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Shalnov [3]
3 years ago
10

Suppose you exert 150 n on your refrigerator and push it across the kitchen floor at constant velocity. what friction force acts

between the refrigerator and the floor?
Physics
2 answers:
Pachacha [2.7K]3 years ago
8 0
Due that the velocity is constant that means that friction force is equal to the force exert by you, otherwise the refrigerator will accelerate or decelerate and in both cases velocity will not be constant.  
So then the friction force between refrigerator and floor is 150 Newtons.
BlackZzzverrR [31]3 years ago
4 0
Friction force is the force arising from two surfaces being pressed together and the force will depend upon the nature of the two surfaces and degree to which they are pressed. Kinetic friction is the force you exert on your refrigerator as you move it across the kitchen floor at constant velocity. 
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A vw beetle goes from 0 to 60 mi/h with an acceleration of 2.35 m/s^2.
Vlad1618 [11]
Part a.
u = 0, the initial velocity
v = 60 mi/h, the final velocity
a = 2.35 m/s², the acceleration.

Note that
1 m = 1609.34 m.
Therefore
v = (60 mi/h)*(1609.34 m/mi)*(1/3600 h/s) = 26.822 m/s
Use the formula
v = u + at
(26.822 m/s) = (2.35 m/s²)*(t s)
t = 26.822/2.35 = 11.4 s

Answer: 11.4 s

Part b.
We already determined that v = 60 mi/h = 26.822 m/s.
t = 0.6 s
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(26.822 m/s) = (a m/s²)*(0.6 s)
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6 0
2 years ago
Model rocket engines are sized by thrust, thrust duration, and total impulse, among other characteristics. A size C5 model rocke
umka2103 [35]

Answer:

v_{f} = 115.95 m / s

Explanation:

This is an exercise of a variable mass system, let's form a system formed by the masses of the rocket, the mass of the engines and the masses of the injected gases, in this case the system has a constant mass and can be solved using the conservation the amount of movement. Which can be described by the expressions

        Thrust = v_{e}  \frac{dM}{dt}

        v_{f}-v₀ = v_{e} ln ( \frac{M_{o} }{M_{f}} )

where v_{e} is the velocity of the gases relative to the rocket

let's apply these expressions to our case

the initial mass is the mass of the engines plus the mass of the fuel plus the kill of the rocket, let's work the system in SI units

       M₀ = 25.5 +12.7 + 54.5 = 92.7 g = 0.0927 kg

     

The final mass is the mass of the engines + the mass of the rocket

      M_{f} = 25.5 +54.5 = 80 g = 0.080 kg

thrust and duration of ignition are given

       thrust = 5.26 N

       t = 1.90 s

Let's start by calculating the velocity of the gases relative to the rocket, where we assume that the rate of consumption is linear

          thrust = v_{e} \frac{M_{f} - M_{o}  }{t_{f} - t_{o}  }

          v_{e} = thrust  \frac{\Delta t}{\Delta M}

          v_{e} = 5.26 \frac{1.90}{0.080 -0.0927}

          v_{e} = - 786.93 m / s

the negative sign indicates that the direction of the gases is opposite to the direction of the rocket

now we look for the final speed of the rocket, which as part of rest its initial speed is zero

            v_{f}-0 = v_{e} ln ( \frac{M_{o} }{M_{f} } )

we calculate

            v_{f} = 786.93 ln (0.0927 / 0.080)

            v_{f} = 115.95 m / s

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