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Diano4ka-milaya [45]
4 years ago
8

QUIK!!!!!!!!!!!!! 1. Compare your scores to your partner’s scores. Explain the reason for any differences. 2. How can improving

your balance and coordination affect your performance in sports or recreational activities? 3. What activity was the most difficult for you to perform? Why? 4. Describe an activity that could help improve your striking and your reaction time skills. 5. Select a physical activity or sport that you have never done but would like to try one day. Identify at least one element of skill-related fitness that is important to being successful in this activity. 6. What power-increasing activities do you need to perform regularly to improve your performance
Physics
1 answer:
kvv77 [185]4 years ago
6 0
2.Improving my balance and coordination can affect my performance in sports or recreational activities by planing better in activities also <span>improving my speed. My activities all need skills that I am really good at.
3.</span>The activity that was most difficult for me to perform was the Stork Stance test because I am not good with balancing and that is something <span>I need to improve.
4.</span>An activity that could help improve my striking and my reaction time.
5.One sport that I have never done but that I would like to try one day is <span>baseball.
</span>One element of skill-related fitness that is important to being successful in the activity is speed and reaction time skills.6. R<span>unning, jogging, sprinting, and jump roping.</span>
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Acceleration on moon is approximately 1/6th of the Earth's; what would be the force of your egg drop package if dropped on the m
cricket20 [7]
  • Acceleration on earth=10m/s^2

Acceleration on moon:-

\\ \rm\Rrightarrow 10\dfrac{1}{6}=1.6m/s^2

Now

  • Mass=0.56kg

\\ \rm\Rrightarrow F=ma

\\ \rm\Rrightarrow F=0.56(1.6)

\\ \rm\Rrightarrow F=0.896N

8 0
3 years ago
algunas fabricas de balones de futbol ubicadas en la costa inflan los balones que van a ser vendidos en ciudades como pasto,tunj
solmaris [256]

Answer:

Debido al cambio de volumen del gas dentro de la bola cuando se somete a una disminución de la presión y un aumento de la temperatura de la atmósfera.

Explanation:

La ubicación de las fábricas que fabrican las bolas que se encuentra en la costa como se indica en la pregunta es a baja altitud y una región más fría donde la presión atmosférica es más alta

Por lo tanto, la presión en la costa = p₁

La temperatura en la costa = T₁

El volumen de la pelota en la costa = V₁

Las condiciones en una ciudad con mayor altitud y temperatura tienen mayor presión atmosférica y temperatura de la siguiente manera;

La presión en la ciudad = p₂ <p₁

La temperatura en la ciudad = T₂> T₁

El volumen de la pelota en la costa = V₂

Por la ley de gas combinada de tenemos;

\dfrac{p_1 \times V_1}{T_1} = \dfrac{p_2 \times V_2}{T_2}

Allí el volumen de la pelota en la ciudad estará dado por la relación;

V_2 = V_1 \times \dfrac{p_1 }{P_2} \times  \dfrac{ T_2}{T_1 }

Dado que p₁> p₂ y T₂> T₁, entonces p₁/p₂> 1 y T₂/T₁> 1 y tenemos;

V₂ = V₁ × p₁/p₂ × T₂/T₁ = V₁ × X, donde X = p₁/p₂ × T₂/T₁> 1

V₂ = V₁ × X (> 1)

Por lo tanto, V₂> V₁ y el volumen ocupado por el gas aumenta cuando la pelota llega a la ciudad y la pelota está más firme.

7 0
3 years ago
A sample of an ideal gas is slowly compressed to one-half its original volume with no change in pressure. If the original root-m
Zigmanuir [339]

Answer:

V_{rms_f}=\sqrt{\frac{3PV}{2nM}} = \sqrt{\frac{1}{2}} \sqrt{\frac{3PV}{M}} = \frac{1}{\sqrt{2}} \sqrt{\frac{3PV}{M}} = \frac{1}{\sqrt{2}} V_{rms_i}

So then the final answer on this case would be:

\frac{V_{rms_i}}{\sqrt{2}}

Explanation:

From the kinetic theory model of gases we know that the velocity rms (speed of gas molecules) is given by:

V_{rms}= \sqrt{\frac{3RT}{M}}  (1)

Where V represent the velocity

R the constant for ideal gases

T the temperature

M the molecular weight of the gas

We also know from the ideal gas law that PV= nRT

If we solve for T we got: T = \frac{PV}{nR}

For the initial state we can replace T into the equation (1) and we got:

V_{rms_i}= \sqrt{\frac{3R (\frac{PV}{nR})}{M}} = \sqrt{\frac{3PV}{M}}

For the final state we know that :V_f = \frac{V}{2} And the pressure not change , so then the final velocity would be:

V_{rms_f}= \sqrt{\frac{3R (\frac{P(V/2)}{nR})}{M}} = \sqrt{\frac{3P(V/2)}{M}}

V_{rms_f}=\sqrt{\frac{3PV}{2nM}} = \sqrt{\frac{1}{2}} \sqrt{\frac{3PV}{M}} = \frac{1}{\sqrt{2}} \sqrt{\frac{3PV}{M}} = \frac{1}{\sqrt{2}} V_{rms_i}

So then the final answer on this case would be:

\frac{V_{rms_i}}{\sqrt{2}}

6 0
3 years ago
How could you increase gravitational force using each factor?
jeka94
The strength of the gravitational force between two objects depends on two factors, mass and distance. the force of gravity the masses exert on each other. If one of the masses is doubled, the force of gravity between the objects is doubled. increases, the force of gravity decreases.
5 0
3 years ago
In a mass spectrometer a particle of mass m and charge q is accelerated through a potential difference V and allowed to enter a
Ghella [55]

Answer:

m = B²qR² / 2 V

Explanation:

If v be the velocity after acceleration under potential difference of V

kinetic energy  = loss of electric potential energy

1/2 m v² = Vq ,

v² = 2 Vq / m ----------------------- ( 1 )

In magnetic field , charged particle comes in circular motion in which magnetic force provides centripetal force

magnetic force = centripetal force

Bqv = mv² / R

v = BqR / m

v² = B²q²R² / m²  ------------------------- (2)

from (1) and (2)

B²q²R² / m²  = 2 Vq / m

m = B²q²R² / 2 Vq

m = B²qR² / 2 V

7 0
3 years ago
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