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2 years ago

Write the meaning of an object has 2 meter length

Physics

1 answer:

Ivahew [28]2 years ago

4 0

Answer:

The object has 2 meter length. This means the length is any quantity with a dimension distance. The definition of the length is how long something is or amount of space. In the given data it is stated that, the object is something that has a length of 2 meter.

Let's take examples to understand.

For example a thread or a table is an object which has a total length of 2 meters.

Another example is something we are measuring it gives us a result of 2 meters of length by using a meter scale or meter tape.

Length is a measure of distance and it is a fundamental quantity. Meter is a international system of units (SI units).

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Water waves in a small tank are .06 m long. They pass a given point at a rate of 14.8 waves every three seconds. What is the spe

snow_lady [41]

Answer:

Speed = 0.296m/2

Period = 0.203 s

Explanation:

If by 'long' you mean the wavelength of the waves, then the wavelength $\lambda=0.06m$ .

The frequency $f$ of the waves is 14.8 waves every 3 seconds or

$f=14.8/3 =4.33Hz$ .

Now the relationship between wavelength $\lambda$ , frequency $f$ and speed $v$ of the waves is:

$v=\lambda f$

We put in the values $\lambda=0.06m$ and $f=4.933Hz$ and get:

$\boxed{v=0.06*4.922=0.296m/s}$

Now the period $T$ is just the inverse of the frequency, or

$T=\frac{1}{f}$

$\boxed{T=\frac{1}{4.933}=0.203\:seconds }$

4 0

2 years ago

Charge q1 is distance s from the negative plate of a parallel-plate capacitor. Charge q2=q1/3 is distance 2s from the negative p

Svetlanka [38]

Answer:

The ratio (U₁/U₂) = 6

Explanation:

U, the potential energy is given as

U = kqQ/r

k = Coulomb's constant

q = charge we're concerned about

Q = charge of the negative plate of the capacitor

r = distance of q from the negative plate of the capacitor.

For charge q₁

U₁ = kq₁Q/s

U₂ = kq₂Q/2s

But q₂ = q₁/3

U₂ becomes U₂ = kq₁Q/6s

U₁ = kq₁Q/s

U₂ = kq₁Q/6s

(U₁/U₂) = 6

5 0

2 years ago

Hurricane katrina, which hit the gulf coast of louisiana and mississippi on august 29, 2005, had the second lowest ever recorded

Simora [160]

The unit 'mb' means millibar which is equivalent to 1/1000 of 1 bar. To convert the units from bar to atmospheres (atm) and to inches Hg (inHg), we need to know the conversion factors.

a.) 1 atm = 1.01325 bar

0.92 mb(1 bar/1000 mbar)(1 atm/1.01325 bar) = 9.08×10⁻⁴ atm

b.) 1 bar = 29.53 inHg

0.92 mb(1 bar/1000 mbar)(29.53 inHg/1 bar) = 0.027 inHg

3 0

2 years ago

A student constructs a simple constant volume gas thermometer and calibrates it using the boiling point of water, 100°C, and the

just olya [345]

Answer:

The pressure corresponding to the absolute zero temperature is 0.997atm.

Explanation:

To solve this question, you draw a straight vertical line with the boiling point temperature and pressure on top of the line and the freezing point temperature and pressure on the lower part. The absolute temperature somewhere in the middle of the line with the pressure to be obtained.

So, we have;

0- (-19) / 100 - (-19) = P - 0.9267 / 1.366 - 0.9267

19 / 119 = P - 0.9267 / 0.4393

Cross multiply, we have

19 * 0.4393 = 119(P -0.9267)

8.3467 = 119P - 110.2773

119P = 118.624

P = 0.997 atm

So at 0°C, the pressure of the thermometer is 0.997atm.

4 0

2 years ago

A resistor with resistance R and an air-gap capacitor of capacitance C are connected in series to a battery (whose strength is "

blsea [12.9K]

Answer:

a) Q = C*emf

b) Reduction in electric field strength and electric potential

c) Initial current through the resistor = emf/R

d) The final charge = K*C*emf

Explanation:

a) The resistors and capacitors are connected in series with the battery

Using Kirchoff's voltage law, sum of all voltages in the circuit is zero

Let $V_{R}$ = Voltage dropped across the Resistor

$V_{c}$ = Voltage dropped across the capacitor

Applying KVL;

$emf - V_{R} - V_{c} = 0\\$ .........................(1)

Since the connection is in series, the same current flow through the circuit

$V_{R} = IR\\Q = CV_{c} \\V_{c} = Q/C$

Putting $V_{c}$ and $V_{R}$ into equation (1)

$emf - IR - Q/C = 0$

At the final charge, the capacitor in fully charged, and current drops to zero due to equilibrium

$I = 0A\\emf = Q/C\\Q = C* emf$

b) Current starts running through the plate because as the sheet of plastic is inserted between the plates both the electric field intensity and the electric potential reduces. The charge also reduces, then current flows

c) The current through the resistor is the current through the entire circuit ( series connection)

$I = I_{o} \exp(\frac{-t}{RC} )\\At time the initial time, t\\t = 0\\ I_{o} = \frac{emf}{R} \\$

Putting the values of t and I₀ into the formula for I written above

$I = \frac{emf}{R} \exp(0)\\I = \frac{emf}{R}$

d) NB: The initial charge on the capacitor = C * emf

The final charge will be:

$Q = K* Q_{initial} \\Q_{initial} = C *emf\\Q_{final} = KCemf$

4 0

2 years ago