All categories

3 years ago

Which type of fitness is not based on performance?

1 answer:

4 0

Fitness can be measured during the actual performance.
Which means that there are a lot of factors that can affect agility, speed, and coordination beside the physical factors itself. Meanwhile, flexibility solely depends on the physical Factors.

The answer is flexibility .

Hope this helps (:

You might be interested in

A 20-kg block is held at rest on the inclined slope by a peg. A 2-kg pendulum starts at rest in a horizontal position when it is

gregori [183]

Complete Question

The diagram of this question is shown on the first uploaded image

Answer:

The distance the block slides before stopping is $d = 0.313 \ m$

Explanation:

The free body diagram for the diagram in the question is shown

From the diagram the angle is $\theta = 25 ^o$

$sin \theta = \frac{h}{d}$

Where $h = h_b - h_a$

So $d sin \theta = h_b - h_a$

From the question we are told that

The mass of the block is $m = 20 \ kg$

The mass of the pendulum is $m_p = 2 \ kg$

The velocity of the pendulum at the bottom of swing is $v_p = 15 m/s$

The coefficient of restitution is $e =0.7$

The coefficient of kinetic friction is $\mu _k = 0.5$

The velocity of the block after the impact is mathematically represented as

$v_2 f = \frac{m_b - em_p}{m_b + m_p} * v_2 i + \frac{[1 + e] m_1}{m_1 + m_2 } v_p$

Where $v_2 i$ is the velocity of the block before collision which is 0

$= \frac{20 - (0.7 * 2)}{(2 + 20)} * 0 + \frac{(1 + 0.7) * 2 }{2 + 20} * 15$

Substituting value

$v_2 f = 2.310\ m/s$

According to conservation of energy principle

The energy at point a = energy at point b

So $PE_A + KE _A = PE_B + KE_B + E_F$

Where

$PE_A$ is the potential energy at A which is mathematically represented as

$PE_A = m_b gh_a$ = 0 at the bottom

$KE _A$ is the kinetic energy at A which is mathematically represented as

$K_A = \frac{1}{2} m_b * v_2f^2$

$PE_B$ is the potential energy at B which is mathematically represented as

$PE_B = m_b gh$

From the diagram $h = h_b -h_a$

$PE_B = m_b g(h_b - h_a)$

$KE _B$ is the kinetic energy at B which is 0 (at the top )

Where is $E_F$ is the workdone against velocity which from the diagram is

$\mu_k m_b g cos 25 *d$

$\frac{1}{2} m_b v_2 f^2 = m_b g h_b + \mu_k m_b g cos \25 * d$

Substituting values

$\frac{1}{2} * 20 * 2.310^2 = 20 * 9.8 * d sin(25) + 0.5* 20 * 9.8 * cos 25 * d$

$d = 0.313 \ m$

6 0

3 years ago

Inappeopiate pratices: what it means? Why it not good idea to use these?

Rina8888 [55]

Answer:

Inappropriate practice is conduct by a practitioner in connection with rendering or initiating services that a practitioner's peers could reasonably conclude was unacceptable to the general body of their profession.

Explanation:

5 0

3 years ago

Mendeleev

Evgesh-ka [11]

Answer:

Explanation:

8 0

3 years ago

Read 2 more answers

An 92-kg football player traveling 5.0m/s in stopped in 10s by a tackler. What is the original kinetic energy of the player? Exp

Artemon [7]

Explanation:

It is given that,

Mass of the football player, m = 92 kg

Velocity of player, v = 5 m/s

Time taken, t = 10 s

(1) We need to find the original kinetic energy of the player. It is given by :

$k=\dfrac{1}{2}mv^2$

$k=\dfrac{1}{2}\times (92\ kg)\times (5\ m/s)^2$

k = 1150 J

In two significant figure, $k=1.2\times 10^3\ J$

(2) We know that work done is equal to the change in kinetic energy. Work done per unit time is called power of the player. We need to find the average power required to stop him. So, his final velocity v = 0

i.e. $P=\dfrac{W}{t}=\dfrac{\Delta K}{t}$

$P=\dfrac{\dfrac{1}{2}\times (92\ kg)\times (5\ m/s)^2}{10\ s}$

P = 115 watts

In two significant figures, $P=1.2\times 10^2\ Watts$

Hence, this is the required solution.

6 0

3 years ago

A fireworks rocket is fired vertically upward. At its maximum height of 90.0 m , it explodes and breaks into two pieces, one wit

Alex73 [517]

Answer:

Ai. Speed of the fragment with mass mA= 1.35 kg is 34.64 m/s

Aii. Speed of the fragment with mass mB = 0.270 kg is 77.46 m/s

B. 475.3 m

Explanation:

A. Determination of the speed of each fragment.

I. Determination of the speed of the fragment with mass mA = 1.35 kg

Mass of fragment (m₁) = 1.35 kg

Kinetic energy (KE) = 810 J

Velocity of fragment (u₁) =?

KE = ½m₁u₁²

810 = ½ × 1.35 × u₁²

810 = 0.675 × u₁²

Divide both side by 0.675

u₁² = 810 / 0.675

u₁² = 1200

Take the square root of both side.

u₁ = √1200

u₁ = 34.64 m/s

Therefore, the speed of the fragment with mass mA = 1.35 kg is 34.64 m/s

II. I. Determination of the speed of the fragment with mass mB = 0.270 kg

Mass of fragment (m₂) = 0.270 kg

Kinetic energy (KE) = 810 J

Velocity of fragment (u₂) =?

KE = ½m₂u₂²

810 = ½ × 0.270 × u₂²

810 = 0.135 × u₂²

Divide both side by 0.135

u₂² = 810 / 0.135

u₂² = 6000

Take the square root of both side.

u₂ = √6000

u₂ = 77.46 m/s

Therefore, the speed of the fragment with mass mB = 0.270 kg is 77.46 m/s

B. Determination of the distance between the points on the ground where they land.

We'll begin by calculating the time taken for the fragments to get to the ground. This can be obtained as follow:

Maximum height (h) = 90.0 m

Acceleration due to gravity (g) = 10 m/s²

Time (t) =?

h = ½gt²

90 = ½ × 10 × t²

90 = 5 × t²

Divide both side by 5

t² = 90/5

t² = 18

Take the square root of both side

t = √18

t = 4.24 s

Thus, it will take 4.24 s for each fragments to get to the ground.

Next, we shall determine the horizontal distance travelled by the fragment with mass mA = 1.35 kg. This is illustrated below:

Velocity of fragment (u₁) = 34.64 m/s

Time (t) = 4.24 s

Horizontal distance travelled by the fragment (s₁) =?

s₁ = u₁t

s₁ = 34.64 × 4.24

s₁ = 146.87 m

Next, we shall determine the horizontal distance travelled by the fragment with mass mB = 0.270 kg. This is illustrated below:

Velocity of fragment (u₂) = 77.46 m/s

Time (t) = 4.24 s

Horizontal distance travelled by the fragment (s₂) =?

s₂ = u₂t

s₂ = 77.46 × 4.24

s₂ = 328.43 m

Finally, we shall determine the distance between the points on the ground where they land.

Horizontal distance travelled by the 1st fragment (s₁) = 146.87 m

Horizontal distance travelled by the 2nd fragment (s₂) = 328.43 m

Distance apart (S) =?

S = s₁ + s₂

S = 146.87 + 328.43

S = 475.3 m

Therefore, the distance between the points on the ground where they land is 475.3 m

3 0

3 years ago