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uranmaximum [27]
1 year ago
7

hello, I need help with a 25 question test! Which statement is an example of heat transfer by convection?Question 4 options:Jane

's feet burned as she walked across the hot beach sand.Hot air from the sand rose and was replaced by cool air blowing in from the ocean as a sea breeze.The black asphalt of the parking lot was much warmer than the green grass surrounding the lot.People were scattered across the beach, sitting on chairs and towels, warming themselves in the bright sunshine.
Physics
1 answer:
garri49 [273]1 year ago
3 0
\begin{gathered} Heat\text{ transfer by }convection\text{ }occurs\text{ on fluids like air or water, hence, an example of heat } \\ \text{transfer by convection is: "Hot air from the sand rose and was replace by cool aire blowing} \\ in\text{ from the ocean as a sea bre}eze." \end{gathered}

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Compare the densities of two objects that have the same volume, but one feels heavier than the other.
motikmotik
We know, density = Mass / Volume
It represents density is directly proportional to mass and indirectly proportional to volume. So, at a constant volume, object with larger density will appear more heavier than that of object with smaller density.

In short, Heavier object will have the higher density than the other.

Hope this helps!
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4 years ago
Substance A has a higher heat capacity than does substance B, and substance B has a higher heat capacity than does substance C.
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3 years ago
Gbenga needs to get glasses to correct his farsightedness. His eyes currently cannot focus on objects that are within 2 ft (or 6
Scilla [17]

Answer:

Focal Length = 38.61cm, Power = 2.59 Diopter, Converging lens.

Explanation:

When an object is placed 25cm from Gbenga's eye, the glasses lens must produce an image 61cm away (Gbenga's eye near point).

An image 61cm from the eye will be (61cm - 1.6cm) from the glasses.

i.e. d_{i}=61cm-1.6cm=59.4cm

and d_{o} = 25cm - 1.6cm = 23.4cm

note d_{i} will be negative because the image is formed on the same side as the object.

finally, d_{i}=-59.4cm\\d_{o}=23.4cm

the formula for finding the focal length f is given as

f=\frac{d_{i}d_{o}  }{d_{o}+d_{i}  }

f=\frac{-59.4*23.4}{23.4-59.4} \\

f=\frac{-1389.96}{-36}

f=38.61cm

The focal length is positive which indicates converging lens

power p=\frac{1}{f}

but f\\ must be in metres

Therefore, f=38.61cm=0.3861m

p=\frac{1}{0.3861}

p=2.59 Diopter

3 0
4 years ago
Start Point: an unlit match. End Point: a lit match.
nalin [4]

Answer:

Ok

Explanation:

5 0
3 years ago
A Hooke's law spring is mounted horizontally over a frictionless surface. The spring is then compressed a distance d and is used
zloy xaker [14]

Answer:

The compression is \sqrt{2} \  d.

Explanation:

A Hooke's law spring compressed has a potential energy

E_{potential} = \frac{1}{2} k (\Delta x)^2

where k is the spring constant and \Delta x the distance to the equilibrium position.

A mass m moving at speed v has a kinetic energy

E_{kinetic} = \frac{1}{2} m v^2.

So, in the first part of the problem, the spring is compressed a distance d, and then launch the mass at velocity v_1. Knowing that the energy is constant.

\frac{1}{2} m v_1^2 = \frac{1}{2} k d^2

If we want to double the kinetic energy, then, the knew kinetic energy for a obtained by compressing the spring a distance D, implies:

2 * (\frac{1}{2} m v_1^2) = \frac{1}{2} k D^2

But, in the left side we can use the previous equation to obtain:

2 * (\frac{1}{2} k d^2) = \frac{1}{2} k D^2

D^2 =  \frac{2 \ (\frac{1}{2} k d^2)}{\frac{1}{2} k}

D^2 =  2 \  d^2

D =  \sqrt{2 \  d^2}

D =  \sqrt{2} \  d

And this is the compression we are looking for

3 0
4 years ago
Read 2 more answers
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