Their magnitude of force must be the same for the tensions to equal out.
An object at equilibrium has zero acceleration so both the magnitude and direction of the object's velocity must be constant. When the body is in equilibrium the forces are balanced. Balanced is a keyword used to describe a balanced state. Therefore the net force is zero and the acceleration is 0 m/s/s. The acceleration of a body in equilibrium must be 0 m/s/s
For a body to be in equilibrium it must not be accelerating. This means that both the net force and net torque on the object must be zero. Clearly when in equilibrium, the net force on the object is zero. According to Newton's second law of motion if the net force is zero the acceleration is also zero. When the acceleration is zero the velocity and thus the velocity is constant by definition. Option d does not apply to equilibrium in this case.
Learn more about Equilibrium here:-https://brainly.com/question/517289
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Hello! So I need some help with this question. I don’t quite understand it and I did have another tutor figure it out and the answer was wrong. It’s not his answer of 58.04. Can you help me?
ANSWER:
58 m/s
STEP-BY-STEP EXPLANATION:
Given:
Initial velocity (u) = 13 m/s
The correct answer is 58 m/sDistance (d) = 400 m
Acceleration (a) = 4 m/s²
We apply the following formula to determine the final velocity:
[tex]\begin{gathered} v^2=u^2+2as \\ \\ \text{ We replacing:} \\ \\ v^2=13^2+2(4)(400) \\ \\ v^2=169+3200 \\ \\ v=\sqrt{3369} \\ \\ v=58.04\cong58\text{ m/s} \end{gathered}[/tex]The correct answer is 58 m/s
Which of the following numbers has the greatest number of significant figures?a.) 144.5b.) 0.009514 c.) 10.507d.) 6.948 ✕ 10
So, the number with the greatest number of significant figures is:
0.009514. [Option B]
A string 0.50 m long is stretched under a tension of 2.0 x 102 N and its fundamental frequency is 400 Hz. If the length if the string is shortened to 0.35 m and the tension is increased to 4.0 x 102 N, what is the new fundamental frequency?
Given,
The initial length of the string, L₁=0.50 m
The tension on the string, T₁=2.0×10² N
The initial fundamental frequency of the string, f₁=400 Hz
The length of the string after it was shortened, L₂=0.35 m
The increased tension on the string, T₂=4.0×10² N
The fundamental frequency of the string before it was shortened is given by,
[tex]f_1=\frac{\sqrt[]{\frac{T_1}{\mu}}}{2L_1}[/tex]Where μ is the mass per unit length of the string.
On rearranging the above equation,
[tex]\begin{gathered} 4L^2_1f^2_1=\frac{T_1}{\mu} \\ \Rightarrow\mu=\frac{T_1}{4L^2_1f^2_1} \end{gathered}[/tex]On substituting the known values,
[tex]\begin{gathered} \mu=\frac{2.0\times10^2}{4\times0.50^2\times400^2} \\ =1.25\times10^{-3}\text{ kg/m} \end{gathered}[/tex]The fundamental frequency after the string was shortened is given by,
[tex]f_2=\frac{\sqrt[]{\frac{T_2}{\mu}}}{2L_2}[/tex]On substituting the known values,
[tex]\begin{gathered} f_2=\frac{\sqrt[]{\frac{4\times10^2}{1.25\times10^{-3}}}}{2\times0.35} \\ =808.1\text{ Hz} \end{gathered}[/tex]Thus the fundamental frequency after the string is shortened and the tension is increased is 808.1 Hz
What voltage was needed for 2 amps to flow through a 600 lamp?
Answer:
I don't kno..................................................w
A 320.-kg reindeer stands in the middle of the railroad tracks, frozen by the lights of an oncoming 10,000.-kg train that is traveling at 10.0 m/s. The engineer sees the reindeer but is unable to stop the train in time and the reindeer rides down the track sitting on the cowcatcher. What is the new combined velocity of the reindeer and train?
Answer:
9.69 m/s
Explanation
By the conservation of momentum, we can write the following equation:
[tex]\begin{gathered} p_i=p_f \\ m_1v_{i1}+m_2v_{i2}=(m_1+m_2)v_f \end{gathered}[/tex]Where m1 is the mass of the reindeer, m2 is the mass of the train, vi1 is the initial velocity of the reindeer, vi2 is the initial velocity of the train and vf is the final velocity of the reindeer and train.
So, replacing m1 = 320 kg, m2 = 10,000 kg, vi1 = 0 m/s and vi2 = 10 m/s, we get
[tex]\begin{gathered} 320(0)+10,000(10)=(320+10000)v_f_{} \\ 0+100,000=(10320)v_f \\ 100,000=10,320v_f \end{gathered}[/tex]Now, solving for the final velocity, we get
[tex]\begin{gathered} v_f=\frac{100,000}{10,320} \\ v_f=9.69\text{ m/s} \end{gathered}[/tex]Therefore, the new combined velocity of the reindeer and train is 9.69 m/s