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3 years ago

Where is heat transferred by conduction in a lava lamp?

Physics

1 answer:

Semmy [17]3 years ago

8 0

It is transferred by direct contact, say you touched it, you would feel the heat

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Freida stands on a hill with a slope of 12 degrees. If her mass is 65 kg, what is the magnitude of the normal force acting on he

aliina [53]

Answer:

Explanation:

7 0

3 years ago

Read 2 more answers

A small balloon is released at a point 150 feet away from an observer, who is on level ground. If the balloon goes straight up a

Elza [17]

Answer:

$\dfrac{dz}{dt}=0.65\ ft/s$

Explanation:

Given that

x= 150 ft

$\dfrac{dy}{dt}= 7\ ft/s$

y= 14 ft

From the diagram

$z^2=x^2+y^2$

When ,x= 150 ft and y= 14 ft

$z^2=150^2+14^2$

$z=\sqrt{150^2+15^2}$

z=150.74 ft

$z^2=x^2+y^2$

By differentiating with respect to time t

$2z\dfrac{dz}{dt}= 2x\dfrac{dx}{dt}+2y\dfrac{dy}{dt}$

$z\dfrac{dz}{dt}= x\dfrac{dx}{dt}+y\dfrac{dy}{dt}$

Here x is constant that is why

$\dfrac{dx}{dt}=0$

$z\dfrac{dz}{dt}= y\dfrac{dy}{dt}$

Now by putting the values in the above equation we get

$150.74\times \dfrac{dz}{dt}=14\times 7$

$\dfrac{dz}{dt}=\dfrac{14\times 7}{150.74}\ ft/s$

$\dfrac{dz}{dt}=0.65\ ft/s$

Therefore the distance between balloon and observer increasing with 0.65 ft/s.

5 0

3 years ago

PLEASE HELP ME (stop putting links ) Two objects m1 and m2, each with a mass of 5 kg and 6 kg separated by a distance. A third o

BabaBlast [244]

Answer:

Explanation:

Newton's Gravitation Law

$\displaystyle \frac{GmM}{d^2}$

where G is a constant, M and M the masses e d the distance betwen masses.

$\displaystyle G\frac{5\cdot2}{x^2}=G\frac{6\cdot 2}{(2x+1)^2} \quad \sqrt{6}x=(2x+1)\sqrt {5} \quad x=\frac{\sqrt{5}}{\sqrt{6}-2\sqrt{5}}$

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3 years ago

How does a 12-month lunar calendar differ from our 12-month solar calendar?.

horrorfan [7]

Answer:

It has about 11 fewer days. It does not have seasons. Its new year always occurs in February instead of on January 1.

Explanation:

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2 years ago

How do you know that forces are balanced when static friction acts on an object?

lyudmila [28]

By looking at the acceleration of the object.
In fact, Netwon's second law states that the resultant of the forces acting on an object is equal to the product between the mass m of the object and its acceleration:
$\sum F = ma$

So, when static friction is acting on the object, if the object is still not moving we know that all the forces are balanced: in fact, since the object is stationary, its acceleration is zero, and so the resultant of the forces (left term in the formula) must be zero as well (i.e. the forces are balanced).

6 0

3 years ago