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Neporo4naja [7]
2 years ago
5

A student conducts an experiment in which a cart is pulled by a variable applied force during a 2sec time interval. In trial 1,

the student exerts the force on a cart of mass M. In trial 2, the student exerts the force on a cart of mass 3M. In trial 3, the student exerts the force on a cart of 5M. In which trial will the cart experience the greatest change in momentum from 0 sec to 2 sec?
Physics
1 answer:
Harlamova29_29 [7]2 years ago
8 0

Answer:

change of momentum does not depend on the mass of the cars, as the force and time are the same all vehicles have the same change of momentum

Explanation:

Let's look for the speed of the car

     F = m a

    a = F / m

We use kinematics to find lips

    v = v₀ + a t

    v = v₀ + (F / m) t

The moment is defined by

    p = m v

The moment change

    Δp = m v - m v₀

Let's replace the speeds in this equation

    Δp = m (v₀ + F / m t) - m v₀

    Δp = m v₀  + F t - m v₀  

    Δp = F t

We see that the change of momentum does not depend on the mass of the cars, as the force and time are the same all vehicles have the same change of momentum

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Answer:

The centripetal acceleration as a multiple of g=9.8 m/s^{2} is 1.020x10^{-3}m/s^{2}

Explanation:

The centripetal acceleration is defined as:

a = \frac{v^{2}}{r}  (1)

Where v is the velocity and r is the radius

Since the person is standing in the Earth surfaces, their velocity will be the same of the Earth. That one can be determined by means of the orbital velocity:

v = \frac{2 \pi r}{T}  (2)

Where r is the radius and T is the period.

For this case the person is standing at a latitude 71.9^{\circ}. Remember that the latitude is given from the equator. The configuration of this system is shown in the image below.

It is necessary to use the radius at the latitude given. That radius can be found by means of trigonometric.

\cos \theta = \frac{adjacent}{hypotenuse}

\cos \theta = \frac{r_{71.9^{\circ}}}{r_{e}} (3)

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r_{71.9^{\circ}} can be isolated from equation 3:

r_{71.9^{\circ}} = r_{e} \cos \theta  (4)

r_{71.9^{\circ}} = (6.37x10^{6}m) \cos (71.9^{\circ})

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Then, equation 2 can be used

v = \frac{2 \pi (1.97x10^{6} m)}{24h}

Notice that the period is the time that the Earth takes to give a complete revolution (24 hours), this period will be expressed in seconds for a better representation of the velocity.

T = 24h . \frac{3600s}{1h} ⇒ 84600s

v = \frac{2 \pi (1.97x10^{6} m)}{84600s}

v = 146.31m/s

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a = \frac{(146.31m/s)^{2}}{(1.97x10^{6}m)}

a = 0.010m/s^{2}

Hence, the centripetal acceleration is 0.010m/s^{2}

To given the centripetal acceleration as a multiple of g=9.8 m/s^{2}​ it is gotten:

\frac{0.010m/s^{2}}{9.8 m/s^{2}} = 1.020x10^{-3}m/s^{2}

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3 years ago
Question 1 of 4 Attempt 4 The acceleration due to gravity, ???? , is constant at sea level on the Earth's surface. However, the
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Answer:

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<em>*At first order on h/R*</em>

Explanation:

Hi!

We can derive this expression for distances h small compared to the earth's radius R.

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F = G \frac{m_1m_2}{r^2}

In the present case m1 will be the mass of the earth.

Additionally, if we remember Newton's second law for the mass m2 (with m2 constant):

F = m_2a

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a(r) = G \frac{m_1}{r^2}

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a(R) =G \frac{m_1}{R^2} \\ \frac{da(r)}{dr}_{r=R} = -2 G\frac{m_1}{R^3}

Therefore:

a(R+h) \approx G\frac{m_1}{R^2} -2G\frac{m_1}{R^2} \frac{h}{R} = g(1-2\frac{h}{R})

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Answer:

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Required

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