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hram777 [196]
3 years ago
7

Slow twitch muscles are needed for?

Physics
2 answers:
kow [346]3 years ago
5 0

The answer to your question is C) Speed or Power Activities

ruslelena [56]3 years ago
4 0

Speed or power activities

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The displacement vector from your house to the library is 740 m long, pointing 40 ∘ north of east. Part A What are the x-compone
Ilia_Sergeevich [38]

The horizontal component of the displacement is 566.9 m

Explanation:

The horizontal (x-) and vertical (y-) components of a vector on the Cartesian plane can be found as follows:

v_x = v cos \theta

v_y = v sin \theta

where

v is the magnitude of the vector

\theta is the angle representing the direction of the vector, measured as above the x-axis.

In this problem, we have:

v = 740 m (magnitude of the vector)

\theta=40^{\circ} (direction of the vector)

Therefore, the two components are

v_x = (740)(cos 40)=566.9 m

v_y = (740)(sin 40)=475.7 m

Learn more about vector components:

brainly.com/question/2678571

#LearnwithBrainly

4 0
3 years ago
A 2.0-kg object moving 5.0 m/s collides with and sticks to an 8.0-kg object initially at rest. Determine the kinetic energy lost
densk [106]

Answer:

20J

Explanation:

In a collision, whether elastic or inelastic, momentum is always conserved. Therefore, using the principle of conservation of momentum we can first get the final velocity of the two bodies after collision. This is given by;

m₁u₁ + m₂u₂ = (m₁ + m₂)v          ---------------(i)

Where;

m₁ and m₂ are the masses of first and second objects respectively

u₁ and u₂ are the initial velocities of the first and second objects respectively

v  is the final velocity of the two objects after collision;

From the question;

m₁ = 2.0kg

m₂ = 8.0kg

u₁ = 5.0m/s

u₂ = 0        (since the object is initially at rest)

<em>Substitute these values into equation (i) as follows;</em>

(2.0 x 5.0) + (8.0 x 0) = (2.0 + 8.0)v

(10.0) + (0) = (10.0)v

10.0 = 10.0v

v = 1m/s

The two bodies stick together and move off with a velocity of 1m/s after collision.

The kinetic energy(KE₁) of the objects before collision is given by

KE₁ = \frac{1}{2}m₁u₁² +  \frac{1}{2}m₂u₂²       ---------------(ii)

Substitute the appropriate values into equation (ii)

KE₁ = (\frac{1}{2} x 2.0 x 5.0²) +  (\frac{1}{2} x 8.0 x 0²)

KE₁ = 25.0J

Also, the kinetic energy(KE₂) of the objects after collision is given by

KE₂ = \frac{1}{2}(m₁ + m₂)v²      ---------------(iii)

Substitute the appropriate values into equation (iii)

KE₂ = \frac{1}{2} ( 2.0 + 8.0) x 1²

KE₂ = 5J

The kinetic energy lost (K) by the system is therefore the difference between the kinetic energy before collision and kinetic energy after collision

K = KE₂ - KE₁

K = 5 - 25

K = -20J

The negative sign shows that energy was lost. The kinetic energy lost by the system is 20J

3 0
3 years ago
Many immigrants lived in​
Fudgin [204]
California I think not sure
7 0
2 years ago
4. Brandon throws a tennis ball vertically upward. The ball returns to the point of release after 4.0 s. What is the
vekshin1

Answer:

-39.2m/s

Explanation:

Using the equation of motion;

v = u + at

Since the ball is thrown upward, the acceleration due to gravity acting on it will be negative, hence a = -g

v = u - gt

Since g = 9.8m/s²

t = 4.0s

u = 0m/s

v = 0 + (-9.8)(4)

v = 0 + (-9.8)(4)

v = -39.2m/s

Hence the speed of the ball before release is -39.2m/s

6 0
2 years ago
How much power is needed to lift the 200-N object to a height of 4 m in 4 s?
Vitek1552 [10]

Answer: 2000 watts

Explanation:

Given that,

power = ?

Weight of object = 200-N

height = 4 m

Time = 4 s

Power is the rate of work done per unit time i.e Power is simply obtained by dividing work by time. Its unit is watts.

i.e Power = work / time

(since work = force x distance, and weight is the force acting on the object due to gravity)

Then, Power = (weight x distance) / time

Power = (200N x 4m) / 4s

Power = 8000Nm / 4s

Power = 2000 watts

Thus, 2000 watts of power is needed to lift the object.

3 0
3 years ago
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