Answer:
a
The radial acceleration is ![a_c = 0.9574 m/s^2](https://tex.z-dn.net/?f=a_c%20%20%3D%200.9574%20m%2Fs%5E2)
b
The horizontal Tension is ![T_x = 0.3294 i \ N](https://tex.z-dn.net/?f=T_x%20%20%3D%200.3294%20i%20%20%5C%20N)
The vertical Tension is ![T_y =3.3712 j \ N](https://tex.z-dn.net/?f=T_y%20%20%3D3.3712%20j%20%20%20%5C%20N)
Explanation:
The diagram illustrating this is shown on the first uploaded
From the question we are told that
The length of the string is ![L = 10.7 \ cm = 0.107 \ m](https://tex.z-dn.net/?f=L%20%3D%20%2010.7%20%5C%20cm%20%20%3D%20%200.107%20%5C%20m)
The mass of the bob is ![m = 0.344 \ kg](https://tex.z-dn.net/?f=m%20%3D%200.344%20%5C%20%20kg)
The angle made by the string is ![\theta = 5.58^o](https://tex.z-dn.net/?f=%5Ctheta%20%20%3D%20%205.58%5Eo)
The centripetal force acting on the bob is mathematically represented as
![F = \frac{mv^2}{r}](https://tex.z-dn.net/?f=F%20%20%3D%20%20%5Cfrac%7Bmv%5E2%7D%7Br%7D)
Now From the diagram we see that this force is equivalent to
where T is the tension on the rope and v is the linear velocity
So
![Tsin \theta = \frac{mv^2}{r}](https://tex.z-dn.net/?f=Tsin%20%5Ctheta%20%20%3D%20%20%20%5Cfrac%7Bmv%5E2%7D%7Br%7D)
Now the downward normal force acting on the bob is mathematically represented as
![Tcos \theta = mg](https://tex.z-dn.net/?f=Tcos%20%5Ctheta%20%3D%20mg)
So
![\frac{Tsin \ttheta }{Tcos \theta } = \frac{\frac{mv^2}{r} }{mg}](https://tex.z-dn.net/?f=%5Cfrac%7BTsin%20%5Cttheta%20%7D%7BTcos%20%5Ctheta%20%7D%20%20%3D%20%20%5Cfrac%7B%5Cfrac%7Bmv%5E2%7D%7Br%7D%20%7D%7Bmg%7D)
=> ![tan \theta = \frac{v^2}{rg}](https://tex.z-dn.net/?f=tan%20%5Ctheta%20%20%3D%20%20%5Cfrac%7Bv%5E2%7D%7Brg%7D)
=> ![g tan \theta = \frac{v^2}{r}](https://tex.z-dn.net/?f=g%20tan%20%5Ctheta%20%20%3D%20%5Cfrac%7Bv%5E2%7D%7Br%7D)
The centripetal acceleration which the same as the radial acceleration of the bob is mathematically represented as
![a_c = \frac{v^2}{r}](https://tex.z-dn.net/?f=a_c%20%20%3D%20%20%5Cfrac%7Bv%5E2%7D%7Br%7D)
=> ![a_c = gtan \theta](https://tex.z-dn.net/?f=a_c%20%20%3D%20gtan%20%5Ctheta)
substituting values
![a_c = 9.8 * tan (5.58)](https://tex.z-dn.net/?f=a_c%20%20%3D%20%209.8%20%20%2A%20%20tan%20%285.58%29)
![a_c = 0.9574 m/s^2](https://tex.z-dn.net/?f=a_c%20%20%3D%200.9574%20m%2Fs%5E2)
The horizontal component is mathematically represented as
![T_x = Tsin \theta = ma_c](https://tex.z-dn.net/?f=T_x%20%20%3D%20Tsin%20%5Ctheta%20%3D%20ma_c)
substituting value
![T_x = 0.3294 \ N](https://tex.z-dn.net/?f=T_x%20%20%3D%200.3294%20%5C%20N)
The vertical component of tension is
![T_y = T \ cos \theta = mg](https://tex.z-dn.net/?f=T_y%20%20%3D%20%20T%20%5C%20cos%20%5Ctheta%20%20%3D%20mg)
substituting value
![T_ y = 0.344 * 9.8](https://tex.z-dn.net/?f=T_%20y%20%20%3D%20%200.344%20%2A%209.8)
The vector representation of the T in term is of the tension on the horizontal and the tension on the vertical is
![T = T_x i + T_y j](https://tex.z-dn.net/?f=T%20%20%3D%20T_x%20i%20%20%2B%20T_y%20%20j)
substituting value
![T = [(0.3294) i + (3.3712)j ] \ N](https://tex.z-dn.net/?f=T%20%20%3D%20%5B%280.3294%29%20i%20%20%2B%20%283.3712%29j%20%5D%20%5C%20%20N)
I = MR^2
The Attempt at a Solution:::
I total = (3M)(0)^2 + (2M)(L/2)^2 + (M)(L)^2
I total = 3ML^2/2
It says the answer is 3ML^2/4 though.
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Answer:
Base units are defined units based on specific objects or events in the physical world. Derived units are defined by combining base units.
Base units are defined by a particular process of measuring a base quantity whereas derived units are defined as algebraic combinations of base units. For example, length is a base quantity in both SI and the English system, but the meter is a base unit in the SI system only.
Answer:
A. The sum of all the forces acting on an object.
Answer:
Explanation:
1) acceleration is the change in velocity of a body with respect to time.
acceleration = velocity/time
Given
velocity = 139m/s
time = 20secs
acceleration = 139/20
acceleration = 6.95m/s²
Hence its average acceleration during the first 20 seconds of the launch is 6.95m/s²
2) Speed is the rate of change of distance with respect to time.
average speed = distance/time
Time = distance/speed
Time = 100/7823
Time = 0.013s
3) Using the equation of motion v² = u²+2as
v² = 0²+2(15)(60)
v² = 30*60
v² =1800
v = √1800
v = 42.43m/s