A lens of focal length 15.0 cm is held 10.0 cm from a page (the object ). Find the magnification .

Answer:

a)

b)

c)

Explanation:

According to the problem, the distance from the building where the ball hits is 16m, and its final elevation exceeds the initial height by 8m.

With this information, we can compute the ball’s starting speed.

a) Let's first assess the horizontal trajectory.

(1)

This gives us our initial equation.

Next, we need to examine the vertical trajectory.

Utilizing in our first equation (1)

Now let’s solve for t.

The ball takes two seconds to reach the adjacent building, allowing us to compute its initial speed.

b) To determine the velocity magnitude just before impact, we must calculate both x and y components.

The computed velocity magnitude is:

c) The ball's angle is:

Response:

Physical well-being

Clarification:

The Pythagorean Theorem can be utilized here: Imagine a car navigating through traffic—when it turns left to travel north, a right angle of 90 degrees is formed. However, the displacement is always the shortest distance connecting the origin and the endpoint, which forms a triangle in this scenario. In a right triangle, the Pythagorean theorem applies: 215^2+45^2=c^2; therefore, v=√(215^2+45^2).

Answer:

force = 6.53× N

Explanation:

Provided data

downward force = 0.60 m

mass m = kg

distance h = 0.40 m

to determine

magnitude of the downward force

solution

we know here mg is apply 0.4 m away from support and

thus applied force is d = 0.6 m from support

therefore

by balancing torque we can compute force

as

force = mass × g × h / d

substituting the values

force = mass × g × h / d

force = ( × 9.81 × 0.4 ) / 0.6

force = 6.53× N

a)

b) 803.4 N

Explanation:

a) At point C (top of the loop), the pilot experiences weightlessness, leading to the normal force from the seat being zero:

N = 0

Consequently, the force balance equation at position C becomes:

where the left term signifies the pilot's weight and the right term represents the centripetal force, with:

= acceleration due to gravity

= jet's velocity at the top

= loop radius

By solving for v,

Thus, this is the jet's speed at C.

The speed at position A (bottom) can be derived from

The distance traveled by the jet corresponds to half the circumference of the circle with radius r, therefore

Given the plane's deceleration is consistent, we can obtain it using the following equation:

b) The pilot experiences a force equal to the normal force from the seat. At point B (half-way through the loop), we find:

- The normal force from the seat, N, directed towards the center of the loop

- As there are no further forces acting toward the central axis, N must equal the centripetal force:

(1)

where

represents the speed at position B.

To deduce the velocity at B, we note that the distance covered by the jet between positions A and B is a quarter of a circle:

With knowledge of the deceleration, we can implement the equation of motion to find the velocity at the midway point B:

Thus, we then apply eq.(1) to determine the normal force acting on the pilot at B: