A weatherman carried an aneroid barometer from the ground floor to his office atop the Sears Tower in Chicago. On the level grou

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.

Upon comparison, it is evident that the student with the highest percent error is A. Student 4, who measured 9.61 m/s². Four students recorded the acceleration of gravity, with the accepted local value being 9.78 m/s². Now let's find out which student's measurement exhibited the greatest percent error.

The height is h = 17 10⁶ meters above the surface of Mars. To determine this, we apply Newton's second law according to the universal law of gravitation, represented by F = m a. The centripetal acceleration a is expressed as v² / r. Applying the gravitational force we have G m M / r² = m v² / r. Given that the speed of the object remains constant, we derive v from d / t, where d is the circumference and t is the orbital period. Substituting gives us d = 2π r and v = 2π r / T. Replacing these values leads to the equation G M / r² = (4π² r² / T) / r, so r³ = G M T² / 4π². Converting time into SI units, T = 24.66 h converts to 88776 seconds. Ultimately, the computed value of r is 2,045 10⁶ m, and after subtracting Mars’ radius of 3.39 10⁶ m, we find the height h to be 17 10⁶ m.

Answer:

35.79 meters

Explanation:

We have an archer, and there is a target. Denote the distance between them as d.

The bowman releases the arrow, which travels the distance d at a velocity of 40 m/s until it hits the target. We establish the equation as:

Right after this, the arrow produces a muffled noise, traveling the same distance d at a speed of 340 m/s in time . Thus, we can derive:

.

Consequently, the sound reaches the archer, precisely 1 second post-firing the bow, resulting in:

.

Using this relationship in the distance formula for sound allows us to write:

.

Substituting the value of d from the first equation yields:

.

Now, after some calculations, we can proceed further:

.

Finally, the value is inserted into the initial equation: