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Lemur [1.5K]
3 years ago
14

The acceleration time plot for a particle (starting from rest) moving on a straight line is shown in figure. For given time inte

rval,
(A) The particle has zero average acceleration
(B) The particle has never turned around.
(C) The particle has zero displacement
(D) The average speed in the interval 0 to 10s is the same as the average speed in the interval 10s to 20s.​

Physics
1 answer:
DanielleElmas [232]3 years ago
4 0

Answer:

Image is not so clear pls send it again.

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If the speed and mass of an object are doubled, which of the following is true?
Harman [31]

Answer:

A is true

Explanation:

Other all are wrong I guess

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3 years ago
I will Mark Brainliest 1. ) Allena and Charissa were discussing whether Cl2 is an element or a compound. Allena said that it is
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3 years ago
An electron in a television tube is accelerated uniformly from rest to a speed of 8.4\times 10^7~\text{m/s}8.4×10 ​7 ​​ m/s over
stich3 [128]

Answer:

P=3.42×10^-6 J/s

Explanation:

From the kinematics of motion with constant acceleration we know that :  

vf^2=vi^2+2*a(xf-xi)

Where :

• vf , vi, are the the final and the initial velocity of the electron  

• a is the acceleration of the electron  

• xf , xi are the final and the initial position of the electron .

Strategy for solving the problem : at first from the given information we calculate the acceleration of the electron.  

Givens: vf = 8.4 x 10^7 m/s , vi, = 0 m/s , xf = 0.025 m and xi = 0 m  

vf^2 =vi^2+2*a(xf-xi)

vf^2-vi^2=2*a(xf-xi)

2*a(xf-xi)= vf^2-vi^2

          a = (vf^2-vi^2)/2(xf-xi)

Pluging known information to get :

a = (vf^2-vi^2)/2(xf-xi)

  = 1.411 × 10^17

From the acceleration and the previous Eq. we can calculate the final velocity of the electron but a new position xf = 0.01 m  

so,

vf^2 =vi^2+2*a(xf-xi)

vf^2 =5.312× 10^7

From the following Eq. we can calculate the time elapsed in this motion .  

xf =xi+vi*t+1/2*a*t

xf =xi+vi*t+1/2*a*t

  t=√2(xf-xi)/a

 t=3.765×10^-10 s

now we can use the power P Eq.  

 P=W/Δt => ΔK/Δt  

Where: the work done W change the kinetic energy K of the electron ,

ΔK=Kf-Ki=>1/2*m*vf^2-1/2*m*vi^2

P=1/2*m*vf^2-1/2*m*vi^2/Δt

P=3.42×10^-6 J/s

6 0
3 years ago
Pls help me guys simple question​
Airida [17]

Answer:

please write neater

Explanation:

can you write neater so I can answer th question but also is a equal to b

7 0
3 years ago
ILL MARK BRAINIEST IF YOU DO THIS CORRECTLY!!!
IRINA_888 [86]

Explanation:

1)5.8m/s

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3)12.69m/s

4)

8 0
3 years ago
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