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juin [17]
2 years ago
12

How does self efficacy relate to sports performance?

Physics
1 answer:
vitfil [10]2 years ago
6 0

Answer: Individuals with high self efficacy will participate readily and more frequently, will put more effort in and persist longer, enhancing performance in sport and exercise

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What will be the average velocity of a body falling in free fall on Earth for 3 s?
SpyIntel [72]

Answer:

29.4m/s

Explanation:

Given parameters:

Time  = 3s

Unknown:

Average velocity  = ?

Solution:

To solve this problem, we use the expression below:

      v  = u + gt

v is the average velocity

u is the initial velocity  = 0m/s

g is the acceleration due to gravity  = 9.8m/s²

t is the time

So;

        v  = 0 + (9.8 x 3)  = 29.4m/s

6 0
3 years ago
Is an example of a physical process that a physical geographer would study.
diamong [38]
D. The environment
Is the right answer
8 0
3 years ago
What are the steps to extract metal from the earths crust
Pani-rosa [81]
<span>Methods of extraction include: extract by electrolysis, extract by reaction with carbon or carbon monoxide, and extracted by various chemical reactions.</span>
4 0
3 years ago
John is a field researcher who studies social interaction within motorcycle groups. He is not an accomplished rider himself, but
uysha [10]
His role as a field research is that of a: Complete participant!
(Option D.)

~Good luck!
7 0
3 years ago
An Atwood machine consists of two masses, mA = 6.8 kg and mB = 8.0 kg , connected by a cord that passes over a pulley free to ro
Lisa [10]

To solve this problem it is necessary to apply the concewptos related to Torque, kinetic movement and Newton's second Law.

By definition Newton's second law is described as

F= ma

Where,

m= mass

a = Acceleration

Part A) According to the information (and as can be seen in the attached graph) a sum of forces is carried out in mass B, it is obtained that,

\sum F = m_b a

m_Bg-T_B = m_Ba

T_B = m_Bg-m_Ba

In the case of mass A,

\sum F = m_A a

T_A = m_Ag-m_Aa

Making summation of Torques in the Pulley we have to

\sum\tau = I\alpha

T_BR_0-T_AR_0=I\alpha

T_B-T_A=I\frac{a}{R^2_0}

Replacing the values previously found,

(m_Bg-m_Ba )-(m_Ag-m_Aa )=I\frac{a}{R^2_0}

(m_B-m_A)g-(m_B+m_A)a=I\frac{a}{R^2_0}

a = \frac{(m_B-m_A)g}{\frac{I}{R_0^2}+(m_B+m_A)}

a = \frac{(m_B-m_A)g}{\frac{MR^2_0^2/2}{R_0^2}+(m_B+m_A)}

a =\frac{(m_B-m_A)g}{\frac{M}{2}+(m_B+m_A)}

Replacing with our values

a =\frac{(8-6.8)(9.8)}{\frac{0.8}{2}+(8+6.8)}

a=0.7736m/s^2

PART B) Ignoring the moment of inertia the acceleration would be given by

a' =\frac{(m_B-m_A)g}{(m_B+m_A)}

a' =\frac{(8-6.8)(9.8)}{(8+6.8)}

a' = 0.7945

Therefore the error would be,

\%error = \frac{a'-a}{a}*100

\%error = \frac{0.7945-0.7736}{0.7736}*100

\%error = 2.7%

8 0
3 years ago
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