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KonstantinChe [14]

3 years ago

7

A person has a mass of 42.0-kg. what is the persons weight on the moon?

1 answer:

Svetlanka [38]3 years ago

4 0

The gpe on the mood = 1.6N
w= m x g
42kg x 1.6N = 67.2N

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The main reason that most professional research telescopes are reflectors is that

GuDViN [60]

<h3><u>Answer;</u></h3>

Large mirrors are easier to build than large lenses.

<h3><u>Explanation;</u></h3>

<em><u>Reflector telescopes have a number of advantages as compared to refracting telescopes and other types of telescopes. </u></em>
<em><u>Reflector telescopes do not suffer from chromatic aberration because all wavelengths will reflect off the mirror in the same way. The support for the objective mirror is all along the back side so they can be made very large.</u></em>
Additionally, reflector telescopes are cheaper to make than refractors of the same size. Also since in reflector telescopes light is reflecting off the objective, rather than passing through it, only one side of the reflector telescope's objective needs to be perfect.

7 0

3 years ago

Lol<br>idk what this is

alex41 [277]

Answer:

I know

Explanation:

Physics= hard

3 0

2 years ago

Read 2 more answers

A drag racing car with a weight of 1600 lbf attains a speed of 270 mph in a quarter-mile race. Immediately after passing the tim

Kaylis [27]

Answer:

15.065ft

Explanation:

To solve this problem it is necessary to consider the aerodynamic concepts related to the Drag Force.

By definition the drag force is expressed as:

$F_D = -\frac{1}{2}\rho V^2 C_d A$

Where

$\rho$ is the density of the flow

V = Velocity

$C_d$ = Drag coefficient

A = Area

For a Car is defined the drag coefficient as 0.3, while the density of air in normal conditions is 1.21kg/m^3

For second Newton's Law the Force is also defined as,

$F=ma=m\frac{dV}{dt}$

Equating both equations we have:

$m\frac{dV}{dt}=-\frac{1}{2}\rho V^2 C_d A$

$m(dV)=-\frac{1}{2}\rho C_d A (dt)$

$\frac{1}{V^2 }(dV)=-\frac{1}{2m}\rho C_d A (dt)$

Integrating

$\int \frac{1}{V^2 }(dV)= - \int\frac{1}{2m}\rho C_d A (dt)$

$-\frac{1}{V}\big|^{V_f}_{V_i}=\frac{1}{2m}(\rho)C_d (\pi r^2) \Delta t$

Here,

$V_f = 60mph = 26.82m/s$

$V_i = 120.7m/s$

$m= 1600lbf = 725.747Kg$

$\rho = 1.21 kg/m^3$

$C_d = 0.3$

$\Delta t=7s$

Replacing:

$\frac{-1}{26.82}+\frac{1}{120.7} = \frac{1}{2(725.747)}(1.21)(0.3)(\pi r^2) (7)$

$-0.029 = -5.4997r^2$

$r = 2.2963m$

$d= r*2 = 4.592m \approx 15.065ft$

4 0

3 years ago

How would convection warm up an iguana cage

harina [27]

Using a basking spot so some sort of heated object, for example heating lamp or heating pad.

3 0

2 years ago

Calculate the kinetic energy in joules of a 1500 kg automobile moving at 19 m/s

Svetlanka [38]

Using
KE = ½mv² = ½×1500×19×19 = 270750 joules

5 0

3 years ago