A bucket of mass m is hanging from the free end of a rope whose other end is wrapped around a drum (radius R, mass M) that can r
1 answer:
Response:
Explanation:
Let T denote the tension.
By employing Newton's second law to analyze the bucket's downward motion, we have:
mg - T = ma
A torque, TR, acts on the drum, inducing an angular acceleration α in it. If I refers to the moment of inertia of the drum, then:
TR = Iα
Rearranging gives: TR = Ia/R
This leads to T = Ia/R²
Substituting this expression for T back into the previous equation yields:
mg - T = ma
mg - Ia/R² = ma
Consequently, we find that mg = Ia/R² + ma
Therefore, a (I/R² + m) = mg
This results in: a = mg / (I/R² + m)
Next, we aim to express T as:
mg - T = ma
which simplifies to mg - ma = T
Rearranging gives mg - m²g / (I/R² + m) = T
Thus, we arrive at: mg - mg / (1 + I / m R²) = T
For part (b), T = Ia/R²
and for part (c), the moment of inertia of a hollow cylinder calculates to:
I = 1/2 M (R² - (R² / 4))
This simplifies to 3/4 x 1/2 MR², yielding 3/8 MR²
Thus, I / R² = 3/8 M
When we substitute, we find a = mg / (3/8 M + m)
and subsequently T = Ia/R²
= 3/8 MR² × mg / (3/8 M + m) × 1/R²
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Answer:
(a). The coefficient of permeability is .
(b). The seepage velocity is 0.0330 cm/s.
(c). The discharge velocity during the test is 0.0187 cm/s.
Explanation:
Given that,
Length = 175 mm
Diameter = 175 mm
Time = 90 sec
Volume= 405 cm³
We need to calculate the discharge
Using formula of discharge
Insert the values into the formula
(a). We will compute the coefficient of permeability
Applying the formula for permeability
Where, Q=discharge
l = length
A = cross-sectional area
h=constant head that causes flow
Plugging the value into the formula
(c). To ascertain the discharge velocity during the testing phase
Utilizing the discharge velocity formula
Substituting the values into the equation
The discharge velocity during the test measures 0.0187 cm/s.
Thus, (a). The coefficient of permeability is .
(b). The seepage velocity computes at 0.0330 cm/s.
(c). The observed discharge velocity during the test equals 0.0187 cm/s.
Answer:
Explanation:
An image of the bond resulting from the search is attached.
Consider the force directed towards thymine as negative.
For the O-H-N combination:
The resulting force from this combination is:
In the case of the N-H-N combination:
The total force acting from this combination is:
The force that thymine applies on adenine is:
When rounded to three significant figures, the net force is
b)
The negative value indicates that the force is attractive, as it is aimed towards thymi.
c)
The force acting on the electron due to the proton is:
Since the electron and proton carry opposite charges, the force on the electron points towards the proton.
d)
The ratio of the above forces is:
Therefore, the bonding strength of the electron in the hydrogen atom is 9.3 times greater than the bonding force between adenine and thymine molecules.
Answer:
La magnitud del momento total es de 21.2 kg m/s y su dirección es de 39.5° respecto al eje x.
Explanation:
¡Hola!
El momento total se calcula como la suma de los momentos de las piezas.
El momento de cada pieza se calcula de la siguiente manera:
p = m · v
Donde:
p = momento.
m = masa.
v = velocidad.
El momento es un vector. La pieza de 200 g se mueve a lo largo del eje x, por lo que su momento será:
p = (m · v, 0)
p = (0.200 kg · 82.0 m/s, 0)
p = (16.4 kg m/s, 0)
La pieza de 300 g se mueve a lo largo del eje y. Su vector momento será:
p =(0, m · v)
p = (0, 0.300 kg · 45.0 m/s)
p = (0, 13.5 kg m/s)
El momento total es la suma de cada momento:
Momento total = (16.4 kg m/s, 0) + (0, 13.5 kg m/s)
Momento total = (16.4 kg m/s + 0, 0 + 13.5 kg m/s)
Momento total = (16.4 kg m/s, 13.5 kg m/s)
La magnitud del momento total se calcula de la siguiente manera:
La dirección del vector de momento se calcula utilizando trigonometría:
cos θ = px/p
Donde px es el componente horizontal del momento total y p es la magnitud del momento total.
cos θ = 16.4 kg m/s / 21.2 kg m/s
θ = 39.3 (39.5° si no redondeamos la magnitud del momento total)
<pFinalmente, la magnitud del momento total es 21.2 kg m/s y su dirección es 39.5° respecto al eje x.