When you apply the torque equation ∑τ = 0 to an object in equilibrium, the axis about which torques are calculated:

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:

Answer:

(A) The tension's magnitude grows to four times the initial value, 4F.

Explanation:

When an object travels in a circular path, a centripetal force is exerted upon it. In this instance, the centripetal force acting on the stone can be represented by .

                   Here, m denotes the mass of the object

                               v is the velocity or speed of the object

                               r signifies the radius of the circular path

Importantly, the tension corresponds to the centripetal force.

Initially, the string completes one revolution each second, and subsequently, it accelerates to perform two revolutions in the same time frame. This signifies that the speed has increased twofold.

Applying our formula:F =

                               where F indicates the tension in the string

assuming the starting speed is v, after doubling it becomes 2v

Maintaining the circle's radius, we arrive at:

F=

From this equation, it's clear that the initial tension has quadrupled.

Consequently, the magnitude of the tension increases to four times its original value, 4F.

Answer:

Explanation:

a) La fuerza neta que actúa sobre la caja en la dirección vertical es:

Fnet=Fg−f−Fp *sin45 °

aquí Fg representa la fuerza gravitacional, f es la fuerza de fricción, y Fp es la fuerza de empuje.

Fnet=ma

ma=Fg−f−Fp *sin45 °

​a=

=0.24 m/s²

Vf =Vi +at

=0.48+0.24*2

Vf=2.98 m/s

b)

Fnet=Fg−f−Fp *sin45 °

=Fg−0.516Fp−Fp *sin45 °

=30-1.273Fp

Fnet=0 (Ya que la velocidad es constante)

Fp=30/1.273

=23.56 N

Answer:

155.38424 K

2.2721 kg/m³

Explanation:

= Reservoir pressure = 10 atm

= Reservoir temperature = 300 K

= Exit pressure = 1 atm

= Exit temperature

= Specific gas constant = 287 J/kgK

= Specific heat ratio = 1.4 for air

Assuming isentropic flow

Flow temperature at exit is 155.38424 K

Density at exit can be derived using the ideal gas equation

Flow density at exit measures 2.2721 kg/m³

Response:

The speed at which the distance from the helicopter to you is changing (in ft/s) after 5 seconds is ft/ sec

Clarification:

Provided:

h(t) = 25 ft/sec

x(t) = 10 ft/ sec

h(5) = 25 ft/sec. 5 = 125 ft

x(5) = 10 ft/sec. 5 = 50 ft

At this point, we can determine the distance between the individual and the helicopter utilizing the Pythagorean theorem

Now, let's calculate the derivative of distance in relation to time

By plugging in the values for h(t) and x(t) and simplifying, we arrive at,

= = ft / sec