What is the equivalent resistance for a series circuit with three resistors : 5.0 ohms, 2.0 ohms, and 12.0 ohms
1 answer:
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Answer:
Its duration is 1.85*10⁻³ s or 1.85 ms
Explanation:
The intensity of electric current I is defined as the amount of electric charge Q (measured in Coulombs) that passes through a section of a conductor in each unit of time. The letter I is used to name the Intensity and its unit is the Ampere (A).
The intensity of electric current is expressed as:
where:
I: Intensity expressed in Amps (A)
Q: Electric charge expressed in Coulombs (C)
t: Time expressed in seconds (s)
Being:
- I= 19500 A
- Q=36 C
- t=?
Replacing:
Solving:
19500 A*t= 36 C
t= 1.85*10⁻³ s= 1.85 ms (being 1 s= 1,000 ms)
<u><em>Its duration is 1.85*10⁻³ s or 1.85 ms</em></u>
Answer:
2.52 × 10⁻² cm
Explanation:
The distance of bright fringe from the center of the screen is given by the formula
Here, wavelength is λ, Distance of the screen from the slits is D, seperation between the
slits is d.
Separation between the slits, d = 0.15 mm
Distance of the screen from the slits = 1.40 m
We have a wavelength, λ1 = 540 nm
=
By substituting all these values in the above equation we get
y1 = mλD/d
We have a wavelength, λ2 = 450 nm
=
By substituting all these values in the above equation we get
According to the problem, these two distance are coincides with each other.
So,
by testing values, the above equation is satisfied only when, m = 5 and m' = 6
Then from the above we have
y1 = y2 = 0.0252 m
= 2.52 × 10⁻² cm
Answer:
the block reaches higher than the sphere
\frac{y_{sphere}} {y_block} = 5/7
Explanation:
We are going to solve this interesting problem
A) in this case a sphere rolls on the ramp, let's find the speed of the center of mass at the exit of the ramp
Let's use the concept of conservation of energy
starting point. At the top of the ramp
Em₀ = U = m g y₁
final point. At the exit of the ramp
Em_f = K + U = ½ m v² + ½ I w² + m g y₂
notice that we include the translational and rotational energy, we assume that the height of the exit ramp is y₂
energy is conserved
Em₀ = Em_f
m g y₁ = ½ m v² + ½ I w² + m g y₂
angular and linear velocity are related
v = w r
the moment of inertia of a sphere is
I = m r²
we substitute
m g (y₁ - y₂) = ½ m v² + ½ ( m r²) (
)²
m g h = ½ m v² (1 + )
where h is the difference in height between the two sides of the ramp
h = y₂ -y₁
mg h = (
m v²)
v = √5/7 √2gh
This is the exit velocity of the vertical movement of the sphere
v_sphere = 0.845 √2gh
B) is the same case, but for a box without friction
starting point
Em₀ = U = mg y₁
final point
Em_f = K + U = ½ m v² + m g y₂
Em₀ = Em_f
mg y₁ = ½ m v² + m g y₂
m g (y₁ -y₂) = ½ m v²
v = √2gh
this is the speed of the box
v_box = √2gh
to know which body reaches higher in the air we can use the kinematic relations
v² = v₀² - 2 g y
at the highest point v = 0
y = vo₀²/ 2g
for the sphere
y_sphere = 5/7 2gh / 2g
y_esfera = 5/7 h
for the block
y_block = 2gh / 2g
y_block = h
therefore the block reaches higher than the sphere