Speed, agility, physical activity
Change in velocity of larger moose: (1/3)v - v = -(2/3)v
<span>change in velocity of small moose: (1/3)v - (-v) = (4/3)v </span>
<span>- (change in velocity of larger moose)/(change in velocity of smaller moose) = 2
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Answer:
d. R4
Explanation:
Generally, the flow of current is always from the positive sign to the negative sign. In the resistors R1, R2, and R3, the direction of flow of current is from the positive sign to the negative sign. However, in the resistor R4, the direction of the flow of current is different from the conventional method. Therefore, the resistor R4 is marked wrongly.
Answer:
B. The elastic portion of a straight-line, downward-sloping demand curve corresponds to the segment above the midpoint.
Explanation:
Elasticity measures the sensitivity of one variable to another. Specifically it is a figure that indicates the percentage variation that a variable will experience in response to a variation of another one percent.
The elasticity of demand measures the reaction of demand when one of the factors that affects it varies.
<u>Elasticity - Price of demand.</u>
easure the sensitivity of the quantity demanded to price variations. It indicates the percentage variation that the quantity demanded of a good will experience if its price rises by 1 percent.
<u>
Elastic Demand
</u>
The demand quantity is relatively sensitive to price variations, so the total expenditure on the product decreases when the price rises, the price elasticity takes value greater than -∞ but less than -1
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