Answer:
Opposition of passing a electric circuit
The object's final velocity, given the data is 10.5 rad/s
<h3>What is acceleration? </h3>
This is defined as the rate of change of velocity which time. It is expressed as
a = (v – u) / t
Where
- a is the acceleration
- v is the final velocity
- u is the initial velocity
- t is the time
<h3>How to determine the final velocity</h3>
The following data were obtained from the question
- Initial velocity (u) = 1.5 rad/s
- Acceleration (a) = 0.75 rad/s²
- Time (t) = 12 s
- Final velocity (v) = ?
The final velocity can be obtained as follow:
a = (v – u) / t
0.75 = (v – 1.5) / 12
Cross multiply
v – 1.5 = 0.75 × 12
v – 1.5 = 9
Collect like terms
v = 9 + 1.5
v = 10.5 rad/s
Thus, the final velocity of the object is 10.5 rad/s
Learn more about acceleration:
brainly.com/question/491732
#SPJ1
Answer:
T’= 4/3 T
The new tension is 4/3 = 1.33 of the previous tension the answer e
Explanation:
For this problem let's use Newton's second law applied to each body
Body A
X axis
T = m_A a
Axis y
N- W_A = 0
Body B
Vertical axis
W_B - T = m_B a
In the reference system we have selected the direction to the right as positive, therefore the downward movement is also positive. The acceleration of the two bodies must be the same so that the rope cannot tension
We write the equations
T = m_A a
W_B –T = M_B a
We solve this system of equations
m_B g = (m_A + m_B) a
a = m_B / (m_A + m_B) g
In this initial case
m_A = M
m_B = M
a = M / (1 + 1) M g
a = ½ g
Let's find the tension
T = m_A a
T = M ½ g
T = ½ M g
Now we change the mass of the second block
m_B = 2M
a = 2M / (1 + 2) M g
a = 2/3 g
We seek tension for this case
T’= m_A a
T’= M 2/3 g
Let's look for the relationship between the tensions of the two cases
T’/ T = 2/3 M g / (½ M g)
T’/ T = 4/3
T’= 4/3 T
The new tension is 4/3 = 1.33 of the previous tension the answer e
Brown dwarf is the first box
White dwarf is the second box
Black dwarf is the third box
Red giant is the fourth box
And
Black hole is the last box
Answer:
The beat frequency when each string is vibrating at its fundamental frequency is 12.6 Hz
Explanation:
Given;
velocity of wave on the string with lower tension, v₁ = 35.2 m/s
the fundamental frequency of the string, F₁ = 258 Hz
<u>velocity of wave on the string with greater tension;</u>
where;
v₁ is the velocity of wave on the string with lower tension
T₁ is tension on the string
μ is mass per unit length
Where;
T₁ lower tension
T₂ greater tension
v₁ velocity of wave in string with lower tension
v₂ velocity of wave in string with greater tension
From the given question;
T₂ = 1.1 T₁
<u>Fundamental frequency of wave on the string with greater tension;</u>
<u />
<u />
Beat frequency = F₂ - F₁
= 270.6 - 258
= 12.6 Hz
Therefore, the beat frequency when each string is vibrating at its fundamental frequency is 12.6 Hz
