There are two different size spherical paintballs and the smaller one has a diameter of 5 cm and the larger one is 9 cm in diame

Answer:

The question is incomplete; refer to the attachment for the diagram

Explanation:

Provided that,

The mass of the car

M = 1500kg

Entering the curve at Point A with a speed of

Va = 100km/hr = 100× 1000/3600

Va = 27.78m/s

The car slows down at a constant rate until it reaches point C at a speed of

Vc = 50km/hr = 50×1000/3600

Vc = 13.89m/s

Radius of curvature at point A

p = 400m

Radius of curvature at point B

p = 80m

The distance from point A to point B as indicated in the attachment is

S=200m

We need to determine the total horizontal forces at points A, B, and C applied by the road on the tire

The constant tangential acceleration can be computed using the equation of motion

Vc² = Va² + 2as

13.89² = 27.78² + 2 × a × 200

192.9 = 771.6 + 400a

400a = 192.9—771.6

400a = -578.7

a = -578.7 / 400

a = —1.45 m/s²

at = —1.45m/s²

The tangential acceleration is -1.45m/s² and it is negative due to the car decelerating

Since the vehicle is decreasing speed at a uniform rate, the tangential acceleration remains the same at every point

At point A

at = -1.45m/s²

At point B

at = -1.45m/s²

At point C

at = -1.45m/s²

Now,

We can compute the normal component of acceleration (centripetal acceleration) at each point since we know the radius of curvature

The centripetal acceleration can be calculated using

ac = v²/ p

At point A ( p = 400)

an = Va²/p = 27.78² / 400

an = 1.93 m/s²

At point B (p = ∞), since point B is the inflection point

Then,

an = Vb²/p =  Vb/∞ = 0

an = 0

At point C ( p = 80m)

an = Vc²/p = 13.89² / 80 = 2.41m/s²

an = 2.41 m/s²

Then,

The tangential force is

Ft = M•at

Ft = 1500 × 1.45

Ft = 2175 N.

Given that the tangential acceleration remains constant, this represents the tangential force at A, B, and C

Next, the normal force

At point A ( an = 1.93m/s²)

Fn = M•an

Fn = 1500 × 1.93

Fn = 2895 N

At point B (an=0)

Fn = M•an

Fn = 0 N

At point C (an= 2.41m/s²)

Fn = M•an

Fn = 1500 × 2.41

Fn = 3615 N.

Thus, the horizontal force acting at each point is

Using the right triangle vector method

F = √(Fn² + Ft²)

At point A

Fa = √(2895² + 2175²)

Fa = √13,111,650

Fa = 3621 N

At point B

Fb = √(0² + 2175²)

Fb = √2175²

Fb = 2175 N

At point C

Fc = √(3615² + 2175²)

Fc = √17,798,850

Fc = 4218.88 N

Fc ≈ 4219N

Answer:

Explanation:

The measurement of pressure is indicated as where p denotes the pressure, signifies density, and h represents height

Given values include pressure , gravity's acceleration , and height =1.163 m

In this scenario, the principles of momentum conservation can be applied since there are no external forces acting on the system. Consequently, the conservation of momentum principle is applicable here. After the bird lands on it, both the bird and the bark will have a unified final speed. Thus, this final speed will be 1 m/s.

Answer: a) t = 1.8 x 10^2 seconds; b) t = 54 seconds; c) t = 49 seconds. Explanation: a) To determine the time of a stationary passenger on the sidewalk, we use the position formula. Given the constant speed of the walkway, we can calculate the time taken for set distances accordingly. This calculation extends into cases where combined velocities for walking are involved in subsequent queries.

Answer:

The initially bent young tree has been straightened by adjusting the tensions of the three guy wires to AB = 7 lb, AC = 8 lb, and AD = 10 lb. Please calculate the force and moment reactions at the trunk's base point O, disregarding the weight of the tree.

C and D are situated 3.1' from the y-axis, while B and C are located 5.4' from the x-axis, and A has a height of 5.2'.

Explanation:

Refer to the attached image.