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Ivenika [448]
2 years ago
6

A car is pulled with a force of 10,000 N. The car's mass is 1267 kg. But, the car covers 394.6 m in 15 seconds

Physics
2 answers:
Lera25 [3.4K]2 years ago
6 0

Answer:

Part a)

a = 7.89 m/s^2

Part b)

a = 3.51 m/s^2

Part c)

\Deltra a = 4.38 m/s^2

Part d)

This difference in acceleration  is due to some frictional force on the surface.

Part e)

F_f = 5552.8 N

Explanation:

Part a)

As we know by newton's II law

F = ma

here we know that

m = 1267 kg

F = 10,000 N

Now we have

a = \frac{F}{m}

a = \frac{10,000}{1267}

a = 7.89 m/s^2

Part b)

distance covered by the car

d = 394.6 m

t = 15 s

now by kinematics we have

d = \frac{1}{2}at^2

394.6 = \frac{1}{2}a(15^2)

a = 3.51 m/s^2

Part c)

Difference of acceleration is given as

\Delta a = a_{expected} - a_{real}

\Delta a = 7.89 - 3.51

\Deltra a = 4.38 m/s^2

Part d)

This difference in acceleration  is due to some frictional force on the surface.

Part e)

Now for magnitude of force is given as

F - F_f = ma

10,000 - F_f = ma

10,000 - F_f = 1267\times 3.51

F_f = 5552.8 N

solong [7]2 years ago
3 0

A. Formula: F=ma or F/m=a

10,000N/1,267kg≈7.9m/s^{2}

B. Formula: a=\frac{V-V_{0} }{t} and s=d/t

speed= 394.6/15

s=26.3m/s

a=\frac{26.3-0}{15}

a=1.75m/s^{2}

C. 7.9-1.75=difference of 6.15m/s^{2}

D. The force that most likely caused this difference is friction forces

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Answer:

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Explanation:

first write the newtons second law:

F_{s}=δma_{s}

Applying bernoulli,s equation as follows:

∑δp+\frac{1}{2} ρδV^{2} +δγz=0\\

Where, δp is the pressure change across the streamline and V is the fluid particle velocity

substitute ρg for {tex]γ[/tex] and g_{0}-cz for g

dp+d(\frac{1}{2}V^{2}+ρ(g_{0}-cz)dz=0

integrating the above equation using limits 1 and 2.

\int\limits^2_1  \, dp +\int\limits^2_1 {(\frac{1}{2}ρV^{2} )} \, +ρ \int\limits^2_1 {(g_{0}-cz )} \,dz=0\\p_{1}^{2}+\frac{1}{2}ρ(V^{2})_{1}^{2}+ρg_{0}z_{1}^{2}-ρc(\frac{z^{2}}{2})_{1}^{2}=0\\p_{2}-p_{1}+\frac{1}{2}ρ(V^{2}_{2}-V^{2}_{1})+ρg_{0}(z_{2}-z_{1})-\frac{1}{2}ρc(z^{2}_{2}-z^{2}_{1})=0\\p+\frac{1}{2}ρV^{2}+ρg_{0}z-\frac{1}{2}ρcz^{2}=constant

there the bernoulli equation for this flow is p+\frac{1}{2}ρV^{2}+ρg_{0}z-\frac{1}{2}ρcz^{2}=constant

note: ρ=density(ρ) in some parts and change(δ) in other parts of this equation. it just doesn't show up as that in formular

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2 years ago
A 4.0-m-diameter playground merry-go-round, with a moment of inertia of 350 kg⋅m2 is freely rotating with an angular velocity of
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Answer:

v = 4.375\,\frac{m}{s}

Explanation:

The situation of the system Ryan - merry-go-round is modelled after the Principle of the Angular Momentum Conservation:

(350\,kg\cdot m^{2})\cdot (1.5\,\frac{rad}{s} ) - (2\,m)\cdot (60\,kg)\cdot v = 0\,kg\cdot \frac{m^{2}}{s}

The initial speed of Ryan is:

v = 4.375\,\frac{m}{s}

5 0
2 years ago
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Suppose your friend claims to have discovered a mysterious force in nature that acts on all particles in some region of space. H
kirill [66]

Answer:

             U = 1 / r²

Explanation:

In this exercise they do not ask for potential energy giving the expression of force, since these two quantities are related

             

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this derivative is a gradient, that is, a directional derivative, so we must have

          dU = - F. dr

the esxresion for strength is

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let's replace

          ∫ dU = - ∫ B / r³  dr

in this case the force and the displacement are parallel, therefore the scalar product is reduced to the algebraic product

let's evaluate the integrals

            U - Uo = -B (- / 2r² + 1 / 2r₀²)

To complete the calculation we must fix the energy at a point, in general the most common choice is to make the potential energy zero (Uo = 0) for when the distance is infinite (r = ∞)

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we substitute the value of B = 2

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2 years ago
Technician a says that using a pressure transducer and lab scope is a similar process to using a vacuum gauge. technician b says
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Answer: Both Technician A and B

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A bird is flying in a room with a velocity field of . Calculate the temperature change that the bird feels after 9 seconds of fl
Korvikt [17]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The temperature change is \frac{dT}{dt} = 1.016 ^oC/m

Explanation:

From the question we are told that

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   The temperature of the room is  T(x, y, u) =  400 -0.4y -0.6z-0.2(5 - x)^2 \  ^o C

    The time considered is  t =  10 \  seconds

    The  distance that the bird flew is  x  =  1 m

 Given that the bird is inside the room then the temperature of the room is equal to the temperature of the bird

Generally the change in the bird temperature with time is mathematically represented as

      \frac{dT}{dt} = -0.4 \frac{dy}{dt} -0.6\frac{dz}{dt} -0.2[2 *  (5-x)] [-\frac{dx}{dt} ]

Here the negative sign in \frac{dx}{dt} is because of the negative sign that is attached to x in the equation

 So

       \frac{dT}{dt} = -0.4v_y  -0.6v_z -0.2[2 *  (5-x)][ -v_x]

From the given equation of velocity field

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     v_z  =  -1.4

So

\frac{dT}{dt} = -0.4[0.2t]  -0.6[-1.4] -0.2[2 *  (5-x)][ -[0.6x]]    

substituting the given values of x and t

\frac{dT}{dt} = -0.4[0.2(10)]  -0.6[-1.4] -0.2[2 *  (5-1)][ -[0.61]]      

\frac{dT}{dt} = -0.8 +0.84 + 0.976  

\frac{dT}{dt} = 1.016 ^oC/m  

5 0
2 years ago
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