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3 years ago

16. A 7500 kg 18-wheeler traveling at 20 m/s exits onto the runaway truck ramp on the freeway.

Physics

1 answer:

miskamm [114]3 years ago

7 0

Answer:

<em>765,000 Joule</em>

Explanation:

<u>Principle of Conservation of Energy </u>

The total energy in an isolated system cannot be created or destroyed, but transformed. Moving objects have kinetic energy, objects placed in some height above a reference level have gravitational potential energy. When they change their motion variables, one energy converts into the other, but if the numbers don't fit, we know there was some other type of energy acting into the system. The most common reason for energy 'losses' is the thermal energy, produced when objects move in rough surfaces or take friction from the air.

The 7,500 kg truck is originally traveling at 20 m/s to a certain height we'll set to 0. Thus, its total energy is

$\displaystyle E_1=\frac{mv^2}{2}$

$\displaystyle E_1=\frac{7,500\ 20^2}{2}$

$E_1=1,500,000\ Joule$

When it comes to a stop, its speed is 0 and its height is 10 m higher than before. It means all the kinetic energy was transformed into other types of energy. The gravitational potential energy is

$U=mgh=(7,500)(9.8)(10)=735,000\ Joule$

Since this number is not equal to the previous value of the energy, the difference is due to thermal energy dissipated by friction

$E_t=1,500,000\ Joule-735,000\ Joule=765,000\ Joule$

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A person exerts a tangential force of 37.7 N on the rim of a disk-shaped merry-go-round of radius 2.75 m and mass 144 kg. If the

Shkiper50 [21]

Answer:

ω = 0.467 rad/s

Explanation:

given,

tangential force exerted by the person = 37.7 N

radius of merry-go-round = 2.75 m

mass of merry-go-round = 144 Kg

angle = 33.2°

moment of inertia

$I = \dfrac{1}{2} m R^2$

$I = \dfrac{1}{2}\times 144 \times 2.75^2$

I = 544.5 kg.m²

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3 years ago

A baseball player hits a 140 g baseball with a force of 2800 N. What is the

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Spacetime interval: What is the interval between two events if in some given inertial reference frame the events are separated b

shepuryov [24]

Answer:

a. $\Delta s ^2 = 8.0888 \ 10^{17} m^2$

b. $\Delta s ^2 = 3.0234 \ 10^{16} m^2$

c. $\Delta s ^2 = 3.0234 \ 10^{20} m^2$

Explanation:

The spacetime interval $\Delta s^2$ is given by

$\Delta s ^2 = \Delta (c t) ^ 2 - \Delta \vec{x}^2$

please, be aware this is the definition for the signature ( + - - - ), for the signature (- + + + ) the spacetime interval is given by:

$\Delta s ^2 = - \Delta (c t) ^ 2 + \Delta \vec{x}^2$ .

Lets work with the signature ( + - - - ), and, if needed in the other signature, we can multiply our interval by -1.

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$\Delta \vec{x}^2 = 5,625 m^2$

$\Delta (c t) ^ 2 = (299,792,458 \frac{m}{s} \ 3 \ s)^2$

$\Delta (c t) ^ 2 = (899,377,374 \ m)^2$

$\Delta (c t) ^ 2 = 8.0888 \ 10^{17} m^2$

$\Delta s ^2 = 8.0888 \ 10^{17} m^2 - 5,625 m^2$

$\Delta s ^2 = 8.0888 \ 10^{17} m^2$

$\Delta \vec{x}^2 = (5 \ 10 \ m)^2$

$\Delta \vec{x}^2 = 2,500 m^2$

$\Delta (c t) ^ 2 = (299,792,458 \frac{m}{s} \ 0.58 \ s)^2$

$\Delta (c t) ^ 2 = (173,879,625.6 \ m)^2$

$\Delta (c t) ^ 2 = 3.0234 \ 10^{16} m^2$

$\Delta s ^2 = 3.0234 \ 10^{16} m^2 - 2,500 m^2$

$\Delta s ^2 = 3.0234 \ 10^{16} m^2$

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$\Delta \vec{x}^2 = 2,500 m^2$

$\Delta (c t) ^ 2 = (299,792,458 \frac{m}{s} \ 58 \ s)^2$

$\Delta (c t) ^ 2 = (1.73879 \ 10^{10} \ m)^2$

$\Delta (c t) ^ 2 = 3.0234 \ 10^{20} m^2$

$\Delta s ^2 = 3.0234 \ 10^{20} m^2 - 2,500 m^2$

$\Delta s ^2 = 3.0234 \ 10^{20} m^2$

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