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aksik [14]
3 years ago
14

A greyhound's velocity changes from rest to 19 m/s in 2 seconds. What is the greyhound's average acceleration?

Physics
2 answers:
Arturiano [62]3 years ago
7 0
We know, a = (v₂-v₁) / (t₂-t₁)
Here, v₂-v₁ = 19 m/s
t₂-t₁ = 2 s

Substitute it into expression, 
a = 19/2
a = 9.5 m/s²

So, your final answer is 9.5 m/s²

Hope this helps!
vfiekz [6]3 years ago
6 0
The correct answer is 9.5 m/s^2. I just took this test.
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Why can videos be streamed from one computer to another with excellent quality?
AleksAgata [21]

Answer:

(d) they are transmitted using digital signals

Explanation:

i just did the quiz an got it right

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2 years ago
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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
A swimmer, capable of swimming at a speed of 1.0 m/s in still water (i.e., the swimmer can swim with a speed of 1.0 m/s relative
Vesnalui [34]
If the current takes him downstream we must find the resultant vector of the velocities: V res= \sqrt{1^{2}+0.91^{2}  } = \sqrt{1.8281}= 1.3520747 Then if the river is 3000 m-wide the swimmer will have to pass:
 1.3520747 · 300 = 4056.14 m                t = 4056.14 m : 1 m/s
a ) It takes 4056.15 seconds ( 1 hour 7 minutes and 36 seconds ) to cross the river.  
b ) 0.91 · 3000 = 2730 m
He will be 2730 m downstream.
5 0
3 years ago
Now review the difference between the temperatures on Earth and Mars. Also look at their distances from the sun.
aksik [14]

Answer: Use Question cove you can get it faster you can get the answer faster! ;) hope this helps ;) but yeah use that and answer is done right away

Explanation: HOPE THIS HELPSS!! ;))

7 0
2 years ago
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If you run 12 m/s for 15 minutes, how far will you go?
vfiekz [6]
10800 m = 10.8 km should be the answer if I am correct
3 0
3 years ago
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