Which of the following is a vector quantity?

PLEASE HELP!! Lab: Electromagnetic Induction

Advocard [28]

I am not sure what this is

5 0

2 years ago

hich muscle fibers are best suited for activities that involve lifting large, heavy objects for a short period of time? cardiac

Temka [501]

Answer:

Dead lifting uses tho muscle fundamentals

Explanation:

6 0

3 years ago

You drop a steel ball bearing, with a radius of 2.40 mm, into a beaker of honey. Note that honey has a viscosity of 6.00 Pa/s an

Stells [14]

Answer:

The “terminal speed” of the ball bearing is 5.609 m/s

Explanation:

Radius of the steel ball R = 2.40 mm

Viscosity of honey η = 6.0 Pa/s

$\text { Viscosity has Density } \sigma=1360 \mathrm{kg} / \mathrm{m}^{3}$

$\text { Steel has a density } \rho=7800 \mathrm{kg} / \mathrm{m}^{3}$

$\left.\mathrm{g}=9.8 \mathrm{m} / \mathrm{s}^{2} \text { (g is referred to as the acceleration of gravity. Its value is } 9.8 \mathrm{m} / \mathrm{s}^{2} \text { on Earth }\right)$

While calculating the terminal speed in liquids where density is high the stokes law is used for viscous force and buoyant force is taken into consideration for effective weight of the object. So the expression for terminal speed (Vt)

$V_{t}=\frac{2 \mathrm{R}^{2}(\rho-\sigma) \mathrm{g}}{9 \eta}$

Substitute the given values to find "terminal speed"

$\mathrm{V}_{\mathrm{t}}=\frac{2 \times 0.0024^{2}(7800-1360) 9.8}{9 \times 6}$

$\mathrm{V}_{\mathrm{t}}=\frac{0.0048 \times 6440 \times 9.8}{54}$

$\mathrm{V}_{\mathrm{t}}=\frac{302.9376}{54}$

$\mathrm{V}_{\mathrm{t}}=5.609 \mathrm{m} / \mathrm{s}$

The “terminal speed” of the ball bearing is 5.609 m/s

7 0

3 years ago

transmission electron microscopes that use high-energy electrons accelerated over a range from 40.0 to 100 kv are employed in ma

Gekata [30.6K]

The spatial limitations in Picometer for the given range of electrons would be around 50 picometers.

What is a transmission electron microscope?

A transmission electron microscope (TEM) is a type of microscope that uses a beam of high-energy electrons to produce detailed images of the structure of materials at the atomic or molecular scale. TEMs work by passing a focused beam of electrons through a thin sample and collecting the transmitted electrons on a fluorescent screen or an electronic detector. The interaction of the sample with the electrons results in the formation of an image that can be magnified and displayed on a computer monitor. TEMs are widely used in the fields of materials science, biology, and nanotechnology and can provide information about the structure, composition, and properties of materials with a high level and resolution.

According to the problem:

The spatial resolution of a transmission electron microscope (TEM) is determined by the size of the electron probe, which is directly related to the energy of the electrons. The higher the energy of the electrons is, the smaller the size of the probe is and the higher the spatial resolution.

At the lower end of the energy range of 40.0 kV, the spatial resolution of the TEM would be on the order of hundreds of nanometers. At the higher end of the range (100 kV), the spatial resolution would be on the order of tens of nanometers.

In general, TEMs with electron energy in the range of 40-100 kV are capable of resolving details down to around 50 picometers (pm). However, the actual spatial resolution will depend on various factors, such as the quality of the electron optics, the stability of the electron beam, and the sample preparation.

It's worth noting that TEMs with even higher electron energies (up to several hundred kV) are available, which can achieve spatial resolutions down to the sub-angstrom level (less than 0.1 pm). However, these instruments are much more expensive and complex to operate than TEMs with lower electron energies.

To know more about de broglie wavelength, visit:

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7 0

10 months ago

In a certain chemical process, a lab technician supplies 292 J of heat to a system. At the same time, 68.0 J of work are done on

vekshin1

Answer:

The increase in the internal energy of the system is 360 Joules.

Explanation:

Given that,

Heat supplied to a system, Q = 292 J

Work done on the system by its surroundings, W = 68 J

We need to find the increase in the internal energy of the system. It can be given by first law of thermodynamics. It is given by :

$dE=dQ+dW\\\\dE=292\ J+68\ J\\\\dE=360\ J$

So, the increase in the internal energy of the system is 360 Joules. Hence, this is the required solution.

3 0

2 years ago