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Zanzabum
4 years ago
11

A ring made of gold is heated and then hammered into a flat, coin-like shape. Is this a physical change? Explain.

Physics
2 answers:
Tamiku [17]4 years ago
7 0
Yes. a physical change is only modifying the structure of it.

a chemical change is permanently changing something.

cutting cheese into slices is a physical change

cooking an egg is a chemical change
andre [41]4 years ago
6 0
Yes,
this would be a physical change because the chemical composition is not being effected. When you change the shape of gold, it will remain gold but have a different look. 
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8 0
3 years ago
Read 2 more answers
If a ball is thrown vertically upward from the roof of 64 foot building with a velocity of 96 ft/sec, then what is the maximum h
masya89 [10]
The height function would be s(t) = -16t^2 + 64t + 32.

Maximum height ---> set s'(t) = 0.

-32t + 64 = 0. ---> t = 2. s(2) = -16(2)^2 + 64(2) + 32 = the maximum height is 96 feet.

To find the velocity when it hits the ground, set s(t) = 0.

-16t^2 + 64t + 32 = 0. Divide by -16.

t^2 - 4t - 2 = 0. The only positive solution is x = 4.45.

s'(4.45) = -32(4.45) + 64 = It's going -78.4 feet/second when it hits the ground.
5 0
3 years ago
2. Why was it important to examine both the color and the streak of your minerals? Think about streak and explain why it’s calle
horrorfan [7]
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3 0
3 years ago
A volume of air increases 0.227 m^3 at a net pressure of 2.07 x 10^7 Pa. How much work is done on the air?
aliya0001 [1]

Answer:

The work done on the air is 4.699 x 10⁶ Joules

Explanation:

Given;

increase in air volume, ΔV = 0.227 m³

net pressure of the air, P = 2.07 x 10⁷ Pa

The work done on the air is given by;

W = PΔV

Where;

W is the work done on the air

P is the net pressure

ΔV  is the increase in air volume

Substitute the given values and solve for work done;

W = (2.07 x 10⁷ Pa) (0.227 m³)

W = 4.699 x 10⁶ Joules

Therefore, the work done on the air is 4.699 x 10⁶ Joules

7 0
3 years ago
What is the velocity of a beam of electrons that goes undeflected when moving perpendicular to an electric and magnetic fields.
Charra [1.4K]

Explanation:

Given that,

Electric field, E = 7900 V/m

Magnetic field, B=9.1\times 10^{-3}\ T

(a) Let v is the velocity of a beam of electrons that goes undeflected when moving perpendicular to an electric and magnetic fields. The magnitude of electric force and the magnetic force balance each other as :

qE=qvB

v=\dfrac{E}{B}

v=\dfrac{7900\ V/m}{9.1\times 10^{-3}\ T}

v=8.68\times 10^5\ m/s

(b) Let r is the radius of the electron orbit. The radius of the motion of electron is given by :

r=\dfrac{mv}{qB}

r=\dfrac{9.1\times 10^{-31}\ kg\times 8.68\times 10^5\ m/s}{1.6\times 10^{-19}\ C\times 9.1\times 10^{-3}\ T}

r=5.4\times 10^{-4}\ m

Hence, this is the required solution.

3 0
3 years ago
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