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3 years ago

How do you find distance on a velocity time graph?

Physics

1 answer:

vekshin13 years ago

8 0

The region under the speed time chart gives the relocation.

In simple to-recall, straightforward terms, remove levels with the result of speed and time. So when you discover the region under a speed time diagram, you are really increasing speed and time. This gives the separation.

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Today’s scientist must search through scientific journals before performing an experiment with making methodical observations wh

Natalija [7]

Answer:

in the lab

Explanation:

cause that is where scientist spend their time doing research ...

4 0

3 years ago

Many biological systems are well-described by the laws of statistical physics. A simple yet often powerful approach is to think

GuDViN [60]

Answer:

z1/z2

Explanation:

we have no quantum effects therefore we can make use of Maxwell Boltzmann distribution in the description of this system.

using the boltzman distribution the probability of finding a particle in energy state

$P_{ei} = \frac{gie^{-ei/kol} }{z}$

we have

gi to be degeneration of the ith state

ei to be energy of ith state

$z=e^{-ei/kbt}$ summation

$P_{ope} = \frac{e^{-ei/kBt} }{z} = \frac{Z_{1} }{Z}$

We have R to be equal to

$\frac{P_{ope} }{P_{Close} } = \frac{Z1}{Z2}$

8 0

3 years ago

What is the relationship between stream and drainage basin

Sergeeva-Olga [200]

Each stream in a drainage system drains into a certain area. In a drainage basin the water falling in the basin drain will fall into the same stream. A drainage divides drawing basin from other drainage basins

5 0

3 years ago

A dentist’s drill starts from rest. After 4.28 s of constant angular acceleration it turns at a rate of 28940 rev/min. Find the

mylen [45]

Answer:

Angular acceleration, is $708.07\ rad/s^2$

Explanation:

Given that,

Initial speed of the drill, $\omega_i=0$

After 4.28 s of constant angular acceleration it turns at a rate of 28940 rev/min, final angular speed, $\omega_f=28940\ rev/min=3030.58\ rad/s$

We need to find the drill’s angular acceleration. It is given by the rate of change of angular velocity.

$\alpha =\dfrac{\omega_f}{t}\\\\\alpha =\dfrac{3030.58\ rad/s}{4.28\ s}\\\\\alpha =708.07\ rad/s^2$

So, the drill's angular acceleration is $708.07\ rad/s^2$ .

4 0

3 years ago

The aurora is caused when electrons and protons, moving in the earth’s magnetic field of ≈5.0×10−5t, collide with molecules of t

andreev551 [17]

The question is incomplete. I can help you by adding the information missing. They want you to calculate a) the radius of the cyclotron orbit for an electron with speed 1.0 * 10^6 m/s^2 and b) the radius of a cyclotron orbit for a proton with speed 5.0 * 10^4 m/s.

The two tasks involve combining the equations of the magnectic force and the centripetal force in a circular motion.

When you do that, you will obtain an expression to find the radius of the circular motion, which is the radius of the cyclotron that impulses the particles.

a)

Magentic force, F = q*v*B

q is the charge of the electron = 1.6 * 10^ -19 C
v is the speed = 1.0 * 10 ^ 6 m/s
B is the magentic field = 5.0 * 10 ^-5 T

Centripetal force, F = m*Ac = m * v^2 / R

where,

Ac = centripetal acceleration
m = mass of the electron = 9.11 * 10 ^-31 kg
R = the radius of the orbit

Now equal the two forces: q*v*B = m * v^2 / R => R = m*v / (q*B)

=> R = (9.11 * 10^31 kg) (1.0*10^6m/s) / [ (1.6 * 10^-19C)* (5.0 * 10^-5T) ]

=> R = 0.114 m

b) The equations are the same, just now use the speed, charge and mass of the proton instead of those of the electron.

R = m*v / (qB) = (1.66*10^-27 kg)(5.0*10^4 m/s) / [(1.6*10^-19C)(5*10^-5T)]

=> R = 10.4 m

4 0

3 years ago