An elevator supported by a single cable descends a shaft at a constant speed. The only forces acting on the elevator are the ten

The required lift force is approximately 866.92 N. To determine this, we first establish the shark's mass at 92 kg and its density at 1040 kg/m³. The volume of the shark is calculated by dividing mass by density, yielding 0.08846 m³. The buoyant force acting on the shark is then determined by multiplying the volume by the density of water and gravity, resulting in a lift force of 866.92 N.

Response:

The car's acceleration magnitude is 35.53 m/s²

Details:

Given;

acceleration of the truck, = 12.7 m/s²

mass of the truck, = 2490 kg

mass of the car, = 890 kg

let the acceleration of the car during the collision =

Using Newton's third law of motion;

The force exerted by the truck equals the force exerted by the car.

The car's force acts in the opposite direction.

Thus, the car's acceleration magnitude is 35.53 m/s²

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

This initial stage represents a right triangle. If the aircraft is positioned 400 km to the east and 300 km to the south of the origin while traveling in a straight path, you can form a right triangle with sides measuring 300, 400, and c. You might notice that these dimensions correspond to multiples of the Pythagorean triple 3, 4, 5, meaning that the length of c is 500 km. Alternatively, you would indicate
.

For the second step, assuming I am correctly understanding "degrees south of east," it involves calculating the angle between the horizontal line indicating east and the trajectory of the aircraft. I created a diagram illustrating this (see attached). You could employ a trigonometric function related to one of the angles to find the solution. I selected
. Therefore, I deduce that the angle is 37° south of east.

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