Answer:
K=24.17 x 10⁻² J s⁻¹c⁻¹m⁻¹
Explanation:
Rate of flow of heat through a material is given by the following expression
where Q is amount of heat flowing in time t through area A and a medium of thickness d having two faces at temperature difference δT . K is thermal conductivity of the medium .
Here Q = 3.34 x 10⁶/6 , t = 24 x 60 x 60 = 86400 s , A = .332 X .332 = .0110224 m² , δT = 104.7
Put these values here
K=24.17 x 10⁻² J s⁻¹c⁻¹m⁻¹
The coefficient of kinetic friction<span> is the force between two objects when one object is moving, or if two objects are moving against each other</span>
Answer:
Explanation:
Rate of flow of liquid through a tube can be expressed by the following expression
V = π P r⁴ / 8ηl
P is pressure difference between end of tube = 618 Pa
r , radius of tube = .5 x 10⁻²
η is viscosity of liquid flowing = .63
l is length of tube = .10 m
V = 3.14 x 618 x ( .5 x 10⁻² )⁴ / (8 x .63 x .10 )
= 240.64 x 10⁻⁸ m³ /s
mass = 240.64 x 1260 x 10⁻⁸ kg / s
= 3.03 x 10⁻³ kg /s
= 3.03 gram /s .
The best name for the ionic bond that forms between them is Beryllium Bromide.
We have been provided with data,
Beryllium charge, q = 2
Bromine charge, q = -1
As we know the valance electron of Be is +2 and the valance electron of bromine is -1. Since one is metallic and the other is non-metallic.
Now, when they combine they exchange valance electron, and bromine change into bromide so they form Beryllium Bromide.
So, the best name for the ionic bond that forms between them is Beryllium Bromide.
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By power transmitted by string, the frequency must be 163.31 Hz.
We need to know about power transmitted to solve this problem. The power transmitted by a wave on string can be determined by this equation
P = 1/2 . μ . ω² . A² . v
where P is the power, μ is mass per unit length of string, ω is angular speed, A is amplitude and v is wave propagation speed.
the wave propagation can be determined as
v = √(F.l/m)
where F is the string tension, l is length and m is the mass.
From the question above, we know that:
l = 2.7 m
m = 260 g = 0.26 kg
F = 36 N
A = 7.7 mm = 0.0077 m
P = 58 W
Find the mass per unit length
μ = m / l
μ = 0.26 / 2.7
μ = 0.096 kg / m
Find the wave propagation speed
v = √(F.l/m)
v = √(36. 2.7 /0.26)
v = √(373.85)
v = 19.34 m/s
Find the angular speed
P = 1/2 . μ . ω² . A² . v
58 = 1/2 . 0.096 . ω² . 0.0077² . 19.34
ω² = 1053777.29
ω = √1053777.29
ω = 1026.54 rad/s
Find the frequency
ω = 2πf
1026.54 = 2 . 3.14 . f
f = 163.31 Hz
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