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2 years ago

Name some devices we use to make measurements

Physics

1 answer:

NikAS [45]2 years ago

4 0

Answer:

Explanation:

Ruler. A steel ruler aids the measurement and layout of straight lines. The ruler, also called "straightedge" or "straight-edged ruler," is a long, thin strip of wood, metal or plastic marked with increments of measurement.

Measuring Tape. The modern measuring tape's roughly palm-sized casing contains a coiled strip of metal marked with increments of measurement. The metal strip, called "tape," attaches to a spring which automatically retracts the tape into the casing following use.

Walking Tape Measure. The walking tape measure, also called "surveyor's measure," records the distance traveled by a wheel. An operator pushes the measure's wheel, similar to a bicycle wheel, by a handle as an attached ticker box displays feet or meters in the same format as a car's odometer.

Laser Measure. The laser measure offers point and shoot distance measurement. In its most basic form, a laser measure is a hand-held electronic device with a digital display.

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Two gliders collide on a frictionless air track that is aligned along the x axis. Glider A has an initial velocity of +4.0 m/s a

DedPeter [7]

Answer:

As collision is elastic,thus we can use conservation of momentum equation

mA=0.2 kg

(vB)1=0 m/s.......................as it is on rest before collision

(vA)1=4 m/s

(vA)2=-1 m/s

(vB)2=2 m/s

using equation

(mA*vA+mB*vB)1= (mA*vA+mB*vB)2

Where 1 and 2 represents before and after collision

(0.2*4)+(mB*0)=(0.2*-1)+(mB*2)

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3 years ago

An electron with speed 2.45 x 10^7 m/s is traveling parallel to a uniform electric field of magnitude 1.18 x 10^4N/C . How much

cupoosta [38]

Answer:

time will elapse before it return to its staring point is 23.6 ns

Explanation:

given data

speed u = 2.45 × $10^{7}$ m/s

uniform electric field E = 1.18 × $10^{4}$ N/C

to find out

How much time will elapse before it returns to its starting point

solution

we find acceleration first by electrostatic force that is

F = Eq

here

F = ma by newton law

ma = Eq

here m is mass , a is acceleration and E is uniform electric field and q is charge of electron

put here all value

9.11 × $10^{-31}$ kg ×a = 1.18 × $10^{4}$ × 1.602 × $10^{-19}$

a = 20.75 × $10^{14}$ m/s²

so acceleration is 20.75 × $10^{14}$ m/s²

and

time required by electron before come rest is

use equation of motion

v = u + at

here v is zero and u is speed given and t is time so put all value

2.45 × $10^{7}$ = 0 + 20.75 × $10^{14}$ (t)

t = 11.80 × $10^{-9}$ s

so time will elapse before it return to its staring point is

time = 2t

time = 2 ×11.80 × $10^{-9}$

time is 23.6 × $10^{-9}$ s

time will elapse before it return to its staring point is 23.6 ns

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2 years ago

What do radio waves and microwaves have in common?

Tamiku [17]

Answer:

I Will say the Answer is A

Explanation:

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2 years ago

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A rectangular coil of wire (a = 22.0 cm, b = 46.0 cm) containing a single turn is placed in a uniform 4.60 T magnetic field, as

frez [133]

Explanation:

Below is an attachment containing the solution.

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3 years ago

A room with dimensions 7.00m×8.00m×2.50m is to be filled with pure oxygen at 22.0∘C and 1.00 atm. The molar mass of oxygen is 32

VashaNatasha [74]

1) 5765 mol

First of all, we need to find the volume of the gas, which corresponds to the volume of the room:

$V=7.00 m\cdot 8.00 m\cdot 2.50 m=140 m^3$

Now we can fidn the number of moles of the gas by using the ideal gas equation:

$pV=nRT$

where

$p=1.00 atm=1.01\cdot 10^5 Pa$ is the gas pressure

$V=140 m^3$ is the gas volume

n is the number of moles

R is the gas constant

$T=22.0^{\circ}+273=295 K$ is the gas temperature

Solving for n,

$n=\frac{pV}{RT}=\frac{(1.01\cdot 10^5 Pa)(140 m^3)}{(8.314 J/molK)(295 K)}=5765 mol$

2) 184 kg

The mass of one mole is equal to the molar mass of the oxygen:

$M_m = 32.0 g/mol$

so if we have n moles, the mass of the n moles will be given by

$m : n = 32.0 g/mol : 1 mol$

since n = 5765 mol, we find

$m=\frac{(5765 mol)(32.0 g/mol)}{1 mol}=1.84\cdot 10^5 g=184 kg$

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3 years ago