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ycow [4]
3 years ago
9

You don't learn any movement concepts until high school.

Physics
2 answers:
stepan [7]3 years ago
5 0

Answer:

yes

Explanation:

you don't learn it until high school

hammer [34]3 years ago
4 0

Answer:

true

Explanation:

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Four 5-gram blocks of metal are sitting out in the sun and absorb the same amount of heat energy. Use the following specific hea
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Answer:

There is NO DATA??

Explanation:

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3 years ago
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Metals are good conductors of electricity because they _________
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Metals are good conductors because the molecules that are inside the metal are tightly packed together. This is why the heat moves through the metal quickly.

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3 years ago
The area of the pond is approximately equal to the area of a circle with radius 297m. Find the mass of the ice. Answer in kilogr
True [87]

Answer:

<em>mass of the ice is 254980463.8T kg</em>

<em>where T is the value of the thickness omitted in the question.</em>

Explanation:

The ice on Walden Pond is .......... thick. The area of the pond is approximately equal to the area of a circle with radius 297 m. Find the mass of the ice.  Answer in kg.

<em>The value of the thickness of the ice T is omitted, but I will show the solution, and the real answer can be gotten by multiplying the final calculated answer here by the thickness of the ice omitted.</em>

Given the radius of the equivalent circle of the ice = 297 m'

the area of the ice can be gotten from area A = \pi r^{2} = 3.142*297^{2} = 277152.678 m^2

recall that the density of ice p ≅ 920 kg/m^3

also,

density of ice p = (mass of ice, m) ÷ (volume of ice, v)

i.e p = m/v

and,

m = pv

substituting the value of the density of water p into the equation, we have,

mass of the ice, m = 920v ....... equ 1

The volume of the ice above will be = (area of the ice, A) x (thickness of the ice, T)

i.e v = AT

substituting the value of area A into the equation, we have

v =  277152.678T  ......equ 2

substitute value of v into equ 1

mass of the ice, m = 920 x (277152.678T)

mass of the ice, m = 254980463.8T kg

where T is the thickness of the ice

NB: To get the mass, multiply this answer with the thickness T given in the question.

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3 years ago
Please help I'm stuck on these 2 problems would help so much!! 10 POINTS<br><br> Thanks!
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Its C because the circuit is connected with battery and light bulb.

4 0
4 years ago
Two round rods, one steel andthe other copper, are joined end to end. Each rod is 0.750 mlong and 1.50 cmin diameter. The combin
nikitadnepr [17]

Answer: a) Strain on Steel rod = 0.0001078

b) elongation on the steel rod = 0.00008085m = 0.008085cm = 0.0081mm.

c) strain on Copper Rod = 0.000189

Explanation: a) To obtain the strain of the steel rod, we invoke Hooke's law which states that, provided the elastic limit of A material isn't exceeded, the stress it undergoes is directly proportional to its strain.

(Stress, σ) ∝ (Strain, ε)

The constant of proportionality is called Young's modulus, E.

σ = Eε

For steel, Younger Modulus as obtained from literature = 210GPa.

Strain = Stress/Young's Modulus

Stress = (Force or Load applied)/ Cross sectional Area.

Force applied For the steel = 4000N

Cross sectional Area = (π(D^2))/4

D = 1.50cm = 0.015m

A = 0.0001767 m2

σ = 4000/0.0001767 = 22637238.257 N/m2

ε = σ/E = 22637238.257/(210 × (10^9)) = 0.0001078.

b) To solve for elongation.

Strain, ε = (elongation, dl)/(original length, lo)

Elongation, dl = strain × original length

dl = 0.0001078 × 0.75 = 0.00008085m = 0.008085cm = 0.0081mm.

c) strain in Copper

ε = σ/E; σ = 22637238.257 N/m2

Young's modulus of Copper, from literature, = 120GPa

ε = 22637238.257/(120 × (10^9)) = 0.000189

4 0
3 years ago
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