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IgorC [24]
3 years ago
8

Please help me .finish this paper​

Physics
1 answer:
Evgesh-ka [11]3 years ago
7 0

<u>S</u><u>o</u><u>l</u><u>u</u><u>t</u><u>i</u><u>o</u><u>n</u><u>-</u><u>1</u><u>:</u><u>-</u>

\boxed{\sf \dfrac{10\times 1000}{60\times 60}}

<u>S</u><u>o</u><u>l</u><u>u</u><u>t</u><u>i</u><u>o</u><u>n</u><u>:</u><u>-</u><u>2</u>

\boxed{\sf Sodium\:and\:Potassium}

<u>S</u><u>o</u><u>l</u><u>u</u><u>t</u><u>i</u><u>o</u><u>n</u><u>:</u><u>-</u><u>3</u>

\boxed{\sf 320m}

<u>S</u><u>o</u><u>l</u><u>u</u><u>t</u><u>i</u><u>o</u><u>n</u><u>:</u><u>-</u><u>4</u>

\boxed{\sf Rough\:tiles\:are\:used\:in\:bathroom}

<u>S</u><u>o</u><u>l</u><u>u</u><u>t</u><u>i</u><u>o</u><u>n</u><u>:</u><u>-</u><u>5</u>

\boxed{\sf Mg_3N_2}

<u>S</u><u>o</u><u>l</u><u>u</u><u>t</u><u>i</u><u>o</u><u>n</u><u>:</u><u>-</u><u>6</u>

\boxed{\sf Grapes\:and\:Rambutan}

<u>S</u><u>o</u><u>l</u><u>u</u><u>t</u><u>i</u><u>o</u><u>n</u><u>:</u><u>-</u><u>7</u>

\boxed{\sf {}^{}_{}N}

<u>S</u><u>o</u><u>l</u><u>u</u><u>t</u><u>i</u><u>o</u><u>n</u><u>:</u><u>-</u><u>8</u>

\boxed{\sf Galactuse}

<u>S</u><u>o</u><u>l</u><u>u</u><u>t</u><u>i</u><u>o</u><u>n</u><u>:</u><u>-</u><u>9</u>

\boxed{\sf Y-X}

<u>S</u><u>o</u><u>l</u><u>u</u><u>t</u><u>i</u><u>o</u><u>n</u><u>:</u><u>-</u><u>10</u>

\boxed{\sf Cell\:wall}

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The standard wave format for any wave is wave. When depicting wave in standard wave format, the direction of motion must be rota
german

Answer:

Transverse wave  and Longitudinal wave  and Electromagnetic wave

Explanation:

  • An inverted wave is a wave in which the vibrations of the particles are perpendicular to the direction of wave motion.
  • Longitudinal waves, on the other hand, are waves in which the vibrations of the particles are parallel to the direction of wave motion.
  • Electromagnetic waves are waves that do not require medium media for transmission, including radio waves, microwaves, UV lights, etc.
  • Most electromagnetic waves are transverse in nature.
4 0
3 years ago
A 75-g bullet is fired from a rifle having a barrel 0.540 m long. choose the origin to be at the location where the bullet begin
lyudmila [28]
Part a) The work done by the gas on the bullet is the integral of the force in dx, where x is the distance covered by the bullet inside the barrel with respect to the origin:
W= \int\limits^{0.540m}_{0} {F} \, dx =  \int\limits^{0.540m}_{0} {(16000+10000x-26000x^2)} \, dx =
=16000x+10000  \frac{x^2}{2} - 26000  \frac{x^3}{3}
By substituting the length of the barrel, L=0.540 m, we find the total work done by the gas on the bullet:
W=16000(0.540m)+10000  \frac{(0.540m)^2}{2} - 26000  \frac{(0.540m)^3}{3}  =
=8733 J=8.73 kJ

part b) The resolution of the problem is the same, we just have to use the new length of the barrel (L=0.95 m) inside the final formula, and we find the new value of the work:
W=16000(0.95m)+10000  \frac{(0.95m)^2}{2} - 26000  \frac{(0.95m)^3}{3}  =
=12280 J=12.28 kJ
5 0
3 years ago
A bullet is fired with a muzzle velocity of 1178 ft/sec from a gun aimed at an angle of 26° above the horizontal. Find the horiz
dalvyx [7]

Answer:

1058.78 ft/sec

Explanation:

Horizontal Component of Velocity; This is the velocity of a body that act on the horizontal axis. I.e Velocity along x-axis

The horizontal velocity of a body can be calculated as shown below.\

Vh = Vcos∅.......................... Equation 1

Where Vh = horizontal component of the velocity, V = The velocity acting between the horizontal and the vertical axis, ∅ = Angle the velocity make with the horizontal.

Given: V = 1178 ft/sec, ∅ = 26°

Substitute into equation 1

Vh = 1178cos26

Vh = 1178(0.8988)

Vh = 1058.78 ft/sec

Hence the horizontal component of the velocity = 1058.78 ft/sec

8 0
3 years ago
Jasmine makes a note card about series circuits. she records the points shown. a 2 column table with 1 row. the column on the le
Bess [88]

Jasmine need to correct the error by switching the headings on the columns adding the title parallel circuits.

<h3>What is a parallel circuit?</h3>

A parallel circuit is a circuit in which the components are connected to a common junction. This implies that if one bulb goes out in a parallel connection, all the bulbs will go out.

As such, Jasmine need to correct the error by switching the headings on the columns adding the title parallel circuits.

Learn more about parallel connection: brainly.com/question/12400458

#SPJ4

7 0
2 years ago
In January 2006, astronomers reported the discovery of a planet comparable in size to the earth orbiting another star and having
Orlov [11]

Answer:

R = 5.28  103 km

Explanation:

The definition of density is

              ρ = m / V

              V = m /ρ

Where m is the mass and V the volume of the body

The volume of a sphere is

            V = 4/3 π r³

Let's replace

             4/3 π r³ = m / ρ

             R =∛ ¾ m / ρ π

The mass of the planet is

              M = 5.5 Me

              R = ∛ ¾ 5.5 Me /ρ π

Let's reduce the density to SI units

             ρ = 1.76 g / cm³ (1 kg / 10³ g) (10² cm / 1 m)³

             ρ = 1.76 10³ kg / m³

Let's calculate

               R = ∛ ¾ 5.5 5.97 10²⁴ / (1.76 10³ pi)

               R = ∛ 0.14723 10²¹

               R = 0.528 10⁷ m

               R = 0.528 104 km

               R = 5.28  103 km

8 0
3 years ago
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