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horrorfan [7]
3 years ago
12

Hardness and density are physical properties. a. True b. False

Physics
2 answers:
antiseptic1488 [7]3 years ago
7 0
a . true hardness and density are physical properties
VMariaS [17]3 years ago
5 0

Answer: The Answer would be True.

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Calculate the gravitational potential energy of the interacting pair of the Earth and a 11 kg block sitting on the surface of th
trasher [3.6K]

Answer:

Gravitational potential energy (GPE) = 107.8J

Explanation:

Gravitational potential energy (GPE) = mgh

Where mass(m) = 11kg

Acceleration due to gravity(g) = 9.8m²/s

height = assumed to be 1m

Force(F) = mg

Force(F) = 11×9.8 = 107.8N

Gravitational potential energy (GPE) = 107.8×1

= 107.8J

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3 years ago
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elixir [45]

Answer:

draw the two atomic structures and then focus in on the electrons either gaining or losing and draw them with arrows and dots

Explanation:

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3 years ago
Example of a static balance
Rufina [12.5K]
Ability to maintain ones balance when not moving.
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8. The force constant of a spring is 150. N/m. (a) How much force is required to stretch the spring 0.25 m? (b) How much work is
Ne4ueva [31]

Answer:

<em>a) 37.5N</em>

<em>b) 9.375Joules</em>

Explanation:

a) According to Hooke's law

F = ke

k is the spring constant

e is the extension;

F = 150 * 0.25

F = 37.5N

b) Work done on the spring = 1/2ke^2

Work done on the spring = 1/2 * 150 * 0.25^2

Work done on the spring  = 75 * 0.0625

Work done on the spring  = 9.375Joules

4 0
3 years ago
A 234.0 g piece of lead is heated to 86.0oC and then dropped into a calorimeter containing 611.0 g of water that initally is at
Vaselesa [24]

Answer:24.70 ^{\circ}C

Explanation:

Given

mass of lead piece m_l=234 gm\approx 0.234 kg

mass of water in calorimeter m_w=611 gm\approx 0.611 kg

Initial temperature of water T_w=24^{\circ}C

Initial temperature of lead piece T_l=24^{\circ}C

we know heat capacity of lead and water are 125.604 J/kg-k and 4.184 kJ/kg-k respectively

Let us take T ^{\circ}C be the final temperature of the system

Conserving energy

heat lost by lead=heat gained by water

m_lc_l(T_l-T)=m_wc_w(T-T_w)

0.234\times 125.604(86-T)=0.611\times 4.184\times 1000(T-24)

86-T=\frac{0.611\times 4.184\times 1000}{29.391}(T-24)

86-T=86.97T-2087.49

T=\frac{2173.491}{87.97}=24.70^{\circ}C

3 0
3 years ago
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