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alexdok [17]
3 years ago
6

Suppose a runner completes one lap around a 400-m track in a time of 50 s. Calculate the average speed of the runner.

Physics
1 answer:
suter [353]3 years ago
6 0

Answer:

8m/s

Explanation:

Average Speed = distance / time = 400/50 = 8m/s

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What is 4 differences between saturated unsaturated and supersaturated solutions
Dennis_Churaev [7]

Answer:

Unsaturated Solution: Less amount of salt in water, clear solution, no precipitation. Saturated Solution: The maximum amount of salt is dissolved in water, Colour of the solution slightly changes, but no precipitation. Supersaturated Solution: More salt is dissolved in water, Cloudy solution, precipitation is visible.

3 0
3 years ago
Why are the nuclei of the heavier elements radioactive and not the lighter elements of nuclei?
KatRina [158]

Answer:

becouse most of nuclear elements are heave

Explanation:

6 0
3 years ago
If x2 = 60, what is the value of x? plus or minus square root 30 plus or minus square root 60 ±30 ±120
Zielflug [23.3K]
X2 = 60
/ 2 / 2
x = 30
Plus or minus square root 60
4 0
3 years ago
What is the frequency for a beam of electrons orbiting in a field of 4.62 x 10^-3 T? Let the mass of an electrons m = 9.31 x 10^
melisa1 [442]

Frequency: 1.27\cdot 10^8 Hz

Explanation:

The force experienced by an electron in a magnetic field is

F=qvB

where

q=1.6\cdot 10^{-19}C is the electron charge

v is the speed of the electron

B is the strength of the magnetic field

Since the force is perpendicular to the direction of motion of the electron, the force acts as centripetal force, so we can write:

qvB=m\frac{v^2}{r}

where

r is the radius of the orbit

m=9.31\cdot 10^{-31}kg is the mass of the electron

Re-arranging the equation,

\frac{v}{r}=\frac{qB}{m} (1)

We also know that in a circular motion, the speed is equal to the ratio between circumference of the orbit and orbital period (T):

v=\frac{2\pi r}{T}

Substituting into (1),

\frac{2\pi}{T}=\frac{qB}{m}

We also know that 1/T is equal to the frequency f, so

f=\frac{qB}{2\pi m}

In this problem,

B=4.62\cdot 10^{-3}T

Therefore, the frequency of the electrons is

f=\frac{(1.6\cdot 10^{-19})(4.62\cdot 10^{-3})}{2\pi(9.31\cdot 10^{-31})}=1.27\cdot 10^8 Hz

4 0
3 years ago
A rock resting high in a cliff is an example of an object with what type of energy?
Elanso [62]

potential energy because it has the ability to do work

Explanation:

A rock resting high in a cliff is an example of potential energy because it has the ability to do work. This form of energy is called potential energy.

  • Energy is defined as the ability to do work. To do work, force must applied using energy to move a body through a particular distance.
  • A body at rest is not doing any work. This is the case with potential energy of the body on a high cliff.
  • At the state of rest, it is not doing any work.
  • The potential energy of this body expresses the ability of such body to do work.

Learn more:

Potential energy brainly.com/question/10770261

#learnwithBrainly

5 0
3 years ago
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