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Papessa [141]
2 years ago
11

At a certain time a particle had a speed of 80 m/s in the positive x direction, and 9.8 s later its speed was 20 m/s in the oppo

site direction. What was the average acceleration of the particle during this 9.8 s interval
Physics
1 answer:
dem82 [27]2 years ago
6 0

The average acceleration of the particle in the time interval of 9.8 s is 10.204 m/s² opposite to the direction of motion.

Given:

Speed of particle, v₁ = 80 m/s (in positive x-direction)

Speed of particle, v₂ = -20 m/s (in opposite direction)

Time interval, Δt = 9.8 s

Calculation:

We know that, the average acceleration is given as:

a_avg = (v₂ - v₁)/ Δt     - ( 1 )

Applying values in above equation we get:

a_avg = (v₂ - v₁)/ Δt

           = (-20 m/s - 80 m/s) / (9.8 s)

           = -10.204 m/s²

Therefore the average acceleration of the particle in the time interval Δt is 10.204 m/s² opposite to the direction of motion.

Learn more about average acceleration here:

<u>brainly.com/question/27184422</u>

#SPJ4

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The correct answer to the question is: A) miles/hour and B) metre/ second.

EXPLANATION:

Before answering this question, first we have to understand speed.

The speed of a body is defined as the rate of distance travelled or the distance travelled by a body per unit time.

Hence, it is a derived quantity which is obtained from distance and time.

The unit of distance can be metre, miles, and the unit of time can be second, minutes or hour.

As speed is the distance covered per unit time, the perfect units will be miles/hour and metre/second.

Hence, the correct options are first and second.

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A billiard ball. unless hit, the balls stay at rest. however when hit into another, the balls do not stop unless acted upon by another force.
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Compressions and rarefactions are characteristic of
Korvikt [17]

Answer:

Of longitudinal waves

Explanation:

Depending on the direction of the oscillation, there are two types of waves:

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- Longitudinal waves: in a longitudinal wave, the oscillations occur parallel to the direction of propagation of the wave. In such a wave, the oscillations are produced by alternating regions of higher density of particles, called compressions, and regions of lower density of particles, called rarefactions. Examples of longitudinal waves are sound waves.

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3 years ago
One electron collides elastically with a second electron initially at rest. After the collision, the radii of their trajectories
ch4aika [34]

Answer:

114.92749 keV

Explanation:

r = Radius of trajectory

m = Mass of electron = 9.11\times 10^{-31}\ kg

B = Magnetic field = 0.044 T

q = Charge of electron = 1.6\times 10^{-19}\ C

The centripetal force and the magnetic forces are conserved

m\frac{v^2}{r}=Bqv\\\Rightarrow v=\frac{Bqr}{m}

Velocity of first electron

v=\frac{Bqr_1}{m}\\\Rightarrow v=\frac{0.044\times 1.6\times 10^{-19}\times 0.01}{9.11\times 10^{-31}}\\\Rightarrow v_1=77277716.79473\ m/s

Velocity of second electron

v=\frac{Bqr_2}{m}\\\Rightarrow v_2=\frac{0.044\times 1.6\times 10^{-19}\times 0.024}{9.11\times 10^{-31}}\\\Rightarrow v_2=185466520.30735\ m/s

Total kinetic energy is given by

K=K_1+K_2\\\Rightarrow K=\frac{1}{2}mv_1^2+\frac{1}{2}mv_2^2\\\Rightarrow K=\frac{1}{2}m(v_1^2+v_2^2)\\\Rightarrow K=\frac{1}{2}\times 9.11\times 10^{-31}(77277716.79473^2+185466520.30735^2)\\\Rightarrow K=1.83884\times 10^{-14}\ J

Converting to eV

1\ J=\frac{1}{1.6\times 10^{-19}}\ eV

1.83884\times 10^{-14}\ J=1.83884\times 10^{-14}\times \frac{1}{1.6\times 10^{-19}}\ eV\\ =114927.49\ ev=114.92749\ keV

The energy of incident electron is 114.92749 keV

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Naddika [18.5K]
The i got answer is 990m
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