A laser emits a light beam with a wavelength of 630 nm. The jet passes a liquid with a refractive index of 1.3.

A video game includes an asteroid that is programmed to move in a straight line across a 17-inch monitor according to the equati

Ainat [17]

Answer:

The asteroid's acceleration at this point is $2.71\ m/s^2$

Explanation:

The equation that governs the trajectory of asteroid is given by :

$x=6.5t-2.3t^3$

The velocity of asteroid is given by :

$v=\dfrac{dx}{dt}\\\\v=\dfrac{d(6.5t-2.3t^3)}{dt}\\\\v=6.5-6.9t^2$

At some point during the trip across the screen, the asteroid is at rest. It means, v = 0

So,

$6.5-6.9t^2=0\\\\t=0.971\ s$

Acceleration,

$a=\dfrac{dv}{dt}\\\\a=\dfrac{d(6.5-6.9t^2)}{dt}\\\\a=-13.8t$

Put t = 0.971 s

$a=-13.8\times 0.197\\\\a=-2.71\ m/s^2$

So, the asteroid's acceleration at this point is $2.71\ m/s^2$ and it is decelerating.

6 0

3 years ago

(a) Calculate the magnitude of the gravitational force exerted by Mars on a 80 kg human standing on the surface of Mars. (The ma

andrew11 [14]

Answer:

a) $F=1.044\times 10^9\ N$

b) $F'=1.044\times 10^9\ N$

c) $F_p=1.0672\times10^{-7}\ N$

d) Treat the humans as though they were points or uniform-density spheres.

Explanation:

Given:

mass of Mars, $M=6.4\times 10^{23}\ kg$

radius of the Mars, $r=3.4\times 10^{6}\ m$

mass of human, $m=80\ kg$

a)

Gravitation force exerted by the Mars on the human body:

$F=G.\frac{M.m}{r^2}$

where:

$G=6.67 \times 10^{-11}\ m^3.kg^{-1}.s^{-2}$ = gravitational constant

$F=6.67\times10^{-11}\times \frac{6.4\times 10^{23}\times 80}{(3.4\times 10^{6})^2}$

$F=1.044\times 10^9\ N$

b)

The magnitude of the gravitational force exerted by the human on Mars is equal to the force by the Mars on human.

$F'=F$

$F'=1.044\times 10^9\ N$

c)

When a similar person of the same mass is standing at a distance of 4 meters:

$F_p=6.67\times10^{-11}\times \frac{80\times 80}{4}$

$F_p=1.0672\times10^{-7}\ N$

d)

The gravitational constant is a universal value and it remains constant in the Universe and does not depends on the size of the mass.

Yes, we have to treat Mars as spherically symmetric so that its center of mass is at its geometric center.

Yes, we also have to ignore the effect of sun, but as asked in the question we have to calculate the gravitational force only due to one body on another specific body which does not brings sun into picture of the consideration.

4 0

3 years ago

The hubbles telescopes orbit is 5.6 x10 ^5 meters above earths suface. the telescope has a mass os 1.1 x10^4 kilograms. earth ex

Andru [333]

(3) 8.3 N/kg. The gravitational field strength at a point is the force per unit mass exerted on a mass placed at that point. So at the point where the Hubble telescope is, it is (9.1 x 10^4)N/(1.1 x 10^4 kg) = 8.3 N/kg

Fam

7 0

3 years ago

You are on an airplane traveling 30° south of due west at 180 m/s with respect to the air. The air is moving with a speed 31 m/s

BlackZzzverrR [31]

1) 211m/s
2)240<span>°
3)759,600m or 759.6 km</span>

6 0

3 years ago

HELP PLS MARKING BRANLIST 100 Pts TAKING TEST RN

vredina [299]

Answer:

0.5 m/s2

Explanation:

accelration formula : final velocty - starting velocity divided by time

5 0

2 years ago