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3 years ago

How does light interact with the surface of a rough stone?

1 answer:

6 0

The only obvious answer from the information provided would have to be B. because a stone is opaque, when there is an opaque object wit a light shining on it, it does not reflect, it absorbs, so the answer is B.

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A bowling (mass = 7.2 kg, radius = 0.11 m) and a billiard ball (mass = 0.38 kg, radius = 0.028 m) may each be treated as uniform

cestrela7 [59]

Hope this helps you!

7 0

3 years ago

Which word best describes this? Particle O Atom O Molecule O Substance

skelet666 [1.2K]

That’s an atom

I hope that helped

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2 years ago

A ball is launched with initial speed v from the ground level up a frictionless hill. The hill becomes steeper as the ball slide

Triss [41]

Answer:

H(max) = (v²/2g)

Explanation:

The maximum height the ball will climb will be when there is no friction at all on the surface of the hill.

Normally, the conservation of kinetic energy (specifically, the work-energy theorem) states that, the change in kinetic energy of a body between two points is equal to the work done in moving the body between the two points.

With no frictional force to do work, all of the initial kinetic emergy is used to climb to the maximum height.

ΔK.E = W

ΔK.E = (final kinetic energy) - (initial kinetic energy)

Final kinetic energy = 0 J, (since the body comes to rest at the height reached)

Initial kinetic energy = (1/2)(m)(v²)

Workdone in moving the body up to the height is done by gravity

W = - mgH

ΔK.E = W

0 - (1/2)(m)(v²) = - mgH

mgH = mv²/2

gH = v²/2

H = v²/2g.

7 0

3 years ago

100 kw of power is delivered to the other side of a city by a pair of power lines with the voltage difference of 13014.1 v.

FinnZ [79.3K]

A) The power delivered to the lines is
$P_{in}= 100 kW=1 \cdot 10^5 W$
And the voltage at which the lines work is
$V=13014.1 V$
Since the power delivered is the product between the voltage and the current:
$P=VI$
We can find the current flowing in the lines:
$I= \frac{P}{V}= \frac{1 \cdot 10^5 W}{13014.1 V}=7.68 A$

b) The voltage change along each line can be found by using Ohm's law:
$\Delta V = IR = (7.68 A)(10 \Omega)=76.8 V$

c) The power wasted as heat along each line is given by:
$P_d = I^2 R = (7.68 A)^2 (10 \Omega) = 590 W$
And since we have 2 lines, the total power wasted as heat in both lines is
$P_d = 2 \cdot 590 W=1180 W$

6 0

3 years ago

List two ways in which heterogeneous mixtures differ from elements

geniusboy [140]

elements cannot be broken down or resolved into simpler materials. A heterogeneous mixture is a mix of elements, compounds, practically anything, and can be resolved into its components parts by physical methods such as filtration, extraction, sublimation, etc

7 0

3 years ago

Read 2 more answers