What collides and creates a movement of heat called conduction?

valina [46]

Answer:

It is generally customary in a free-body diagram to represent the object by a box and to draw the force arrow from the center of the box outward in the direction that the force is acting. An example of a free-body diagram is shown at the right. T he free-body diagram above depicts four forces acting upon the object.

Explanation:

3 0

2 years ago

An automobile spare tire is inflated to a certain pressure. When the tire is placed on a car, the weight of the car causes the p

Eduardwww [97]

When the tire is on the car, the pressure is greatest INSIDE the tire,

in that dark chamber where the captive air is trapped. It is the same

pressure at every point inside there, and if it isn't greater than the

pressure OUTSIDE the tire, then you've got a flat.

4 0

3 years ago

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At the beginning of 2016, robotics inc. acquired a manufacturing facility for $13.1 million. $10.1 million of the purchase price

ki77a [65]

$ 376153 is depreciation on the building for 2018.

<u>Explanation:</u>

Given data:

Original cost = $13.1 million

Residual value = $2.1 million

Useful life = 20 -year

Depreciation on the building for 2016 = $=\frac{\text {original cost-residual value}}{\text { Useful life }}$

By substituting the given values, we get,

$=\frac{(\$ 13.1 \text { million }-\$ 2.1\text { million) } }{20 \text { years }}=\frac{11.1 \text { million }}{20 \text { years }}$ = 5500000

Depreciation for 2016 $5500000

Depreciation for 2017 $5500000

Depreciation at Dec 31, 2017 $ 5500000

In 2018, the estimates of useful life and residual value were changed to 15 years and $610000, but 2 yrs have already passed, so there're 13 more yrs to go.

Depreciation on the building for 2018,

$=\frac{\text { book value at the end of } 2016-\text {residual value}}{\text { useful life }}$

= $=\frac{\$ 5500000-\$ 610000}{13}=\frac{4890000}{13}=\$ 376153$

4 0

2 years ago

A 13-cm-diameter cd has a mass of 25 g . part a what is the cd's moment of inertia for rotation about a perpendicular axis throu

faltersainse [42]

<span>Answer: For a disc, the moment of inertia about the perpendicular axis through the center is given by 0.5MR^2. where M is the mass of the disc and R is the radius of the disc. For the axis through the edge, use parallel axis theorem. I = I(axis through center of mass) + M(distance between the axes)^2 = 0.5MR^2 + MR^2 (since the axis through center of mass is the axis through the center) = 1.5 MR^2</span>

8 0

2 years ago

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Calculate the force of gravity between planet X and planet y if both planets are 3.75 X 10^11 m apart, planet X has a mass of 1.

GenaCL600 [577]

So, the force of gravity that the asteroid and the planet have on each other approximately $\boxed{\sf{2.9 \times 10^{17} \: N}}$

<h3>Introduction</h3>

Hi ! Now, I will help to discuss about the gravitational force between two objects. The force of gravity is not affected by the radius of an object, but radius between two object. Moreover, if the object is a planet, the radius of the planet is only to calculate the "gravitational acceleration" on the planet itself,does not determine the gravitational force between the two planets. For the gravitational force between two objects, it can be calculated using the following formula :

$\boxed{\sf{\bold{F = G \times \frac{m_1 \times m_2}{r^2}}}}$