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maksim [4K]
3 years ago
8

There are_____________ centimeters in 1 meter.

Physics
2 answers:
Setler [38]3 years ago
8 0
There is 100 centimeters in a meter
ivanzaharov [21]3 years ago
6 0

Answer: 100

Explanation:

just multiply the value by 100 :)

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Light can travel from the sun to ________ in less than ten minutes.
Solnce55 [7]
To the Earth in less than ten minutes. 
5 0
3 years ago
A simple series circuit consists of a 150 Ω resistor, a 29 V battery, a switch, and a 2.1 pF parallel-plate capacitor (initially
Rufina [12.5K]
Find the electric flux and the disp at t=0.50ns 
<span>Given: </span>
<span>Resistor R = 160 Ω </span>
<span>Voltage ε = 22.0 V </span>
<span>Capacitor C = 3.10 pF = 3.10 * 10^-12 F </span>
<span>time t = 0.5 ns = 0.5 * 10^-9 s </span>
<span>ε0 = 8.85 * 10^-12 </span>
<span>Solution: </span>
<span>ELECTRIC FLUX: </span>
<span>Φ = Q/ε0 </span>
<span>we have ε0, we need to find Q the charge </span>
<span>STEP 1: FIND Q </span>
<span>Q = C ε ( 1 - e^(-t/RC) ) </span>
<span>Q = { 3.10 * 10^-12 } { 22.0 } { 1 - e^(- 0.5 * 10^-9 / 160 *3.10 * 10^-12 ) } </span>
<span>Q = { 3.10 * 10^-12 } { 22.0 } { 1 - 0.365 } </span>
<span>Q = { 3.10 * 10^-12 } { 22.0 } { 0.635 } </span>
<span>Q = 43.31 * 10^-12 C </span>
<span>STEP 2: WE HAVE Q AND ε0 > >>> SOLVE FOR ELECTRIC FLUX >>> </span>
<span>Φ = Q/ε0 </span>
<span>Φ = { 43.31 * 10^-12 C } / { ε0 = 8.85 * 10^-12 } </span>
<span>Φ = 4.8937 = 4.9 V.m </span>
<span>DISPLACEMENT CURRENT </span>
<span>we use the following equation: </span>
<span>I = { ε / R } { e^(-t/RC) } </span>
<span>I = { 22 / 160 } { e^(- 0.5 * 10^-9 / 160 *3.10 * 10^-12 ) } </span>
<span>I = { 0.1375 } { 0.365 } </span>
<span>I = 0.0502 A = 0.05 A </span>
8 0
3 years ago
A liquid is used to make a mercury-type barometer. The barometer is intended for space-faring astronauts. At the surface of the
Anarel [89]

Answer:

Density of liquid = 4730 kg/m³

Atmospheric pressure on planet X = 8401.7 N/m²

Explanation:

Pressure, P = ρgh where ρ = density of liquid, g =9.8 m/s² and h = height of column at earth's surface = 2185 mm. Since P = atmospheric pressure, for mercury, P = ρ₁gh₁ where ρ₁ = 13.6 g/cm³ and h₁ = 760 mm

So, ρgh = ρ₁gh₁

ρ = ρ₁h₁/h = 13.6 g/cm³ × 760/2185 = 4.73 g/cm³ = 4730 kg/m³

The atmospheric pressure on planet X

P = ρg₁h₃     g₁ = g/4 and h₃ = 725 mm = 0.725 m

on planet X

P = ρg₁h₃ = (4730 kg/m³ × 9.8 m/s² × 0.725 m)/4 = 8401.7 N/m²

6 0
3 years ago
A golf ball is hit with a golf club. While the ball flies through the air, which forces act on the ball? Neglect air resistance.
34kurt

To solve this problem it is necessary to apply the equation related to the Gravitational Force, the equation describes that

F = \frac{GMm}{r^2}

Where,

G = Gravitational Universal Constant

M = Mass of Earth (or Bigger star)

m = Mass of Object  (or smallest star)

r = Radius

From the statement we know that once the impact is made, the golf ball is subjected to the forces that are exerted in nature. Since the air resistance, which would represent the drag force, is ignored. Only the forces related to gravity remain.

The gravitational force carries 'pushes' or 'attracts' the body towards the earth, while the speed decreases as it reaches its maximum height.

When the ball has reached its maximum height only the force of gravity begins to act on it, generating the attraction to the earth in parabolic motion.

Therefore the correct answer is B.

6 0
3 years ago
Read 2 more answers
An ideal gas is sealed in a container at constant volume. if the temperature t is increased to 4t, the pressure will be:
Westkost [7]
I believe that it would increase to 4p
6 0
3 years ago
Read 2 more answers
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