Answer:
v1 = 15.90 m/s
v2 = 8.46 m/s
mechanical energy before collision = 32.4 J
mechanical energy after collision = 32.433 J
Explanation:
given data
mass m = 0.2 kg
speed = 18 m/s
angle = 28°
to find out
final velocity and mechanical energy both before and after the collision
solution
we know that conservation of momentum remain same so in x direction
mv = mv1 cosθ + mv2cosθ
put here value
0.2(18) = 0.2 v1 cos(28) + 0.2 v2 cos(90-28)
3.6 = 0.1765 V1 + 0.09389 v2 ................1
and
in y axis
mv = mv1 sinθ - mv2sinθ
0 = 0.2 v1 sin28 - 0.2 v2 sin(90-28)
0 = 0.09389 v1 - 0.1768 v2 .......................2
from equation 1 and 2
v1 = 15.90 m/s
v2 = 8.46 m/s
so
mechanical energy before collision = 1/2 mv1² + 1/2 mv2²
mechanical energy before collision = 1/2 (0.2)(18)² + 0
mechanical energy before collision = 32.4 J
and
mechanical energy after collision = 1/2 (0.2)(15.90)² + 1/2 (0.2)(8.46)²
mechanical energy after collision = 32.433 J
Answer:
The pressure of the tire will be 4.19 atm
Explanation:
The ideal gas equation is:
p0 * V0 / T0 = p1 * V1 / T1
Since V0 = V1 because the volume of the tire doesn't change
p0 / T0 = p1 / T1
p1 = p0 * T1 / T0
We need absolute temperatures for this equation
16 C = 289 K
45 C = 318 K
Now we replace the values:
p1 = 3.71 * 318 / 289 = 4.19 atm
Answer:
a) The speed of the slider is 4.28 in/s
b) The velocity vector is 2.33 in/s
Explanation:
a) According to the diagram 1 in the attached image:
Also:
![v_{C} =v_{A}+w_{AC}*r_{C/A}\\v_{Ci}=-3j+\left[\begin{array}{ccc}i&j&k\\0&0&w_{AC} \\6.883&-9.829&0\end{array}\right]\\v_{Ci}=-3j+(0+9.829w_{AC} i-(0-6.883w_{AC})j\\v_{Ci}=9.829w_{AC}i+(-3+6.883w_{AC})j](https://tex.z-dn.net/?f=v_%7BC%7D%20%3Dv_%7BA%7D%2Bw_%7BAC%7D%2Ar_%7BC%2FA%7D%5C%5Cv_%7BCi%7D%3D-3j%2B%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Di%26j%26k%5C%5C0%260%26w_%7BAC%7D%20%5C%5C6.883%26-9.829%260%5Cend%7Barray%7D%5Cright%5D%5C%5Cv_%7BCi%7D%3D-3j%2B%280%2B9.829w_%7BAC%7D%20i-%280-6.883w_%7BAC%7D%29j%5C%5Cv_%7BCi%7D%3D9.829w_%7BAC%7Di%2B%28-3%2B6.883w_%7BAC%7D%29j)
If we comparing both sides of the expression:
b) According to the diagram 2 in the attached image:
![v_{C}=v_{A}+w_{AC}r_{C/A}\\v_{C}=-3j+\left[\begin{array}{ccc}i&j&k\\0&0&w_{AC}\\7.713&-9.192&0\end{array}\right] \\v_{Ci}=-3j+(9.192w_{AC})i+7.713w_{AC}j\\v_{Ci}=9.192w_{AC}i+(7.713w_{AC}-3)j](https://tex.z-dn.net/?f=v_%7BC%7D%3Dv_%7BA%7D%2Bw_%7BAC%7Dr_%7BC%2FA%7D%5C%5Cv_%7BC%7D%3D-3j%2B%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Di%26j%26k%5C%5C0%260%26w_%7BAC%7D%5C%5C7.713%26-9.192%260%5Cend%7Barray%7D%5Cright%5D%20%5C%5Cv_%7BCi%7D%3D-3j%2B%289.192w_%7BAC%7D%29i%2B7.713w_%7BAC%7Dj%5C%5Cv_%7BCi%7D%3D9.192w_%7BAC%7Di%2B%287.713w_%7BAC%7D-3%29j)
Comparing both sides of the expression:
![v_{B}=v_{C}+w_{AC}r_{B/C}\\v_{B}=3.57i+\left[\begin{array}{ccc}i&j&k\\0&0&0.388\\-3.856&4.59&0\end{array}\right] \\v_{B}=3.57i+(0-1.78)i-(0+1.499)j\\v_{B}=1.787i-1.499j\\|v_{B}|=\sqrt{1.787^{2}+1.499^{2} } =2.33in/s](https://tex.z-dn.net/?f=v_%7BB%7D%3Dv_%7BC%7D%2Bw_%7BAC%7Dr_%7BB%2FC%7D%5C%5Cv_%7BB%7D%3D3.57i%2B%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Di%26j%26k%5C%5C0%260%260.388%5C%5C-3.856%264.59%260%5Cend%7Barray%7D%5Cright%5D%20%20%5C%5Cv_%7BB%7D%3D3.57i%2B%280-1.78%29i-%280%2B1.499%29j%5C%5Cv_%7BB%7D%3D1.787i-1.499j%5C%5C%7Cv_%7BB%7D%7C%3D%5Csqrt%7B1.787%5E%7B2%7D%2B1.499%5E%7B2%7D%20%20%7D%20%3D2.33in%2Fs)
Answer:
Explanation:
Equation for translational kinetic energy = 1/2 m V² where m is mass and V is velocity .
In first case let mass of car be m
Translational kinetic energy = 1/2 m x 40² = 800 m .
In the second case ,
Translational kinetic energy = 1/2 m x 20² = 200 m
So , in former case kinetic energy of car is 4 times that of second case.
Ferromagnetic, paramagnetic, and diamagnetic
