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3 years ago

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85. A 1500 kg car accelerates forward due to an applied force of 25001 while experiencing 175 N of friction a. What is the magni

tude of the weight of the car? b. What is the magnitude of the force that causes the acceleration? c. What is the acceleration of the car?

1 answer:

klio [65]3 years ago

6 0

Answer:

See answers below

Explanation:

a.

F = mg,

F = (1500 kg)(9.8 m/s²) = 14,700 N

b.

2500 N - 175 N = 2325 N

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a. The ball's horizontal and vertical positions at time $t$ are given by

$x=\left(27.4\,\dfrac{\rm m}{\rm s}\right)t$

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The ball reaches the net when $x=12.0\,\rm m$ :

$12.0\,\mathrm m=\left(27.4\,\dfrac{\rm m}{\rm s}\right)t\implies t=0.438\,\rm s$

At this time, the ball is at an altitude of

$2.34\,\mathrm m-\dfrac g2\left(0.438\,\mathrm s\right)^2=1.40\,\mathrm m$

which is 1.40 m - 0.900 m = 0.500 m above the net.

b. The change in angle gives the ball the new position functions

$x=\left(27.4\,\dfrac{\rm m}{\rm s}\right)\cos(-5.00^\circ)t$

$y=2.34\,\mathrm m+\left(27.4\,\dfrac{\rm m}{\rm s}\right)\sin(-5.00^\circ)t-\dfrac g2t^2$

The ball reaches the net at time $t$ such that

$\left(27.4\,\dfrac{\rm m}{\rm s}\right)\cos(-5.00^\circ)t=12.0\,\mathrm m\implies t=0.440\,\mathrm s$

at which point the ball's vertical position would be

$2.34\,\mathrm m+\left(27.4\,\dfrac{\rm m}{\rm s}\right)\sin(-5.00^\circ)\left(0.440\,\mathrm s\right)-\dfrac g2\left(0.440\,\mathrm s\right)^2=0.343\,\mathrm m$

so that the ball does not clear the net with 0.343 m - 0.900 m = -0.557 m.

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3 years ago

When a hot and cold object are placed in contact, the hot one loses energy. Does this violate energy conservation? Why or why no

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Answer:

This does not violate the conservation of energy.

Explanation:

This does not violate the conservation of energy because the hot body gives energy in the form of heat to the colder body, this second absorbs energy. This will be the case until both bodies reach the same temperature, reaching thermal equilibrium and reducing the transfer of thermal energy. In this way the energy was only transferred from one body to another but the total energy of the system (body 1 plus body 2) will be the same as in the beginning, respecting the principle of conservation of energy or also called the first principle of thermodynamics .

The part of physics that studies these processes is in turn called heat transfer or heat transfer or thermal transfer. Heat transfer occurs whenever there is a thermal gradient or when two systems with different temperatures come into contact. The process persists until thermal equilibrium is reached, that is, until temperatures are equalized. When there is a temperature difference between two objects or regions close enough, the heat transfer cannot be stopped, it can only be slowed down.

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4 years ago

A Boulder drops in the water and creates a wave with a period of 2s/cycle and a wavelength of .75 m/cycle. How fast is the wave

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Answer:

v = 0.375 m/s

Explanation:

Given that,

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The period of a wave is 2s/cycle

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3 years ago

Two floors in a building are separated by 4.1 m. People move between the two floors on a set of stairs. (a) Determine the change

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Answer:

a) The change in potential energy of a 3.0 kilograms backpack carried up the stairs.

b) The change in potential energy of a persona with weight 650 newtons that descends the stairs is -2665 joules.

Explanation:

Let consider the bottom of the first floor in a building as the zero reference ( $z = 0\,m$ ). The change in potential energy experimented by a particle ( $\Delta U_{g}$ ), measured in joules, is:

$\Delta U_{g} = m\cdot g\cdot (z_{f}-z_{o})$ (1)

Where:

$m$ - Mass, measured in kilograms.

$g$ - Gravitational acceleration, measured in meters per square second.

$z_{o}$ , $z_{f}$ - Initial and final height with respect to zero reference, measured in meters.

Please notice that $m\cdot g$ is the weight of the particle, measured in newtons.

a) If we know that $m = 3\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ , $z_{o} = 0\,m$ and $z_{f} = 4.1\,m$ , then the change in potential energy is:

$\Delta U_{g} = (3\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (4.1\,m-0\,m)$

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The change in potential energy of a 3.0 kilograms backpack carried up the stairs.

b) If we know that $m\cdot g = 650\,N$ , $z_{o} = 4.1\,m$ and $z_{f} = 0\,m$ , then the change in potential energy is:

$\Delta U_{g} = (650\,N)\cdot (0\,m-4.1\,m)$

$\Delta U_{g} = -2665\,J$

The change in potential energy of a persona with weight 650 newtons that descends the stairs is -2665 joules.

5 0

3 years ago