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3 years ago

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If the road becomes wet or crowded, you should ____. slow down and increase your following distance All choices are incorrect. m

aintain your speed and following distance speed up and decrease your following distance Submit answer

1 answer:

Vaselesa [24]3 years ago

7 0

Answer:

The first one.

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The diagram is an illustration of<br> NO LINKS

V125BC [204]

Answer:

A longitudinal wave

Explanation:

6 0

3 years ago

What was the result of the Mexican army's victory over the Texan garrison at the Alamo?

sasho [114]

The result of the Mexican victory was that fallen defenders became heroes to the cause of Texan independence.<span> The Battle of the Alamo took place between February 23 and March 6, 1836 and became the central episode of the Texas Revolution . After this thirteen-day battle, the Mexican troops of General President Antonio Lopez de Santa Anna began an attack on San Antonio de Bexar, the current San Antonio in Texas. The Battle of the Alamo fought the army of Mexico against a group of Texan rebels, mostly American settlers. More than four thousand men from Santa Ana stood in front of the Alamo Fort , the last stronghold of the rebels, which barely reached 187. The Alamo was not a fortress prepared to withstand a siege. It is believed that all the rebels of the Alamo died in the siege, but Santa Anna came to lose up to about 900 men during the days that lasted the fight. However, the worst result for Santa Ana was precisely the resistance that the Texan rebels had in the Alamo, which fostered the fighting spirit of the Texans. A few days later, on March 14, 1836, Texas became independent from Mexico and a month later, Santa Ana was imprisoned.</span>

3 0

3 years ago

Read 2 more answers

Two uniform, solid cylinders of radius R and total mass M are connected along their common axis by a short, light rod and rest o

sveta [45]

Explanation:

A) To prove the motion of the center of mass of the cylinders is simple harmonic:

System diagram for given situation is shown in attached Fig. 1

We can prove the motion of the center of mass of the cylinders is simple harmonic if

$a_{x} = -\omega^{2} x$

where aₓ is acceleration when attached cylinders move in horizontal direction:

<h3>PROOF:</h3>

rotational inertia for cylinders is given as:

$I=\frac{1}{2}MR^{2}$ -----(1)

Newton's second law for angular motion is:

∑τ = Iα ------(2)

For linear motion in horizontal direction it is:

∑Fₓ = Maₓ ------ (3)

By definition of torque:

τ = RF --------(4)

Put (4) and (1) in (2)

$RF=\frac{1}{2}MR^{2}\alpha$

$RF=\frac{1}{2}MR^{2}\alpha$

from Fig 3 it can be seen that fs is force by which the cylinders roll without slipping as they oscillate

So above equation becomes

$f_{s}=\frac{1}{2}MR\alpha$ ------ (5)

As angular acceleration is related to linear by:

$a= R\alpha$

Eq (5) becomes

$f_{s}=\frac{1}{2}Ma_{x}$ ---- (6)

aₓ shows displacement in horizontal direction

From (3)

∑Fₓ = Maₓ

Fₓ is sum of fs and restoring force that spring exerts:

$\sum F_{x} = f_{s} - kx$ ----(7)

Put (7) in (3)

$f_{s} - kx = Ma_{x}$ [/tex] -----(8)

Using (6) in (8)

$\frac{1}{2}Ma_{x} - kx =Ma_{x}$

$a_{x} = \frac{2k}{3M} x$ --- (9)

For spring mass system

$a= -\omega^{2} x$ ----- (10)

Equating (9) and (10)

$\omega^{2} = \frac{2k}{3M}$

$\omega = \sqrt{ \frac{2k}{3M}}$

then (9) becomes

$a_{x} = - \omega^{2}x$

(The minus sign says that x and aₓ have opposite directions as shown in fig 3)

This proves that the motion of the center of mass of the cylinders is simple harmonic.

<h3 /><h3>B) Time Period</h3>

Time period is related to angular frequency as:

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

$T = 2\pi \sqrt{\frac{3M}{2k}$

5 0

3 years ago

1. A small light bulb is shining light on a basketball (diameter is 23 cm or 9 inches). The light bulb is 3 m from the closest s

siniylev [52]

Answer:

The size (diameter) of the basketball's shadow on the wall is approximately 53.38 cm

Explanation:

The given parameters of the basketball are;

The diameter of the basketball = 23 cm (9 inches)

The distance of the light bulb from the closest side of the basketball = 3 m

The distance from the ball to the wall = 4 m

The distance from the light source to the center of the ball, d = 3 m + 0.23/2 m = 3.115 m

The angle the light ray makes with the edge of the ball, θ = arctan(0.115/3.115)

Therefore, the ratio of the shadow width divided by 2 to the distance from the light from the wall = 0.115/3.115

The distance from the light from the wall = 3 m + 4 m + 0.23 m = 7.23 m

Therefore;

((The width of the shadow)/2)/(The distance from the light from the wall) = 0.115/3.115

∴ ((The width of the shadow)/2)/(7.23 m) = 0.115/3.115

((The width of the shadow)/2) = 7.23 m × 0.115/3.115 = 16629/62300 m ≈ 0.2669 m = 26.69 cm

The width (diameter) of the shadow on the wall = 2 × 16629/62300 m ≈ 0.5338 m = 53.38 cm

The size (diameter) of the basketball's shadow on the wall ≈ 53.38 cm

4 0

3 years ago

The electron affinity of thulium has been measured by a technique known as laser photodetachment electron spectroscopy. In this

antoniya [11.8K]

Answer:

ΔE = 1.031 eV

Explanation:

For this exercise let's calculate the energy of the photons using Planck's equation

E = h f

wavelength and frequency are related

c = λ f

f = c /λ

let's substitute

E = h c /λ

let's calculate

E = 6.63 10⁻³⁴ 3 10⁸/1064 10⁻⁹

E = 1.869 10⁻¹⁹ J

let's reduce to eV

E = 1.869 10⁻¹⁹ J (1 eV / 1.6 10⁻¹⁹ J)

E = 1.168 eV

therefore the electron affinity is

ΔE = E - 0.137

ΔE = 1.168 - 0.137

ΔE = 1.031 eV

3 0

3 years ago