\frac{1}{4}mv_f^2 +mgh\\\\\frac{3}{4}mv_i^2 = \frac{3}{4}mv_f^2 +mgh\\\\mgh = \frac{3}{4}mv_i^2 - \frac{3}{4}mv_f^2\\\\gh = \frac{3}{4}v_i^2 - \frac{3}{4}v_f^2\\\\$