Answer: 5.5 ×10^-2 ×3R. Non metals that are solid at room temperature has mono atomic atoms that has only 3 translational modes.
This is why their constant volume molar heat capacities are measured in multiples of 3R.
Explanation: see attachment
Ummmmmmmmmm is this what subject??????? Because I hv no idea unless u tell me a subject
Answer:
a). linear acceleration of the hoop and block = 3.2895 <u>m</u>
s²
c). magnitude of the angular acceleration of the disk pulley = 41.119 <u>rad</u>
s²
d). tensions in the string between the block and disk pulley = 11.842 N
tensions in the string between the hoop and disk pulley = 15.625 N
check the pictures below for further explanation and for the remaining answers. I hope it helps you. Thank you
Explanation:
Start by writing "F=ma" equations for each of the things that moves. Also, since some of the objects (the pulley and the orange sphere) rotate, you should write "τ = Iα" equations (net torque = moment of inertia × angular acceleration) for those. In the end, you should have enough equations that you can combine them and solve for the desired quantities.
First, the hoop. There's no indication that it rotates, so we don't need a "τ = Iα" equation for it; just do "F=ma". The hoop has gravity ((mhoop)g) pulling down, and the tension in the vertical string (call it "T_v") pulling up.
Fnet = ma
Answer:
The angle of the launch is 17.09 degrees.
Explanation:
Given that,
The initial speed of a cannon is 100 m/s
It is launched at some angle and it lands after being in the air for 6 s.
We need to find the angle of the launch.
The time taken by the projectile to reach the ground is called the time of flight. It is given by the formula as follows :

Here,
= launch angle

Hence, the angle of the launch is 17.09 degrees.
Answer:
a = - 3.14[m/s^2]
Explanation:
To solve this problem we must use the following kinematic equation:
![v_{f} ^{2}= v_{o} ^{2}+2*a*dx\\where:\\v_{f} = final velocity = 0 [m/s]\\v_{o}= initial velocity = 48.8 [m/s]\\ a = acceleration [m/s^2]\\dx = displacement [m]\\](https://tex.z-dn.net/?f=v_%7Bf%7D%20%5E%7B2%7D%3D%20v_%7Bo%7D%20%5E%7B2%7D%2B2%2Aa%2Adx%5C%5Cwhere%3A%5C%5Cv_%7Bf%7D%20%3D%20final%20velocity%20%3D%200%20%5Bm%2Fs%5D%5C%5Cv_%7Bo%7D%3D%20initial%20velocity%20%3D%2048.8%20%5Bm%2Fs%5D%5C%5C%20a%20%3D%20acceleration%20%5Bm%2Fs%5E2%5D%5C%5Cdx%20%3D%20displacement%20%5Bm%5D%5C%5C)
Now replacing:
![0 = 48.8^{2}+2*a*(378.7) \\-2381.44 = 757.4*a\\a = -3.14[m/s^2]](https://tex.z-dn.net/?f=0%20%3D%2048.8%5E%7B2%7D%2B2%2Aa%2A%28378.7%29%20%5C%5C-2381.44%20%3D%20757.4%2Aa%5C%5Ca%20%3D%20-3.14%5Bm%2Fs%5E2%5D)
The negative sign means the vehicle slows down.