All categories

3 years ago

Describe the basis concept of the atwood’s machine

1 answer:

8 0

Answer:The Atwood Machine is a device that demonstrates the basic principles of acceleration and dynamics. You'll mostly see Atwood machines in Physics laboratories and classrooms. It consists of two objects with different masses that hang vertically from a frictionless pulley that has a very small, negligible mass.

Explanation:

You might be interested in

Which title best reflects the main idea of the passage? The Role of Convection in the Distribution of Earth's Energy The Role of

Leto [7]

Answer:

The Role of Heat Transfer Methods in the Distribution of Earth's Energy

Explanation:

8 0

3 years ago

Read 2 more answers

At what time was the person at a position of 0m?

Anni [7]

Answer: The person was not at a position of "0" at any time. The person started at 10 metres from the starting line. The explanation below shows how to use the standard formula for position when the initial position is not "0". It is noteworthy that the standard expression of the formula for distance travelled does not include a variable (e.g. "d") for distance at the start (when t(time) = 0)

Explanation: At time = 0, the start, the person was at 10m distance from the starting line. Therefore, to use the standard equation, "s + ut + 1/2att (t squared, that is), distance from starting line = 10 + s, that is, total distance from starting line equals initial position, 10 metres, plus "s" (distance travelled from t = 0 to t = 1) in metres.

for the section of the graph from "0" seconds (t = 0) to 1 second (t = 1):

s = ut + 1/2att

the initial position is 10 metres.

s = 10

the distance is constant from t = 0 to t = 1, therefore the velocity for the whole of that section of graph must be 0.

u = 0

there is no change in the velocity from t = 0 to t= 1, therefore the acceleration for the first section of the graph must be 0.

a = 0

s = ut + 1/2att

= (0 x 1) + 1/2 (0 x 1 x 1)

= (0) + 1/2 (0)

= 0

total distance from starting line (position) equals initial position plus change in position (distance travelled).

at t = 1,

position = 10 + 0

= 10 metres

The whole of the graph can be analysed using this process for each straight section of the graph separately, adding "s" for each section to the previous total of distance from starting line.

using "d" for initial distance from starting line ( position ), d1 for distance from starting line at t = 1, d2 for distance from starting line at t = 2, etcetera:

section 1, t = 0 to t = 1:

d1 (t=0 to t=1) = 10 + s (t=0 to t=1).

section 2, t= 1 to t = 2:

d2 (t=0 to t=2) = 10 + s (t=0 to t=1) + s (t=1 to t=2).

etcetera.

8 0

3 years ago

Review. An electron moves in a circular path perpendicular to a constant magnetic field of magnitude 1.00 mT . The angular momen

Ghella [55]

The speed of an electron when it moves in a circular path perpendicular to a constant magnetic field is 8.88 x 10^7 m/s.

The angular momentum(L) of an electron moving in a circular path is given by the formula,

L = mvr ........(i)

We know that the radius of the path of an electron in a magnetic field is

r = mv/qB

Putting this value in equation (i),

L = mv x mv/qB

or L = (mv)^2/qB

Putting the given values in the above equation,

4 x 10^-25 = (9.1x10^-31)^2 x v^2/ 1.6 x 10^-19 x 1 x 10^-3

v comes out to be 8.88 x 10^7 m/s.

Hence, the speed of an electron when it moves in a circular path perpendicular to a constant magnetic field is 8.88 x 10^7 m/s.

To know more about "angular momentum", refer to the following link:

brainly.com/question/15104254?referrer=searchResults

#SPJ4

5 0

1 year ago

When an electron falls from a higher to a lower energy level in an atom, the photon released has a wavelength of 121.6 nm. What

yarga [219]

Answer:

$\Delta E=1.64*10^{-18}J$

Explanation:

The energy difference between the energy levels involved in the transition of the electron is directly proportional to the frequency of the emitted photon:

$\Delta E=h\nu(1)$

Where h is the Planck constant. The photon's frequency is inversely proportional to its wavelegth:

$\nu=\frac{c}{\lambda}(2)$

Here c is the speed of light. Replacing (2) in (1):

$\Delta E=\frac{hc}{\lambda}\\\Delta E=\frac{(6.63*10^{-34}J\cdot s)(3*10^8\frac{m}{s})}{121.6*10^{-9}m}\\\Delta E=1.64*10^{-18}J$

6 0

3 years ago

A rock is thrown vertically upward from some height above the ground. It rises to some maximum height and falls back to the grou

True [87]

Answer:

At the highest point the velocity is zero, the acceleration is directed downward.

Explanation:

This is a free-fall problem, in the case of something being thrown or dropped, the acceleration is equal to -gravity, so -9.80m/s^2. So, the acceleration is never 0 here.

I attached an image from my lecture today, I find it to be helpful. You can see that because of gravity the acceleration is pulled downwards.

At the highest point the velocity is 0, but it's changing direction and that's why there's still an acceleration there.

6 0

4 years ago