Answer:
The specific heat for the metal is 0.466 J/g°C.
Explanation:
Given,
Q = 1120 Joules
mass = 12 grams
T₁ = 100°C
T₂ = 300°C
The specific heat for the metal can be calculated by using the formula
Q = (mass) (ΔT) (Cp)
ΔT = T₂ - T₁ = 300°C - 100°C = 200°C
Substituting values,
1120 = (12)(200)(Cp)
Cp = 0.466 J/g°C.
Therefore, specific heat of the metal is 0.466 J/g°C.
Answer:No, it doesn't move easily downward because it will try to resist the movement ,due to a resistance force of inertia that it possess at rest.
Explanation:when an object has higher or larger mass it tends to resist any motion given to it unlike the one with lower mass.
The larger the mass the more resistance force an object has.
Answer:
a.6.5025 J
b.6.5025 J
Explanation:
We are given that
Mass of pellet,m=0.27 g=
1 kg=1000 g
Spring constant,k=1800 N/m
x=8.5 cm=
1m=100 cm
a.Potential energy stored in the compressed spring is given by
P.E=
b.By using law of conservation of energy
P.E of spring=K.E of the pellet
K.E of the pellet=6.5025 J
In fact, entropy of an isolated system never decreases (2nd law of thermodynamics), unless some external energy is provided in order to "restore" order in the system and decrease its entropy.
(note that when external energy is added to the system, it is no longer "isolated").
*This is only true if the question is referring to a certain system within the universe. If we are considering the universe itself as the system, then this option is no longer correct, because no external energy can be provided to the universe, and since the universe is an isolated system, its entropy can never decrease. If we are considering the universe itself as the system, none of the options is true.
Answer:
5 × 10¹⁰ or 5 billion times louder is the intensity of sound at a rock concert in comparison with that of a whisper
Explanation:
Given that at a rock concert;
the intensity of sound is 110 dB compared to a whisper of 3 dB
Now; how many times louder will that of the whisper be compared to the sound of the rock concert
Using the formula for calculating decibels (dB);
For 110 dB
dB = 10log₁₀(W)
110 dB = 
110 dB = 10¹¹
For 3dB
dB = 10log₁₀(W)
3 dB = 
3 dB = 2
Now:
(110/3)dB = 10¹¹/2
(110/3)dB = 5 × 10¹⁰ or 5 billion times louder is the intensity of sound at a rock concert in comparison with that of a whisper
