A wire with a resistance of 20 Ω is connected to a 12-V battery. What is the current flowing through the wire?
2 answers:
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Newton's second law allows us to find the results for the string tensions are:
- T₁ = 6.7 N
- T₂ = 16.54 N
Newton's second law gives a relationship between force, mass and acceleration of bodies
∑ F = ma
Where the bold letters indicate vectors, F is the force, m the mass and the acceleration.
Free-body diagrams are representations of the forces applied to bodies without the details of them.
The reference system is a coordinate system with respect to which the forces decompose, in this case the x-axis is parallel to the plane and the positive direction in the direction of movement, the y-axis is perpendicular to the plane.
In the attachment we see a free-body diagram of the three-block system.
Let's apply Newton's second law to each body.
Block C
Y-axis
Block A
X axis
Y axis
Block B
X axis
Y axis
Let's use trigonometry to find the components of the weight.
Block A
cos θ =
sin θ =
Block B
cos θ =
sin θ =
Let's write our system of equations.
Let's find the acceleration of the bodies, adding the equations.
The weight is
W = mg
Let's substitute
Indicate ma mass of the block a ma = 1.00 kg, the mass of the block b mb = 2.2 kg and the weight of the block c Wc = 16.2 N, let's find the mass of block c.
m_c = Wc / g
m_c = 16.2 / 9.8
m_c = 1.65 kg
we substitute the values
The negative sign indicates that the system is descending, to be able to give a specified value an angle is needed, they assume that the angle of the ramp is 45º
a = - 0.3196 sin 45
a = -0.226 m / s
Taking the acceleration we are going to look for the tensions.
From the equation of block C
T₂ = 16.54 N
From the equation of block B
T₁ = 6.7 N
In conclusion using Newton's second law we can find the results for the string tensions are:
- T₁ = 6.7 N
- T₂ = 16.54 N
Learn more here: brainly.com/question/20575355
Answer:
fundamental frequency= 20Hz
length= 4.3m
Explanation:
resonating frequencies = 60 Hz,100 Hz,140 Hz
speed of sound =344 m/s
as well behaves as organ pipe with are end closed
only odd harmonics are allowed
, so the vibration will corresponds to n=1,3,5,7....
(2n−1) λ
/4 = l (∵λ
= V/f)
(2n-1) V / 4f = l
(2m-1)/6= (2n-1)/10= (2p-1)/14 =A
The given frequencies happen to be 3rd,5th & 7th harmonics of the fundamental frequency n=20Hz.
Therefore, The fundamental frequency will be 20 Hz because 60 Hz is corresponding 3rd harmonic
In order to find length,
(2n-1) V / 4f = l
= 4.3m
therefore, the well is 4.3m deep