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Anna11 [10]
3 years ago
15

A student is holding a stone at a certain height. The stone has 50 joules of potential energy and 0 joules of kinetic energy. Th

e stone is released from that height. How much energy does the stone have
right before it hits the ground?
Select one:
a) The stone will have no energy
b) The stone will have about 25 joules each of kinetic and potential energy
c) The stone will have about 50 joules of potential energy and ovogres van kinetic
energy
d) The stone will have about 50 joules of kinetic energy and ovoules of potencia
energy​
Physics
1 answer:
Hitman42 [59]3 years ago
6 0

Answer:

d) The stone will have about 50 joules of kinetic energy and 0 joules of potential energy​ .

Explanation:

Given :

Initial Potential energy , P_i=50\ J .

Initial Kinetic energy , K_i=0\ J . ( because ball is in rest )

Now , we know , kinetic energy is maximum when an object reaches ground .

Also , potential energy is zero when an object is in ground .

We know , by conservation of energy :

Initial total energy = Final total energy

P_i+K_i=P_f+K_f\\\\50+0=0+K_f\\\\K_f=50 \ J

Therefore , option d) is correct .

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One uniform ladder of mass 30 kg and 10 m long rests against a frictionless vertical wall and makes an angle of 60o with the flo
yuradex [85]

Answer:

   μ = 0.37

Explanation:

For this exercise we must use the translational and rotational equilibrium equations.

We set our reference system at the highest point of the ladder where it touches the vertical wall. We assume that counterclockwise rotation is positive

let's write the rotational equilibrium

           W₁  x/2 + W₂ x₂ - fr y = 0

where W₁ is the weight of the mass ladder m₁ = 30kg, W₂ is the weight of the man 700 N, let's use trigonometry to find the distances

             cos 60 = x / L

where L is the length of the ladder

              x = L cos 60

            sin 60 = y / L

           y = L sin60

the horizontal distance of man is

            cos 60 = x2 / 7.0

            x2 = 7 cos 60

we substitute

         m₁ g L cos 60/2 + W₂ 7 cos 60 - fr L sin60 = 0

         fr = (m1 g L cos 60/2 + W2 7 cos 60) / L sin 60

let's calculate

         fr = (30 9.8 10 cos 60 2 + 700 7 cos 60) / (10 sin 60)

         fr = (735 + 2450) / 8.66

         fr = 367.78 N

the friction force has the expression

         fr = μ N

write the translational equilibrium equation

         N - W₁ -W₂ = 0

         N = m₁ g + W₂

         N = 30 9.8 + 700

         N = 994 N

we clear the friction force from the eucacion

        μ = fr / N

        μ = 367.78 / 994

        μ = 0.37

3 0
3 years ago
A planned high-speed train between Houston and Dallas will travel a distance of 386 kilometers in 5.40 × 10^3 seconds. What is t
Mazyrski [523]

¡Hellow!

For this problem, first, lets convert the seconds in hours:

5,4x10³\rightarrow 5400

h = sec / 3600

h = 5400 s / 3600

h = 1,5

Let's recabe information:

d (Distance) = 386 km

t (Time) = 1,5 h

v (Velocity) = ?

For calculate velocity, let's applicate formula:

                                                    \boxed{\boxed{\textbf{d = v * t} } }

Reeplace according we information:

386 km = v * 1,5 h

v = 386 km / 1,5 h

v = 257,33 km/h

The velocity of the train is of <u>257,33 kilometers for hour.</u>

<u></u>

Extra:

For convert km/h to m/s, we divide the velocity of km/h for 3,6:

m/s = km/h / 3,6

Let's reeplace:

m/s = 257,33 km/h / 3,6

m/s = 71,48

¿Good Luck?

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