Which of these statements best describes how energy is transferred when an object experiences friction?

slader the cross section of a 5-ft long trough is an isosceles trapezoid with a 2 foot lower base, a 3-foot upper base, and an a

Ostrovityanka [42]

Answer:

0.08 ft/min

Explanation:

To get the speed at witch the water raising at a given point we need to know the area it needs to fill at that point in the trough (the longitudinal section), which is given by the height at that point.

So we need to get the lenght of the sides for a height of 1 foot. Given the geometry of the trough, one side is the depth d and the other (lets call it l) is given by:

$l=\frac{3-2}{2}\,ft+2\,ft\\l=2.5\,ft$

since the difference between the upper and lower base is the increase in the base and we are only at halft the height.

Now we can calculate the longitudinal section A at that point:

$A=d\times l\\A=5\,ft \times 2.5\, ft\\A=12.5\, ft^{2}$

And the raising speed v of the water is given by:

$v=\frac{q}{A}\\v=\frac{1\, \frac{ft^3}{min}}{12.5\, ft^2}\\v=0.08\, \frac{ft}{min}$

where q is the water flow (1 cubic foot per minute).

7 0

3 years ago

Adhesion is when forces of the molecules are more strongly drawn to each other than the molecules in other substances true or fa

SSSSS [86.1K]

Answer:

False

Explanation:

that is cohesion. adhesion is force between dissimilar molecules of a body

6 0

3 years ago

IBM has a fast computer that it calls the Blue Gene/L that can do '136.8

dmitriy555 [2]

Answer:

138.6 megacalculations

Explanation:

This is a pretty straightforward one.

All it needs is to convert the degree of measurement.

Dimensions in physics are attributed names, which state the power to which they're are raised. Just as how

Kilo and Mega means the numbers are raised to the power of 3 and 6 respectively. There also exists the ones that indicates how small, such as milli and micro, which are to the powers of -3 & -6.

The question says the IBM computer calculates at an astonishing 136.8 teracalculations.

Tera in physics means it's raised to the power of 12. Thus, the IBM calculates at an astonishing rate of

136.8*10^12 calculations per second.

We're then asked how many calculations it does in 1 micro second. Like I had highlighted earlier, 1 micro second is 1 raised to the power of -6. Or succinctly put,

1 micro second = 1*10^-6.

If the IBM does

138.6*10^12 = 1 second,

Then it does

x = 1*10^-6 second.

When we cross multiply, we have

138.6*10^12 * 1*10^-6, and that is

138.6*10^6 calculations, or say, 138.6 megacalculations.

The IBM does 138.6 megacalculations in 1 micro second, which is still astonishing, by the way

6 0

2 years ago

Suppose that the height of the incline is h = 14.7 m. Find the speed at the bottom for each of the following objects. 1.solid sp

tensa zangetsu [6.8K]

Answer:

1. 14.4 m/s 2. 13.2 m/s 3. 12.0 m/s 4. 13.9 m/s

Explanation:

Assuming no friction present, the different objects roll without slipping, so there is a constant relationship between linear and angular velocity, as follows:

ω= v/r

If no friction exists, the change in total kinetic energy must be equal in magnitude to the change in the gravitational potential energy:

∆K = -∆U

½ *m*v² + ½* I* ω² = m*g*h

Simplifying and replacing the value of the angular velocity:

½ * v² + ½ I *(v/r)² = g*h (1)

In order to answer the question, we just need to replace h by the value given, and I (moment of inertia) for the value for each different object, as follows:

Solid Sphere I = 2/5* m *r²

Replacing in (1):

½ * v² + ½ (2/5 *m*r²) *(v/r)² = g*h

Replacing by the value given for h, and solving for v:

v = √(10/7*9.8 m/s2*14.7 m) = 14. 4 m/s

Spherical shell I=2/3*m*r²

Replacing in (1):

½ * v² + ½ (2/3 *m*r²) *(v/r)² = g*h

Replacing by the value given for h, and solving for v:

v = √(6/5*9.8 m/s2*14.7 m) = 13.2 m/s

Hoop I= m*r²

Replacing in (1):

½ * v² + ½ (m*r²) *(v/r)² = g*h

Replacing by the value given for h, and solving for v:

v = √(9.8 m/s2*14.7 m) = 12.0 m/s

Cylinder I = 1/2 * m* r²

Replacing in (1):

½ * v² + ½ (1/2 *m*r²) *(v/r)²= g*h

Replacing by the value given for h, and solving for v:

v = 2*√(1/3*9.8 m/s2*14.7 m) = 13.9 m/s

5 0

3 years ago

A surveyor is using a magnetic compass 6.4 m below a power line in which there is a steady current of 120 A. (a) What is the mag

Mademuasel [1]

Answer:

Explanation:

Magnetic field due to a long current carrying conductor

μ₀ / 4π x 2i / r ( i = current , r = distance of point from wire )

= 10⁻⁷ x 2 x 120 / 6.4 ( i = 120 A , r = 6.4m )

= 37.5 x 10⁻⁷ T .

= 3. 75 X 10⁻⁶ T .

= 3.75 µT.

b )

The direction of this field will be horizontal hence it will affect magnetic needle.

6 0

3 years ago

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