Answer:
R=V/I=6/2=3 ohm
time = 5 minutes = 5*60=300 seconds
I=2 A
Energy = I²Rt=(2)²*3*300=4*900=3600 J
Answer: Her velocity magnitude (v) relative to the shore is 5.70 km/h.
Explanation:
Let Q be the speed of the boat, and P be the speed of the river flow.
R represents the resultant velocity combining boat velocity and river current.
According to vector addition using the law of triangles:
From the diagram:
P = 3.5 km/h, Q = 4.5 km/h
Therefore, her velocity magnitude relative to the shore is 5.70 km/h.
The work done can be calculated using the equation:
Work = Force x Distance = Change in kinetic energy
The kinetic energy is derived using the following formula: KE = (1/2)*m*v^2
Thus, the change in kinetic energy is calculated as (1/2)*m*(Vf)^2 - (1/2)*m*(Vo)^2
Where:
Vf represents the final speed = 90 kph = 25 m/s
Vo denotes the initial speed = 72 kph = 20 m/s
By substituting in the given values:
Work = (1/2)*2500*(25^2) - (1/2)*2500*(20^2) = 281250 J, which can also be represented as 2.8 x 10^5 Joules.
The correct choice among the options is A.
Speed is defined as distance over time. Hence, to determine the distance, we use d = V * t. Plugging in the values yields d = (72 Km / h) * (1h / 3600s) * (4.0 s) = 0.08Km. Thus, during this distracted period, a distance of 0.08Km was covered.
