A river flows due east at 12 km/h. A boat crosses the 720 m wide river by maintaining a constant velocity of 72 mi/h due north r
elative to the water. If no correction is made for the current, how far downstream does the boat move by the time it reaches the far shore?
1 answer:
Answer:
The boat will be 74 .17 meters downstream by the time it reaches the shore.
Explanation:
Consider the vector diagrams for velocity and distance shown below.
converting 72 miles per hour to km/hr
we have 72 miles per hour 72 × 1.60934 = 115.83 km/hr
The velocity vectors form a right angled triangle, and can be solved using simple trigonometric laws
This is the vector angle with which the ship drifts away with respect to its northward direction.
<em>From the sketch of the displacement vectors, we can use trigonometric ratios to determine the distance the boat moves downstream.</em>
Let x be the distance the boat moves downstream.d
∴The boat will be 74 .17 meters downstream by the time it reaches the shore.
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Answer:
The minimum coefficient of friction is 0.544
Solution:
As per the question:
Radius of the curve, R = 48 m
Speed of the car, v = 16 m/s
To calculate the minimum coefficient of static friction:
The centrifugal force on the box is in the outward direction and is given by:
where
= coefficient of static friction
The net force on the box is zero, since, the box is stationary and is given by:
Answer:
final velocity = 0.08585m/s
Explanation:
We are taking train cars as our system. In this system no external force is acting. So we can apply the law of conservation of linear momentum.
The law of conservation of linear momentum states that the total linear momentum of a system remains constant if there is no external force acting on the system. That is total linear momentum before = total linear momentum after
total linear momentum before = linear momentum of first train car + linear momentum of second train car
We know that linear momentum = mv
where,
m = mass
v = velocity
thus,
total linear momentum before = m₁v₁ + m₂v₂
m₁ = mass of first train car = 135,000kg
v₁ = velocity of first train car = 0.305m/s
m₂ = mass of first second car = 100,000kg
v₂ = velocity of second train car = −0.210m/s
Note: Momentum is a vector. So while adding momentum we should take account of its direction too. Here since second train car is moving in a direction opposite to that of the first one, we have taken its velocity as negative.
total linear momentum before = m₁v₁ + m₂v₂
= 135,000x0.305 + 100,000x(−0.210)
= 135,000x0.305 - 100,000x0.210
= 20,175 kgm/s
Now we have to find total linear momentum after bumping. After the bumping both the train cars will be moving together with a common velocity(say v).
Therefore, total linear momentum after = mv
m = m₁ + m₂ = 135,000 + 100,000 = 235,000
total linear momentum before = total linear momentum after
235,000v = 20,175
v =
= 0.08585m/s