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3 years ago

Which of the following is not a health hazard associated with driving on mountain roads?

1 answer:

5 0

Among the following options, the answer would be A. Carbon monoxide poisoning. This usually is from breathing carbon monoxide too much and symptoms may include headache,weakness and confusion. Hope this is the right answer and would be of help then.

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pav-90 [236]

Answer:

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Explanation:

7 0

3 years ago

Read 2 more answers

If the electron has half the speed needed to reach the negative plate, it will turn around and go towards the positive plate. Wh

In-s [12.5K]

Answer:

v = -v₀ / 2

Explanation:

For this exercise let's use kinematics relations.

Let's use the initial conditions to find the acceleration of the electron

v² = v₀² - 2a y

when the initial velocity is vo it reaches just the negative plate so v = 0

a = v₀² / 2y

now they tell us that the initial velocity is half

v’² = v₀’² - 2 a y’

v₀ ’= v₀ / 2

at the point where turn v = 0

0 = v₀² /4 - 2 a y '

v₀² /4 = 2 (v₀² / 2y) y’

y = 4 y'

y ’= y / 4

We can see that when the velocity is half, advance only ¼ of the distance between the plates, now let's calculate the velocity if it leaves this position with zero velocity.

v² = v₀² -2a y’

v² = 0 - 2 (v₀² / 2y) y / 4

v² = -v₀² / 4

v = -v₀ / 2

We can see that as the system has no friction, the arrival speed is the same as the exit speed, but with the opposite direction.

7 0

3 years ago

A 24 kg child sits on a 2.0-m-long rope swing. You are going to give the child a small, brief push at regular intervals.

Bess [88]

time should you wait between pushes is 2.83 sec.

the question is incomplete, full statement is-

A 24 kg child sits on a 2.0-m-long rope swing. You are going to give the child a small, brief push at regular intervals. If you want to increase the amplitude of her motion as quickly as possible, how much time should you wait between pushes?

<h3>What is Amplitude?</h3>

In physics, amplitude refers to the greatest displacement or distance that a point on a vibrating body or wave may move relative to its equilibrium location. It is equivalent to the vibration path's half-length.

regular interval - at similarly spaced intervals: having the same interval of time between occurrences From 4 a.m. to midnight, the buses operate at regular intervals. The boards are positioned at regular intervals, with an equal amount of space between each.

The length of swing, l = 2.1 m

The time between the pushes is nothing but the Time period

and is given by the formula,

$T = 2\pi ( \frac{l}{g} )^{\frac{1}{2} }$

= 2 * 3.14 ( 2.0/ 9.8 ) ^ (1/2)

= 2.83 sec

to learn more about Amplitude go to - brainly.com/question/3613222

#SPJ4

3 0

1 year ago

A swimmer is capable of swimming 0.42 m/s in still water. part a if she aims her body directly across a 66-m-wide river whose cu

satela [25.4K]

<span>If the swimmer is swimming perpendicular to the current, it will take her 66m / 0.42 m/s = 157.14 seconds to cross the river. At the same time, the current will be taking her downstream at a rate of 0.32 m/s. So, when she reaches the opposite bank, her total downstream distance traveled will have been 0.32*157.14 = 50.28 meters.</span>

3 0

3 years ago

Two objects of the same size are both perfect blackbodies. One has a temperature of 3000 K, so its frequency of maximum emission

bija089 [108]

Answer:

a) The colder body (3000k), b) hearter body c) 12000K body

Explanation:

This exercise should know the power emitted by the objects and the distribution of this emission in the energy spectrum, for this we will use Stefan's laws and that of Wien's displacement

Stefan's Law P = σ A e T⁴

Wien displacement law λ T = 2,898 10⁻³ m K

Let's calculate the power emitted for each object.

As they are perfect black bodies e = 1, they also indicate that they have the same area

T = 3000K

P₁ = σ A T₁⁴

T = 12000K

P₂ = σ A T₂⁴

P₂ / P₁ = T₂⁴ / T₁⁴

P₂ / P₁ = (12000/3000)⁴

P₂ / P₁ = 256

This indicates that the hottest body emission is 256 times the coldest body emission.

Let's calculate the maximum emission wavelength

Body 1

T = 3000K

λ T = 2,898 10-3

λ₁ = 2.89810-3 / T

λ₁ = 2,898 10-3 / 3000

λ₁ = 0.966 10-6 m

λ₁ = 966 nm

T = 12000K

λ₂ = 2,898 10-3 / 12000

λ₂ = 0.2415 10-6 m

λ₂ = 214 nm

a) The colder body (3000k) emits more light in the infrared, since the emission of the hot body is at a minimum (emission tail)

b) The two bodies have emission in this region, the body of 3000K in the part of rise of the emission and the body to 12000K in the descent of the emission even when this body emits 256 times more than the other, so this body should have the highest broadcast in this area

c) The emission of the hottest 12000K body is mainly in UV

d) The hottest body emits more energy in UV and visible

e) No body has greater emission in all zones

5 0

3 years ago