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EastWind [94]
4 years ago
8

A car parked on level pavement exerts a force of 10,000 newtons on the ground. What force does the pavement exert back on the ca

r ?
Physics
1 answer:
natka813 [3]4 years ago
7 0

Answer:

Normal force of 10,000N

Explanation:

From the question, the weight the car exerts on the pavement is 10,000N.

The pavement exerts upward and perpendicular contact force called normal force on the car to support its weight. Also, the normal force is equal and opposite to the weigh of the car.

Hence the pavement exerts normal force of 10,000N back on the car to prevent it from passing through it.

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Difference between 3.15 m and 2.0 m with the correct number of significant figures?
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How do you find the speed of an object given its mass and kinetic energy (what is the formula)?
madam [21]
   v  =   √ { 2*(KE) ] / m } ; 

Now, plug in the known values for "KE" ["kinetic energy"] and "m" ["mass"] ; 
        
and solve for "v".

______________________________________________________
Explanation:
_____________________________________________________
The formula is:  KE = (½) * (m) * (v²) ;
_____________________________________
  
"Kinetic energy" = (½) * (mass) * (velocity , "squared")
________________________________________________
Note:  Velocity is similar to speed, in that velocity means "speed and direction";  however, if you "square" a negative number, you will get a "positive"; since:  a "negative" multiplied by a "negative" equals a "positive".
____________________________________________
So, we have the formula:
___________________________________
KE = (½) * (m) * (v²) ;  to solve for "(v)" ; velocity, which is very similar to                                          the "speed"; 
___________________________________________________
we arrange the formula ;
__________________________________________________
(KE) = (½) * (m) * (v²) ;  ↔  (½)*(m)* (v²) = (KE) ; 
___________________________________________________

→ We have:  (½)*(m)* (v²) = (KE)  ; we isolate, "m" (mass) on one side of the equation:
______________________________________________________
   
→ We divide each side of the equation by: "[(½)* (m)]" ; 
___________________________________________________
    
           →   [ (½)*(m)*(v²) ] /  [(½)* (m)]  = (KE) / [(½)* (m)]<span> ;
</span>______________________________________________________
 to get: 
______________________________________________________
                           →   v²     =   (KE) / [(½)* (m)]
                     
                           →   v²     = 2 KE / m
_______________________________________________________
Take the "square root" of each side of the equation ;
_______________________________________________________
                          →  √ (v²)  =  √ { 2*(KE) ] / m }
________________________________________________________

                          →     v  =   √ { 2*(KE) ] / m } ; 

Now, plug in the known values for "KE" ["kinetic energy"] and "m" ["mass"]; 
       
and solve for "v".

______________________________________________________
8 0
3 years ago
A physics student swings a tennis ball connected to a rope in a vertical circle with a constant speed of 6.29 m/s. The ball has
Alex777 [14]

Answer:

r = 0.5 m

Explanation:

First we find the angular speed of the ball by using its period:

ω = θ/t

For the time period:

ω = angular speed = ?

θ = angular displacement = 2π rad

t = time period = 0.5 s

Therefore,

ω = 2π rad/0.5 s

ω = 12.56 rad/s

Now, for the radius:

v = rω

r = v/ω

where,

v = linear speed = 6.29 m/s

r = radius = ?

r = (6.29 m/s)/(12.56 rad/s)

<u>r = 0.5 m</u>

8 0
3 years ago
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