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maxonik [38]
3 years ago
5

A clamp-type measuring instrument operates on the principle of

Physics
1 answer:
Hitman42 [59]3 years ago
8 0

<em>A clamp-type measuring instrument operates on the principle of; </em>

A. induction

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Two mechanical devices typically used in laboratories to accurately measure small objects or distances are the _____ and _____.
Lera25 [3.4K]
Micrometers and calipers

3 0
3 years ago
Modeled after the two satellites problem (slide 12 in Lecture13, Universal gravity), imagine now if the period of the satellite
ikadub [295]

Answer:

  R = 6.3456 10⁴  mile

Explanation:

For this exercise we will use Newton's second law where force is gravitational force

      F = m a

The satellite is in a circular orbit therefore the acceleration is centripetal

      a = v² / r

Where the distance is taken from the center of the Earth

     G m M / r² = m v² / r

     G M / r = v²

The speed module is constant, let's use the uniform motion relationships, with the length of the circle is

     d = 2π  r

     v = d / t

The time for a full turn is called period (T)

Let's replace

     G M / r = (2π r / T)²

     r³ = G M T²²2 / 4π²

     r = ∛ (G M T² / 4π²)

We have the magnitudes in several types of units

      T = 88.59 h (3600 s / 1h) = 3.189 10⁵ s

      Re = 6.37 10⁶ m

Let's calculate

     r = ∛ (6.67 10⁻¹¹ 5.98 10²⁴ (3,189 10⁵)²/4π²)

     r = ∛ (1.027487 10²⁴)

     r = 1.0847 10⁸ m

This is the distance from the center of the Earth, the distance you want the surface is

     R = r - Re

     R = 108.47 10⁶ - 6.37 10⁶

     R = 102.1 10⁶ m

Let's reduce to miles

      R = 102.1 10⁶ m (1 mile / 1609 m)

     

      R = 6.3456 10⁴  mile

6 0
3 years ago
A basketball player grabbing a rebound jumps 76.0 cm vertically. How much total time (ascent and descent) does the player spend
Viefleur [7K]

a) we can answer the first part of this by recognizing the player rises 0.76m, reaches the apex of motion, and then falls back to the ground we can ask how

long it takes to fall 0.13 m from rest: dist = 1/2 gt^2 or t=sqrt[2d/g] t=0.175

s this is the time to fall from the top; it would take the same time to travel

upward the final 0.13 m, so the total time spent in the upper 0.15 m is 2x0.175

= 0.35s

b) there are a couple of ways of finding thetime it takes to travel the bottom 0.13m first way: we can use d=1/2gt^2 twice

to solve this problem the time it takes to fall the final 0.13 m is: time it

takes to fall 0.76 m - time it takes to fall 0.63 m t = sqrt[2d/g] = 0.399 s to

fall 0.76 m, and this equation yields it takes 0.359 s to fall 0.63 m, so it

takes 0.04 s to fall the final 0.13 m. The total time spent in the lower 0.13 m

is then twice this, or 0.08s

5 0
4 years ago
Read 2 more answers
when a 4.25 kg object is placed on top of a vertical spring the spring compress a distance of 2.62 cm what is the spring force c
Alina [70]

Answer:

425n

Explanation:

4 0
3 years ago
Estimate the change in the equilibrium melting point of copper caused by a change in pressure of 10 kbar. The molar volume of co
mr Goodwill [35]

Answer:

The change in the equilibrium melting point is 4.162 K.

Explanation:

Given that,

Pressure = 10 kbar

Molar volume of copperV=8.0\times10^{-6}\ m^3

Volume of liquid V=7.6\times10^{-6}\ m^3

Latent heat of fusion L= 13.05 kJ

Melting point =1085°C

We need to calculate the change temperature

Using Clapeyron equation

\dfrac{\Delta P}{\Delta T}=\dfrac{\Delta H}{T\Delta V}

Put the value into the formula

\dfrac{1000\times10^{5}}{\Delta T}=\dfrac{13050}{(1085+273)\times(8.0-7.6)\times10^{-6}}

\Delta T=\dfrac{1000\times10^{-5}\times(1085+273)\times(8.0-7.6)\times10^{-6}}{13050}

\Delta T=4.162\ K

Hence, The change in the equilibrium melting point is 4.162 K.

5 0
3 years ago
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