HELP Due right now

Physics

1 answer:

xz_007 [3.2K]2 years ago

7 0

Answer:

so here the initial momentum was all in the X direction. This is Equalling the mass. I'm someone multiplied by visa one initial Equalling 1345 kilograms multiplied by 15.7 meters per second. This is Equalling 2.11 times 10 to the fourth kilograms meters per second. Now the car stick together and we can say that the final momentum is Equalling. I'm someone plus himself to multiplied by the final velocity. This is gonna be equal to 1345 kilograms plus 1900 23 kilograms. This is multiplied by 14 0.5 meters per second. And we find that the final month Thomas Equalling 4.74 times 10 to the fourth kilograms meters per second. Now the y component of the final momentum, we can say that he's sub final. Why would be equaling two 4.74 times 10 to the fourth kilograms meters per second? This would be equally rather multiplied by sign of 63.5 degrees. This is Equalling 4.24 times 10 to the fourth kilograms meters per second. By law, of the conservation of momentum. We can say that the Y component of the final momentum equals the why component of the initial momentum. This is equaling EMS up to visa to initial and be some to initial is found to be equal to the final. The white component of final momentum divided by the second cars mass and so this would be 4.24 times tend to the fourth kilograms meters per second, divided by M sub 2 1923 kilograms and this is giving us 22.0 meters per second. And of course this is greater than 20 whether 20.1 meters per second. And so we can say that yes, this car is exceeding the speed limit.

Explanation:

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Read 2 more answers

An 85.0-N crate of apples sits at rest on a ramp that runs from the ground to the bed of a truck. The ramp is inclined at 28.0°

irina [24]

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A 75.1 N and a direction of 152° to the vertical.

B 85.0 N at 0° to the vertical.

Explanation:

A) The interaction partner of this normal force has what magnitude and direction?

The interaction partner of this normal force is the component of the weight of the crate perpendicular to the ramp. It has a magnitude of 85cos28° = 75.1 N and a direction of 180° - 28° = 152° to the vertical(since it is directed downwards perpendicular to the ramp).

B) The normal and frictional forces are perpendicular components of the contact force exerted on the crate by the ramp. What is the magnitude and direction of the contact force?

Since this force has to balance the weight of the crate, its magnitude is 85.0 N. Its direction has to be vertically opposite to that of the weight.

Since the weight is 180° to the vertical (since it is directed downwards), this force is 0° to the vertical.

So, this force has a magnitude of 85.0 N and a direction of 0° to the vertical.

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