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svetoff [14.1K]

3 years ago

7

What is the z coordinate for the above vector?

1 answer:

Hitman42 [59]3 years ago

4 0

120 m would be he answer

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Complete the statement by filling in the Blanks:

Anika [276]

Answer:

induced current

Explanation:

intentionally manipulated.

5 0

3 years ago

(15 points) (Asap!!)<br>In what two ways can you increase the elastic potential energy of a spring?

adelina 88 [10]

Hello There

Answers: T<span>he elastic potential energy can be increased by: </span>

<span>1) Getting a spring with a higher spring constant</span>

<span>2) Increasing the length at which the spring is compressed.

Reasons: Getting a stronger spring makes it stronger which equals more energy. While increasing the compression on the spring, increases the stored energy which makes it more powerful when its released

I hope this helps
-Chris</span>

5 0

2 years ago

Read 2 more answers

A 1.30-m string of weight 0.0125 N is tied to the ceiling at its upper end, and the lower end supports a weight W. Neglect the v

atroni [7]

Answer:

1. t = 0.0819s

2. W = 0.25N

3. n = 36

4. y(x , t)= Acos[172x + 2730t]

Explanation:

1) The given equation is

$y(x, t) = Acos(kx -wt)$

The relationship between velocity and propagation constant is

$v = \frac{\omega}{k}=\frac{2730rad/sec}{172rad/m}\\\\$

v = 15.87m/s

Time taken, $t = \frac{\lambda}{v}$

$= \frac{1.3}{15.87}\\\\=0.0819 sec$

t = 0.0819s

2)

The velocity of transverse wave is given by

$v = \sqrt{\frac{T}{\mu}}$

$v = \sqrt{\frac{W}{\frac{m}{\lambda}}}$

mass of string is calculated thus

mg = 0.0125N

$m = \frac{0.0125N}{9.8N/s}$

m = 0.00128kg

$\omega = \frac{v^2m}{\lambda}$

$\omega = \frac{(15.87^2)(0.00128)}{1.30}$

$\omega =$ 0.25N

3)

The propagation constant k is

$k=\frac{2\pi}{\lambda}$

hence

$\lambda = \frac{2\pi}{k}\\\\\lambda = \frac{2 \times 3.142}{172}$

$\lambda =$ 0.036 m

No of wavelengths, n is

$n = \frac{L}{\lambda}\\\\n = \frac{1.30m}{0.036m}\\$

n = 36

4)

The equation of wave travelling down the string is

$y(x, t)=Acos[kx -wt]\\\\becomes\\\\y(x , t)= Acos[(172 rad.m)x + (2730 rad.s)t]$

$without, unit\\\\y(x , t)= Acos[172x + 2730t]$

7 0

3 years ago

A merry-go-round with a a radius of R = 1.9 m and moment of inertia I = 209 kg-m2 is spinning with an initial angular speed of ω

RideAnS [48]

Answer:

340.67 kgm²/s

Explanation:

R = Radius of merry-go-round = 1.9 m

I = Moment of inertia = 209 kgm²

$\omega_i$ = Initial angular velocity = 1.63 rad/s

m = Mass of person = 73 kg

v = Velocity = 4.8 m/s

Initial angular momentum is given by

$L=I\omega_i\\\Rightarrow L=209\times 1.63\\\Rightarrow L=340.67\ kgm^2/s$

The initial angular momentum of the merry-go-round is 340.67 kgm²/s

8 0

3 years ago

Which direction will the object move?

kap26 [50]

Answer: it will move to the left

Explanation: i remeber doing this

3 0

3 years ago

Read 2 more answers