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drek231 [11]
3 years ago
9

(a) Calculate the linear acceleration of a car, the 0.220-m radius tires of which have an angular acceleration of 11.0 rad/s2. A

ssume no slippage.
Physics
1 answer:
Alisiya [41]3 years ago
8 0

Answer:

The value is  a_t =  2.42 \  m/s^2

Explanation:

From the question we are told that

  The radius of the tires is  r =  0.22 \  m

   The  angular acceleration is  \alpha  =  11.0 \ rad/s^2

Generally the linear acceleration is mathematically represented as

     a_t =  r *  \alpha

=>  a_t =  0.22  *  11

=>  a_t =  2.42 \  m/s^2

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A student who takes a multiple-choice test by reading the stem of each item, generating the correct response before looking at t
Ludmilka [50]

Answer:

(A) a heuristic

Explanation:

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3 years ago
NEED HELP FAST ILL MARK BRAINLIEST AND RATE 5 STARS AND SAY THANK YOU
daser333 [38]

Acceleration = (change in speed)/(time for the change)

Change in speed = (end speed) - (start speed)

Change in speed = (10 m/s) - (20 m/s) = -10 m/s

Time for the change = 5.00 seconds

Acceleration = (-10 m/s) / (5 sec)

<em>Acceleration = -2 m/s²</em>

That's choice-A .

8 0
3 years ago
Read 2 more answers
Please answer the number 5 6 and 7. I don't know what to do its our hw in physics. (new lesson as well)
Aleks04 [339]

From the picture, I see that you had no trouble at all with #4.
Well, #5, 6, and 7 are easily handled in exactly the same way.

Just as you did with #4, please sketch these on paper
as I walk you through the solutions.  That'll help you see
immediately what's going on.

#5.b).
Traveling east at 3 m/s for 4 seconds,
he covers  (3 m/s) x (4 sec) = 12 meters.

Traveling south at 5 m/s for 2 seconds,
he covers (5 m/s) x (2 sec) = 10 meters.

The total distance he covers is  (12m + 10m) = 22 meters.

#5.c).
Average speed (scalar)

                           = (distance covered)/(time to cover the distance)

                           = (22 meters)/(6 sec) = 3-2/3 m/s .

#5.d).
Displacement (vector)

                       = distance between the start-point and the end-point,
                          regardless of the route traveled,
                      
  in the direction from the start-point to the end-point.

Distance from the start-point to the end-point =

               √(12² + 10²) = √(144 + 100) = √(244) = 15.62 meters

in the direction of  arctan(10/12) south of east

                             =  39.8° south of east.
 
#5.e).
Average velocity (vector) =

             (displacement vector) / (time)

         =  15.62 meters directed 39.8° south of east / 6 seconds

         =  2.603 m/s directed 39.8° south of east.

 #6).
Magnitude = √(5.2² + 2.1²) = √(27.04 + 4.41) = √31.45 = 5.608 km.

Direction = arctan(5.2/2.1) south of east

                =   68° south of east  =  158° bearing . 

#7).
Magnitude = √(39² + 57²) = √(1521 + 3249) = √( 4770)

                                                                         =  69.07 m/s .

Direction = arctan (57/39) south of west

               =   55.6° south of west

                    Bearing = 214.4°

                    Compass: 0.65° past "southwest by south".  


I'm grateful for the privilege and opportunity to practice my math,
and I shall cherish the bounty of 5 points that came with it.

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A person standing 102 meter from the foot of high building claps his hand and hears the echo 0.6 seconds later. what is the spee
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Answer:

Explanation:Let me tell you the logic you need for this: If the boy hears an echo, this means ... 100 m from a tall building claps his hands and hears an echo 0.6 seconds later. ... you can find the rate of travel (meters travelled per second) dividing the total ... the speed of sound in a hypothetical silent craft

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