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s344n2d4d5 [400]

2 years ago

8

As the wavelength increases, the frequency (2 points) decreases and energy decreases. increases and energy increases. decreases

and energy increases. increases and energy decreases

2 answers:

lora16 [44]2 years ago

4 0

Answer:

I TOOK THE TEST! the answer is INCREASES AND ENERGY INCREASES!

frozen [14]2 years ago

3 0

Bohr's equation for the change in energy is
$\Delta E= \frac{hc}{\lambda}$
where
h = Planck's constant
c == the velocity of light
λ = wavelength.

The velocity is related to wavelength and frequency, f, by
c = fλ

Let us examine the given answers on the basis of the given equations.

a. As λ increases, f decreases and ΔE decreases.
TRUE

b. As λ increases, f increases and ΔE increases.
FALSE

c. As λ increases, f increases and ΔE decreases.
FALSE

Answer:
As the wavelength increases, the frequency decreases and energy decreases.

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In mammals, the weight of the heart is approximately 0.5% of the total body weight. Write a linear model that gives the heart we

jekas [21]

Answer:

The weight of heart of a human is 0.93 lbs.

Explanation:

Given that,

Approximately weight of heart is 0.5 % of the total body weight.

Weight of human = 185 lbs

Let the the weight of total body is w and weight of heart is $w_{h}$ .

We need to calculate the weight of heart of a human

Using given data

$w_{h}=0.5\times w$

Where, h = weight of heart

w = weight of human

$w_{h}=\dfrac{0.5}{100}\times 185$

$w_{h}=0.93\ lbs$

Hence, The weight of heart of a human is 0.93 lbs.

8 0

2 years ago

A child is playing with a spring toy, first stretching and then compressing it.

Reika [66]

Sorry, I’m only in 6th Grade, I don’t know the answer to this question.

5 0

2 years ago

A helicopter is traveling at 86.0 km/h at an angle of 35° to the ground. What is the value of Ax? Round your answer to the neare

jasenka [17]

Answer: 70.5 km/h

Justification:

The question is not clearly stated but it seems you are asking for the x - component of the velocity of the helicopter.

You can find the x and y - components of the velocity using the trigonometric ratios sine and cosine.

The sine ratio relates the y-component and the velocity by:

sin(angle) = y-component of velocity / velocity

The cosine ratio related the x-component and the velocity by:

cos(angle) = x-component of velocity / velocity.

Since you have the angle and the velocity and are asked by the x-component of the velocity, you need to use the cosine ratio:

cos(35°)= x-component / 86.0 km/h

=> x -component = 86.0 km/h * cos(35°) = 70.5 km/h

6 0

2 years ago

Read 2 more answers

ery large accelerations can injure the body, especially if they last for a considerable length of time. One model used to gauge

IgorC [24]

Answer:

2.98 second

Explanation:

The severity index is defined by :

$S=a^{5/2}t$

a is dimensionless constant that equals the number of multiples of g

Conditions are given as :

Initial velocity, u = 0

Acceleration, a = 34 m/s²

Final velocity, v = 16.4 km/h = 4.56 m/s

We can find t from the above data as follows :

$t=\dfrac{v-u}{a}\\\\t=\dfrac{4.56-0}{34}\\\\t=0.134\ s$

As a is the acceleration that is multiple of g.

So,

$a=\dfrac{34}{9.8}=3.46$

So,

Severity index,

$S=a^{5/2}t\\\\S=(3.46)^{5/2}\times 0.134\\\\S=2.98\ s$

Hence, the severity index for the collision is 2.98 seconds.

6 0

2 years ago

1. Do alto de uma plataforma com 15m de altura, é lançado horizontalmente um projéctil. Pretende-se atingir um alvo localizado n

sveta [45]

Answer:

(a). The initial velocity is 28.58m/s

(b). The speed when touching the ground is 33.3m/s.

Explanation:

The equations governing the position of the projectile are

$(1).\: x =v_0t$

$(2).\: y= 15m-\dfrac{1}{2}gt^2$

where $v_0$ is the initial velocity.

(a).

When the projectile hits the 50m mark, $y=0$ ; therefore,

$0=15-\dfrac{1}{2}gt^2$

solving for $t$ we get:

$t= 1.75s.$

Thus, the projectile must hit the 50m mark in 1.75s, and this condition demands from equation (1) that

$50m = v_0(1.75s)$

which gives

$\boxed{v_0 = 28.58m/s.}$

(b).

The horizontal velocity remains unchanged just before the projectile touches the ground because gravity acts only along the vertical direction; therefore,

$v_x = 28.58m/s.$

the vertical component of the velocity is

$v_y = gt \\v_y = (9.8m/s^2)(1.75s)\\\\{v_y = 17.15m/s.$

which gives a speed $v$ of

$v = \sqrt{v_x^2+v_y^2}$

$\boxed{v =33.3m/s.}$

4 0

1 year ago