Answer:
1) F = 24 N
2) Distance = 1 m
Explanation:
We are given;
Mass; m = 120 g = 0.12 kg
Initial velocity; u = 20 m/s
Final velocity; v = 0 m/s since it came to rest.
Time; t = 0.1 s
We can calculate acceleration from Newton's first equation of motion;
a = (v - u)/t
a = (0 - 20)/0.1
a = -200 m/s²
1) magnitude of the resistance will be;
F = ma
F = 0.12 × (-200)
F = -24 N
Since, we are dealing with the magnitude, we will take the absolute value. Thus, F = 24 N
2) To find the distance moved by the bullet, we know that;
Distance = Average speed × time
Thus;
Distance = ((v + u)/2) × t
Distance = ((0 + 20)/2) × 0.1
Distance = 1 m
Answer:
Red light
Explanation:
The energy emitted during an electron transition in an atom of hydrogen is given by
where
is the energy of the lowest level
n1 and n2 are the numbers corresponding to the two levels
Here we have
n1 = 3
n2 = 2
So the energy of the emitted photon is
Converting into Joules,
And now we can find the wavelength of the emitted photon by using the equation
where h is the Planck constant and c is the speed of light. Solving for 
,
And this wavelength corresponds to red light.
Answer:
1.92 J
Explanation:
From the question given above, the following data were obtained:
Mass (m) = 200 Kg
Spring constant (K) = 10⁶ N/m
Workdone =?
Next, we shall determine the force exerted on the spring. This can be obtained as follow:
Mass (m) = 200 Kg
Acceleration due to gravity (g) = 9.8 m/s²
Force (F) =?
F = m × g
F = 200 × 9.8
F = 1960 N
Next we shall determine the extent to which the spring stretches. This can be obtained as follow:
Spring constant (K) = 10⁶ N/m
Force (F) = 1960 N
Extention (e) =?
F = Ke
1960 = 10⁶ × e
Divide both side by 10⁶
e = 1960 / 10⁶
e = 0.00196 m
Finally, we shall determine energy (Workdone) on the spring as follow:
Spring constant (K) = 10⁶ N/m
Extention (e) = 0.00196 m
Energy (E) =?
E = ½Ke²
E = ½ × 10⁶ × (0.00196)²
E = 1.92 J
Therefore, the Workdone on the spring is 1.92 J
Explanation:
Sucrose is a disaccharide which is composed of fructose and glucose. Sucrose molecule has oxygen atoms bonded to hydrogen atoms (O-H bonds - Polar groups) on all ends of its double 6-Carbon ring. The areas near the oxygen atoms are slightly negative, and the areas near the hydrogen atoms are slightly positive that is, the O-H bonds are polar. They bond with the neighbouring Oxygen and Hydrogen atoms because of their
dipole - dipole attractions and hence hydrogen bonds are formed.
However, the covalent bonds within the molecule aren't broken. But rather, the hydrogen bonds holding the sucrose molecules in the crystalline lattice.
