Question

A 60lb Belgian is moving in the x-direction at 14 m/s. Just 0.34s later it is moving 16.8m/s at 34 degrees to the out-of-bounds line (call it the x-axis) of the French Ring stadium. What magnitude and direction did each paw's force apply to the ground to make this move if slow motion video showed at full stretch, the dog's legs made a 30-degree angle to the ground?

Answer 1

do ur school

Answer 2

home word

Answer 3

To solve this problem, we can use the conservation of momentum equation, which states that the momentum before an event is equal to the momentum after the event.

Let's assume that the Belgian dog is moving in the positive x-direction before the event, and after the event, it is moving at an angle of 34 degrees with the x-axis. The momentum before the event is:

p1 = m*v1 = 60 lb * 14 m/s = 840 lb m/s

where m is the mass of the dog and v1 is its velocity before the event.

The momentum after the event is:

p2 = m*v2 = 60 lb * 16.8 m/s = 1008 lb m/s

where v2 is the velocity of the dog after the event.

Since momentum is conserved, we can set p1 equal to p2 and solve for v2:

p1 = p2

m*v1 = m*v2

v2 = v1 * (m2/m1)

where m2 is the mass of the dog and m1 is the mass of the ground.

Since the mass of the ground is much larger than the mass of the dog, we can assume that m2/m1 is negligible and v2 is approximately equal to v1. Therefore, the velocity of the dog after the event is approximately 14 m/s.

To find the magnitude and direction of each paw's force, we can use the following equations:

F = ma

a = (v2 - v1)/t

where F is the force applied by each paw, m is the mass of the dog, a is the acceleration of the dog, and t is the time interval between the two velocities.

Substituting the values, we get:

a = (16.8 m/s - 14 m/s)/0.34 s = 8.24 m/s^2

Now, we can use the angle between the dog's legs and the ground to calculate the magnitude and direction of each paw's force. Since the legs are at a 30-degree angle to the ground, the component of the force in the vertical direction is:

F_vertical = m*a*sin(30) = 60 lb * 8.24 m/s^2 * sin(30) = 247.2 lb

The component of the force in the horizontal direction is:

F_horizontal = m*a*cos(30) = 60 lb * 8.24 m/s^2 * cos(30) = 507.4 lb

Therefore, each paw applied a force of approximately 247.2 lb in the vertical direction and 507.4 lb in the horizontal direction to make the move. The direction of the force is at a 30-degree angle with the ground.

Answer 4

wth

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