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Novay_Z [31]
2 years ago
6

conductors of resistance 2 ohm 3 ohm and 6 ohm are connected (1) in series and (2) in parallel. calculate the equivalent resista

nce of each combination​
Physics
1 answer:
Georgia [21]2 years ago
4 0
Answer
For conductors in series 1/2+1/3+1/6=1
And in parallel 2+3+6= 11
For the combination of series and parallel you get 2+11=13
Explanation:

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A block of mass m = 4.8 kg slides from left to right across a frictionless surface with a speed Vi=7.3m/s. It collides with a bl
nirvana33 [79]

Answer:

The speed of the 11.5kg block after the collision is V≅4.1 m/s

Explanation:

ma= 4.8 kg

va1= 7.3 m/s

va2= - 2.5 m/s

mb= 11.5 kg

vb1= 0 m/s

vb2= ?

vb2= ( ma*va1 - ma*va2) / mb

vb2= 4.09 m/s ≅ 4.1 m/s

7 0
3 years ago
Why a drop of spirit on the hand feels colder than a drop of water at the same temperature​
brilliants [131]

Answer:

This is because spirit has a lower boiling point when compared to water

Explanation:

spirit has a lower boiling point when compared to water which means it has the capacity to pull more heat from your hand and also it can do this very fast. This is why our hand feels colder.

5 0
2 years ago
The driver sees that the road is empty and accelerates at 1.0 m/s2 for 5.0 s. what can you determine about the truck's motion us
vekshin1

kinematic equation

v=u+at

v-u=at

v-u = 1x5

the driver will have increased speed by 5 m/s. actual speeds unknown

7 0
3 years ago
you just saw a magazine ad that shows several people having fun at a football game. They are all wearing a specific shoe brand.
Scilla [17]
The answer for the question is bandwagon. Hope it helped :)

6 0
3 years ago
A phonograph record accelerates from rest to 28.0 rpm in 5.73 s.
Arturiano [62]

Answer:

a) \alpha=0.5117\ rad.s^{-2}

b) \theta=8.4\ rad

Explanation:

Given:

  • initial rotational speed of phonograph, \omega_i=0\ rad.s^{-1}
  • final rotational speed of phonograph, N_f=28\ rpm \Rightarrow \omega_f=2\pi\times\frac{28}{60} =2.932\ rad.s^{-1}
  • time taken for the acceleration, t=5.73\ s

a)

Now angular acceleration:

\alpha=\frac{\omega_f-\omega_i}{t}

\alpha=\frac{2.932-0}{5.73}

\alpha=0.5117\ rad.s^{-2}

b)

Using eq. of motion:

\theta=\omega_i.t+\frac{1}{2} \alpha.t^2

\theta=0+\frac{1}{2}\times 0.5117\times 5.73^2

\theta=8.4\ rad

5 0
3 years ago
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