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kenny6666 [7]
3 years ago
10

Which is NOT a property of an acid

Physics
1 answer:
nikdorinn [45]3 years ago
5 0
What are the options?
You might be interested in
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
Need a little help here :(
Goshia [24]

Answer:

The output out be 200

Explanation:

Hope this helps :))

8 0
3 years ago
Read 2 more answers
A ball is thrown vertically upwards with an initial velocity of 20.00 m/s. Neglecting air resistance, how long is the ball in th
algol [13]

Answer:

1) The greatest height attained by the ball equals 20.387 meters.

2) The time it takes for the ball to reach 15 meters approximately equals 1 second.

Explanation:

The greatest height will be attained when the ball stop's in the air and starts falling back to the earth.

thus using third equation of kinematics we obtain the height attained as

v^2=u^2+2as

where

'v' is the final speed of the ball

'u' is the initial speed of the ball

'a' is the acceleration that the ball is under which in this case equals 9.81 m/s^{2}

's' is the distance it covers

Thus for maximum height applying the values in the equation we get

0=20^{2}-2\times 9.81\times h\\\\\therefore h=\frac{20^{2}}{2\times 9.81}=20.387meters

Using the same equation we can find the speed of the ball when it reaches 15 meters of height as  

v^2=20^{2}-2\times 9.81\times 15\\\\v^{2}=105.7\\\\\therefore v=10.28m/s

the time it takes to reduce the velocity to this value can be found by first equation of kinematics as

v=u+at\\\\t=\frac{v-u}{a}\\\\t=\frac{10.28-20}{-9.81}=0.991seconds\approx 1second

4 0
3 years ago
Which of the following is associated with fission but not fusion?
kodGreya [7K]

Answer:

D. Uranium

Explanation:

I just got the answer right on my quiz.

8 0
3 years ago
You are presently taking a weather observation. The sky is full of wispy cirrus clouds estimated to be about 10 km overhead. If
marshall27 [118]

Answer:

x = 2000 Km

Explanation:

Given

y = 10 km

Slope: 1 : 200

x = ?

We can apply the formula

y / x = 1 / 200   ⇒     x = 200*y = 200*10 Km

⇒     x = 2000 Km

7 0
3 years ago
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