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3 years ago

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Infer whether a system can have kinetic energy and potential energy at the same time

1 answer:

trapecia [35]3 years ago

7 0

A system can have kinetic and potential energy, for example, an object falling but it hasn't reached the ground. It's still in the air, and has "potential" to go faster or hit something on it's way down, but since it's moving it also has kinetic energy.

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The engine oil at 150 degree Celsius is cooled to 80 degree Celsius in a parallel flow heat exchanger by water entering at 25 de

Setler [38]

Answer:

Explanation:

Attach is the solution

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3 years ago

what equastion do you use to solve Riders in a carnival ride stand with their backs against the wall of a circular room of diame

Hitman42 [59]

Answer:

μsmín = 0.1

Explanation:

There are three external forces acting on the riders, two in the vertical direction that oppose each other, the force due to gravity (which we call weight) and the friction force.
This friction force has a maximum value, that can be written as follows:

$F_{frmax} = \mu_{s} *F_{n} (1)$

where μs is the coefficient of static friction, and Fn is the normal force,

perpendicular to the wall and aiming to the center of rotation.

This force is the only force acting in the horizontal direction, but, at the same time, is the force that keeps the riders rotating, which is the centripetal force.
This force has the following general expression:

$F_{c} = m* \omega^{2} * r (2)$

where ω is the angular velocity of the riders, and r the distance to the

center of rotation (the radius of the circle), and m the mass of the

riders.

Since Fc is actually Fn, we can replace the right side of (2) in (1), as

follows:

$F_{frmax} = m* \mu_{s} * \omega^{2} * r (3)$

When the riders are on the verge of sliding down, this force must be equal to the weight Fg, so we can write the following equation:

$m* g = m* \mu_{smin} * \omega^{2} * r (4)$

(The coefficient of static friction is the minimum possible, due to any value less than it would cause the riders to slide down)

Cancelling the masses on both sides of (4), we get:

$g = \mu_{smin} * \omega^{2} * r (5)$

Prior to solve (5) we need to convert ω from rev/min to rad/sec, as follows:

$60 rev/min * \frac{2*\pi rad}{1 rev} *\frac{1min}{60 sec} =6.28 rad/sec (6)$

Replacing by the givens in (5), we can solve for μsmín, as follows:

$\mu_{smin} = \frac{g}{\omega^{2} *r} = \frac{9.8m/s2}{(6.28rad/sec)^{2} *2.5 m} =0.1 (7)$

5 0

2 years ago

What happens with the plates at this Juan de fuca

djverab [1.8K]

Answer:

Subduction of the Juan de Fuca plate causes melting and magma generation in the mantle which rises to the surface to create the Cascade volcanoes.

Explanation:

8 0

3 years ago

A 3-kg rock is thrown upward with a force of 200 N at a location where the local gravitational acceleration is 9.79 m/s2 . Deter

expeople1 [14]

Answer: $56.87m/s^{2}$

Explanation:

If we make an analysis of the net force $F_{net}$ of the rock that was thrown upwards, we will have the following:

$F_{net}=F_{up}-W$ (1)

Where:

$F_{up}=200N$ is the force with which the rock was thrown

$W$ is the weight of the rock

Being the weight the relation between the mass $m=3kg$ of the rock and the acceleration due gravity $g=9.79m/s^{2}$ :

$W=m.g=(3kg)(9.79m/s^{2})$ (2)

$W=29.37 N$ (3)

Substituting (3) in (1):

$F_{net}=200N-29.37 N$ (4)

$F_{net}=170.63 N$ (5) This is the net Force on the rock

On the other hand, we know this force is equal to the multiplication of the mass with the acceleration, according to Newton's 2nd Law:

$F_{net}=m.a$ (6)

Finding the acceleration $a$ :

$a=\frac{F_{net}}{m}$ (7)

$a=\frac{170.63 N}{3kg}$ (8)

Finally:

$a=56.87m/s^{2}$

3 0

2 years ago

What should you do before each laboratory exercise?

liraira [26]

Answer:

OA. . review all safety procedures and the lab activity procedure

Explanation:

3 0

2 years ago

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