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3 years ago

Calculate the total resistance in a series circuit made up of resistances of 3Ω, 4Ω, and 5Ω.

1 answer:

5 0

The total resistance in a series circuit is equal to the sum of all resistors (R total = ΣRi).

R total = R1 + R2 + R3 = (3 + 4 + 5) Ω = 12 Ω

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You're driving a vehicle of mass 1350 kg and you need to make a turn on a flat road. The radius of curvature of the turn is 71 m

sergey [27]

Answer:

$v=12.65\ m.s^{-1}$

Explanation:

Given:

mass of vehicle, $m=1350\ kg$

radius of curvature, $r=71\ m$

coefficient of friction, $\mu=0.23$

<u>During the turn to prevent the skidding of the vehicle its centripetal force must be equal to the opposite balancing frictional force:</u>

$m.\frac{v^2}{r} =\mu.N$

where:

$\mu=$ coefficient of friction

$N=$ normal reaction force due to weight of the car

$v=$ velocity of the car

$1350\times \frac{v^2}{71} =0.23\times (1350\times 9.8)$

$v=12.65\ m.s^{-1}$ is the maximum velocity at which the vehicle can turn without skidding.

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3 years ago

Which best describes how combustion works?

jekas [21]

The answer is A, it breaks down and releases thermal energy.

3 0

2 years ago

Read 2 more answers

What is the momentum of a 12 kg condor flying at 6 m/s?

rusak2 [61]

Answer:

Explanation:

Formula

$p=mv\\=12*6\\=72$

5 0

3 years ago

A groups beliefs, values, and patterns of behavior as known as its

Nady [450]

I believe the answer is Nonmaterial Culture.

8 0

2 years ago

A ball of radius r rolls on the inside of a track of radius R. If the ball starts from rest at the vertical edge of the track, f

meriva

Answer:

$v=\sqrt{\dfrac{10g(R-r)}{7}}$

Explanation:

Given that

Radius of track = R

Radius of ball = r

The ball can be treated as solid sphere, so

The moment of inertia of ball

$I=\dfrac{2}{5}mr^2$

When the ball reach at the lowest position then it will have both angular and linear speed.

Condition for rolling without slipping v= ωr

Form energy conservation

$mgR=mgr+\dfrac{1}{2}mv^2+\dfrac{1}{2}I\omega^2$

v= ωr

$I=\dfrac{2}{5}mr^2$

$mgR=mgr+\dfrac{1}{2}mv^2+\dfrac{1}{2}\times \dfrac{2}{5}mr^2\omega^2$

$mg(R-r)=\dfrac{1}{2}mv^2+\dfrac{1}{2}\times \dfrac{2}{5}mv^2$

$2mg(R-r)=mv^2+\dfrac{2}{5}mv^2$

$2g(R-r)=\dfrac{7}{5}v^2$

$v=\sqrt{\dfrac{10g(R-r)}{7}}$

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3 years ago