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3 years ago

The initial speed of a body is 3.28 m/s. What is its speed after 2.32 s if it accelerates

1 answer:

6 0

Final velocity = initial velocity + acceleration * time

v = u + at

v = 3.28 + 2.32 * 2.08

v = 3.28 + 4.83

v = 8.11 m/s

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3. What are some questions that philosophers ask about knowledge?

Mariulka [41]

Hello there,

What are some questions that philosophers ask about knowledge?

I hope I would be understanding your question right. But I believe that on question they could ask is " What part in the human brain could store up all this knowledge that any human can store".

I truly hope this helps.

~Jurgen

5 0

3 years ago

A cable with a linear density of \mu=0.109~\text{kg/m}μ=0.109 kg/m is hung from telephone poles. The tension in the cable is 572

RUDIKE [14]

Answer:

$f=16.46 Hz$

Explanation:

The equation of the speed of a mechanical wave in terms of the tension and linear density, of the cable in our case, is given by:

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

Where:

T is the tension of the cable (T = 572 N)
μ is the linear density of the cable (μ = 0.109 kg/m)

And we know that v = λ*f

λ is the wavelength
f is the frequency

Because a standing waves pattern is produced that has 4.5 wavelengths between the two poles and the distance between poles is 19.9 meters, the value of the wavelength is: λ = 19.9/4.5 = 4.4 m.

Therefore, the frequency will be:

$\lambda f=\sqrt{\frac{T}{\mu}}$

$f=\frac{1}{\lambda}\sqrt{\frac{T}{\mu}}$

$f=\frac{1}{4.4}\sqrt{\frac{572}{0.109}}$

$f=16.46 Hz$

I hope it helps you!

6 0

3 years ago

3) A small, 30 gram pith ball carrying 3 nC of charge is dangling from an insulating string. Another charged pith ball carrying

ludmilkaskok [199]

Answer:

q2 = 5.1μC

Explanation:

From a forces diagram we can stablish that:

Fe = T*sin(16) (eq1)

T*cos(16) = m*g (eq2)

Where Fe is the electric force and T is the tension of the string. Solving for Fe:

Fe = m*g*tan(16)

The magnitude of the electric force is calculated as:

$Fe = \frac{K*q1*q2}{d^2} = m*g*tan(16)$ Solving for q2:

$q2 = \frac{m*g*tan(16)*d^2}{K*q1}=5.1 \mu C$

7 0

2 years ago

Read 2 more answers

A fellow student with a mathematical bent tells you that the wave function of a traveling wave on a thin rope is y(x,t)= 2.30mmc

MakcuM [25]

y(x,t)= 2.30mmcos[(6.98rad/m)x + (742 rad/s)t]
A) Amplitude is 2.30mm
B) Frequency 1/2.30mm
C) Wavelength is 6.98rad/m
D) Wave Speed is 742 rad/s
E) Direction the wave is traveling

7 0

3 years ago

A particle position as a function of time t is given by r=(5.0t+6.0t^2)mi+(7.0-3.0t^3)mj. At t=5s find the magnitude and directi

horsena [70]

Horizontal component:

x(t) = 5t + 6t²

x(5) = 25 + 6(25) = 175 m

Vertical component:

y(t) = 7 - 3t³

y(5) = 7 - 3(125) = -368 m

Components combined:

r(5) = (175 m) i - (368 m) j

Taken relative to the point r(0) = (0 i + 7 j) m ,
the displacement vector when t=5 is

   (175 m) i - (375 m) j .

Its magnitude is √(175² + 375²)

   = √(30,625 + 140,625)

   = √171,250 = 413.823... m    (rounded)

Its direction is tan⁻¹(-375/175) = tan⁻¹(-2.14285...) = - 64.98°

6 0

3 years ago