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3 years ago

Which statement accurately describes a sample of water during parts a and c of the heating curve

Physics

2 answers:

vivado [14]3 years ago

5 0

Answer:

A and C is about 12 cm away from each other.

Explanation:

Olin [163]3 years ago

3 0

Explanation:

A and C is about 25 ccentimetre away

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natali 33 [55]

As waves get closer to the beach they increase in energy

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2 years ago

Suppose you have thrown a rock nearly straight up at a coconut in a palm tree,and the rock misses on the way up but hits the coc

Margarita [4]

Most likely it would dislodge the coconut on the way down due to gravity because on the way up the gravity would slow down the rock but on they down the gravity pulls the rock

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2 years ago

Sonja [21]

Answer

2) 1.5×10-2 m

Explanation

The potential difference is related to the electric field by:

$\Delta V=Ed$ (1)

where

$\Delta V$ is the potential difference

E is the electric field

d is the distance

We want to know the distance the detectors have to be placed in order to achieve an electric field of

$E=1 V/cm=100 V/m$

when connected to a battery with potential difference

$\Delta V=1.5 V$

Solving the equation (1) for d, we find

$d=\frac{\Delta V}{E}=\frac{1.5 V}{100 V/m}=0.015 m=1.5 \cdot 10^{-2} m$

5 0

2 years ago

a cone is constructed by cutting a sector from a circular sheet of metal with radius . the cut sheet is then folded up and welde

Svet_ta [14]

The expression for the radius and height of the cone can be obtained from

the property of a function at the maximum point.

$The \ radius, \ of \ the \ base \ of \ the \ cone \ is \ \sqrt{ \dfrac{3}{4}} \times radius \ of \ circular \ sheet \ metal$

The height of the cone is half the length of the radius of the circular sheet metal.

Reasons:

The part used to form the cone = A sector of a circle

The length of the arc of the sector = The perimeter of the circle formed by the base of the cone.

$Volume \ of \ a \ cone = \dfrac{1}{3} \cdot \pi \cdot r^3 \cdot h$

$Volume \ of \ a \ cone, \, V = \dfrac{1}{3} \cdot \pi \cdot r^3 \cdot \sqrt{(s^2- r^2)}$

θ/360·2·π·s = 2·π·r

Where;

s = The radius of he circular sheet metal

h = s² - r²

$\dfrac{dV}{dr} = \dfrac{d}{dr} \left(\dfrac{1}{3} \cdot \pi \cdot r^3 \cdot \sqrt{(s^2- r^2)}\right) = \dfrac{\pi \cdot (3 \cdot r^2 \cdot s^2 - 4 \cdot r^4)}{\sqrt{(s^2- r^2)}} = 0$