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anyanavicka [17]

3 years ago

14

If an electron is accelerated from rest through a potential difference of 1200V find its approximate velocity at the end of this

process. (qe = 1.6 x 10^-19 C; me = 9.1 x 10^-11kg)

1 answer:

kolbaska11 [484]3 years ago

8 0

Answer: 2.1 × 10^7 m/s

Explanation:

Please see the attachments below

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A ball at rest starts rolling down a hill with a constant acceleration of 3.2 meters/second2. What is the final velocity of the

leva [86]

Used an app called Mephyso to do the calculation.

8 0

3 years ago

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The per-unit impedance of a single-phase electric load is 0.3. The base power is 500 kVA, and the base voltage is 13.8 kV. a. Fi

Leto [7]

Answer:

114.26

Explanation:

a)Formula for per unit impedance for change of base is

Zpu2= Zpu1×(kV1/kV2)²×(kVA2/kVA1)

Zpu2: New per unit impedance

Zpu1: given per unit impedance

kV1: give base voltage

kV2: New bas votlage

kVA1: given bas power

kVA2: new base power

In the question

Zpu2=??

Zpu1= 0.3

kV2=24kV

kV1= 13.8 kV

kVA2= 1MVA ×1000= 1000 kVA

kVA1=500kVA

Zpu2= 0.3(13.8/24)²×(1000/500)

Zpu2= 0.198

b) to find ohmic impedance we will first calculate base value of impedance(Zbase). So,

Zbase= kV²/MVA

Zbase= 13.8²/(500/1000)

Zbase=380.88

Now that we have base value of impedance, Zbase, we can calculate actual ohmic value of impedance(Zactual) by using the following formula:

Zpu=Zactual/Zbase

0.3= Zactual/380.88

Zactual= 114.26 ohms

8 0

3 years ago

Relate a real life phenomenon with each branch of physics

anastassius [24]

Answer:

Branches of physics with real life examples

In measuring and understanding nuclear fission (a real life phenomenon), all branches of theoretical and experimental physics have to be employed. Physics branches needed in it are, radiation detection and measurement, nuclear physics, statistical physics, thermodynamics, and almost all others.

Explanation:

4 0

2 years ago

An object thrown vertically upward from the surface of a celestial body at a velocity of 36 m/s reaches a height of sequalsminu

Anarel [89]

Answer:

v = -1.8t+36

20 seconds

360 m

40 seconds

36 m/s

The object speed will increase when it is coming down from its highest height.

Explanation:

$s=-0.9t^2+36t$

Differentiating with respect to time we get

$\frac{ds}{dt}=-1.8t+36\\\Rightarrow v=-1.8t+36$

a) Velocity of the object after t seconds is v = -1.8t+36

At the highest point v will be 0

$0=-1.8t+36\\\Rightarrow t=\frac{-36}{-1.8}\\\Rightarrow t=20\ s$

b) The object will reach the highest point after 20 seconds

$s=-0.9t^2+36t\\\Rightarrow s=-0.9\times 20^2+36\times 20\\\Rightarrow s=360\ m$

c) Highest point the object will reach is 360 m

$s=-0.9t^2+36t\\\Rightarrow 360=-0.9t^2+36t\\\Rightarrow -0.9t^2+36t-360=0\\\Rightarrow -9t^2+360t-3600=0$

$\frac{-360\pm \sqrt{0}}{2\left(-9\right)}\\\Rightarrow t=20\ s$

d) Time taken to strike the ground would be 20+20 = 40 seconds

$[tex]v=u+at\\\Rightarrow v=0+0.9\times 2\times 20\\\Rightarrow v=36\ m/s$

Acceleration will be taken as positive because the object is going down. Hence, the sign changes. 2 is multiplied because the expression is given in the form of $s=ut+\frac{1}{2}at^2$

e) The velocity with which the object strikes the ground will be 36 m/s

f) The speed will increase when the object has gone up and for 20 seconds and falls down for 20 seconds. The object speed will increase when it is coming down from its highest height.

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2 years ago

At a particular instant the magnitude of the momentum of a planet is 2.05 × 10^29 kg·m/s, and the force exerted on it by the sta

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