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Rus_ich [418]
3 years ago
5

Bob is pushing a box across the floor at a constant speed of 1.4 m/s, applying a horizontal force whose magnitude is 65 N. Alice

is pushing an identical box across the floor at a constant speed of 2.8 m/s, applying a horizontal force. (a) What is the magnitude of the force that Alice is applying to the box? F = N (b) With the two boxes starting from rest, explain qualitatively what Alice and Bob did to get their boxes moving at different constant speeds. In order to keep the box moving twice as fast, Alice had to apply a constant force that was twice as large as the force that Bob applied. Each initially applied a force bigger than static friction to get the box moving and accelerating, then when the desired final speed was achieved they reduced the force to make the net force zero.
Physics
1 answer:
bija089 [108]3 years ago
5 0

Answer:

a) 65 N

b) Each initially applied a force bigger than static friction to get the box moving and accelerating, then when the desired final speed was achieved they reduced the force to make the net force zero.

Explanation:

a) Here, both boxes are identical. They have different velocities but the velocities are constant hence the force Alice is applying is 65 N. Constant velocity means there is no acceleration.

b) The acceleration that Alice provided must have been different than that of Bob. But after reaching their desired speed they stopped accelerating. This would make the net force zero.

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A uniform stationary ladder of length L and mass M leans against a smooth vertical wall, while its bottom legs rest on a rough h
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Answer:A uniform ladder of mass and length leans at an angle against a frictionless wall .If the coefficient of static friction between the ladder and the ground is , determine a formula for the minimum angle at which the ladder will not slip.

Explanation:A uniform ladder of mass and length leans at an angle against a frictionless wall .If the coefficient of static friction between the ladder and the ground is , determine a formula for the minimum angle at which the ladder will not slip.

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3 years ago
A 1200 kg car is being driven down a road. If it has 101 kJ of kinetic energy, what is its speed?
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Answer:

The solved problem is in the photo. Hope it helps.

5 0
3 years ago
Un automovil parte del reposo y acelera uniformemente hasta alcanzar una rapidez de 0,255km/h en un tiempo de 3/4 Minutos determ
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Answer:

a = 1.5*10^-3 m/s^2

x = 0.033m = 3.3cm

Explanation:

To calculate the acceleration and the distance traveled by the car you use the following formulas:

v=v_o+at    (1)

x=v_ot+\frac{1}{2}at^2   (2)

v: final velocity = 0,255 km/h

vo: initial velocity = 0 m/s

t: time = 3/4 min

a: acceleration = ?

x: distance

In order to use the equations (1) and (2) you first convert the units of the final velocity to m/s, and the time to seconds.

v=0,255\frac{km}{h}*\frac{1000m}{1km}*\frac{1h}{3600s}\\\\v=0.07m/s\\\\t=\frac{3}{4}min*\frac{60s}{1min}=45s

Next, you solve the equation (1) for the acceleration a:

a=\frac{v}{t}=\frac{0.07m/s}{45s}=1.5*10^{-3}\frac{m}{s^2}

With this value of a you can calculate the distance traveled by the car, by using the equation (2):

x=\frac{1}{2}(1.5*10^{-3}m/s^2)(45s)^2=0.033m=3.3cm

hence, the acceleration of the car is 1.5*10^-3 m/s^2 and the distance traveled in 3/4 min is 0.033m

5 0
3 years ago
A diesel engine does not use spark plugs to ignite the fuel and air in the cylinders. Instead, the temperature required to ignit
Klio2033 [76]

Answer:

Compression ratio = 24.42

Explanation:

From Thermodynamic,

Adiabatic Equation ⇒ TV^{γ-1} = Constant.

⇒T₁V₁^{γ-1}  = T₂V₂^{γ-1} ..................(1)

Where T₁= initial Temperature, V₁ = Initial Volume, T₂ = final Temperature, V₂ = Final Volume.

Given:  T₁= 18°C we convert to  Kelvin(K) by adding 273.

∴      T₁= 18°C + 273 = 291K

        T₂ = 733° C  also,we convert to  Kelvin(K) by adding 273

 ∴      T₂ = 733° C +273 =1046K

          γ = 7/5

 ∴ Rearranging equation(1), we have

    (T₁/T₂) = (V₂/V₁)^{γ-1}................(2)

also rearranging equation(2) we have

   (V₁/V₂)^{γ-1} = (T₂/T₁).

Where (V₁/V₂) = Compression ratio.

∴ (V₁/V₂)^{(7/5)-1} =( 1046/291)

simplifying the index in the equation

I.e (7/5)-1 = (7-5 )/5 = 2/5.

(V₁/V₂)^2/5 =(1046/291)^2/5

Multiplying the power on both side of the equation by 5/2.

∴(V₁/V₂)^(2/5)×(5/2) = (1046/291)^(5/2)

⇒ V₁/V₂= (1046/291)^2.5=( 3.59)^2.5

    V₁/V₂ = 24.42.

∴ Compression ratio = 24.42

7 0
3 years ago
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