Answer:
Adding heat makes the particles move faster so the particles have more kinetic energy when more thermal energy is added
Explanation:
Answer:
1) the new power coming from the amplifier is 19.02 W
2) The distance away from the amplifier now is 5.50 m
3) u₁ = 69.24 m
Therefore have to move u₁ - u ( 69.24 - 5.50) = 63.74 farther
Explanation:
Lets say that I am at a distance "u" from the TV,
Let I₁ be the corresponding intensity of the sound at my location when sound level is 125dB
SO
S(indB) = 10log (I₁/1₀)
we substitute
125 = 10(I₁/10⁻¹²)
12.5 = log (I₁/10⁻¹²)
10^12.5 = I₁/10^-12
I₁ = 10^12.5 × 10^-12
I₁ = 10^0.5 W/m²
Now I₂ will be intensity of sound when corresponding sound level is 107 dB
107 = 10log(I₂/10⁻²)
10.7 = log(I₂/10⁻¹²)
10^10.7 = I₂ / 10^-12
I₂ = 10^10.7 × 10^-12
I₂ = 10^-1.3 W/m²
Now since we know that
I = P/4πu² ⇒ p = 4πu²I
THEN P₁ = 4πu²I₁ and P₂ =4πu²I₂
Therefore
P₁/P₂ = I₁/I₂
WE substitute
P₂ = P₁(I₂/I₁) = 1200 × ( 10^-1.3 / 10^0.5)
P₂ = 19.02 W
the new power coming from the amplifier is 19.02 W
2)
P₁ = 4πu²I₁
u =√(p₁/4πI₁)
u = √(1200/4π × 10^0.5)
u = 5.50 m
The distance away from the amplifier now is 5.50 m
3)
Let I₃ be the intensity corresponding to required sound level 85 dB
85 = 10log(I₃/10⁻¹²)
8.5 = log (I₃/10⁻¹²)
10^8.5 = I₃ / 10^-12
I₃ = 10^8.5 × 10^-12
I₃ = 10^-3.5 w/m²
Now, I ∝ 1/u²
so I₂/I₃ = u₁²/u²
u₁ = √(I₂/I₃) × u
u₁ = √(10^-1.3 / 10^-3.5) × 5.50
u₁ = 69.24 m
Therefore have to move u₁ - u ( 69.24 - 5.50) = 63.74 farther
Answer:
28 miles
Explanation:
3,223 meters/day x 14 days = 45,122 meters
1 meter = 3.28 feet
45,122 meters = 148,000 feet
5280 feet = 1 mile
148,000 feet = 28 miles
Answer:
D) the two spheres remain of equal size.
Explanation:
- Since the body of the sphere is made up of both the same material. Thus the orientation will not affect the expansion. That is solid upon solid and hollow upon the hollow sphere. Hence it can be said that both the sphere expands and is due to the material used for making both of them is the same.
Answer:
v = - 1,715 m / s
, x = 0.0156 m
Explanation:
This is an oscillatory movement exercise, which is described by the expression
x = A cos (wt + Ф)
we can assume that the block is released from its maximum elongation, so the phase constant (Ф) is zero
As we are told that the stone does not affect the movement of the spring mass system, the amplitude and angular velocity do not change, in the upward movement the stone is attached to the mass, but in the downward movement the mass has an acceleration greater than g leave the stone behind, let's look for time, for this we use the definition of speed and acceleration
v = dx / dt
v = - A w sin wt
a = - Aw² cos wt
a = -g
-g = - Aw² cos wt
wt = cos⁻¹ (g / Aw²)
t = 1 / w cos⁻¹ (g / Aw²)
angular velocity and frequency are related
w = 2π f
w = 2π 4
w = 8π rad / s
remember that the angles are in radians
t = 1 / 8π cos⁻¹ (9.8 / (0.07 64π²))
t = 0.039789 1.3473
t = 0.0536 s
let's find the speed for this time
v = - A w sin wt
v = - 0.07 8π sin (8π 0.0536)
v = - 1,715 m / s
the distance is
x = A cos wt
x = 0.07 cos (8π 0.0536)
x = 0.0156 m
