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sattari [20]
2 years ago
13

7. A particle is forced to move in a straight line path. It returns to the starting point after 10 s. The total distance covered

by the particle is 20 m. Which of the following statements is true regarding the motion of the particle? a The average velocity of the particle is zero b The displacement of the particle is zero c The average speed of the particle is 2 m/s d All of the above​
Physics
1 answer:
lutik1710 [3]2 years ago
8 0

d. All three statements are true.

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The answer is B. False
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which of the following cannot be determined by looking at the phase diagram? A. melting point, B. boiling point, C. subilmation
bulgar [2K]
 I am pretty sure that <span>the following whihc cannot be determined by looking at the phase diagram is definitely </span>D. system pressure.  I consider this one to be correct because only this point is not included into<span> phase diagram and can't be determined itself. Hope it will help! Regards!</span>
7 0
3 years ago
If the sun's mass is about average, how many stars are there in the milky way galaxy? the mass of the sun is of the order of 103
ra1l [238]
Just assume that the sun has the average mass of all the stars
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4 0
4 years ago
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The car has a constant deceleration of 4.20 m/s^2. If its initial velocity was 24.0 m/s, how long does it take to come to a stop
nalin [4]

Answer:

The time is 5.71 sec.

Explanation:

Given that,

Acceleration a= -4.20 m/s^2

Initial velocity = 24.0 m/s

We need to calculate the time

Using equation of motion

v = u+at[/tex]

Where, v = final velocity

u = inital velocity

t = time

a = acceleration

Put the value into the formula

0 =24.0 +(-4.20)\times t

t = \dfrac{-24.0}{-4.20}

t=5.71\ sec

Hence, The time is 5.71 sec.

5 0
4 years ago
The Chernobyl reactor accident in what is now Ukraine was the worst nuclear disaster of all time. Fission products from the reac
Anna35 [415]

Answer:

The correct answer is "53.15 days".

Explanation:

Given that:

Half life of 131_{I},

T_{\frac{1}{2} }= 8 \ days

  • Let the initial activity be "R_o".
  • and, activity to time t be "R".

To find t when R will be "1%" of R_o, then

⇒ R=\frac{1}{100}R_o

As we know,

⇒ R=R_o e^{-\lambda t}

or,

∴ e^{\lambda t}=\frac{R_o}{R}

By putting the values, we get

        =\frac{R_o}{\frac{R}{100} }

        =100

We know that,

Decay constant, \lambda = \frac{ln2}{T_{\frac{1}{2} }}

hence,

⇒ \lambda t=ln100

     t=\frac{ln100}{\lambda}

        =\frac{ln100}{\frac{ln2}{8} }

        =53.15 \ days  

5 0
3 years ago
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