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2 years ago

How to find the coefficient of kinetic friction on an incline?

Physics

1 answer:

PolarNik [594]2 years ago

7 0

Answer:

An object slides down an inclined plane at a constant velocity if the net force on the object is zero. We can use this fact to measure the coefficient of kinetic friction between two objects.

Explanation:

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Can someone help me with this?

oksian1 [2.3K]

Answer:

Yes

Explanation:

5 0

2 years ago

A solid sphere has a radius of 0.200 m and a mass of 150.0 kg. how much work is required to get the sphere rolling with an angul

Allisa [31]

Here in this case we can use work energy theorem

As per work energy theorem

Work done by all forces = Change in kinetic Energy of the object

Total kinetic energy of the solid sphere is ZERO initially as it is given at rest.

Final total kinetic energy is sum of rotational kinetic energy and translational kinetic energy

$KE = \frac{1}{2}Iw^2 +\frac{1}{2} mv^2$

also we know that

$I = \frac{2}{5}mR^2$

$w= \frac{v}{R}$

Now kinetic energy is given by

$KE = \frac{1}{2}(\frac{2}{5}mR^2)w^2 +\frac{1}{2} m(Rw)^2$

$KE = \frac{1}{5}mR^2w^2 +\frac{1}{2} mR^2w^2$

$KE = \frac{7}{10}mR^2w^2$

$KE = \frac{7}{10}*150*(0.200)^2(50)^2$

$KE = 10500 J$

Now by work energy theorem

Work done = 10500 - 0 = 10500 J

So in the above case work done on sphere is 10500 J

7 0

2 years ago

Renewable energy sources are ____________________?

Nina [5.8K]

Answer:

Eco-friendly

Explanation:

6 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:

brainly.com/question/17295250

#SPJ4

7 0

7 months ago

A parallel-plate capacitor has a plate area of 0.2 m^2 and a plate separation of 0.1 mm. If the charge on each plate has a magni

barxatty [35]

Answer:

The potential difference across the plates is 226 V.

Explanation:

Given;

area of the capacitor plate, A = 0.2 m²

separation, d = 0.1 mm = 0.1 x 10⁻³ m

charge on each plate, Q = 4 x 10⁻⁶ C

Charge on the capacitor is given by;

Q = CV

Where;

C is the capacitance of the capacitor, given as;

C = ε₀A / d

Then, the potential difference across the plates is given by;

$V = \frac{Q}{C} = \frac{Qd}{\epsilon_o A} = \frac{(4*10^{-6})(0.1*10^{-3})}{(8.85*10^{-12})(0.2)}\\\\V = 226 \ V$

Therefore, the potential difference across the plates is 226 V.

5 0

2 years ago