If the number of particles of a gas is increased, the gas pressure will do what?

Sam heaves a 16lb shot straight upward, giving it a constant upward acceleration from rest of 35 m/s^2 for 64.0 cm. He releases

makvit [3.9K]

Answer:

6.69 m/s

4.483 m

1.42s

Explanation:

Given that:

Initial Velocity, u = 0

Final velocity, v =?

Acceleration, a = 35m/s²

1.) using the relation :

v² = u² + 2as

v² = 0 + 2(35) * 64*10^-2m

v² = 70 * 0.64

v = sqrt(44.8)

v = 6.693

v = 6.69 m/s

B.) height from the ground, h0 = 2.2

How high ball went , h:

Using :

v² = u² + 2as

Upward motion, g = - ve

0 = 6.69² + 2(-9.8)*(h - 2.2)

0= 6.69² - 19.6(h - 2.2)

44.7561 + 43.12 - 19.6h = 0

19.6h = 44.7561 - 43.12

h = 87.8761 / 19.6

h = 4.483 m

C.)

vt - 0.5gt² = h - h0

6.69t - 0.5(9.8)t²

6.69t - 4.9t² = 1.83 - 2.2

-4.9t² + 6.69t + 0.37 = 0

Using the quadratic equation solver :

Taking the positive root:

1.4185 = 1.42s

5 0

2 years ago

Minerals that form from magma process will tend to form ?

NISA [10]

igneous rocks, and can either be entrusive or extrusive.

3 0

3 years ago

Electromagnetic waves polarized in the x direction propagate towards an ideal polarizer whose axis is aligned with the y-axis. H

Digiron [165]

Answer:

there is ZERO TRANSMISSION through the polarizer.

Explanation:

Given the data in the question;

waves was polarized in the x direction propagate towards an ideal polarizer whose axis is aligned with the y-axis.

so θ = 90° { aligned with the y-axis }

now, we know that for polarized light going at angle θ to the polarizer;

I = I₀ × cos²θ

Where I₀ is the intensity of unpolarized light.

I is the intensity after polarization.

so we substitute the value of θ

I = I₀ × cos²( 90° )

I = I₀ × 0

I = 0

Therefore, there is ZERO TRANSMISSION through the polarizer.

6 0

2 years ago

A century ago, swords were made from various metals, especially steel. The property that makes steel a good choice for sword mak

tigry1 [53]

The answer should be hardness and strength
~emily

8 0

3 years ago

Read 2 more answers

What is the orbital period of a spacecraft in a low orbit near the surface of mars? The radius of mars is 3.4×106m.

valkas [14]

<h2>Answer: 56.718 min</h2>

Explanation:

According to the Third Kepler’s Law of Planetary motion<em> </em><em>“The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”. </em>

In other words, this law states a relation between the orbital period $T$ of a body (moon, planet, satellite) orbiting a greater body in space with the size $a$ of its orbit.

This Law is originally expressed as follows:

$T^{2}=\frac{4\pi^{2}}{GM}a^{3}$ (1)

Where;

$G$ is the Gravitational Constant and its value is $6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}$

$M=6.39(10)^{23}kg$ is the mass of Mars

$a=3.4(10)^{6}m$ is the semimajor axis of the orbit the spacecraft describes around Mars (assuming it is a <u>circular orbit </u>and a <u>low orbit near the surface </u>as well, the semimajor axis is equal to the radius of the orbit)

If we want to find the period, we have to express equation (1) as written below and substitute all the values:

$T=\sqrt{\frac{4\pi^{2}}{GM}a^{3}}$ (2)

$T=\sqrt{\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(6.39(10)^{23}kg)}(3.4(10)^{6}m)^{3}}$ (3)

$T=\sqrt{11581157.44 s^{2}}$ (4)

Finally:

$T=3403.1099s=56.718min$ This is the orbital period of a spacecraft in a low orbit near the surface of mars

6 0

3 years ago