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ale4655 [162]
3 years ago
13

The Lamborghini Huracan has an initial acceleration of 0.75g. Its mass, with a driver, is 1510 kg.

Physics
1 answer:
Cerrena [4.2K]3 years ago
4 0

Answer:

11109.825 N

Explanation:

Given Data:

total mass =m=1510 kg

initial acceleration (a) =0.75g                  ( g=9.81 m/s² )

F=ma

  = (1510)*( 0.75*9.81)

  = 11109.825 N

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C. Both force and displacement

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A large truck collides with a small car. True or False: The truck exerted a greater magnitude force on the car than the car exer
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Answer:

False.

Explanation:

The forces on the car and truck are equal and opposite. The equal forces cause accelerations of the truck and car inversely proportional to their mass. That is, If the Truck A exerts a force FAB on car B, then the car will exert a force FBA on the truck. Therefore,

FBA = −FAB

However, this can be explained by Newton's second law. Let's say the truck has mass M and the car has mass m. If the magnitude of the force that both vehicles experience is F, then the magnitudes of their respective accelerations are:

atruck = F/M

acar = F/m

and combining these we get:

atruck/acar = m/M

So if the mass of the car is a lot less than the mass of the truck, then the acceleration of the truck is much smaller than the acceleration of the car, and if you were to watch the collision, the truck would pretty much seem like it's motion was unaffected, but the car's motion will change quite a bit.

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3 years ago
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3 years ago
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A 3.5-kg object placed on an inclined plane (angle 30? above the horizontal) is connected by a string going over a pulley to a 1
Blababa [14]

Answer:

a= 0.22 m/s²

Explanation:

Given that

M = 3.5 kg

θ = 30°

m = 1 kg

μ= 0.3

The force due to gravity

F₁= M g sinθ

F₁=3.5 x 10 x sin 30

F₁= 17.5 N

F₂ = m g

F₂ = 1 x 10 = 10 N

The maximum value of the friction force on the incline plane

Fr = μ M g cosθ

Fr = 0.3 x 2.5 x 10 cos30°

Fr= 6.49 N

Lets take acceleration of the system is a  m/s²

F₁ - F₂  - Fr =  (M+m) a

17.5 - 10 - 6.49 = (3.5+1)a

a= 0.22 m/s²

7 0
3 years ago
Two 22.7 kg ice sleds initially at rest, are placed a short distance apart, one directly behind the other, as shown in Fig. 1. A
boyakko [2]

Newton's third law of motion sates that force is directly proportional to the rate of change of momentum produced

(a) The final speeds of the ice sleds is approximately 0.49 m/s each

(b) The impulse on the cat is 11.0715 kg·m/s

(c) The average force on the right sled is 922.625 N

The reason for arriving at the above values is as follows:

The given parameters are;

The masses of the two ice sleds, m₁ = m₂ = 22.7 kg

The initial speed of the ice, v₁ = v₂ = 0

The mass of the cat, m₃ = 3.63 kg

The initial speed of the cat, v₃ = 0

The horizontal speed of the cat, v₃ = 3.05 m/s

(a) The required parameter:

The final speed of the two sleds

For the first jump to the right, we have;

By the law of conservation of momentum

Initial momentum = Final momentum

∴ m₁ × v₁ + m₃ × v₃ = m₁ × v₁' + m₃ × v₃'

Where;

v₁' = The final velocity of the ice sled on the left

v₃' = The final velocity of the cat

Plugging in the values gives;

22.7 kg × 0 + 3.63 × 0 = 22.7 × v₁' + 3.63 × 3.05

∴  22.7 × v₁'  = -3.63 × 3.05

v₁' =  -3.63 × 3.05/22.7 ≈ -0.49

The final velocity of the ice sled on the left, v₁' ≈ -0.49 m/s (opposite to the direction to the motion of the cat)

The final speed ≈ 0.49 m/s

For the second jump to the left, we have;

By conservation of momentum law,  m₂ × v₂ + m₃ × v₃ = m₂ × v₂' + m₃ × v₃'

Where;

v₂' = The final velocity of the ice sled on the right

v₃' = The final velocity of the cat

Plugging in the values gives;

22.7 kg × 0 + 3.63 × 0 = 22.7 × v₂' + 3.63 × 3.05

∴  22.7 × v₂'  = -3.63 × 3.05

v₂' =  -3.63 × 3.05/22.7 ≈ -0.49

The final velocity of the ice sled on the right = -0.49 m/s (opposite to the direction to the motion of the cat)

The final speed ≈ 0.49 m/s

(b) The required parameter;

The impulse of the force

The impulse on the cat = Mass of the cat × Change in velocity

The change in velocity, Δv = Initial velocity - Final velocity

Where;

The initial velocity = The velocity of the cat before it lands = 3.05 m/s

The final velocity = The velocity of the cat after coming to rest =

∴ Δv = 3.05 m/s - 0 = 3.05 m/s

The impulse on the cat = 3.63 kg × 3.05 m/s = 11.0715 kg·m/s

(c) The required information

The average velocity

Impulse = F_{average} × Δt

Where;

Δt = The time of collision = The time it takes the cat to finish landing = 12 ms

12 ms = 12/1000 s = 0.012 s

We get;

F_{average} = \mathbf{\dfrac{Impulse}{\Delta \ t}}

∴ F_{average} = \dfrac{11.0715 \ kg \cdot m/s}{0.012 \ s}  = 922.625 \ kg\cdot m/s^2 = 922.625 \ N  

The average force on the right sled applied by the cat while landing, \mathbf{F_{average}} = 922.625 N

Learn more about conservation of momentum here:

brainly.com/question/7538238

brainly.com/question/20568685

brainly.com/question/22257327

8 0
2 years ago
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