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Leokris [45]
3 years ago
15

Based on this simulation, how would you describe the relationship between potential energy and kinetic energy? Why does this rel

ationship occur?
Physics
1 answer:
Fudgin [204]3 years ago
6 0

Answer:

The total energy of the skater is the sum of his potential energy and kinetic energy. As potential energy decreases, kinetic energy increases, and vice versa. This occurs because kinetic energy is the energy of motion and potential energy is stored energy.

Explanation:

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Tired of being chased by a jaguar, you set a trap. Hoping to drop it on the jaguar, you try to push a
Alenkasestr [34]

Answer: acceleration = 3.27m/s^2

Explanation:

Given that the

Mass M = 44kg

Angle Ø = 15 degree

Coefficient of friction ų = 0.7

Force F = 222N

F - Fr = ma ...... (1)

Where Fr = frictional force

Fr = ųN

N = normal reaction = mg

Fr = ųmgsinØ

Fr = 0.7 × 44 × 9.81 × sin 15

Fr = 78.2N

Substitutes Fr, F and M into equation one.

222 - 78.2 = 44a

143.79 = 44a

Make a the subject of formula

a = 143.79/44

Acceleration a = 3.27 m/s^2

6 0
4 years ago
Un depósito de gran superficie se llena de agua hasta una altura de 0,3 m. En el fondo del depósito hay un orificio de 5 cm2 de
ahrayia [7]

Answer:

a) El caudal de salida del chorro es 1.213\times 10^{-3}\,\frac{m^{3}}{s}.

Explanation:

a) Asúmase que el tanque se encuentra a presión atmósferica y que la sima del tanque tiene una altura de 0 metros. La rapidez de salida del chorro del depósito se determined a partir del Principio de Bernoulli, cuya línea de corriente entre la cima y la sima del tanque queda descrita por la siguiente ecuación:

\Delta z = \frac{v_{out}^{2}}{2\cdot g}

Donde:

\Delta z - Diferencia de altura, medida en metros.

g - Constante gravitacional, medida en metros por segundo al cuadrado.

v_{out} - Rapidez de salida del chorro, medida en metros por segundo.

Se despeja la rapidez de salida del chorro:

v_{out} = \sqrt{2\cdot g \cdot \Delta z}

Si g = 9.807\,\frac{m}{s^{2}} y \Delta z = 0.3\,m, entonces la rapidez de salida del chorro es:

v_{out} = \sqrt{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.3\,m)}

v_{out} \approx 2.426\,\frac{m}{s}

Ahora, la cantidad de líquido que sale del depósito por unidad de tiempo se obtiene al multiplicar la rapidez de salida del chorro por el área transversal del orificio. Esto es:

\dot V_{out} = v_{out}\cdot A_{t}

Donde:

v_{out} - Rapidez de salida del chorro, medida en metros por segundo.

A_{t} - Área transversal del orificio, medido en metros cuadrados.

\dot V_{out} - Caudal de salida del chorro, medido en metros cúbicos por segundo.

Dado que v_{out} = 2.426\,\frac{m}{s} y A_{t} = 5\,cm^{2}, el caudal de salida del chorro es:

\dot V_{out} = \left(2.426\,\frac{m}{s} \right)\cdot (5\,cm^{2})\cdot \left(\frac{1}{10000}\,\frac{m^{2}}{cm^{2}}  \right)

\dot V_{out} = 1.213\times 10^{-3}\,\frac{m^{3}}{s}

El caudal de salida del chorro es 1.213\times 10^{-3}\,\frac{m^{3}}{s}.

5 0
3 years ago
The United States and South Korean soccer teams are playing in the first round of the World Cup. An American kicks the ball, giv
garik1379 [7]

Answer:

Vf = 3.67 [m/s]

Explanation:

To solve this problem we must use the following equation of kinematics.

v_{f} ^{2} =v_{i} ^{2} -(2*a*x)

where:

Vf = final velocity [m/s]

Vi = initial velocity = 4.3 [m/s]

a = acceleration or desacceleration = 0.5 [m/s²]

x = distance = 5 [m]

Note: The negative sign in the above equation means that the velocity of the ball is decreasing (desacceleration).

Now replacing:

Vf² = (4.3)² - (2*0.5*5)

Vf² = 18.49 - 5

Vf² = 13.49

using the square root, we have.

Vf = 3.67 [m/s]

8 0
3 years ago
An instrument is thrown upward with a speed of 15 m/s on the surface of planet X where the acceleration due to gravity is 2.5 m/
Katen [24]
<h2>Answer: 12 s</h2>

Explanation:

The situation described here is parabolic movement. However, as we are told <u>the instrument is thrown upward</u> from the surface, we will only use the equations related to the Y axis.

In this sense, the main movement equation in the Y axis is:

y-y_{o}=V_{o}.t-\frac{1}{2}g.t^{2}    (1)

Where:

y  is the instrument's final position  

y_{o}=0  is the instrument's initial position

V_{o}=15m/s is the instrument's initial velocity

t is the time the parabolic movement lasts

g=2.5\frac{m}{s^{2}}  is the acceleration due to gravity at the surface of planet X.

As we know y_{o}=0  and y=0 when the object hits the ground, equation (1) is rewritten as:

0=V_{o}.t-\frac{1}{2}g.t^{2}    (2)

Finding t:

0=t(V_{o}-\frac{1}{2}g.t^{2})   (3)

t=\frac{2V_{o}}{g}   (4)

t=\frac{2(15m/s)}{2.5\frac{m}{s^{2}}}   (5)

Finally:

t=12s

3 0
4 years ago
There are five different types of muscles in exercise physiology.
9966 [12]

Answer: The answer is B) False

Explanation: There are only three muscle types in the exercise physiology.


Skeletal Muscle

Smooth Muscle

Cardiac Muscle

3 0
3 years ago
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