Answer:
The equilibrium temperature is
21.97°c
Explanation:
This problem bothers on the heat capacity of materials
Given data
specific heat capacities
copper is Cc =390 J/kg⋅C∘,
aluminun Ca = 900 J/kg⋅C∘,
water Cw = 4186 J/kg⋅C∘.
Mass of substances
Copper Mc = 235g
Aluminum Ma = 135g
Water Mw = 825g
Temperatures
Copper θc = 255°c
Water and aluminum calorimeter θ1= 16°c
Equilibrium temperature θf =?
Applying the principle of conservation of heat energy, heat loss by copper equal heat gained by aluminum calorimeter and water
McCc(θc-θf) =(MaCa+MwCw)(θf-θ1)
Substituting our data into the expression we have
235*390(255-θf)=
(135*900+825*4186)(θf-16)
91650(255-θf)=(3574950)(θf-16)
23.37*10^6-91650*θf=3.57*10^6θf- +57.2*10^6
Collecting like terms and rearranging
23.37*10^6+57.2*10^6=3.57*10^6θf+91650θf
8.2*10^6=3.66*10^6θf
θf=80.5*10^6/3.6*10^6
θf =21.97°c
A person lifting a chair is converting chemical energy to mechanical energy.
Answer:
Explanation:
Small grains are negatively charged by the wind while big grains is positively charged and remains at the ground . This process creates an electric field due to the presence of oppositely charged particles.
When ever electric field exists it is directed from a positive charge to a negative charge so the here electric field is towards an upwards direction.
Complete Question
An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field in the wire changes with time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is measured in seconds.
I = 1.2 A at time 5 secs.
Find the charge Q passing through a cross-section of the conductor between time 0 seconds and time 5 seconds.
Answer:
The charge is 
Explanation:
From the question we are told that
The diameter of the wire is 
The radius of the wire is 
The resistivity of aluminum is 
The electric field change is mathematically defied as
Generally the charge is mathematically represented as
Where A is the area which is mathematically represented as
So
Therefore
substituting values
![Q = 120 \int\limits^{t}_{0} { [ 0.0004t^2 - 0.0001t +0.0004] } \, dt](https://tex.z-dn.net/?f=Q%20%3D%20120%20%5Cint%5Climits%5E%7Bt%7D_%7B0%7D%20%7B%20%5B%200.0004t%5E2%20-%200.0001t%20%2B0.0004%5D%20%7D%20%5C%2C%20dt)
![Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | t} \atop {0}} \right.](https://tex.z-dn.net/?f=Q%20%3D%20120%20%5B%20%5Cfrac%7B0.0004t%5E3%20%7D%7B3%7D%20-%20%5Cfrac%7B0.0001%20t%5E2%7D%7B2%7D%20%2B0.0004t%5D%20%7D%20%20%5Cleft%20%7C%20t%7D%20%5Catop%20%7B0%7D%7D%20%5Cright.)
From the question we are told that t = 5 sec
![Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | 5} \atop {0}} \right.](https://tex.z-dn.net/?f=Q%20%3D%20120%20%5B%20%5Cfrac%7B0.0004t%5E3%20%7D%7B3%7D%20-%20%5Cfrac%7B0.0001%20t%5E2%7D%7B2%7D%20%2B0.0004t%5D%20%7D%20%20%5Cleft%20%7C%205%7D%20%5Catop%20%7B0%7D%7D%20%5Cright.)
![Q = 120 [ \frac{0.0004(5)^3 }{3} - \frac{0.0001 (5)^2}{2} +0.0004(5)] }](https://tex.z-dn.net/?f=Q%20%3D%20120%20%5B%20%5Cfrac%7B0.0004%285%29%5E3%20%7D%7B3%7D%20-%20%5Cfrac%7B0.0001%20%285%29%5E2%7D%7B2%7D%20%2B0.0004%285%29%5D%20%7D)
Answer:
625000 N/ m
Explanation:
m= 20 kg
v= 30 m/s
x= 12 cm
k = ?
Here when the mass when hits at spring its speed is
Vi= 30 m/s
Finally it comes to rest after compressing for 12 cm
i-e Vf = 0 m/s
Distance= S= 12 cm = 0.12 m
using
2aS= Vf2 - Vi2
==> 2a ×0.12 = o- 30 × 30
==> a = 900 ÷ 0.24 = 3750 m/sec2
Now we know;
F = ma
F= -Kx
==> ma= -kx
==> 20 × 3750 = -K × 0.12
==> k = 625000 N/ m
