Answer:
L = - 1361.591 k Kgm/s
Explanation:
Given
mA = 55.2 Kg
vA = 3.45 m/s
rA = 6.00 m
mB = 62.4 Kg
vB = 4.23 m/s
rB = 3.00 m
mC = 72.1 Kg
vC = 4.75 m/s
rC = - 5.00 m
then we apply the equation
L = (mv x r)
⇒ LA = mA*vA x rA = 55.2 *(3.45 i)x(6 j) = (1142.64 k) Kgm/s
⇒ LB = mB*vB x rB = 62.4 *(4.23 j)x(3 i) = (- 791.856 k) Kgm/s
⇒ LC = mC*vC x rC = 72.1 *(- 4.75 j)x(- 5 i) = (- 1712.375 k) Kgm/s
Finally, the total counterclockwise angular momentum of the three joggers about the origin is
L = LA + LB + LC = (1142.64 - 791.856 -1712.375) k Kgm/s
L = - 1361.591 k Kgm/s
Answer:
The distance traveled during its acceleration, d = 214.38 m
Explanation:
Given,
The object's acceleration, a = -6.8 m/s²
The initial speed of the object, u = 54 m/s
The final speed of the object, v = 0
The acceleration of the object is given by the formula,
a = (v - u) / t m/s²
∴ t = (v - u) / a
= (0 - 54) / (-6.8)
= 7.94 s
The average velocity of the object,
V = (54 + 0)/2
= 27 m/s
The displacement of the object,
d = V x t meter
= 27 x 7.94
= 214.38 m
Hence, the distance the object traveled during that acceleration is, a = 214.38 m
Answer:
a) 4.2m/s
b) 5.0m/s
Explanation:
This problem is solved using the principle of conservation of linear momentum which states that in a closed system of colliding bodies, the sum of the total momenta before collision is equal to the sum of the total momenta after collision.
The problem is also an illustration of elastic collision where there is no loss in kinetic energy.
Equation (1) is a mathematical representation of the the principle of conservation of linear momentum for two colliding bodies of masses 
and 
whose respective velocities before collision are 
and 
;
where 
and 
are their respective velocities after collision.
Given;
Note that =0 because the second mass was at rest before the collision.
Also, since the two masses are equal, we can say that 
so that equation (1) is reduced as follows;
m cancels out of both sides of equation (2), and we obtain the following;
a) When 
, we obtain the following by equation(3)
b) As stops moving 
, therefore,
Answer:
QC = 122 KJ
QH = 2.64 x 122 = 322 KJ
Explanation:
TH = 500 Degree C = 500 + 273 = 773 K
TC = 20 degree C = 20 + 273 = 293 K
W cycle = 200 KJ
Use the formula for the work done in a cycle
Wcycle = QH - QC
200 = QH - QC ..... (1)
Usse
TH / TC = QH / QC
773 / 293 = QH / QC
QH / QC = 2.64
QH = 2.64 QC Put it in equation (1)
200 = 2.64 QC - QC
QC = 122 KJ
So, QH = 2.64 x 122 = 322 KJ
Hello There!
The resistance of a conductor depends on all of the following except mass.
Mass wouldn't affect the resistance in any way.
Hope This Helps You!
Good Luck :)
- Hannah ❤
