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love history [14]
3 years ago
12

Not only did Skid get shot out of a cannon when he was a clown in the circus, but he was also launched into the air by a vertica

lly compressed spring. Skid of 55 kg is standing on a massless platform attached to a spring of constant 8700 N/m that is compressed 125 cm. Skid is launched from rest. He eventually leaves the platform and reaches a peak high above the platform. (a) What is the velocity of Skid when he leaves the platform
Physics
1 answer:
tiny-mole [99]3 years ago
8 0

Answer:

v = 16.11 m / s

Explanation:

For this exercise we must use the principle of conservation of energy. We set a reference system on the part of the platform without elongation

starting point. When the spring is compressed

        Em₀ = K_e + U = ½ k x² + m g x ’

final point. The point where it leaves the platform

        Em_f = K = ½ m v²

energy is conserved

         Em₀ = Em_f

         ½ k x² + m g x ’= ½ m v²

         v² = \frac{k}{m}  x² + g x

let's calculate

         v² = \frac{8700}{55}  1.25² + 9.8 1.25

         v² = 247.159 + 12.25 = 259.409

         v = 16.11 m / s

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A. the gas particles move faster and collide more frequently, which causes an increase in pressure.

Increasing temperature increases the energy of the gas, which causes the kinetic energy of the molecules to increase.
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3 years ago
When a vapor condenses to a liquid, an amount of thermal energy (Q
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Answer:

true

Explanation:

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2 years ago
Your office has a 0.025 m^3 cylindrical container of drinking water. The radius of the container is about 13 cm. Required:a. Whe
uranmaximum [27]

Answer:

<em>a) 105935.7 Pa</em>

<em>b) 103630.35 Pa</em>

Explanation:

The volume of the container = 0.025 m^3

The radius of the container = 13 cm = 0.13 m

We have to find the height of the tank

From the equation for finding the volume of the cylinder,

V = \pi r^2h

where

V is the volume of the cylinder

h is the height of the cylinder

substituting values, we have

0.025 = 3.142 x 0.13^2 x h

0.025 = 0.0531h

h = 0.025/0.0531 = 0.47 m

Pressure at the bottom of the tank P = ρgh

where

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

g is the acceleration due to gravity = 9.81 m/s^2

h is the depth of water which is equal to the height of the tank

substituting values, we have

P = 1000 x 9.81 x 0.47 = 4610.7 Pa

atmospheric pressure = 101325 Pa

therefore, the pressure in the tank bottom above atmospheric pressure = 101325 Pa + 4610.7 Pa = <em>105935.7 Pa</em>

b) For half way down the container, depth of water will be = 0.47/2 = 0.235 m

pressure P = 1000 x 9.81 x 0.235 = 2305.35 Pa

This pressure above atmospheric pressure = 101325 Pa + 2305.35 Pa = <em>103630.35 Pa</em>

7 0
3 years ago
Your cousin Jannik skis down a blue square ski slope, with an initial speed of 3.6 m/s. He travels 15 m down the mountain side b
fenix001 [56]

Answer: The loss of energy due to friction is equal to 1,253 J.

Explanation:

The problem tells us that the skier has an initial speed of 3.6 m/s, which means that his initial kinetic energy is as follows:

K₁ = 1/2 m v₁² = 1/2 . 58.0 Kg. (3.6)² (m/s)² =  376 J

After coming to a  flat landing, his final speed is 7.8 m/s, so the final kinetic energy is as follows:

K₂ = 1/2 m v₂² = 1/2. 58.0 Kg. (7.8)² (m/s)² = 1,764 J

Now, when skying down the slope the increase in kinetic energy only can come from another type of energy, in this case, gravitational potential energy.

If we take the ground flat level as a Zero reference, the initial gravitational potential energy, can be written as follows, by definition:

U₁ = m.g. h (1)

Now, we don't know the value of the height h, but we know that the incline has a 18º angle above the horizontal, and that the distance travelled along the incline is 15 m.

By definition, the sinus of an angle, is equal to the proportion between the height and the hypotenuse , so we can write the following equation:

sin 18º = h / 15 m ⇒ h = 15 m. sin 18º = 4.6 m

Replacing in (1), we get:

U₁ = 58.0 Kg. 9.8 m/s². 4.6 m = 2,641 J

So, we can get the total initial mechanical energy, as follows:

E₁ = K₁ + U₁ = 376 J + 2,641 J = 3,017 J

After arriving to the flat zone, all potential energy has become in kinetic energy, even though not completely, due to the effect of friction.

This remaining kinetic energy can be written as follows:

E₂ = K₂ = 1,764 J

The difference E₂-E₁, is the loss of energy due to friction forces acting during the travel along the 15 m path, and is as follows:

ΔE= E₂ - E₁ = 1,764 J - 3,017 J = -1,253 J

8 0
3 years ago
A 9.0-V battery moves 20 mC of charge through a circuit running from its positive terminal to its negative terminal. How much en
dem82 [27]

Answer:

E = 0.18 J

Explanation:

given,

Potential of the battery,V = 9 V

Charge on the circuit, Q = 20 m C

                                        = 20 x 10⁻³ C

energy delivered in the circuit

E = Q V

E = 20 x 10⁻³ x 9

E =  180 x 10⁻³

E = 0.18 J

Energy delivered in the circuit is equal to E = 0.18 J

7 0
4 years ago
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