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Tasya [4]
3 years ago
6

23. Milo created the data table based on his science experiment. What was the velocity of the

Physics
1 answer:
Tems11 [23]3 years ago
8 0
6 m/s because the position of the object was increasing 6 m every second
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The Mars Curiosity rover was required to land on the surface of Mars with a velocity of 1 m/s. Given the mass of the landing veh
Aliun [14]

Answer:

The value is      A   = 39315 \  m^2

Explanation:

From the question we are told that

    The velocity which the rover is suppose to land with is  v  =  1 \ m/s

    The  mass of the rover and the parachute is  m  =  2270 \ kg

     The  drag coefficient is  C__{D}}  =  0.5

      The atmospheric density of Earth  is  \rho =  1.2 \  kg/m^3

     The acceleration due to gravity in Mars is  g_m  =  3.689 \  m/s^2

     

Generally the Mars  atmosphere density is mathematically represented as

          \rho_m  =  0.71 *  \rho

=>        \rho_m  =  0.71 *  1.2

=>        \rho_m  = 0.852 \  kg/m^3

Generally the drag force on the rover and the parachute  is mathematically represented as

          F__{D}} =  m  *  g_{m}

=>       F__{D}} =  2270   *  3.689  

=>       F__{D}} =  8374 \ N  

Gnerally this drag force is mathematically represented as

         F__{D}} =   C__{D}} *  A *  \frac{\rho_m * v^2 }{2}

Here A is the frontal area

So  

         A   =  \frac{2 *  F__D }{ C__D}  *  \rho_m  * v^2   }

=>       A   =  \frac{2 * 8374 }{ 0.5 *  0.852    *  1 ^2   }

=>       A   = 39315 \  m^2

8 0
3 years ago
NASA scientists suggest using rotating cylindrical spacecraft to replicate gravity while in a weightless environment. Consider s
dusya [7]

Answer:

Explanation:

Given

diameter of spacecraft d=148\ m

radius r=74\ m

Force of gravity F_g=mg

where m =mass of object

g=acceleration due to  gravity on earth

Suppose v is the speed at which spacecraft is rotating so a net centripetal  acceleration is acting on spacecraft which is given by

F_c=\frac{mv^2}{r}

F_c=F_g

\frac{mv^2}{r}=mg

\frac{v^2}{r}=g

v=\sqrt{gr}

v=\sqrt{1450.4}

v=38.08\ m/s    

8 0
3 years ago
Some superconductors are capable of carrying a very large quantity of current. If the measured current is 1.00 ´ 105 A, how many
Zielflug [23.3K]

Answer:

The 6.25 \times 10^{23} electrons are moving through the superconductor per second.

Explanation:

Given :

Current I = 1 \times 10^{5} A

Charge of electron e = 1.6 \times 10^{-19} C

Time t = 1 sec

From the formula of current,

Current is the number of charges flowing per unit time.

   I = \frac{ne}{t}

Where n = number of charges means in our case number of electrons

   n = \frac{It}{e}

   n = \frac{1 \times 10^{5} }{1.6 \times 10^{-19} }

   n = 6.25 \times 10^{23}

Therefore, 6.25 \times 10^{23} electrons are moving through the superconductor per second.

5 0
3 years ago
Which one of the following temperatures (in °C) is equivalent to 294 K?
vaieri [72.5K]

Answer:

21

Explanation: its actually 20.85 but i guess they round to 21

5 0
3 years ago
A(an) ________ complex refers to the feeling that one lacks worth and doesn't measure up to the standards of others or of societ
saw5 [17]

Answer:

Explanation:

Inferiority Complex

3 0
3 years ago
Read 2 more answers
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