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d1i1m1o1n [39]
1 year ago
14

Why do women typically tend to have slightly greater stability than men?

Physics
1 answer:
Pani-rosa [81]1 year ago
3 0

Women have lower centers of gravity, and lower centers of gravity provide more stability.

The correct option is C.

<h3>What is centre of gravity with example?</h3>

The location in a system or body where the weight is equally distributed and all sides are in balance is known as the center of gravity. The middle of a seesaw is an illustration of a center of gravity. the center of mass of an object in a gravitational field that is uniform.

<h3>How can you improve your center of gravity?</h3>

Training your legs and lower body is essential since they contain the biggest muscles in your body, as any competent bodybuilder or strength athlete will tell you. Exercises like squats, lunges, and calf raises should be a regular part of your workout routine if you want to reduce your center of gravity.

To know more about center of gravity visit:

brainly.com/question/17409320

#SPJ9

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A jumbo jet must reach a speed of 360 km/h on the runway for takeoff. What is the lowest constant acceleration needed for takeof
weeeeeb [17]

The lowest constant acceleration needed for takeoff from a 1.80 km runway is 2.8 m/s².

To find the answer, we need to know about the Newton's equation of motion.

<h3>What's the Newton's equation of motion to find the acceleration in term of initial velocity, final velocity and distance?</h3>
  • The Newton's equation of motion that connects velocity, distance and acceleration is V² - U²= 2aS
  • V= final velocity, U= initial velocity, S= distance and a= acceleration
<h3>What's the acceleration, if the initial velocity, final velocity and distance are 0 m/s, 360km/h and 1.8 km respectively?</h3>
  • Here, S= 1.8 km or 1800 m, V= 360km/h or 100m/s , U= 0 m/s
  • So, 100²-0= 2×a×1800

=> 10000= 3600a

=> a= 10000/3600 = 2.8 m/s²

Thus, we can conclude that the lowest constant acceleration needed for takeoff from a 1.80 km runway is 2.8 m/s².

Learn more about the Newton's equation of motion here:

brainly.com/question/8898885

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6 0
2 years ago
A compressor receives air at 290 K, 100 kPa and a shaft work of 5.5 kW from a gasoline engine. It is to deliver a mass flow rate
Sladkaya [172]

Answer:

P_2=4091\ KPa

Explanation:

Given that

T₁ = 290 K

P₁ = 100 KPa

Power P =5.5 KW

mass flow rate

\dot{m}= 0.01\ kg/s

Lets take the exit temperature = T₂

We know that

P=\dot{m}\ C_p (T_2-T_1)

5.5=0.01\times 1.005(T_2-290})\\T_2=\dfrac{5.5}{0.01\times 1.005}+290\ K\\\\T_2=837.26\ K

If we assume that process inside the compressor is adiabatic then we can say that

\dfrac{T_2}{T_1}=\left(\dfrac{P_2}{P_1}\right)^{0.285}

\dfrac{837.26}{290}=\left(\dfrac{P_2}{100}\right)^{0.285}\\2.88=\left(\dfrac{P_2}{100}\right)^{0.285}\\

2.88^{\frac{1}{0.285}}=\dfrac{P_2}{100}

P_2=40.91\times 100 \ KPa

P_2=4091\ KPa

That is why the exit pressure will be 4091 KPa.

4 0
3 years ago
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