A rock is launched vertically upward by a slingshot. The elastic band of the slingshot accelerates the rock from rest to 40 m/s

Assuming that the mass of the empty wagon is "M," according to Newton's second law, we can derive the following relationships. Given that the empty wagon accelerates at 1.4 m/s², we proceed with this information. If a child weighing three times the mass of the wagon is on it, we can establish the relevant equations.

Answer:

H = 109.14 cm

Explanation:

Given,                                                            

Assume that the total energy equals 1 unit.                                

Energy remaining after the first collision = 0.78 x 1 unit

Balance after the first impact = 0.78 units

Remaining energy after the second impact = 0.78 ^2 units

Balance after the second impact = 0.6084 units

Remaining energy after the third impact = 0.78 ^3 units

Balance after the third impact = 0.475 units

The height reached after the third collision is equivalent to the remaining energy.

Let H denote the height achieved after three bounces.

0.475 (m g h) = m g H                  

H = 0.475 x h                                    

H = 0.475 x 2.3 m                          

H = 1.0914 m                      

H = 109.14 cm                      

Response:

Clarification:

We need an expression that shows how much water has been drained from the tub. This is represented by v, which indicates how many gallons have flowed out since the plug was taken out. Each gallon removed equates to 8.345 pounds of water, so the weight of the drained water Q in pounds as a function of v can be expressed as:

Where v signifies the number of gallons emptied from the tub.

Have a great day! Let me know if there's anything else I can assist with.

Answer:

(A) = 3.57 m

Explanation:

According to the question, the information provided is:

diameter (d) = 3.2 m

mass (m) = 42 kg

angular speed (ω) = 4.27 rad/s

Using the conservation of energy principle, we have

mgh = 0.5 mv² + 0.5Iω²...equation 1

where

Inertia (I) = 0.5mr²

ω = v/r

Revising equation 1, it turns into

mgh = 0.5 mv² + 0.5(0.5mr²)(v/r)²

resulting in gh = 0.5 v² + 0.5(0.5)v²

This simplifies to 4gh = 2v² + v²

thus h = 3v² ÷ 4g... equation 2

Given ω = v/r, we find v = ωr = 4.27 × (3.2 ÷ 2)

which yields v = 6.8 m/s

Next, substituting the value of v into equation 2 gives us

h = 3v² ÷ 4g

h = 3 × (6.8)² ÷ (4 × 9.8)

h = 3.57 m