is 468km per minute describing speed or velocity

Explanation:

a) E= fs

b) 2E = mv²

c) E= Mc ²

Answer:

Motion at one end of the beam results in motion at the other end in the opposite direction.

Explanation:

The location of the fulcrum can magnify (or reduce) the force applied at one end at the expense (or advantage) of the distance over which the other end travels. Sorry if I get this wrong! I am in 5th grade. ♥

Answer:

Sodium phosphide

Explanation:

Answer:

sodium phosphide

Explanation:

i just took the test. Good luck!! Have a great day!!

Answer:

vasporisation

Explanation:

process used to change liquid to gass

Answer:

I believe it's Evaportation or another word would be Vaporisation.

Explanation:

hope this helps! ^^

Answer:

I'm fairly certain it's 1,800

Explanation:

The efficiency of the wedge with force F=900 N displaces the wedge 0.10 m is 55.55%  or 56%.

The efficiency of the machine equals the ratio of output energy and input energy. It is also defined as the ratio of work done on a load by the machine to the work done on the effort by the machine.

A load is defined as the weight lifted by the machine. The effort is defined as the force applied to bring the desired change in the load. Work done is defined as the product of force and displacement.

The efficiency, η = Output power / Input power, and the highest efficiency of the machine is equal to one. The efficiency of the machine has no unit. The efficiency η = (Actual work done / Ideal work done) × 100.

From the given,

Force = 900 N

Displacement = 0.1 m

Work done = F × s = 900×0.1 = 90 J

Actual work done = 50 J

Efficiency,η = (50/90) × 100

           η  = 55.55 %

Thus, the efficiency of the wedge is 55.56%.

To learn more about the Efficiency of a machine:

https://brainly.com/question/11752408

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Answer:

 F = 99 v₂₀

v₂₀ = 1 m / s,        F = 99 N

Explanation:

In this exercise it is asked to find the force during the collision, for this we use the relationship between the momentum and the momentum of car 1

            I = Δp

            F t = p_f- p₀

            F t = m (v_f -v₀)                        (1)

We must find the final speed of car 1, for this we define a system formed by the two cars, in this case the forces during the collision are internal and the moment is conserved

initial instant. Before the crash

        p₀ = 0 + m₂ v₂₀

final instant. After the crash

        p_f = m₁ v₁ + m₂ v_{2f}

the moment is preserved

        p₀ = p_f

        m₂ v₂₀ = m₁ v_{1f} + m₂ v_{2f}           (2)

        m₂ (v₂₀ - v_2f}) = m₁ v_{1f}

as the collision is elastic the kinetic energy is also conserved

        K₀ = K_f

        ½ m₂ v₂₀² = ½ m₁ v_{1f}² + ½ m₂ v_{2f}²

        m₂ (v₂₀² -v_{2f}²) = m₁ v_{1f}²

let's write our system of equations, using

         a² - b² = (a + b) (a-b)

         m₂ (v₂₀ - v_{2f}) = m₁ v_{1f}

         m₂ (v₂₀ -v_{2f}) (v₂₀ + v_{2f}) = m₁ v_{1f}²

to solve we divide the equations

       v₂₀ + v_{2f} = v_{1f}

with this we substitute in equation 2 and find the speed of each car, in this case we need the speed of car 1

         m₂ v₂₀ = m₁ v_{1f} + m₂ (v_{1f}-v₂₀)

         2m₂ v₂₀ = (m₁ + m₂) v_{1f}

          v_{1f} = [tex]\frac{2m_2}{m_1+m_2} v_{2o}[/tex]

We substitute in the drive ratio of car 1

            F t = m (v_f -v₀)

            F = m₁ (\frac{2m_2}{m_1+m_2}  v_{2o} - 0) / t

            F = [tex]\frac{2m_1 m_2 }{m_1+m_2} \ \frac{v_{2o}}{t}[/tex]

the mass of each car is the mass of the car plus the mass of the boy

           m₁ = 50 +40 = 90 kg

           m₂ = 50 +60 = 110 kg

time is t = 1

we substitute the values

           F = [tex]\frac{ 2\ 90 \ 110}{90+110} \ \frac{v_{2o}}{1}[/tex]2 90 100/90 + 110 vo2 / 1

           F = 99 v₂₀

The value of the initial velocity of car 2 is not indicated in the problem, if this velocity is known it can be included and the force value is obtained, suppose that the initial velocity v₂₀ = 1 m / s

           F = 99 N