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Anna35 [415]
3 years ago
6

A rod of propoer length l0 is at rest in frame S'. it ilies in the x',y' plane and amkes an angle of sin^-1(3/5) What must be th

e value of v if as measured in S the rod is at a 45 degree)
Physics
1 answer:
Blizzard [7]3 years ago
5 0

Answer:

v = 1.98*10^8 m/s

Explanation:

Given:

- Rod at rest in S' frame

- makes an angle Q = sin^-1 (3/5) in reference frame S'

- makes an angle of 45 degree in frame S

Find:

What must be the value of v if as measured in S the rod is at a 45 degree)

Solution:

- In reference frame S'

                x' component = L*cos(Q)

                y' component = L*sin(Q)

- Apply length contraction to convert projected S' frame lengths to S frame:

                x component = L*cos(Q) / γ           (Length contraction)

                y component = L*sin(Q)                  (No motion)

- If the rod is at angle 45° to the x axis, as measured in F, then the x and y components must be  equal:

                L*sin(Q) = L*cos(Q) / γ    

Given:       γ = c / sqrt(c^2 - v^2)

                 c / sqrt(c^2 - v^2) = cot(Q)

                 1 - (v/c)^2 = tan(Q)

                 v = c*sqrt( 1 - tan^2 (Q))

For the case when Q = sin^-1 (3/5)::

                 tan(Q) = 3/4

                 v = c*sqrt( 1 - (3/4)^2)

                 v = c*sqrt(7) / 4 = 1.98*10^8 m/s

 

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Levart [38]

Answer:

1.3823 rad/s

20.7345 m/s

28.66129935 m/s²

a=2.92164g

2006.29095 N radially outward

Explanation:

r = Radius = 15 m

m = Mass of person = 70 kg

g = Acceleration due to gravity = 9.81 m/s²

Angular velocity is given by

\omega=13.2\times \dfrac{2\pi}{60}\\\Rightarrow \omega=1.3823\ rad/s

Angular velocity is 1.3823 rad/s

Linear velocity is given by

v=r\omega\\\Rightarrow v=15\times 1.3823\\\Rightarrow v=20.7345\ m/s

The linear velocity is 20.7345 m/s

Centripetal acceleration is given by

a_c=r\omega^2\\\Rightarrow a_c=15\times 1.3823^2\\\Rightarrow a_c=28.66129935\ m/s^2

The centripetal acceleration is 28.66129935 m/s²

Acceleration in terms of g

\dfrac{a}{g}=\dfrac{28.66129935}{9.81}\\\Rightarrow a=2.92164g

a=2.92164g

Centripetal force is given by

F_c=ma_c\\\Rightarrow F_c=70\times 28.66129935\\\Rightarrow F_c=2006.29095\ N

The centripetal force is 2006.29095 N radially outward

The torque will be experienced when the centrifuge is speeding up of slowing down i.e., when it is accelerating and decelerating.

3 0
4 years ago
Write an expression for a transverse harmonic wave that has a wavelength of 2.5 m and propagates to the right with a speed of 13
lyudmila [28]

Answer:

y = 0.14 Cos\left ( 2.512x-34.66t \right )

Explanation:

wavelength, λ = 2.5 m

speed, v = 13.8 m/s

Amplitude, A = 0.14 m

The general equation of the transverse harmonic wave which is travelling right is given by

y = A Sin\left ( \frac{2\pi }{\lambda } (x - vt)+\phi \right  )

where, Ф is phase

At t = 0, x = 0 , y = 0.14 m

0.14 = 0.14 Sin Ф

Ф = π/2

So, the equation is

y = 0.14Sin\left ( \frac{2\pi }{2.5 } (x - 13.8t)+\frac{\pi }{2} \right  )

y = 0.14 Cos\left ( 2.512x-34.66t \right )

3 0
3 years ago
Five identical quintuplets leave earth when they reach the age of 21, in the year 2121. Each quintuplet goes on a spaceship jour
Elena-2011 [213]

Answer:

Explanation:

This is a problem based on time dilation , a theory given by Albert Einstein .

The formula of time dilation is as follows .

t₁ = \frac{t}{\sqrt{1-\frac{v^2}{c^2} } }

t is time measured on the earth and t₁ is time measured by man on ship .

A ) Given t = 20 years , t₁ = ? v = .4c

\frac{20}{\sqrt{1-\frac{.16c^2}{c^2} } }

=1.09 x 20

t₁= 21.82 years

B ) Given t = 5 years , t₁ = ? v = .2c

\frac{5}{\sqrt{1-\frac{.04c^2}{c^2} } }

=1.02 x 5

t₁= 5.1 years

C ) Given t = 10 years , t₁ = ? v = .8c

\frac{10}{\sqrt{1-\frac{.64c^2}{c^2} } }

=1.67 x 10

t₁= 16.7  years

D ) Given t = 10 years , t₁ = ? v = .4c

\frac{10}{\sqrt{1-\frac{.16c^2}{c^2} } }

=1.09 x 10

t₁= 10.9  years

E ) Given t = 20 years , t₁ = ? v = .8c

\frac{20}{\sqrt{1-\frac{.64c^2}{c^2} } }

=1.67 x 20

t₁= 33.4   years

7 0
3 years ago
Two racing boats set out from the same dock and speed away at the same constant speed of 104 km/h for half an hour (0.500 h), th
Zinaida [17]

Answer:

The blue boat traveled 6.1 km farther west than the green boat

The green boat traveled 10.7 km farther south than the blue boat

Explanation:

The equation for linear uniform speed movement is

X(t) = X0 + v * t

Since we have two coordinates (X, Y) we use

X(t) = X0 + vx * t

Y(t) = Y0 + vy * t

The dock will be the origin of coordinates so X0 and Y0 will be zero. The X axis will be pointing west and the Y axis south.

The blue boat moves with a direction 24° south of west, so it will have speeds:

vxb = 104 * cos(24) = 95 km/h

vyb = 104 * sin(24) = 42.3 km/h

And the green boat:

vxg = 104 * cos(37.7) = 82.3 km/h

vyg = 104 * sin(37.7) = 63.6 km/h

After half hour the boats will have arrived at positions

Xb = 95 * 0.5 = 47.5 km

Yb = 42.3 * 0.5 = 21.1 km

And

Xg = 82.3 * 0.5 = 41.4 km

Yg = 63.6 * 0.5 = 31.8 km

The difference in positions of the boats

47.5 - 41.4 = 6.1 km

31.8 - 21.1 = 10.7 km

5 0
3 years ago
Please answer! thank you!
lesantik [10]

Answer:

Maybe A.... I'm not sure

Explanation:

8 0
3 years ago
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