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sergey [27]
2 years ago
6

How much kinetic energy is required to break through?

Physics
1 answer:
Ksju [112]2 years ago
7 0

Explanation:

The natural environment or natural world encompasses all living and non-living things occurring naturally, meaning in this case not artificial. The term is most often applied to the Earth or some parts of Earth

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What is the acceleration of a car that moves at a steady velocity of 100 km/h for 100 seconds? Explain your answer.
RUDIKE [14]

Answer:

a = 0 m/s²

Explanation:

given,

car moving at steady velocity = 100 Km/h

         1 km/h = 0.278 m/s

      100 Km/h = 27.8 m/s

time of acceleration = 100 s

acceleration is equal to change in velocity per unit time.

 a=\dfrac{dv}{dt}

change in velocity of the car is 27.8 - 27.8 = 0

 a=\dfrac{0}{100}

        a = 0 m/s²

If the car is moving with steady velocity then acceleration of the car is zero.

Hence, the acceleration of the car is equal to a = 0 m/s²

3 0
3 years ago
Before a rifle is fired, the linear momentum of the bullet-rifle system is zero.
marusya05 [52]

Hi there!

II. Linear momentum of the system is zero.

This is an example of a RECOIL collision. With the Law of Conservation of Momentum, momentum remains constant before and after the collision.

Thus, the total momentum would also be equivalent to zero after the collision.

3 0
2 years ago
Two point charge
timama [110]

Answer:

Showing results for Two point charge q, separated by 1.5cm have change value of +2.0 and -4.0AND/C respectively what is the magnitude of the Electric force midway between them?

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4 0
1 year ago
A pyrotechnical releases a 3 kg firecracker from rest. at t=0.4 s, the firecracker is moving downward with a speed 4 m/s. At the
olga2289 [7]

Answer:

a) F = 30 N, b)   I = 12 N s , c)  I = -12 N s , d) ΔI = 0 N s

Explanation:

This exercise is a case at the moment, let's define the system formed by the firecracker and its two parts, in this case the forces during the explosion are internal and the moment is conserved

Initial, before the explosion

     p₀ = m v

The speed can be found by kinematics

     v = v₀ - g t

     v = 0 - 10 0.4

     v = -4.0 m / s

Final after division

     pf = m₁ v₁f + m₂ v₂f

    p₀ = pf

    M v = m₁ v₁f + m₂ v₂f

Where M is the initial mass (M = 3 kg), m₁ is the mass mtop (m₁ = 1 kg) and m₂ in the mass m botton (m₂ = 2kg) and the piece that moves up (v₁f = 6m/s )

a) before the explosion the only force acting on the body is gravity

     F = mg

     F = 3 10 = 30 N

b) The expression for momentum is

     I = Ft

Before the explosion the only force that acts is the weight

    I = mg t

    I = 3 10 0.4

    I = 12 N s

c) To calculate this part we use the conservation of the moment and calculate the speed of the body that descends body 2

    M v = m₁ v₁f + m₂ v₂f

    v₂f = (M v - m₁ v₁f) / m₂

    v₂f = (3 (-4) - 1 6) / 2

   v₂f = - 9 m / 2

The negative sign indicates that body 2 (botton) is descending

Now we can use the momentum and momentum relationship for the body during the explosion

    I = F t = Dp

   F t = pf –po)

   F t= [m₁ v₁f + m₂ v₂f]

   

   I = [1 6 + 2 (-9) -0]

   I = -12 N s

This is the impulse during the explosion the negative sign indicates that it is headed down

d) impulse change

I₀ = Mv

I₀ = 3 *4

I₀ =-12 N s

 ΔI =If – I₀  

ΔI = - 12 – (-12)

ΔI = -0 N s

3 0
3 years ago
All of the members of a community belong to the same species tf
Anna35 [415]
That is false

I hope this helps!
3 0
3 years ago
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