Answer:
(A) 60 J
Explanation:
At state 1
KE₁=100 J
At state 2
KE₂ = 0
U₂=80 J
Given that surface is rough so friction force will act in opposite to the direction of motion
Lets take work done by friction = Wfr
From work power energy
Work done by all forces = Change in kinetic energy
Wfr + U₂=ΔKE
Wfr+80 = 100
Wfr= 20 J
Now when book slides from top position then
Wfr+ U = KEf - KEi
-20 + 80 = KEf-0
KEf= 60 J
(A) 60 J
Multiply it by a fraction equal to ' 1 ', like this:
(14.8 cm) x (1 meter/100 cm) = 14.8/100 = 0.148 meter
Answer:
(a) 5.43 x 10⁵ J
(b) 3.07 x 10⁵ J
(c) 45 °C
Explanation:
(a)
= Latent heat of fusion of ice to water = 3.33 x 10⁵ J/kg
m = mass of ice = 1.63 kg
= Energy required to melt the ice
Energy required to melt the ice is given as
= m
= (1.63) (3.33 x 10⁵)
= 5.43 x 10⁵ J
(b)
E = Total energy transferred = 8.50 x 10⁵ J
Q = Amount of energy remaining to raise the temperature
Using conservation of energy
E = + Q
8.50 x 10⁵ = 5.43 x 10⁵ + Q
Q = 3.07 x 10⁵ J
(c)
T₀ = initial temperature = 0°C
T = Final temperature
m = mass of water = 1.63 kg
c = specific heat of water = 4186 J/(kg °C)
Q = Amount of energy to raise the temperature of water = 3.07 x 10⁵ J
Using the equation
Q = m c (T - T₀)
3.07 x 10⁵ = (1.63) (4186) (T - 0)
T = 45 °C
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:
Explanation:
Given that the grand stone has initial angular velocity of
w(ini)= 6rad/
And it has a final angular velocity of
w(fin)=12.20rad/sec
Time taken is t=16s
Using equation of angular motion
To get angular acceleration (α)
w(fin)=w(ini)+αt
12.20=6+16α
16α=12.20-6
16α=6.2
α=6.2/16
α=0.3875rad/sec²
The angular acceleration is 0.39rad/s²
Angle that he turn using
w(fin)²=w(ini)²+2αθ
12.2²=6²+2×0.3875θ
12.2²-6²=0.775θ
0.775θ=112.84
Then, θ=112.84/0.775
θ=145.6radian
The angular displacement is 145.6rad
