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3 years ago

9

Think about the pencil-dropping activity that you did in the introduction. What did the target finally look like?

1 answer:

s344n2d4d5 [400]3 years ago

3 0

Answer:

By dropping a pencil from a certain fixed height again and again it will make the target super messay with marks of dot everywhere on the target and some even out side the target.

Explanation:

You might be interested in

What wavelength of radiation has photons of energy 2.87×x10−18 j?

Masja [62]

The energy of a photon is given by
$E=hf$
where h is the Planck constant and f is the frequency of the light.
Given the relationship between the frequency f, the wavelength $\lambda$ and the speed of light c for an electromagnetic wave:
$f= \frac{c}{\lambda}$
we can rewrite the energy of the photon as
$\lambda = \frac{hc}{E}$
and by substituting $h=6.6 \cdot 10^{-34 J}$ , $c=3 \cdot 10^8 m/s$ and $E=2.87 \cdot 10^{-18}J$ , we get the wavelength of the photon:
$\lambda = \frac{hc}{E}=6.9 \cdot 10^{-8} m = 69 nm$

4 0

3 years ago

A baseball is hit from a height of 4 feet above the ground with an initial velocity of 110 feet per second and at an angle of 35

mrs_skeptik [129]

Answer:

Part a)

$h = 62.2 ft$

Part b)

It will not able to cross the pole

Explanation:

As we know that ball is hit with speed of 110 ft/s at an angle of 35 degree

so here we will say

$v_y = 110 sin35$

$v_x = 110 cos35$

now at the maximum height the vertical velocity will become zero

so here we can use kinematics

$v_f^2 - v_y^2 = 2 a h$

here we have

$a = -32 ft/s^2$

$v_f = 0$

$v_y = 63.1 ft/s$

now we have

$0 - 63.1^2 = 2(-32)h$

$h = 62.2 ft$

Part b)

now the height of ball is related to the distance from point of projection is given as

$y = xtan\theta - \frac{gx^2}{2v^2cos^2\theta}$

now we know that

$x = 375 ft$

$y = 375(tan35) - \frac{(32)(375^2)}{2(110^2)(cos35)^2}$

$y = 262.6 - 277.12 = -14.5 ft$

since its coming negative so it will not able to cross the pole

3 0

3 years ago

Se usa una bombona de He para inflar globos, de volumen 15L con una presión de 200 atm. Si los globos tienen un volumen de 1.5L

Bingel [31]

Answer:

La cantidad de globos llenos es de aproximadamente 1.974 globos

Explanation:

Los parámetros dados son;

El gas utilizado para inflar el globo = Él

El volumen del gas He, V₁ = 15 L

La presión del gas He, P₁ = 200 atm = 20,265,000 Pa

El volumen de cada globo = 1,5 L

La presión del He dentro del globo, P₂ = 770 mmHg = 102,641 Pa

Sea 'V₂', que represente el volumen total que el gas ocupa en todos los globos, y 'x', que represente el número de globos;

Por lo tanto, tenemos;

1,5 · x = V₂

Según la ley de Boyle, tenemos;

P₁ · V₁ = P₂ · V₂

∴ V₂ = P₁ · V₁ / (P₂)

Al introducir los valores conocidos, obtenemos;

V₂ = 20,265,000 Pa × 15 L / (102,641 Pa) = 2,961.53548 L

1,5 L × x = V₂

∴ 1,5 L × x = 2,961,53548 L

x = 2,961.53548 L / (1.5 L) = 1,974.35699 redondeamos hacia abajo para obtener un número entero de globos de la siguiente manera;

El número de globos, x ≈ 1,974 globos.

8 0

3 years ago

Can I have some help

earnstyle [38]

I think is D cuz all Yoou really need to do is look at the graph

4 0

3 years ago

HELP ASAP !!!!!!!!!!!!!!

kompoz [17]

The Big Bang theory supports that belief. If everything exploded from one point, then it is continuing to expand outward now

5 0

3 years ago

Read 2 more answers