Answer:
<em>i</em><em> </em><em>gu</em><em>ess</em><em> </em><em>it</em><em>'s</em><em> </em><em>gender</em>
<em>hop</em><em>e</em><em> this</em><em> helps</em>
Stomata is the lay of the leaf found below the skin
]To calculate the amount of heat entering or leaving a system, the equation
Q
=
m
c
Δ
T
is used.
Explanation:
m = mass (in grams)
c = specific heat capacity (J/g°C)
ΔT = change in temperature (°C)
Here, we will use the specific heat capacity for liquid water which is 4.19 J/g°C.
The mass given is 25.0 grams.
As for the change in temperature, I will assume that it start off at room temperature, 25°C.
25
°
C
−
0
°
C
=
25
°
C
Q
=
m
c
Δ
T
Q
=
25
g
r
a
m
s
⋅
4.19
J
g
°
C
⋅
25
°
C
Q
=
2618.75
J
Take into account significant figures and the answer should be
2.6
⋅
10
3
J
Answer link
Answer:
A. 21 rad/s
B. 200.5 rpm
C. 26.46 rad/s2
D. 0.299 Hz
Explanation:
Parameters given are:
Tangential velocity,v = 1.26m/s
Diameter (outer edge) = 0.120m
Outer edge radius, ro = 0.12/2
= 0.06m
A. Calculate Angular velocity
Angular velocity, w = v/r
= v/ro
= 1.26/0.06 = 21 rad/s
In rpm,
2pi/60 rad/s = 1 rpm
1 rad/s = 60/2pi rpm
Sp, 21 rad/s = 60 * 21/2pi rpm
= 200.5 rpm
B. Calculate the inner edge radius, ri given w = 500 rpm
Converting rpm to rad/s,
= 2pi/60 * 500 rad/s
= 52.34 rad/s
ri = v/w
= 1.26/ 52.36
= 0.024m
C. Calculating centripetal acceleration, a from the outer edge, ro
a = v2/r = w2r
= (1.26)2/0.06
= 26.46 rad/s
D. Calculate the frequency, f of the outer edge angular velocity, w = 21 rad/s
f = 2pi/w
= 2pi/21
= 0.299 Hz (or per second)