Answer:
335°C
Explanation:
Heat gained or lost is:
q = m C ΔT
where m is the mass, C is the specific heat capacity, and ΔT is the change in temperature.
Heat gained by the water = heat lost by the copper
mw Cw ΔTw = mc Cc ΔTc
The water and copper reach the same final temperature, so:
mw Cw (T - Tw) = mc Cc (Tc - T)
Given:
mw = 390 g
Cw = 4.186 J/g/°C
Tw = 22.6°C
mc = 248 g
Cc = 0.386 J/g/°C
T = 39.9°C
Find: Tc
(390) (4.186) (39.9 - 22.6) = (248) (0.386) (Tc - 39.9)
Tc = 335
Answer:
0.5 m
Explanation:
Givens:
ym1 = 2.5 mm
ym2 = 4.5 mm
Ф_1=π / 4
Ф_2=π / 2
We have 2 ways to solve this problem. The first one given that the 2 waves have the frequency then we know that the resultant wave amplitude is
Ym = (ym1 + ym2)cos(Ф_2/2)
By substitution we have
Ym= (0.025 + 0.045)cos(π/4) = 0.496 m
The second one is it treat them as Phasors where the phase between them is Ф_2=π / 2 Therefore
Ym^2=(ym1^2+ym2^2)
So we have Ym=√0.025^2+0.045^2
= 0.5 m
That prediction is not correct because Xenon is extremely stable; column 18 of the periodic table contains the noble gasses, which are stable because their outer-most energy levels are completely filled. Having the octet (8) of valence electrons means that the element no longer needs to lose or gain electrons to gain stability.
The column 17 elements are unstable because they only have one valence electron short of the stable octet configuration of the noble gasses.
Answer:
Hz
Explanation:
We know that
1 cm = 0.01 m
= Length of the human ear canal = 2.5 cm = 0.025 m
= Speed of sound = 340 ms⁻¹
= First resonant frequency
The human ear canal behaves as a closed pipe and for a closed pipe, nth resonant frequency is given as
for first resonant frequency, we have n = 1
Inserting the values
Hz
Answer:
I = 16 kg*m²
Explanation:
Newton's second law for rotation
τ = I * α Formula (1)
where:
τ : It is the moment applied to the body. (Nxm)
I : it is the moment of inertia of the body with respect to the axis of rotation (kg*m²)
α : It is angular acceleration. (rad/s²)
Kinematics of the wheel
Equation of circular motion uniformly accelerated :
ωf = ω₀+ α*t Formula (2)
Where:
α : Angular acceleration (rad/s²)
ω₀ : Initial angular speed ( rad/s)
ωf : Final angular speed ( rad
t : time interval (rad)
Data
ω₀ = 0
ωf = 1.2 rad/s
t = 2 s
Angular acceleration of the wheel
We replace data in the formula (2):
ωf = ω₀+ α*t
1.2= 0+ α*(2)
α*(2) = 1.2
α = 1.2 / 2
α = 0.6 rad/s²
Magnitude of the net torque (τ )
τ = F *R
Where:
F = tangential force (N)
R = radio (m)
τ = 80 N *0.12 m
τ = 9.6 N *m
Rotational inertia of the wheel
We replace data in the formula (1):
τ = I * α
9.6 = I *(0.6
)
I = 9.6 / (0.6
)
I = 16 kg*m²
