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3 years ago

Escape speed from the surface of earth is about

2 answers:

8 0

<span>Escape velocity is the speed that an object needs to be traveling to break free of a planet or moon's gravity well and leave it without further propulsion. For example, a spacecraft leaving the surface of Earth needs to be going 7 miles per second, or nearly 25,000 miles per hour to leave without falling back to the surface or falling into orbit</span>

dezoksy [38]3 years ago

6 0

<span>On the surface of the Earth, the escape velocity is about 11.2 km/s, which is approximately 33 times the speed of sound (Mach 33) and several times the muzzle velocity of a rifle bullet (up to 1.7 km/s). However, at 9,000 km altitude in "space", it is slightly less than 7.1 km/s.</span>

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A laser beam is incident at an angle of 30.0° from the vertical onto a solution of corn syrup in water. The beam is refracted to

dimaraw [331]

Answer with Explanation:

We are given that

Angle of incidence, $i=30^{\circ}$

Angle of refraction, $r=19.24^{\circ}$

a.Refractive index of air, $n_1=1$

We know that

$n_2sinr=n_1sini$

$n_2=\frac{n_1sin i}{sin r}=\frac{sin30}{sin19.24}=1.517$

b.Wavelength of red light in vacuum, $\lambda=632.8nm=632.8\times 10^{-9} m$

$1nm=10^{-9} m$

Wavelength in the solution, $\lambda'=\frac{\lambda}{n_2}$

$\lambda'=\frac{632.8}{1.517}=417nm$

c.Frequency does not change .It remains same in vacuum and solution.

Frequency, $\nu=\frac{c}{\lamda}=\frac{3\times 10^8}{632.8\times 10^{-9}}$

Where $c=3\times 10^8 m/s$

Frequency, $\nu=4.74\times 10^{14}Hz$

d.Speed in the solution, $v=\frac{c}{n_2}$

$v=\frac{3\times 10^8}{1.517}=1.98\times 10^8m/s$

5 0

3 years ago

A wheel accelerates from rest to 34.7 rad/s at a rate of 47.0 rad/s^2. Through what angle (in radians) did the wheel turn while

dem82 [27]

12.8 rad

Explanation:

The angular displacement $\theta$ through which the wheel turned can be determined from the equation below:

$\omega^2 = \omega_0^2 + 2\alpha\theta$ (1)

where

$\omega_0 = 0$

$\omega = 34.7\:\text{rad/s}$

$\alpha = 47.0\:\text{rad/s}^2$

Using these values, we can solve for $\theta$ from Eqn(1) as follows:

$2\alpha\theta = \omega^2 - \omega_0^2$

$\theta = \dfrac{\omega^2 - \omega_0^2}{2\alpha}$

$\:\:\:\:= \dfrac{(34.7\:\text{rad/s})^2 - 0}{2(47.0\:\text{rad/s}^2)}$

$\:\:\:\:= 12.8\:\text{rad}$

7 0

2 years ago

Thin Layer Chromatography consists of three parts: The analyte, the stationary phase, and mobile phase. Match each of these term

Margaret [11]

Answer:

Analyte⇒ one of analgesics

stationery phase⇒ silica

mobile phase⇒ solvent

Explanation:

during the thin layer chromatography non volatile mixtures are separated.The technique is performed on the plastic or aluminum foil that is coated with a thin layer.

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2 years ago

Will give 30 points!!!

AysviL [449]

The answer is C I believe

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2 years ago

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B) A skilled jet fighter flies a stunt plane in a vertical circle of 1200 ft

vodka [1.7K]

Answer:I need the ans to this one too

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3 years ago