Answer: The answer is density
Hello User,
Approximately 32 electrons can be fit in the fourth energy level.
Solution:
2+4+6+10+10=32
Answer:
new temperature of the tire will be 278.76 K
Explanation:
when the temperature increases, the particles will have greater kinetic energy and also the pressure will be increase for the gas particles.
so when the temperature increases, pressure will also increase and vice versa
T is directly proportional to P
T1 = initial temperature= 303 k
P1= Initial pressure = 325000 pa
T2= Final temperature= ?
P2= Final pressure = 299000 pa
mathematically
P1/T1= P2/T2
T2= P2 x T1/P1
T2 = 299000 x 303/ 325000= 278.76 k
Answer:
Weight = 966 Newton.
Explanation:
Given the following data;
Length = 1.2 m
Width = 2.3 m
Pressure = 350 Pa
To find the weight of the tank;
We know that weight is the force of gravity acting on an object multiplied by its mass.
Weight = mg = force
Hence, we would determine the force using the parameters that were given.
But we would first determine the area of the rectangular tank.
Area of rectangle, A = length * width
A = 1.2 * 2.3
A = 2.76 m²
Mathematically, pressure is given by the formula;
Pressure = force/area
Force = pressure * area
Substituting into the formula, we have;
Force = 2.76 * 350
Force = 966 Newton
Therefore, the weight of the tank is 966 Newton.
Answer:
L = 0 m
Therefore, the cricket was 0m off the ground when it became Moe’s lunch.
Explanation:
Let L represent Moe's height during the leap.
Moe's velocity v at any point in time during the leap is;
v = dL/dt = u - gt .......1
Where;
u = it's initial speed
g = acceleration due to gravity on Mars
t = time
The determine how far the cricket was off the ground when it became Moe’s lunch.
We need to integrate equation 1 with respect to t
L = ∫dL/dt = ∫( u - gt)
L = ut - 0.5gt^2 + L₀
Where;
L₀ = Moe's initial height = 0
u = 105m/s
t = 56 s
g = 3.75 m/s^2
Substituting the values, we have;
L = (105×56) -(0.5×3.75×56^2) + 0
L = 0 m
Therefore, the cricket was 0m off the ground when it became Moe’s lunch.
