(G)-->Iron and steel industry is called a heavy industry because all the raw material as well as finished goods are heavy and bulky entailing heavy transportation costs. Iron ore, coking coal and limestone are required in the ratio of 4:2:1 approximately. Some quantity of manganese is also required to harden the steel.
(H)-->Sodium is very reactive in nature. When exposed in air, it automatically forms Na2O. When it is put in water it reacts vigorously and starts burning on water. Due to the above reasons Sodium is called an active metal.
(I)-->Down the group, the effective nuclear charge experienced by valence electrons is decreasing because the outermost electrons are far away from the nucleus. Thus, these electrons can be lost easily by the element to form positive ions. Hence, the chemical reactivity of metals increases on going down a group.
(J)-->While moving from top to bottom in a group of the periodic table, the reactivity of non- metals decreases. While moving from top to bottom in a group of non- metals, the atomic size increases with the additional number of shells and the force of attraction between the nucleus and valence shell decreases.
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
Answer:
After 1 sec = 4.9 m
After 2 sec = 19.6 m
After 3 sec = 44.1 m
After 4 sec = 78.4 m
After 5 sec = 122.5 m
Explanation:
After 1 sec:
<em>u=0m/s t=1 s a=9.8m/s²</em>
s = ut + (1/2)at²
=0(1) + (1/2)(9.8)(1²) = 4.9m
After 2 sec:
<em>u=0m/s t=2 s a=9.8m/s²</em>
s = ut + (1/2)at²
=0(2) + (1/2)(9.8)(2²) = 19.6m
After 3 sec:
<em>u=0m/s t=3 s a=9.8m/s²</em>
s = ut + (1/2)at²
=0(3) + (1/2)(9.8)(3²) = 44.1m
After 4 sec:
<em>u=0m/s t=4 s a=9.8m/s²</em>
s = ut + (1/2)at²
=0(4) + (1/2)(9.8)(4²) = 78.4m
After 5 sec:
<em>u=0m/s t=5 s a=9.8m/s²</em>
s = ut + (1/2)at²
=0(5) + (1/2)(9.8)(5²) = 122.5m
To solve this problem we will apply the concepts of equilibrium and Newton's second law.
According to the description given, it is under constant ascending acceleration, and the balance of the forces corresponding to the tension of the rope and the weight of the elevator must be equal to said acceleration. So
Here,
T = Tension
m = Mass
g = Gravitational Acceleration
a = Acceleration (upward)
Rearranging to find T,
Therefore the tension force in the cable is 10290.15N
1) In the reference frame of one electron: 0.38c
To find the relative velocity of one electron with respect to the other, we must use the following formula:
where
u is the velocity of one electron
v is the velocity of the second electron
c is the speed of light
In this problem:
u = 0.2c
v = -0.2c (since the second electron is moving towards the first one, so in the opposite direction)
Substituting, we find:
2) In the reference frame of the laboratory: -0.2c and +0.2c
In this case, there is no calculation to be done. In fact, we are already given the speed of the two electrons; we are also told that they travel in opposite direction, so their velocities are
+0.2c
-0.2c
