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Roman55 [17]
2 years ago
10

MY PAPER IS DUE IN AN HOUR!!!

Physics
2 answers:
Doss [256]2 years ago
7 0

Answer:

Your kinetic energy is always proportional to the square of your speed,

no matter whether you're on a bicycle, a roller coaster, or a baby carriage.

The faster you're moving, the more kinetic energy you have.

Your potential energy doesn't depend on speed at all, only on height.

On a roller coaster, the minimum speeds are usually at the top of a hill

or a loop, so the potential energy would generally correspond to the

slowest speeds. But again, that's not because of the speed. It's because

of the height.

Explanation:

Delvig [45]2 years ago
6 0

Answer:

njjnbhkjgjbvggvn

Explanation:

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Using this illustration, what is the chemical formula
forsale [732]

Answer:la primera o la tercera

Explanation:

Porque son 3 átomos en uno

4 0
2 years ago
Explain how surface and subsurface events are integral parts of the rock cycle.
vagabundo [1.1K]
Example of surface events are erosion and weathering. Erosion is the carrying of a particle from one place to the other and weathering is the breaking down of particles. These processes help in rock formation because this allows physical changes (grouping together or breaking down) on a certain substance. Subsurface events are those which happened underground such as the flow of underground water which subsequently allow the deposition of minerals, etc. 
3 0
3 years ago
PLS SOMEONE HELP QUICK
SVETLANKA909090 [29]

Answer:

the rates of rock formation are similar. i could be wrong tho.....

Explanation:

6 0
2 years ago
Read 2 more answers
A girl drops a stone from the top a tower 45m tall. At the same time, a boy standing at the base of the tower, projects another
Advocard [28]

Answer:

(i) The stones meet at 1.8 second

(ii) The point at which the stones meet, is 28.8 m above the base of the building and 16.2 m below the top of the building.

Explanation:

(i)

First we consider the stone dropped by the girl. We have data:

Vi = Initial Velocity of Stone = 0 m/s   (Since the stone was initially at rest)

t = Time Period

g = 10 m/s²

s₁ = Distance Covered by Stone

Using 2nd equation of motion, we get:

s₁ = Vi t + (0.5)gt²

s₁ = (0)(t) + (0.5)(10)t²

s₁ = 5t²   ----- equation (1)

Now, we consider the stone throne vertically upward by the boy. We have data:

Vi = Initial Velocity of Stone = 25 m/s

t = Time Period

g = - 10 m/s²   (negative sign due to upward motion)

s₂ = Distance Covered by Stone

Using 2nd equation of motion, we get:

s₂ = Vi t + (0.5)gt²

s₂ = (25)(t) + (0.5)(-10)t²

s₂ = 25t - 5t²   ----- equation (2)

At, the point where both the stones meet, the sum of distances covered by both stones must be equal to the height of building (i.e 45 m).

s₁ + s₂ = 45

using values from equation (1) and equation (2)

5t² + 25t - 5t² = 45

25t = 45

t = 45/25

<u>t =  1.8 sec</u>

(ii)

using this value of of t in equation (2)

s₂ = (25)(1.8) - (5)(1.8)²

<u>s₂ = 28.8 m</u>

using this value of of t in equation (1)

s₁ = (5)(1.8)²

<u>s₁ = 16.2 m</u>

<u>Hence, the point at which the stones meet, is 28.8 m above the base of the building and 16.2 m below the top of the building.</u>

6 0
3 years ago
1) Si un mango cae a una velocidad de 75m/s y tarda 26 seg. en caer. ¿ Cuál habrá sido la velocidad con qué el mango llegó al su
Lyrx [107]

Answer:

El mango llega al suelo a una velocidad de 329.982 metros por segundo.

Explanation:

El mango experimenta un movimiento de caída libre, es decir, un movimiento uniformemente acelerado debido a la gravedad terrestre, despreciando los efectos de la viscosidad del aire y la rotación planetaria. Entonces, la velocidad final del mango, es decir, la velocidad con la que llega al suelo, se puede determinar mediante la siguiente fórmula cinemática:

v = v_{o}+g\cdot t (1)

Donde:

v_{o} - Velocidad inicial, en metros por segundo.

v - Velocidad final, en metros por segundo.

g - Aceleración gravitacional, en metros por segundo al cuadrado.

t - Tiempo, en segundos.

Si sabemos que v_{o} = -75\,\frac{m}{s}, g = -9.807\,\frac{m}{s^{2}} y t = 26\,s, entonces la velocidad final del mango es:

v = v_{o}+g\cdot t

v = -75\,\frac{m}{s}+\left(-9.807\,\frac{m}{s} \right)\cdot (26\,s)

v = -329.982\,\frac{m}{s}

El mango llega al suelo a una velocidad de 329.982 metros por segundo.

8 0
2 years ago
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