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netineya [11]
3 years ago
6

What is inelastic deformation?

Physics
1 answer:
EleoNora [17]3 years ago
7 0
The answer is A. Deformation is where the object doesnt return to its original shape.
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Explain how characteristic and traits are related<br>​
jok3333 [9.3K]
Traits are basically your phenotype. They include things like hair color, height, and eye color. Alleles are versions of genes. ... This is a pretty basic idea of how traits and alleles are related.
4 0
3 years ago
What will be the final temperature if a 4.00 g silver ring at 41.0◦C if it gives off 18.0 J of heat to the surroundings? The spe
Eduardwww [97]

Answer:

Final temperature, T_f=21.85^{\circ}

Explanation:

Given that,

Mass of silver ring, m = 4 g

Initial temperature, T_i=41^{\circ}C

Heat released, Q = -18 J (as heat is released)

Specific heat capacity of silver, c=0.235\ J/g\ C

To find,

Final temperature

Solution,

The expression for the specific heat is given by :

Q=mc\Delta T

Q=mc(T_f-T_i)

T_f=\dfrac{-Q}{mc}+T_i

T_f=\dfrac{-18}{4\times 0.235}+41

T_f=21.85^{\circ}

So, the final temperature of silver is 21.85 degrees Celsius.

5 0
3 years ago
Which telescope would be better viewing a faint, distant star? Why?
grin007 [14]
Reflecting telescope. Reflecting telescopes tend to have larger objective (due to the use of mirrors, mirrors are a lot cheaper than lenses) and have the ability to collect more light, while refracting telescopes are limited to objective lenses with smaller diameters due to their structural limitations (chromatic abbreviation, for example). Therefore, reflecting telescopes should be better at viewing faint distant stars
4 0
3 years ago
Two moles of an ideal monatomic gas are at a temperature of 345 K. Then, 2250 J of heat is added to the gas, and 870 J of work i
Lapatulllka [165]

Answer: The final temperature is 470K

Explanation: Using the relation;

Q= ΔU +W

Given, n = 2mol

Initial temperature T1= 345K

Heat =Q= 2250J

Workdone=W=-870J(work is done on gas)

T2 =Final temperature =?

ΔU =3/2nR(T2-T1)

ΔU=3/2 × 2 ×8.314 (T2 - 345)

ΔU=24.942(T2-345)

Therefore Q = 24.942(T2-345)+ (-870)

2250=24.942(T2-345)+ (-870)

125.09=(T2-345)

T2 =470K

Therfore the final temperature is 470K

5 0
3 years ago
An ion accelerated through a potential difference of 115 V experiences an increase in kinetic energy of 7.37 x 1017 J. Calculate
riadik2000 [5.3K]

Answer: 6.408(10)^{-19} C

Explanation:

This problem can be solved by the following equation:

\Delta K=q V

Where:

\Delta K=7.37(10)^{-17} J is the change in kinetic energy

V=115 V is the electric potential difference

q is the electric charge

Finding q:

q=\frac{\Delta K}{V}

q=\frac{7.37(10)^{-17} J}{115 V}

Finally:

q=6.408(10)^{-19} C

4 0
3 years ago
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