Name the force between the brakes and the wheels that does work to slow down the vehicle when the brakes are applied.

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

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3 years ago

How light is channelled down an optical fibre

coldgirl [10]

Explanation:

Suppose you want to shine a flashlight beam down a long, straight hallway. Just point the beam straight down the hallway -- light travels in straight lines, so it is no problem. What if the hallway has a bend in it? You could place a mirror at the bend to reflect the light beam around the corner. What if the hallway is very winding with multiple bends? You might line the walls with mirrors and angle the beam so that it bounces from side-to-side all along the hallway. This is exactly what happens in an optical fiber.

The light in a fiber-optic cable travels through the core (hallway) by constantly bouncing from the cladding (mirror-lined walls), a principle called total internal reflection. Because the cladding does not absorb any light from the core, the light wave can travel great distances.

However, some of the light signal degrades within the fiber, mostly due to impurities in the glass. The extent that the signal degrades depends on the purity of the glass and the wavelength of the transmitted light (for example, 850 nm = 60 to 75 percent/km; 1,300 nm = 50 to 60 percent/km; 1,550 nm is greater than 50 percent/km). Some premium optical fibers show much less signal degradation -- less than 10 percent/km at 1,550 nm.

1

3 0

3 years ago

What is the resistance force when you walk up an inclined plane?<br> Please help quick!

Dominik [7]

Friction
Friction also affects the movement of an object on a slope. Friction is a force that offers resistance to movement when one object is in contact with another. Imagine now that you were on the downside of the object and applying force to keep the object in the same place (not moving)

5 0

2 years ago

Back in their laboratory, the team of scientists carefully measure the changes in density of a freshwater sample as they decreas

BabaBlast [244]

Answer: As the temperature of the water decreases from 20 degrees celsius to 4 degrees celsius, the density increases.

Explanation:

7 0

2 years ago

Read 2 more answers

1. Two forces act on a box as follows: F1 = 100 N at 01 = 170° and F2 = 75 N

lilavasa [31]

Answer:

a) F = 64.30 N, b) θ = 121.4º

Explanation:

Forces are vector quantities so one of the best methods to add them is to decompose each force and add the components

let's use trigonometry

Force F1

sin 170 = F_{1y} / F₁

cos 170 = F₁ₓ / F₁

F_{1y} = F₁ sin 170

F₁ₓ = F₁ cos 170

F_{1y} = 100 sin 170 = 17.36 N

F₁ₓ = 100 cos 170 = -98.48 N

Force F2

sin 30 = F_{2y} / F₂

cos 30 = F₂ₓ / F₂

F_{2y} = F₂ sin 30

F₂ₓ = F₂ cos 30

F_{2y} = 75 sin 30 = 37.5 N

F₂ₓ = 75 cos 30 = 64.95 N

the resultant force is

X axis

Fₓ = F₁ₓ + F₂ₓ

Fₓ = -98.48 +64.95

Fₓ = -33.53 N

Y axis

F_y = F_{1y} + F_{2y}

F_y = 17.36 + 37.5

F_y = 54.86 N

a) the magnitude of the resultant vector

let's use Pythagoras' theorem

F = Ra Fx ^ 2 + Fy²

F = Ra 33.53² + 54.86²

F = 64.30 N

b) the direction of the resultant

let's use trigonometry

tan θ’= F_y / Fₓ

θ'= $tan^{-1} \frac{F_y}{F_x}$

θ'= tan⁻¹ (54.86 / (33.53)

θ’= 58.6º

this angle is in the second quadrant

The angle measured from the positive side of the x-axis is

θ = 180 -θ'

θ = 180- 58.6

θ = 121.4º

5 0

2 years ago

Name the force between the brakes and the wheels that does work to slow down the vehicle when the brakes are applied. ​

Name the force between the brakes and the wheels that does work to slow down the vehicle when the brakes are applied.