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3 years ago

An object is moving in the plane according to these parametric equations:

Physics

2 answers:

morpeh [17]3 years ago

8 0

Answer:

.a vx = -3π

b.vy = 0

c.c. m = sin(4πt + π/2) / [πt + cos(4πt + π/2)]

d.m = sin(4π/6 + π/2) / [π/6 + cos(4π/6 + π/2)]

e. t = -1.0

f.f. t = -0.35

g.vx = π - 4πsin (4π(0.124) + π/2)

h.vmax=4π cos (4π(0.045) + π/2)

i.s(t) = [x(t)^2 + y(t)^2]^(1/2)

s'(t) = d [x(t)^2 + y(t)^2]^(1/2) / dt

Explanation:

x(t) = πt + cos(4πt + π/2)

differentiating te orizontal distance wit respect to time t, will give horizontal velocity

change in displacement per change in time is velocity

vx = dx/dt = π - 4πsin (4πt + π/2)

vx = π - 4π sin (0 + π/2)

at t =0, substituting te value of t into the above

vx = π - 4π (1)

vx = -3π

y(t) = sin(4πt + π/2)

differentiate wit respect to t

dy/dt=4π cos (4πt + π/2)

π/2=90

when t=0

b. vy=dy/dt = 4π cos (4πt + π/2)

vy = 0

c. slope of te tanent line

y(t)/x(t)

c. m = sin(4πt + π/2) / [πt + cos(4πt + π/2)]

d. at t=1/6, we substitute into answer gotten in c

m = sin(4π/6 + π/2) / [π/6 + cos(4π/6 + π/2)]

e. t = -1.0

f. t = -0.35

g. Solve for t

vx = π - 4πsin (4πt + π/2) = 0

at maximum v=0

(4πt + π/2)=0.0043

t=-π/2+0.0043/(4π)

t=-0.124

vx = π - 4πsin (4π(0.124) + π/2)

h. Solve for t

vy = 4π cos (4πt + π/2) = 0

(4πt + π/2=1

t=0.57/4π

t=0.045

vmax=4π cos (4π(0.045) + π/2)

i. resultant of te displacement

s(t) = [x(t)^2 + y(t)^2]^(1/2)

h. s'(t) = d [x(t)^2 + y(t)^2]^(1/2) / dt

s'(t)=

k and l. Solve for the values of t

d [x(t)^2 + y(t)^2]^(1/2) / dt = 0

And substitute to determine the maximum and minimum speeds.

aniked [119]3 years ago

5 0

A. The horizontal velocity is
vx = dx/dt = π - 4πsin (4πt + π/2)
vx = π - 4π sin (0 + π/2)
vx = π - 4π (1)
vx = -3π

b. vy = 4π cos (4πt + π/2)
vy = 0

c. m = sin(4πt + π/2) / [πt + cos(4πt + π/2)]

d. m = sin(4π/6 + π/2) / [π/6 + cos(4π/6 + π/2)]

e. t = -1.0

f. t = -0.35

g. Solve for t
vx = π - 4πsin (4πt + π/2) = 0
Then substitute back to solve for vxmax

h. Solve for t
vy = 4π cos (4πt + π/2) = 0
The substitute back to solve for vymax

i. s(t) = [x(t)^2 + y(t)^2]^(1/2)

h. s'(t) = d [x(t)^2 + y(t)^2]^(1/2) / dt

k and l. Solve for the values of t
d [x(t)^2 + y(t)^2]^(1/2) / dt = 0
And substitute to determine the maximum and minimum speeds.

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Fynjy0 [20]

Answer:

6.13428 rev/s

Explanation:

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In this system the angular momentum is conserved

$L_f=L_i\\\Rightarrow I_f\omega_f=I_i\omega_i\\\Rightarrow \omega_f=\dfrac{I_i\omega_i}{I_f}\\\Rightarrow \omega_f=\dfrac{(5.7+2\times 2.5\times 0.76^2)3}{4.2}\\\Rightarrow \omega_f=6.13428\ rev/s$

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An intergalactic rock star bangs his drum every 1.50 s. A person on earth measures that the time between beats is 2.70 s. How fa

sineoko [7]

Answer:

v = 83.1 % of speed of light

Explanation:

given,

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T_s is the ship time = 1.5 s

we know,

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$1.5= 2.7\times \sqrt{1-\dfrac{v^2}{c^2}}$

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If the average speed of a person was 1.2 meters/second, does this mean that their speed was exactly 1.2 meters/second the whole

alexandr1967 [171]

Average speed is defined by the following formula

$v = \frac{D}{t}$

here

D = total distance that an object move from its initial position to final position

t = total time of the motion

so here we will say that there is no such relation between initial or final speed or we can say maximum or minimum speed of object with average speed of object.

We only need to find the total distance and total time of motion in order to find the average speed

here we can see many examples like let say an object moves with speed v1 for time t1 and then with speed v2 for time t2 then here average speed is given as

$d = v_1 t_1 + v_2t_2$

since we know that distance covered is product of speed and time

that's why we used above equation for finding total distance

now the average speed will be

$v_{avg} = \frac{v_1t_1 + v_2 t_2}{t_1 + t_2}$

so this is how we can find the average speed for above motion

so average speed is always between maximum and minimum speed any value in-between.

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Ulleksa [173]

Answer:

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Initial angular momentum is given by

$L=I\omega_i\\\Rightarrow L=196\times 1.53\\\Rightarrow L=299.88\ kgm^2/s$

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Angular momentum is given by

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The angular momentum of the person 2 meters before she jumps on the merry-go-round is 499.758 kgm²/s

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