What test are included in the Physical Fitness Test?

What does addition of two vectors give you?

mash [69]

Answer:

Explanation:

To add or subtract two vectors, add or subtract the corresponding components. Let →u=⟨u1,u2⟩ and →v=⟨v1,v2⟩ be two vectors. The sum of two or more vectors is called the resultant. The resultant of two vectors can be found using either the parallelogram method or the triangle method .

4 0

4 years ago

(a) What is the potential between two points situated 10 cm and 20 cm from a 3.0-μC point charge? (b) To what location should th

julia-pushkina [17]

Answer:

(a) 135 kV

(b) The charge chould be moved to infinity

Explanation:

(a)

The potential at a distance of r from a point charge, Q, is given by

$V = -\dfrac{kQ}{r}$

where $k = 9\times 10^9 \text{ F/m}$

Difference in potential between the points is

$kQ\left[-\dfrac{1}{0.2\text{ m}} -\left( -\dfrac{1}{0.1\text{ m}}\right)\right] = \dfrac{kQ}{0.2\text{ m}} = \dfrac{9\times10^9\text{ F/m}\times3\times10^{-6}\text{ C}}{0.2\text{ m}}$

$PD = 135\times 10^3\text{ V} = 135\text{ kV}$

(b)

If this potential difference is increased by a factor of 2, then the new pd = 135 kV × 2 = 270 kV. Let the distance of the new location be x.

$270\times10^3 = kQ\left[-\dfrac{1}{x}-\left(-\dfrac{1}{0.1\text{ m}}\right)\right]$

$10 - \dfrac{1}{x} = \dfrac{270000}{9\times10^9\times3\times10^{-6}} = 10$

$\dfrac{1}{x} = 0$

$x = \infty$

The charge chould be moved to infinity

7 0

3 years ago

At what point in its motion is the KE of a pendulum bob a maximum? 1. The KE does not change. 2. at the lowest point 3. at the h

Andrew [12]

Answer:

2. at the lowest point

Explanation:

The motion of the pendulum is a continuous conversion between kinetic energy (KE) and gravitational potential energy (GPE). This is because the mechanical energy of the pendulum, which is sum of KE and GPE, is constant:

E = KE + GPE = const.

Therefore, when KE is maximum, GPE is minimum, and viceversa.

So, the point of the motion where the KE is maximum is where the GPE is minimum: and since the GPE is directly proportional to the heigth of the bob:

$GPE=mgh$

we see that GPE is minimum when the bob is at the lowest point,so the correct answer is

2. at the lowest point

3 0

4 years ago

You are designing a flywheel. It is to start from rest and then rotate with a constant angular acceleration of 0.200 rev/s^2. Th

Rama09 [41]

Answer:

I = 8.75 kg m

Explanation:

This is a rotational movement exercise, let's start with kinetic energy

K = ½ I w²

They tell us that K = 330 J, let's find the angular velocity with kinematics

w² = w₀² + 2 α θ

as part of rest w₀ = 0

w = √ 2α θ

let's reduce the revolutions to the SI system

θ = 30.0 rev (2π rad / 1 rev) = 60π rad

let's calculate the angular velocity

w = √(2 0.200 60π)

w = 8.683 rad / s

we clear from the first equation

I = 2K / w²

let's calculate

I = 2 330 / 8,683²

I = 8.75 kg m

4 0

3 years ago

A 25,000 kg traveling east collides with a 2,000 kg truck standing still on the tracks. After the collision the train and truck

Elis [28]

Answer:

24.084 m/s

Explanation:

From the law of conservation of linear momentum

Total momentum before collision equals to the total momentum after collision

Since momentum=mv where m is mass and v is velocity

$M_{truck}V_{truck}=V_{common}*(M_{truck} +M_{standing})$ where $M_{truck}$ is the mass of the truck, $V_{truck}$ is velocity of the truck, $V_{common}$ is the common velocity of moving and standing truck after collision and $M_{standing}$ is the mass of the standing truck

Making $V_{truck}$ the subject we obtain

$V_{truck}=\frac { V_{common}*(M_{truck} +M_{standing})}{M_{truck}}$

Substituting $M_{truck}$ as 25000 Kg, $V_{common}$ as 22.3 m/s, $M_{standing}$ as 2000 Kg we obtain

$V_{truck}=\frac { 22.3 m/s *(25000 Kg +2000 Kg)}{25000}= 24.084 m/s$

Therefore, assuming no friction and considering that after collision they still move eastwards hence common velocity and initial truck velocities are positive

The truck was moving at 24.084 m/s

3 0

3 years ago