Which observation would be evidence that heat was transferred by radiation?

When you voice the vowel sound in "hat," you narrow the opening where your throat opens into the cavity of your mouth so that yo

xz_007 [3.2K]

Answer: 0.1 m, 0.0583 m

Explanation:

We are given that:

Frequency for throat= 800 Hz

Frequency for mouth= 1500 Hz

Sound speed= 350 m/s

We have to find the corresponding lengths.

We have

f= $\frac{v}{4L}$

or L= $\frac{v}{4f}$

For the throat= L= $\frac{350}{4*800\\}$ = 0.1 m

For the mouth= L= $\frac{350}{4*1500}$ = 0.0583 m

3 0

3 years ago

Projectile <br> SHOW WORK<br> WILL MARK BRANLIEST <br> (Draw Picture and Label)

m_a_m_a [10]

a) The horizontal distance covered by the projectile is 600 m

b) The projectile reaches its maximum height after 3.00 s

c) The altitude of the highest point is 44.1 m

Explanation:

a)

The motion of the projectile consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction

- A uniformly accelerated motion, with constant acceleration (acceleration of gravity) in the downward direction

In this part A, we just need to analyze the horizontal motion. We know that:

The projectile travels horizontally with a constant velocity ov $v_x = 100 m/s$
The total time of flight of the projectile is $t=6.00 s$

Therefore, the horizontal distance covered by the projectile is given by

$x=v_x t$

And substituting, we find

$x=(100)(6.0) = 600 m$

b)

For this part, we need to analyze the vertical motion of the projectile.

First, we want to find the initial vertical velocity. We can do it by using the equation for the vertical displacement:

$s=u_y t + \frac{1}{2}at^2$

where:

$u_y$ is the initial vertical velocity

$a=g=-9.8 m/s^2$ is the acceleration of gravity (negative because it is downward)

t is the time

s is the vertical displacement

We know that the total time of flight is t = 6.00 s: this means that when t=6, the projectile returns to its initial vertical position, so s = 0. Substituting and solving for $u_y$ , we get

$u_y = - \frac{1}{2}at=-\frac{1}{2}(-9.8)(6)=29.4 m/s$

The vertical velocity then as a function of t is given by

$v_y = u_y + at$

And at the maximum height, it becomes zero: $v_y = 0$ . Substituting and solving for t, we find the time at which the projectile reaches the maximum height:

$t=-\frac{u_y}{a}=-\frac{29.4}{-9.8}=3.00 s$

c)

To find the altitude of the highest point in the path, we use again the equation:

$s=u_y t + \frac{1}{2}at^2$

where

$u_y = 29.4 m/s$ is the initial vertical velocity

t = 3.00 s is the time at which the projectile reaches the highest point

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

Substituting the values, we find

$s=(29.4)(3.00)+\frac{1}{2}(-9.8)(3.00)^2=44.1 m$

So, the highest point is at 44.1 m above the ground.

Learn more about projectiles:

brainly.com/question/8751410

#LearnwithBrainly

8 0

3 years ago

The distance between the first and fifth minima of a single-slit diffraction pattern is 0.500 mm with the screen 37.0 cm away fr

WARRIOR [948]

In the single-slit experiment, the displacement of the minima of the diffraction pattern on the screen is given by
$y_n= \frac{n \lambda D}{a}$ (1)
where
n is the order of the minimum
y is the displacement of the nth-minimum from the center of the diffraction pattern
$\lambda$ is the light's wavelength
D is the distance of the screen from the slit
a is the width of the slit

In our problem,
$D=37.0 cm=0.37 m$
$\lambda=530 nm=5.3 \cdot 10^{-7} m$
while the distance between the first and the fifth minima is
$y_5-y_1 = 0.500 mm=0.5 \cdot 10^{-3} m$ (2)

If we use the formula to rewrite $y_5, y_1$ , eq.(2) becomes
$\frac{5 \lambda D}{a} - \frac{1 \lambda D}{a} =\frac{4 \lambda D}{a}= 0.5 \cdot 10^{-3} m$
Which we can solve to find a, the width of the slit:
$a= \frac{4 \lambda D}{0.5 \cdot 10^{-3} m}= \frac{4 (5.3 \cdot 10^{-7} m)(0.37 m)}{0.5 \cdot 10^{-3} m}= 1.57 \cdot 10^{-3} m=1.57 mm$

7 0

3 years ago

A car traveling at 23m/s starts to decelerate steadily. It comes to a complete stop in 7 seconds. What is the acceleration?

Lena [83]

The car has an initial velocity $v_0$ of 23 m/s and a final velocity $v$ of 0 m/s. Recall that for constant acceleration,

$v=v_0+at$

The car stops in 7 s, so the acceleration is

$0\,\dfrac{\mathrm m}{\mathrm s}=23\,\dfrac{\mathrm m}{\mathrm s}+a(7\,\mathrm s)$

$\implies a\approx-3.29\,\dfrac{\mathrm m}{\mathrm s^2}$

7 0

3 years ago

A total solar eclipse is a rare event. Although they occur somewhere on earth every 18 months on average, it is estimated that t

Setler79 [48]

Because the tip of the moon's shadow ... the area of "totality" ... is never more than a couple hundred miles across, It never covers a single place for more than 7 minutes, and can never stay on the Earth's surface for more than a few hours altogether during one eclipse.

If you're not inside that small area, you don't see a total eclipse.

3 0

3 years ago

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