8 a Name the bones that articulate (join together) in the knee joint

How easy is it to die in quick sand (I know the answer to this one too i just want to see who understands)

matrenka [14]

There have been 580 cases!! Wow

8 0

3 years ago

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A 3.50 cm tall object is held 24.8 cm from a lens of focal length 16.0 cm. What is the image height?

Mandarinka [93]

Answer:

Explanation:

Using the formula to first get the image distance

1/f = 1/u+1/v

f = focal length of the lens

u = object distance

v = image distance

Given f = 16.0 cm, u = 24.8 cm

1/v = 1/16 - 1/24.8

1/v = 0.0625-0.04032

1/v = 0.02218

v = 1/0.02218

v = 45.09 cm

To get the image height, we will us the magnification formula.

Mag = v/u = Hi/H

Hi = image height = ?

H = object height = 3.50 cm

45.09/24.8 = Hi/3.50

Hi = (45.09*3.50)/24.8

Hi = 6.36 cm

The image height is 6.36 cm

6 0

3 years ago

FIGURE 2 shows a 1.5 kg block is hung by a light string which is wound around a smooth pulley of radius 20 cm. The moment of ine

Sindrei [870]

Answer:

At t = 4.2 s

Angular velocity: 6. 17 rad /s

The number of revolutions: 2.06

Explanation:

First, we consider all the forces acting on the pulley.

There is only one force acting on the pulley, and that is due to the 1.5 kg mass attached to it.

Therefore, the torque on the pulley is

$\tau=Fd=mg\cdot R$

where m is the mass of the block, g is the acceleration due to gravity, and R is the radius of the pulley.

Now we also know that the torque is related to angular acceleration α by

$\tau=I\alpha$

therefore, equating this to the above equation gives

$mg\cdot R=I\alpha$

solving for alpha gives

$\alpha=\frac{mgR}{I}$

Now putting in m = 1.5 kg, g = 9.8 m/s^2, R = 20 cm = 0.20 m, and I = 2 kg m^2 gives

$\alpha=\frac{1.5\cdot9.8\cdot0.20}{2}$ $\boxed{\alpha=1.47s^{-2}}$

Now that we have the value of the angular acceleration in hand, we can use the kinematics equations for the rotational motion to find the angular velocity and the number of revolutions at t = 4.2 s.

The first kinematic equation we use is

$\theta=\theta_0+\omega_0t+\frac{1}{2}\alpha t^2$

since the pulley starts from rest ω0 = 0 and theta = 0; therefore, we have

$\theta=\frac{1}{2}\alpha t^2$

Therefore, ar t = 4.2 s, the above gives

$\theta=\frac{1}{2}(1.47)(4.2)^2$

$\boxed{\theta=12.97}$

So how many revolutions is this?

To find out we just divide by 2 pi:

$\#\text{rev}=\frac{\theta}{2\pi}=\frac{12.97}{2\pi}$ $\boxed{\#\text{rev}=2.06}$

Or about 2 revolutions.

Now to find the angular velocity at t = 4.2 s, we use another rotational kinematics equation:

$\omega^2=w^2_0+2\alpha(\Delta\theta)_{}$

Since the pulley starts from rest, ω0 = 0. The change in angle Δθ we calculated above is 12.97. The value of alpha we already know to be 1.47; therefore, the above becomes: