Given that,
Atmospheric Pressure = 14.7 psi
Cooking Pressure = 14.7 +11.1 = 25.8 psi
Take, Atmospheric Temperature = 25 °C
Cooking Temperature = ??
Since, we know that Gas equation is given by:
PV = nRT
or
P ∝ T
P1 / T1 = P2 / T2
14.7/ 25 = 25.8/ T2
T2 = 25*25.8/14.7
T2 = 43.87 °C
The cooking pressure will be 43.87 °C.
With an acceleration of 4.0 m/s², the bottle attains a speed of
(4.0 m/s²) * (9.0 s) = 36 m/s
so that the slide has length ∆<em>x</em> such that
(36 m/s)² - 0² = 2 * (4.0 m/s²) * ∆<em>x</em>
==> ∆<em>x</em> = 162 m ≈ 160 m
Alternatively, we know the bottle covers a distance ∆<em>x</em> with acceleration <em>a</em> at time <em>t</em> according to
∆<em>x</em> = 1/2 <em>a</em> <em>t</em>²
so that
∆<em>x</em> = 1/2 * (4.0 m/s²) * (9.0 s)² = 162 m ≈ 160 m
Answer:
The percentage of mass recover from recrystallization = 24.69 %
Explanation:
Given that
Initial mass of crude acetanilide = 166 gm
The mass after recrystallization = 125 gm
The mass recover from recrystallization= 166 - 125 gm
The mass recover from recrystallization= 41 gm
The percentage of mass recover from recrystallization can be find as
The percentage of mass recover from recrystallization = 24.69 %
Answer:
a) 38.3mA
b) 109.396A/m^2
c) 1.283cm/s
d) 226.582V/m
Explanation:
The equation for current is given by:
I = V/R = 35.6/ 935 = 0.03829 A Approximately 38.3mA
b) The equation to find the magnitude is given by:
J = I/A = 0.03829/ 0.000350m^2
J = 109.396A/m^2
c) The equation to calculate drift velocity of the electron is given by: Vd = J/ ne = 109.396/( 5.33×10^23 )(1.60×10^-19)
Vd = 0.01283 approximately 1.283cm/s
d) The magnitude of electric field in the block, E = V/L = 35.8/ 0.158m = 226.582V/m
Answer:
1.17 m
Explanation:
From the question,
s₁ = vt₁/2................ Equation 1
Where s₁ = distance of the reflecting object for the first echo, v = speed of the sound in air, t₁ = time to dectect the first echo.
Given: v = 343 m/s, t = 0.0115 s
Substitute into equation 1
s₁ = (343×0.0115)/2
s₁ = 1.97 m.
Similarly,
s₂ = vt₂/2.................. Equation 2
Where s₂ = distance of the reflecting object for the second echo, t₂ = Time taken to detect the second echo
Given: v = 343 m/s, t₂ = 0.0183 s
Substitute into equation 2
s₂ = (343×0.0183)/2
s₂ = 3.14 m
The distance moved by the reflecting object from s₁ to s₂ = s₂-s₁
s₂-s₁ = (3.14-1.97) m = 1.17 m