Answer:
The velocity at B is
The velocity at C is ![v_c =10.80 \ m/s](https://tex.z-dn.net/?f=v_c%20%3D10.80%20%5C%20m%2Fs)
Explanation:
From the question we are told that
The mass of the bead is ![m_b = 0.368 \ kg](https://tex.z-dn.net/?f=m_b%20%20%3D%200.368%20%5C%20kg)
The first height is ![h_1 = 4.20 \ m](https://tex.z-dn.net/?f=h_1%20%3D%204.20%20%5C%20m)
The second height is ![h_2 = 2.14 \ m](https://tex.z-dn.net/?f=h_2%20%3D%202.14%20%5C%20m)
The initial speed of the bead is ![u = 1.96 \ m/s](https://tex.z-dn.net/?f=u%20%20%3D%201.96%20%5C%20m%2Fs)
According to the law of energy conservation
![KE _ A + PE_A = KE_B + PE_B](https://tex.z-dn.net/?f=KE%20_%20A%20%2B%20PE_A%20%3D%20%20KE_B%20%2B%20PE_B)
Now
is the kinetic energy at A and it is mathematically represented as
![KE_A = \frac{1}{2} mu^2](https://tex.z-dn.net/?f=KE_A%20%20%3D%20%20%5Cfrac%7B1%7D%7B2%7D%20mu%5E2)
is the potential energy at A which is mathematically represented as
is the kinetic energy at B which is mathematically represented as
Where v is the speed of the bead at B
is the potential energy at B which equal to 0 because height is 0 at B
So
![\frac{1}{2} mu^2 + mg h_1 = \frac{1}{2} mv^2](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%20mu%5E2%20%2B%20mg%20h_1%20%3D%20%20%5Cfrac%7B1%7D%7B2%7D%20mv%5E2)
Making v the subject
![v = \sqrt{u ^2 + 2gh_1}](https://tex.z-dn.net/?f=v%20%3D%20%5Csqrt%7Bu%20%5E2%20%2B%202gh_1%7D)
substituting values
According to the law of energy conservation
![KE _ B + PE_B = KE_C + PE_C](https://tex.z-dn.net/?f=KE%20_%20B%20%2B%20PE_B%20%3D%20%20KE_C%20%2B%20PE_C)
So
![KE_B = \frac{1}{2} m v^2](https://tex.z-dn.net/?f=KE_B%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20m%20v%5E2)
![PE_B = 0](https://tex.z-dn.net/?f=PE_B%20%20%3D%200)
is the kinetic energy at c which is mathematically represented as
![KE_C = \frac{1}{2} * m v_c^2](https://tex.z-dn.net/?f=KE_C%20%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%20%2A%20%20%20m%20v_c%5E2)
is the potential energy at C which is mathematically represented as
![PE_C = mg h_2](https://tex.z-dn.net/?f=PE_C%20%20%3D%20mg%20h_2)
So
![\frac{1}{2} m v^2 = \frac{1}{2} * m v_c^2 + mg h_2](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%20m%20v%5E2%20%20%3D%20%20%20%5Cfrac%7B1%7D%7B2%7D%20%20%2A%20%20%20m%20v_c%5E2%20%2B%20%20mg%20h_2)
![v^2 = v_c^2 + 2g h_2](https://tex.z-dn.net/?f=v%5E2%20%20%3D%20%20%20%20v_c%5E2%20%2B%20%202g%20h_2)
making
the subject
![v_c = \sqrt{12.6^2 - 2* 9.8 * 2.14}](https://tex.z-dn.net/?f=v_c%20%3D%20%5Csqrt%7B12.6%5E2%20-%202%2A%209.8%20%2A%202.14%7D)
![v_c =10.80 \ m/s](https://tex.z-dn.net/?f=v_c%20%3D10.80%20%5C%20m%2Fs)
Answer: D. The force and displacement are in the same direction.
Explanation:
The Work
done by a Force
refers to the release of potential energy from a body that is moved by the application of that force to overcome a resistance along a path with distance
.
Work is a scalar magnitude, and its unit in the International System of Units is the Joule (like energy).
Now, when the applied force is constant and the direction of the force and the direction of the displacement are <u>parallel</u>, the equation to calculate it is:
(1)
When they are not parallel, both directions form an angle, let's call it
. In that case the expression to calculate the Work is:
(2)
When the force and displacement are perpendicular to each other,
and <u>no work is done</u>.
Answer:
12 N
Explanation:
Given,
Mass ( m ) = 3 kg
Acceleration ( a ) = 4 m/s^2
To find : Force ( F ) = ?
Formula : -
F = ma
F
= 3 x 4
= 12 N
Therefore, the force acting on the object is 12 N.
Stan looga
this is correct answer
It is false it is a metle core