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4 years ago

14

Perform calculations using the circuit illustrated. Round all the numerical answers to the tenths place.

2 answers:

Nina [5.8K]4 years ago

8 0

Answer:

1. 5.9.

2. 1

3. D

Explanation:

DochEvi [55]4 years ago

3 0

The total resistance is 5.9.

The expected current is 1.

The current will be the lowest at point D.

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Which waves require a medium for transmission

uysha [10]

I believe it is Mechanical Waves. they are waves that require a material medium for propagation

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4 years ago

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A candy-filled piñata is hung from a tree for Elia's birthday. During an unsuccessful attempt to break the 4.4-kg piñata, Tonja

Klio2033 [76]

Answer: v = 0.6 m/s

Explanation: <u>Momentum</u> <u>Conservation</u> <u>Principle</u> states that for a collision between two objects in an isolated system, the total momentum of the objects before the collision is equal to the total momentum of the objects after the collision.

Momentum is calculated as Q = m.v

For the piñata problem:

$Q_{i}=Q_{f}$

$m_{p}v_{p}_{i}+m_{s}v_{s}_{i}=m_{p}v_{p}_{f}+m_{s}v_{s}_{f}$

Before the collision, the piñata is not moving, so $v_{p}_{i}=0$ .

After the collision, the stick stops, so $v_{s}_{f}=0$ .

Rearraging, we have:

$m_{s}v_{s}_{i}=m_{p}v_{p}_{f}$

$v_{p}_{f}=\frac{m_{s}v_{s}_{i}}{m_{p}}$

Substituting:

$v_{p}_{f}=\frac{(0.54)(4.8)}{(4.4)}$

$v_{p}_{f}=$ 0.6

Immediately after being cracked by the stick, the piñata has a swing speed of 0.6 m/s.

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3 years ago

A solid, uniform sphere of mass 2.0 kg and radius 1.8 m rolls from rest without slipping down an inclined plane of height 7.5 m.

Molodets [167]

Answer:

$w^2=5.5rads/s$

Explanation:

From the question we are told that:

Mass $m=2.0kg$

Radius $r=1.8m$

Height $h=7.5m$

Generally the equation for Potential energy is mathematically given by

Potential energy=Kinetic energy+Rotational energy

$mgh=\frac{1}{2}mv^2+\frac{1}{2}Iw^2$

Since there is no slipping

$v=rw$

Therefore

$mgh=\frac{1}{2}mr^2w^2+\frac{1}{2}Iw^2$

Where

$I=\frac{1}{2}mr^2$

$l=3.24m$

$2*9.81*7.5=\frac{1}{2}(2)(1.8)^2w^2+\frac{1}{2}(3.24)w^2\\\\$

$147.15=3.24w^2+1.62w^2$

$w^2=\frac{147.15}{4.86}$

$w^2=\sqrt{\frac{147.15}{4.86}}$

$w^2=5.5rads/s$

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3 years ago

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NeTakaya

Answer:

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3 years ago

Under ideal conditions (no atmospheric interference of any kind), if I hit a golf ball at an angle of 25 degrees at an initial s

g100num [7]

Answer:

The required angle is (90-25)° = 65°

Explanation:

The given motion is an example of projectile motion.

Let 'v' be the initial velocity and '∅' be the angle of projection.

Let 't' be the time taken for complete motion.

Let 'g' be the acceleration due to gravity

Taking components of velocity in horizontal(x) and vertical(y) direction.

$v_{x}$ = v cos(∅)

$v_{y}$ = v sin(∅)

We know that for a projectile motion,

t = $\frac{2vsin(∅)}{g}$

Since there is no force acting on the golf ball in horizonal direction.

Total distance(d) covered in horizontal direction is -

d = $v_{x}$ ×t = vcos(∅)× $\frac{2vsin(∅)}{g}$ = $\frac{v^{2}sin(2∅) }{g}$ .

If the golf ball has to travel the same distance 'd' for same initital velocity v = 23m/s , then the above equation should have 2 solutions of initial angle 'α' and 'β' such that -

α +β = 90° as-

d = $\frac{v^{2}sin(2α) }{g}$ = $\frac{v^{2}sin(2[90-β]) }{g}$ = $\frac{v^{2}sin(180-2β) }{g}$ = $\frac{v^{2}sin(2β) }{g}$ .

∴ For the initial angles 'α' or 'β' , total horizontal distance 'd' travelled remains the same.

∴ If α = 25° , then

β = 90-25 = 65°

∴ The required angle is 65°.

5 0

3 years ago

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