Ask question

Ask question

All categories

Makovka662 [10]

3 years ago

14

Neglecting the effects of air resistance, explain what the thrower would have to do to catch the ball at the end of the trajecto

ry. Why?

1 answer:

kap26 [50]3 years ago

3 0

Answer:

the sising of neviage

Explanation:

You might be interested in

Qqqqqqqqqqqqqqqqqqqqq

rodikova [14]

Answer:

flip the cards over

Explanation:

7 0

3 years ago

Read 2 more answers

What does critical mean?

iren2701 [21]

expressing adverse or disapproving comments or judgments.

5 0

3 years ago

Read 2 more answers

I need answers and solvings to these questions

den301095 [7]

1) The period of a simple pendulum depends on B) III. only (the length of the pendulum)

2) The angular acceleration is C) $15.7 rad/s^2$

3) The frequency of the oscillation is C) 1.6 Hz

4) The period of vibration is B) 0.6 s

5) The diameter of the nozzle is A) 5.0 mm

6) The force that must be applied is B) 266.7 N

Explanation:

1)

The period of a simple pendulum is given by

$T=2\pi \sqrt{\frac{L}{g}}$

where

T is the period

L is the length of the pendulum

g is the acceleration of gravity

From the equation, we see that the period of the pendulum depends only on its length and on the acceleration of gravity, while there is no dependence on the mass of the pendulum or on the amplitude of oscillation. Therefore, the correct option is

B) III. only (the length of the pendulum)

2)

The angular acceleration of the rotating disc is given by the equation

$\alpha = \frac{\omega_f - \omega_i}{t}$

where

$\omega_f$ is the final angular velocity

$\omega_i$ is the initial angular velocity

t is the time elapsed

For the compact disc in this problem we have:

$\omega_i = 0$ (since it starts from rest)

$\omega_f = 300 rpm \cdot \frac{2\pi rad/rev}{60 s/min}=31.4 rad/s$ is the final angular velocity

t = 2 s

Substituting, we find

$\alpha = \frac{31.4-0}{2}=15.7 rad/s^2$

3)

For a simple harmonic oscillator, the acceleration and the displacement of the system are related by the equation

$a=-\omega^2 x$

where

a is the acceleration

x is the displacement

$\omega$ is the angular frequency of the system

For the oscillator in this problem, we have the following relationship

$a=-100 x$

which implies that

$\omega^2 = 100$

And so

$\omega = \sqrt{100}=10 rad/s$

Also, the angular frequency is related to the frequency $f$ by

$f=\frac{\omega}{2\pi}$

Therefore, the frequency of this simple harmonic oscillator is

$f=\frac{10}{2\pi}=1.6 Hz$

4)

When the mass is hanging on the sping, the weight of the mass is equal to the restoring force on the spring, so we can write

$mg=kx$

where

m is the mass

$g=9.8 m/s^2$ is the acceleration of gravity

k is the spring constant

x = 8.0 cm = 0.08 m is the stretching of the spring

We can re-arrange the equation as

$\frac{k}{m}=\frac{g}{x}=\frac{9.8}{0.08}=122.5$

The angular frequency of the spring is given by

$\omega=\sqrt{\frac{k}{m}}=\sqrt{122.5}=11.1 Hz$

And therefore, its period is

$T=\frac{2\pi}{\omega}=\frac{2\pi}{11.1}=0.6 s$

5)

According to the equation of continuity, the volume flow rate must remain constant, so we can write

$A_1 v_1 = A_2 v_2$

where

$A_1 = \pi r_1^2$ is the cross-sectional area of the hose, with $r_1 = 5 mm$ being the radius of the hose

$v_1 = 4 m/s$ is the speed of the petrol in the hose

$A_2 = \pi r_2^2$ is the cross-sectional area of the nozzle, with $r_2$ being the radius of the nozzle

$v_2 = 16 m/s$ is the speed in the nozzle

Solving for $r_2$ , we find the radius of the nozzle:

$\pi r_1^2 v_1 = \pi r_2^2 v_2\\r_2 = r_1 \sqrt{\frac{v_1}{v_2}}=(5)\sqrt{\frac{4}{16}}=2.5 mm$

So, the diameter of the nozzle will be

$d_2 = 2r_2 = 2(2.5)=5.0 mm$

6)

According to the Pascal principle, the pressure on the two pistons is the same, so we can write

$\frac{F_1}{A_1}=\frac{F_2}{A_2}$

where

$F_1$ is the force that must be applied to the small piston

$A_1 = \pi r_1^2$ is the area of the first piston, with $r_1= 2 cm$ being its radius

$F_2 = mg = (1500 kg)(9.8 m/s^2)=14700 N$ is the force applied on the bigger piston (the weight of the car)

$A_2 = \pi r_2^2$ is the area of the bigger piston, with $r_2= 15 cm$ being its radius

Solving for $F_1$ , we find

$F_1 = \frac{F_2A_1}{A_2}=\frac{F_2 \pi r_1^2}{\pi r_2^2}=\frac{(14700)(2)^2}{(15)^2}=261 N$

So, the closest answer is B) 266.7 N.

Learn more about pressure:

brainly.com/question/4868239

brainly.com/question/2438000

#LearnwithBrainly

5 0

3 years ago

The normal eye, myopic eye and old age

yanalaym [24]

Answer:

1) f’₀ / f = 1.10, the relationship between the focal length (f'₀) and the distance to the retina (image) is given by the constructor's equation

2) the two diameters have the same order of magnitude and are very close to each other

Explanation:

You have some problems in the writing of your exercise, we will try to answer.

1) The equation to be used in geometric optics is the constructor equation

$\frac{1}{f} = \frac{1}{p} + \frac{1}{q}$

where p and q are the distance to the object and the image, respectively, f is the focal length

* For the normal eye and with presbyopia

the object is at infinity (p = inf) and the image is on the retina (q = 15 mm = 1.5 cm)

$\frac{1}{f'_o} = 1/ inf + \frac{1}{1.5}$

f'₀ = 1.5 cm

this is the focal length for this type of eye

* Eye with myopia

the distance to the object is p = 15 cm the distance to the image that is on the retina is q = 1.5 cm

1 / f = 1/15 + 1 / 1.5

1 / f = 0.733

f = 1.36 cm

this is the focal length for the myopic eye.

In general, the two focal lengths are related

f’₀ / f = 1.5 / 1.36

f’₀ / f = 1.10

The question of the relationship between the focal length (f'₀) and the distance to the retina (image) is given by the constructor's equation

2) For this second part we have a diffraction problem, the point diameter corresponds to the first zero of the diffraction pattern that is given by the expression for a linear slit

a sin θ= m λ

the first zero occurs for m = 1, as the angles are very small

tan θ = y / f = sin θ / cos θ

for some very small the cosine is 1

sin θ = y / f

where f is the distance of the lens (eye)

y / f = lam / a

in the case of the eye we have a circular slit, therefore the system must be solved in polar coordinates, giving a numerical factor

y / f = 1.22 λ / D

y = 1.22 λ f / D

where D is the diameter of the eye

D = 2R₀

D = 2 0.1

D = 0.2 cm

the eye has its highest sensitivity for lam = 550 10⁻⁹ m (green light), let's use this wavelength for the calculation

* normal eye

the focal length of the normal eye can be accommodated to give a focus on the immobile retian, so let's use the constructor equation

\frac{1}{f} = \frac{1}{p} + \frac{1}{q}

sustitute

$\frac{1}{f} = \frac{1}{25} + \frac{1}{1.5}$

$\frac{1}{f}$ = 0.7066

f = 1.415 cm

therefore the diffraction is

y = 1.22 550 10⁻⁹ 1.415 / 0.2

y = 4.75 10⁻⁶ m

this is the radius, the diffraction diameter is

d = 2y

d_normal = 9.49 10⁻⁶ m

* myopic eye

In the statement they indicate that the distance to the object is p = 15 cm, the retina is at the same distance, it does not move, q = 1.5 cm

$\frac{1}{f} = \frac{1}{15} + \frac{1}{ 1.5}$

$\frac{1}{f}$ = 0.733

f = 1.36 cm

diffraction is

y = 1.22 550 10-9 1.36 10-2 / 0.2 10--2

y = 4.56 10-6 m

the diffraction diameter is

d_myope = 2y

d_myope = 9.16 10-6 m

$\frac{d_{normal}}{d_{myope}}$ = 9.49 /9.16

\frac{d_{normal}}{d_{myope}} = 1.04

we can see that the two diameters have the same order of magnitude and are very close to each other

8 0

3 years ago

Explain why you can only change one variable in an experiment

ohaa [14]

The purpose of an experiment is to LEARN the EFFECT of something.

The way you do that is to CHANGE the thing and see what happens.

You can change as many things as you want to. But If you change
TWO things and observe the result, then you don't know which one
of them caused the effect you see.

Or maybe BOTH of them working together caused it. You don't know.

So your experiment is not really much good. You need to do it again.

5 0

3 years ago