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nexus9112 [7]
2 years ago
5

4. What type of assessments are based on repeatable, measurable data?

Physics
1 answer:
Harman [31]2 years ago
8 0

Answer:

  • OBJECTIVE ASSESSMENTS
  • SUBJECTIVE ASSESSMENTS
  • BODY COMPOSITION ASSESSMENTS
  • COGNITIVE ASSESSMENTS

Explanation:

PLS MAKE ME BRAINLIEST

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How high would you have to lift a 1000kg car to give it a potential energy of:
Elza [17]

Given parameters:

Mass of the car = 1000kg

Unknown:

Height  = ?

To find the heights for the different amount potential energy given, we need to understand what potential energy is.

Potential energy is the energy at rest due to the position of a body.

 It is mathematically expressed as:

          P.E  = mgh

m is the mass

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

h is the height of the car

Now the unknown is h, height and we make it the subject of the expression to make for easy calculation.

               h = \frac{P.E }{mg}

<u>For 2.0 x 10³ J;</u>

                  h  = \frac{2000}{1000 x 9.8}   = 0.204m

<u>For 2.0 x 10⁵ J;</u>

                  h  = \frac{200000}{9.8 x 1000}   = 20.4m

<u>For 1.0kJ  = 1 x 10³J; </u>

                  h  = \frac{1000}{9.8 x 1000}   = 0.102m

   

5 0
3 years ago
The x-component of vector R is Rx = −28.2 units and its y-component is Ry = 19.6 units. What are its magnitude and direction? Gi
ikadub [295]

Answer:

Explanation:

Rx = -28.2 units

Ry = 19.6 units

magnitude of R = √  [( - 28.2 )² + ( 19.6 ) ]

= √ ( 795.24 + 384.16 )

= 34.34 units

If θ  be the angle measured counterclockwise from the +x-direction

Tanθ = 19.6 / - 28.2 = -0.695

θ = 180 - 34.8

= 145.2° .

7 0
3 years ago
A heavy solid disk rotating freely and slowed only by friction applied at its outer edge takes 120 seconds to come to a stop.
alisha [4.7K]

Answer:

The time is 16 min.

Explanation:

Given that,

Time = 120 sec

We need to calculate the moment of inertia

Using formula of moment of inertia

I=\dfrac{1}{2}MR^2

If the disk had twice the radius and twice the mass

The new moment of inertia

I'=\dfrac{1}{2}\times2M\times(2R)^2

I'=8I

We know,

The torque is

\tau=F\times R

We need to calculate the initial rotation acceleration

Using formula of acceleration

\alpha=\dfrac{\tau}{I}

Put the value in to the formula

\alpha=\dfrac{F\times R}{\dfrac{1}{2}MR^2}

\alpha=\dfrac{2F}{MR}

We need to calculate the new rotation acceleration

Using formula of acceleration

\alpha'=\dfrac{\tau}{I'}

Put the value in to the formula

\alpha=\dfrac{F\times R}{8\times\dfrac{1}{2}MR^2}

\alpha=\dfrac{2F}{8MR}

\alpha=\dfrac{\alpha}{8}

Rotation speed is same.

We need to calculate the time

Using formula angular velocity

\Omega=\omega'

\alpha\time t=\alpha'\times t'

Put the value into the formula

\alpha\times120=\dfrac{\alpha}{8}\times t'

t'=960\ sec

t'=16\ min

Hence, The time is 16 min.

5 0
3 years ago
A rifle has a mass of 7-kg and the bullet has a mass of 0.7-kg. If the velocity of the bullet is 350-m/s after the rifle is fire
nevsk [136]

Answer:

-35 m/s

Explanation:

Momentum is conserved.

Momentum before firing = momentum after firing

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

Before the bullet is fired, the bullet and rifle have no velocity, so u₁ and u₂ are 0.

0 = m₁v₁ + m₂v₂

Given m₁ = 0.7 kg, v₁ = 350 m/s, and m₂ = 7 kg:

0 = (0.7 kg) (350 m/s) + (7 kg) v

v = -35 m/s

The rifle recoils at 35 m/s in the opposite direction.

8 0
3 years ago
The equation of damped oscillations is given in the form x=0.05e^-0.25sin½πt (m). Find the velocity of an oscillating point at t
zhannawk [14.2K]

Explanation:

The logarithmic damping decrement of a mathematical pendulum is DeltaT=0.5. How will the amplitude of oscillations decrease during one full oscillation of the pendulum

8 0
3 years ago
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