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2 years ago

Which device is used to measure interior diameter of a water pipe and in which units are the callibrations

Physics

2 answers:

Rzqust [24]2 years ago

7 0

Answer:

The instrument is  micrometer screw gauge .

Units : millimetres

Explanation:

$.$

ahrayia [7]2 years ago

3 0

We can use a vernier calliper

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Two capacitors have the same size of plates and the same distance 4 mm between the plates The potentials of the two plates in ca

OverLord2011 [107]

Answer:

E = -4000 N / C

Explanation:

The potential and electric field are related

V = - E s

E = - V / s

we reduce the magnitudes to the SI system

s = 4 mm (1 m / 1000 mm) = 0.004 m

we calculate

E = - 16 /0.004

E = -4000 N / C

8 0

3 years ago

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Lelechka [254]

Answer:

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2 years ago

A roller coaster travels around a vertical 8-m radius loop. Determine the speed at the top of the loop if the normal force exert

strojnjashka [21]

Answer:

$v=10m/sec$

Explanation:

From the question we are told that

Radius of vertical r= 8m

Force exerted by passengers is 1/4 of weight

Generally the net force acting on top of the roller coaster is give to be

$F_N+Fg$

where

$F_N =forceof the normal$

$Fg= force due to gravity$

Generally the net force is given to be $FC(force towards center)$

$F_C =F_N + Fg$

$F_N =F_C -Fg$

$F_N=F_C-F_g$

$F_N=\frac{mv^2}{R} -mg$

Mathematical we can now derive V

$m_g + \frac{8m}{4}= \frac{mv^2}{8}$

$\frac{5mg}{4} =\frac{mv^2}{8}$

$v^2 =\frac{40*10}{4}$

$v=10m/sec$

Therefore the speed of the roller coaster is given ton be $v=10m/sec$

5 0

3 years ago

You are given two unknown point charges, Q1, and Q2. At a point on the line joining them, one third of the way from Q1 to Q2 the

podryga [215]

The formula that is applicable here is E = kQ/r^2 in which the energy of attraction is proportional to the charges and inversely proportional to the square of the distance. In this case,
kQ1/(r1)^2 = kQ2/(r2)^2 r1=l/3, r2=2l/3solve Q1/Q2
kQ1/(l/3)^2 = kQ2/(2l/3)^2 kQ1/(l^2/9) = kQ2/(4l^2/9)Q1/Q2 = 1/4

3 0

2 years ago

Read 2 more answers

The radius of a planet is 2400 km, and the acceleration due to gravity at its surface is 3.6 m/s2.

kiruha [24]

Answer:

$3.1\cdot10^{23}\:\mathrm{kg}$

Explanation:

We can use Newton's Universal Law of Gravitation to solve this problem:

$g_P=G\frac{m}{r^2}$ ., where $g_P$ is acceleration due to gravity at the planet's surface, $G$ is gravitational constant $6.67\cdot 10^{-11}$ , $m$ is the mass of the planet, and $r$ is the radius of the planet.

Since acceleration due to gravity is given as $m/s^2$ , our radius should be meters. Therefore, convert $2400$ kilometers to meters:

$2400\:\mathrm{km}=2,400,000\:\mathrm{m}$ .

Now plugging in our values, we get:

$3.6=6.67\cdot10^{-11}\frac{m}{(2,400,000)^2}$ ,

Solving for $m$ :

$m=\frac{2,400,000^2\cdot3.6}{6.67\cdot 10^{-11}},\\m=\fbox{$3.1\cdot10^{23}\:\mathrm{kg}$}$ .

6 0

2 years ago

Which device is used to measure interior diameter of a water pipe and in which units are the callibrations​

Which device is used to measure interior diameter of a water pipe and in which units are the callibrations