For balancing the lever, force on both the sides shall be equal. so,
Force on 3 m end = m × a = 3 × 98.1 = 294.3
Now, on 6 m end, it would be: = 294.3/6 = 49.05
After rounding-off to the nearest hundredth value, it would be: 49 N
Finally, Option A would be your correct answer.
Hope this helps!
Answer:
a = 3.61[m/s^2]
Explanation:
To find this acceleration we must remember newton's second law which tells us that the total sum of forces is equal to the product of mass by acceleration.
In this case we have:
![F = m*a\\\\m=mass = 3.6[kg]\\F = force = 13[N]\\13 = 3.6*a\\a = 3.61[m/s^2]](https://tex.z-dn.net/?f=F%20%3D%20m%2Aa%5C%5C%5C%5Cm%3Dmass%20%3D%203.6%5Bkg%5D%5C%5CF%20%3D%20force%20%3D%2013%5BN%5D%5C%5C13%20%3D%203.6%2Aa%5C%5Ca%20%3D%203.61%5Bm%2Fs%5E2%5D)
Answer:
it is 27 degree.
Explanation:
because angle of reflection is equal to angle of Inc dence
Complete Question:
When specially prepared Hydrogen atoms with their electrons in the 6f state are placed into a strong uniform magnetic field, the degenerate energy levels split into several levels. This is the so called normal Zeeman effect.
Ignoring the electron spin what is the largest possible energy difference, if the magnetic field is 2.02 Tesla?
Answer:
ΔE = 1.224 * 10⁻²² J
Explanation:
In the 6f state, the orbital quantum number, L = 3
The magnetic quantum number, 
The change in energy due to Zeeman effect is given by:
Magnetic field B = 2.02 T
Bohr magnetron, 
ΔE = 1.224 * 10⁻²² J
(a) The moment of inertia of the wheel is 78.2 kgm².
(b) The mass (in kg) of the wheel is 1,436.2 kg.
(c) The angular speed (in rad/s) of the wheel at the end of this time period is 3.376 rad/s.
<h3>
Moment of inertia of the wheel</h3>
Apply principle of conservation of angular momentum;
Fr = Iα
where;
- F is applied force
- r is radius of the cylinder
- α is angular acceleration
- I is moment of inertia
I = Fr/α
I = (200 x 0.33) / (0.844)
I = 78.2 kgm²
<h3>Mass of the wheel</h3>
I = ¹/₂MR²
where;
- M is mass of the solid cylinder
- R is radius of the solid cylinder
- I is moment of inertia of the solid cylinder
2I = MR²
M = 2I/R²
M = (2 x 78.2) / (0.33²)
M = 1,436.2 kg
<h3>Angular speed of the wheel after 4 seconds</h3>
ω = αt
ω = 0.844 x 4
ω = 3.376 rad/s
Thus, the moment of inertia of the wheel is 78.2 kgm².
The mass (in kg) of the wheel is 1,436.2 kg.
The angular speed (in rad/s) of the wheel at the end of this time period is 3.376 rad/s.
Learn more about moment of inertia here: brainly.com/question/14839816
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