![\huge\underline{\underline{\boxed{\mathbb {EXPLANATION}}}}](https://tex.z-dn.net/?f=%5Chuge%5Cunderline%7B%5Cunderline%7B%5Cboxed%7B%5Cmathbb%20%7BEXPLANATION%7D%7D%7D%7D)
The heat capacity is given by the expression:
![\longrightarrow \sf{\triangle Q= m \triangle C \triangle T}](https://tex.z-dn.net/?f=%5Clongrightarrow%20%5Csf%7B%5Ctriangle%20Q%3D%20m%20%5Ctriangle%20C%20%20%5Ctriangle%20%20%20T%7D)
![\longrightarrow \sf{Q= \: Heat}](https://tex.z-dn.net/?f=%5Clongrightarrow%20%5Csf%7BQ%3D%20%5C%3A%20Heat%7D)
![\longrightarrow \sf{M= \: Mass}](https://tex.z-dn.net/?f=%5Clongrightarrow%20%5Csf%7BM%3D%20%5C%3A%20Mass%7D)
![\longrightarrow \sf{C= \: Specific \: Heat}](https://tex.z-dn.net/?f=%5Clongrightarrow%20%5Csf%7BC%3D%20%5C%3A%20Specific%20%5C%3A%20Heat%7D)
![\longrightarrow \sf{T= \: Temperature}](https://tex.z-dn.net/?f=%5Clongrightarrow%20%5Csf%7BT%3D%20%5C%3A%20Temperature%7D)
![\huge\underline{\underline{\boxed{\mathbb {ANSWER:}}}}](https://tex.z-dn.net/?f=%5Chuge%5Cunderline%7B%5Cunderline%7B%5Cboxed%7B%5Cmathbb%20%7BANSWER%3A%7D%7D%7D%7D)
When the
is measured in the calorimeter, we obtain a value, and since we know the mass of the material and we control the change in
, we can then determine the specific heat "C" by simply remplazing in the expression.
The amount of matter in an object is called Mass.
The answer is
2.5 N
B
this needs to be 20 letters long so this part doesn't matter
Hello! So Ik this can be wrong but I worked hard for it(: It took 50 joules to push a shopping cart 5 meters with what force was the shopping cart pushed. blinded him to enable him and his men to escape. the answer is: round. Done!
(a) 3.5 Hz
The angular frequency in a spring-mass system is given by
![\omega=\sqrt{\frac{k}{m}}](https://tex.z-dn.net/?f=%5Comega%3D%5Csqrt%7B%5Cfrac%7Bk%7D%7Bm%7D%7D)
where
k is the spring constant
m is the mass
Here in this problem we have
k = 160 N/m
m = 0.340 kg
So the angular frequency is
![\omega=\sqrt{\frac{160 N/m}{0.340 kg}}=21.7 rad/s](https://tex.z-dn.net/?f=%5Comega%3D%5Csqrt%7B%5Cfrac%7B160%20N%2Fm%7D%7B0.340%20kg%7D%7D%3D21.7%20rad%2Fs)
And the frequency of the motion instead is given by:
![f=\frac{\omega}{2\pi}=\frac{21.7 rad/s}{2\pi}=3.5 Hz](https://tex.z-dn.net/?f=f%3D%5Cfrac%7B%5Comega%7D%7B2%5Cpi%7D%3D%5Cfrac%7B21.7%20rad%2Fs%7D%7B2%5Cpi%7D%3D3.5%20Hz)
(b) 0.021 m
The block is oscillating up and down together with the upper end of the spring. The block will lose contact with the spring when the direction of motion of the spring changes: this occurs when the spring is at maximum displacement, so at
x = A
where A is the amplitude of the motion.
The maximum displacement is given by Hook's law:
![F=kA](https://tex.z-dn.net/?f=F%3DkA)
where
F is the force applied initially to the spring, so it is equal to the weight of the block:
![F=mg=(0.340 kg)(9.81 m/s^2)=3.34 N](https://tex.z-dn.net/?f=F%3Dmg%3D%280.340%20kg%29%289.81%20m%2Fs%5E2%29%3D3.34%20N)
k = 160 N/m is the spring constant
Solving for A, we find
![A=\frac{F}{k}=\frac{3.34 N}{160 N/m}=0.021 m](https://tex.z-dn.net/?f=A%3D%5Cfrac%7BF%7D%7Bk%7D%3D%5Cfrac%7B3.34%20N%7D%7B160%20N%2Fm%7D%3D0.021%20m)