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ollegr [7]
3 years ago
12

two toy cars can roll without friction on a horizontal surface. Car A is a 5 kg. Car B is 0.5 kg. Both cars start from rest. Equ

al sized forces are used to push each cart forward for a time of 1 second. After 1 second the force is removed. What is the difference in magnitude of the momenta of the two cars, right after the force is removed?Choices:A. The momentum of car A(5kg) is EQUAL to that of car B(0.5)B. The momentum of car A (5kg) is LESS than that car B (0.5)C. The momentum of car A (5kg) is MORE than that car B (0.5)D. 5.5 kg m/s
Physics
1 answer:
Llana [10]3 years ago
8 0

Answer:

A. The momentum of car A(5kg) is EQUAL to that of car B(0.5)

Explanation:

The moment, or impulse formula of the same forces acting on both car within 1 second is

\Delta p = F * \Delta t

In our case the forces are the same, the time duration of force acting on the cars are the same. Therefore, their momentum right after the force must also be the same.

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Answer:

v= s/t

Explanation:

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A right triangle ABC has sides /AB/ =9cm, /BC/=12cm find /AC/ and angles ACB and ABC if angle BAC= 90°​
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Two teams of nine members each engage in a tug of war. Each of the first team’s members has an average mass of 68 kg and exerts
olga2289 [7]

Answer:

a)  a = - 0.106 m/s^2  (←)

b) T = 12215.1064 N

Explanation:

If

F₁ = 9*1350 N = 12150 N   (→)

F₂ = 9*1365 N = 12285 N  (←)

∑Fx = M*a = (M₁  +M₂)*a           (→)

F₁ - F₂ = (M₁  +M₂)*a        

→       a = (F₁ - F₂) / (M₁  +M₂ ) = (12150-12285)N/(9*68+9*73)Kg

→       a = - 0.106 m/s^2            (←)

(b) What is the tension in the section of rope between the teams?

If we apply  ∑Fx = M*a   for the team 1

F₁ - T = - M₁*a  ⇒   T = F₁ + M₁*a  

⇒   T = 12150 N + (9 * 68 Kg)*(0.106 m/s^2)

⇒ T = 12215.1064 N

If we choose the team 2 we get

- F₂ + T = - M₂*a  ⇒   T = F₂ - M₂*a  

⇒   T = 12285 N - (9 * 73 Kg)*(0.106 m/s^2)

⇒ T = 12215.1064 N

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