The average dissipated power in a resistor in a ac circuit is:
where R is the resistance, and
is the root mean square current, defined as
where
is the peak value of the current. Substituting the second formula into the first one, we find
and if we re-arrange this formula and use the data of the problem, we can find the value of the peak current I0:
Answer:
D
Explanation:
The answer is Niels Bohr's planetary model, the difference between this model and all of the other models is that the Bohr's PM Is more of layers of
Nucleus - Protons and Neutrons
Electron Orbital - Period 1 Elements
2 electrons
Electron Orbital - Period 2 Elements
8 electrons
Electron Orbital - Period 3 Elements
8 electrons
If that made sense-
The force acting on the object is constant, so the acceleration of the object is also constant. By definition of average acceleration, this acceleration was
<em>a</em> = ∆<em>v</em> / ∆<em>t</em> = (6 m/s - 0) / (1.7 s) ≈ 3.52941 m/s²
By Newton's second law, the magnitude of the force <em>F</em> is proportional to the acceleration <em>a</em> according to
<em>F</em> = <em>m a</em>
where <em>m</em> is the object's mass. Solving for <em>m</em> gives
<em>m</em> = <em>F</em> / <em>a</em> = (10 N) / (3.52941 m/s²) ≈ 2.8 kg
Answer:
μ=0.151
Explanation:
Given that
m= 3.5 Kg
d= 0.96 m
F= 22 N
v= 1.36 m/s
Lets take coefficient of kinetic friction = μ
Friction force Fr=μ m g
Lets take acceleration of block is a m/s²
F- Fr = m a
22 - μ x 3.5 x 10 = 3.5 a ( take g =10 m/s²)
a= 6.28 - 35μ m/s²
The final speed of the block is v
v= 1.36 m/s
We know that
v²= u²+ 2 a d
u= 0 m/s given that
1.36² = 2 x a x 0.96
a= 0.963 m/s²
a= 6.28 - 35μ m/s²
6.28 - 35μ = 0.963
μ=0.151
