Hi,Find answers from Task 5
1.(X+4)+(X)+(X+4)+(X)=50cm
4x+8=50cm
4x=42
X=10.5cm
Length=10.5+4=14.5cm
Width=10.5cm
Area= length × width=(10.5/100) × (14.5/100) =0.0152m2
2. Volume of a sphere= 4/3 ×π×r³
4/3 ×π×r³=3.2×10^-6 m³
r³=3.2×10^-6 m³/1.33×π
r³=7.64134761e-7
r=0.00914m
Surface area of the blood drop= 4πr²
=4×3.142×0.00914×0.00914=0.00105m²
3.
Equation of an ideal gas = PV =n RT
Equation for pressure, = P= n RT/V
Equation for the volume of an ideal gas= V= n RT/P
If the volume of gas doubles ,V(new)= 2n RT/P
Equation for temperature of an ideal gas, T = PV/n R
If temperature of gas triples, T (new)= 3PV/n R
New Equation for Pressure, = n× R× (3PV/n R)/(2n RT/P)
Pressure factor increase= P(new)/P(old) ={ n× R× (3PV/n R)/(2n RT/P)}/{ n RT/V}
=3PV²/2n RT
What information can scientists obtain from tree rings?
Answer's <u>I chose</u>:
<h3>how narrow the rings are</h3><h3>how the climate changed in the tree’s life</h3><h3>how wide the rings are</h3>
Please <u>correct</u> me if there are <em>more </em>or <em>less</em>
Please give a brainliest and a thanks.
<h2>❣</h2>
N2 = 3*n1
T2 = 2*T1
V1 = V2
(n2 * T2)/P2 = (n1 * T1)/P1
3 n1 * 2 T1 / P2 = n1 *T1 / P1
P2 = 6*P1
Since P2 is 6P1, it is 6 times greater than original pressure
<span>-- the product of the net charges on the objects;. -- the distance between the centers of their net charges. (Pretty much identical to the formula for gravitational force)</span>
Answer:
4.7 m³
Explanation:
We'll use the gas law P1 • V1 / T1 = P2 • V2 / T2
* Givens :
P1 = 101 kPa , V1 = 2 m³ , T1 = 300.15 K , P2 = 40 kPa , T2 = 283.15 K
( We must always convert the temperature unit to Kelvin "K")
* What we want to find :
V2 = ?
* Solution :
101 × 2 / 300.15 = 40 × V2 / 283.15
V2 × 40 / 283.15 ≈ 0.67
V2 = 0.67 × 283.15 / 40
V2 ≈ 4.7 m³
