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Snowcat [4.5K]
2 years ago
12

A ball is launched with an initial horizontal velocity of 10.0 meters per second. It takes 500 milliseconds for the ball to reac

h its maximum height.
Determine the maximum horizontal distance that the ball will travel.
Calculate the initial vertical velocity of the ball.
Physics
1 answer:
hichkok12 [17]2 years ago
5 0

Answer:

500ms times 2 would be when the ball reaches the max horizontal distance.

Then to find the angle, use the formula of time to reach max height t = u sin theta / g . With t being the max height time 500ms, u being 10m/s

For initial vertical velocity just use u sin theta.

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Consider a system of a cliff diver and the earth. the gravitational potential energy of the system decreases by 28,000 j as the
olga2289 [7]

The decrease in gravitational potential energy of the system is given by

\Delta U = (mg) \Delta h

where

m is the mass, g is the gravitational acceleration, and \Delta h is the variation of height of the system.


(mg) also corresponds to the weight of the diver, therefore if we rearrange the equation and we use \Delta U=28000 J and \Delta h=32.0 m, we can find her weight:

mg=\frac{\Delta U}{\Delta h}=\frac{28000 J}{32 m}=875 N

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A solid object has a density of 2.85 g/cm3. According to Archimedes, what can be done to make this object float in water?
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3 years ago
Friends who are your age and share the same interests are called
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Read 2 more answers
A 15 m uniform ladder weighing 500 N rests against a frictionless wall. The ladder makes a 60° angle with horizontal. (a) Find t
scoray [572]

Answer:

a)    F₁ = 267.3 N,   N₁ = 1300 N,  b)    μ = 0.324

Explanation:

For this exercise we use the rotational equilibrium condition, we have a reference system is the floor and the anticlockwise rotations as positive, in the adjoint we can see a diagram of the forces

           

let's use subscript 1 for the ladder and 2 for the firefighter

            ∑ τ = 0

          -W₁ x₁ - W₂ x₂ + N₁ y = 0

           N₁ = \frac{W_1 x_1 + W_2 x_2}{y}          (1)

the center of mass of the ladder is at its geometric center,

d = L / 2 = 15/2 = 7.5 m

         cos 60 = x₁ / d₁

         x₁ = d₁ cos 60

         x₁ = 7.5 cos 60

         x₁ = 3.75 m

for the firefighter d₂ = 4 m

         cos 60 = x₂ / d₂

         x₂ = d₂ cos 60

          x₂ = 4 cos 60 = 2 m

for the fulcrum d₃ = 15 m

         sin 60 = y / d₃

         y = d₃ sin 60

         y = 15 sin 60

         y = 13 m

we look for the Normal by substituting in equation 1

         N₂ = \frac{500 \ 3.75 \ + 800 \ 2}{13}

         N₂ = 267.3 N

now let's use the translational equilibrium relations

 X axis

           F₁ - N₂ = 0

           F₁ = N₂

           F₁ = 267.3 N

Axis y

          N₁ - W₁ -W₂ = 0

          N₁ = W₁ + W₂

          N₁ = 500 + 800

          N₁ = 1300 N

b) for this case change the firefighter's distance d₂ = 9 m

          x₂ = 9 cos 60

          x₂ = 4.5 m

we substitute in 1

          N₂ = \frac{500 \ 3.75 \ + 800 \ 4.5}{13}  

          N₂ = 421.15 N

of the translational equilibrium equation on the x-axis

          fr = F₁ = N₂

          fr = 421.15 N

friction force has the expression

          fr = μ N

in this case the reaction of the Earth to the support of the ladder is N1 = 1300N

          μ = fr / N₁

          μ = 421.15 / 1300

          μ = 0.324

8 0
2 years ago
A rock with a mass of 540 g in air is found to have an apparent mass of 342 g when submerged in water. (a) What mass of water is
AleksandrR [38]

(a) 198 g

When the rock is submerged into the water, there are two forces acting on the rock:

- its weight, equal to W=mg (m=mass, g=acceleration of gravity), downward

- the buoyant force, equal to B=m_w g (m_w=mass of water displaced), upward

So the resultant force, which is the apparent weight of the rock (W'), is

W'=W-B

which can be rewritten as

m'g = mg-m_w g

where m' is the apparent mass of the rock. Using:

m = 540 g

m' = 342 g

we find the mass of water displaced

m_w = m-m'=540 g-342 g=198 g

(b) 1.98\cdot 10^{-4} m^3

If the rock is completely submerged, the volume of the rock corresponds to the volume of water  displaced.

The volume of water displaced is given by

V_w = \frac{m_w}{\rho_w}

where

m_w = 198 g = 0.198 kg is the mass of the water displaced

\rho_w = 1000 kg/m^3 is the density of the water

Substituting,

V_w = \frac{0.198}{1000}=1.98\cdot 10^{-4} m^3

And so this is also the volume of the rock.

(c) 2727 kg/m^3

The average density of the rock is given by

\rho = \frac{m}{V}

where

m = 540 g = 0.540 kg is the mass of the rock

V=1.98\cdot 10^{-4} m^3 is its volume

Substituting into the equation, we find

\rho = \frac{0.540 kg}{1.98\cdot 10^{-4}}=2727 kg/m^3

3 0
3 years ago
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