10. Josie and Trey were working on their physics project and both built catapults. Trey's catapult shot a ball

ella [17]

C : I did this already

6 0

3 years ago

Meg goes swimming on a hot afternoon. When she comes out of the pool, her foot senses that the prevement is unbearably hot. Supp

mart [117]

1.Record her observation with the time it was hot.
2. Gather info about the pavement and its surroundings. Find out what it's made of and what its temp. is at different times of the day.
3. Come up with a hypothesis about why it is hot.
4. Design an experiment to test the hypothesis. If she thinks the Sun is responsible (which she should b smart enough to know), keep it covered during the day time and check it's temp.
5. Come up with a conclusion. If her hypothesis is not supported, design a new experiment or gather more info.

7 0

3 years ago

Two polarizing sheets have their transmission axes crossed so that no light is transmitted. A third sheet is inserted so that it

jek_recluse [69]

Answer:

a) I= I₀ (cos²θ - cos⁴θ) b) 75.5º

Explanation:

a) For this exercise we must use Malus's law

I = I₀ cos² θ

where tea is the angle between the two polarizers.

We apply this expression to our case

* Polarizer 1 suppose that it is vertical and polarizer 2 (intermediate) is at an angle θ with respect to the vertical

I₁ = I₀ cos² θ

* We analyze for the polarity 2 and the last polarizer 3 which indicate that it must be at 90º from the first one, therefore it must be horizontal.

The angle of polarizers 2 and 3 is θ' measured from the horizontal, if we measure with respect to the vertical

θ₂ = 90- θ’ = θ

fiate that in the exercise we must take a reference system and measure everything with respect to this system.

I = I₁ cos² θ'

we substitute

I = (I₀ cos² tea) cos² (θ - 90)

cos (θ -90) = cos θ cos 90 + sin θ sin 90 = sin θ

I = Io cos² θ sin² θ

1= cos²θ+ sin²θ

sin²θ = 1 - cos²θ

I= I₀ (cos²θ - cos⁴θ)

b) to find when the intensity is maximum,

we can use that we have an extreme point when the drift is zero

$\frac{dI}{d \theta}$ = 0

\frac{dI}{d \theta}= Io (2 cos θ - 4 cos³θ) = 0

whereby

cos θ - 2 cos³ θ = 0

cos θ ( 1 - 2 cos² θ) = 0

The zeros of this function are in

θ = 90º

1-2cos²θ =0 cos θ = 0.25 θ = 75.5º

Let's analyze this two results for the angle of 90º the intnesidd is zero with respect to the first polarizer, so it is not an acceptable solution.

Consequently, the angle that allows the maximum intensity to pass is 75.5º

5 0

2 years ago

Starting from rest, a disk rotates about its central axis with constant angular acceleration. In 1.00 s, it rotates 21.0 rad. Du

ELEN [110]

With constant angular acceleration $\alpha$ , the disk achieves an angular velocity $\omega$ at time $t$ according to

$\omega=\alpha t$

and angular displacement $\theta$ according to

$\theta=\dfrac12\alpha t^2$

a. So after 1.00 s, having rotated 21.0 rad, it must have undergone an acceleration of

$21.0\,\mathrm{rad}=\dfrac12\alpha(1.00\,\mathrm s)^2\implies\alpha=42.0\dfrac{\rm rad}{\mathrm s^2}$

b. Under constant acceleration, the average angular velocity is equivalent to

$\omega_{\rm avg}=\dfrac{\omega_f+\omega_i}2$

where $\omega_f$ and $\omega_i$ are the final and initial angular velocities, respectively. Then

$\omega_{\rm avg}=\dfrac{\left(42.0\frac{\rm rad}{\mathrm s^2}\right)(1.00\,\mathrm s)}2=42.0\dfrac{\rm rad}{\rm s}$

c. After 1.00 s, the disk has instantaneous angular velocity

$\omega=\left(42.0\dfrac{\rm rad}{\mathrm s^2}\right)(1.00\,\mathrm s)=42.0\dfrac{\rm rad}{\rm s}$

d. During the next 1.00 s, the disk will start moving with the angular velocity $\omega_0$ equal to the one found in part (c). Ignoring the 21.0 rad it had rotated in the first 1.00 s interval, the disk will rotate by angle $\theta$ according to

$\theta=\omega_0t+\dfrac12\alpha t^2$

which would be equal to

$\theta=\left(42.0\dfrac{\rm rad}{\rm s}\right)(1.00\,\mathrm s)+\dfrac12\left(42.0\dfrac{\rm rad}{\mathrm s^2}\right)(1.00\,\mathrm s)^2=63.0\,\mathrm{rad}$

5 0

3 years ago

Which feature of an object affects its weight? Select three options.

Vadim26 [7]

Answer:mass of the object,how much force earth exerts on the object,and shape of the object

Explanation:

6 0

2 years ago