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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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Answer:

(a) E=0 : 0 cm from the center of the sphere

(b) E= 227.8*10³ N/C : 10.0 cm from the center of the sphere

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Explanation:

If we have a uniform charge sphere we can use the following formulas to calculate the Electric field due to the charge of the sphere

$E=\frac{K*Q}{r^{2} }$ : Formula (1) To calculate the electric field in the region outside the sphere r ≥ a

$E=k*\frac{Q}{a^{3} } *r$ :Formula (2) To calculate the electric field in the inner region of the sphere. r ≤ a

Where:

K: coulomb constant

a: sphere radius

Q: Total sphere charge

r : Distance from the center of the sphere to the region where the electric field is calculated

Equivalences

1μC=10⁻⁶C

1cm= 10⁻²m

Data

k= 9*10⁹ N*m²/C²

Q=16.2 μC=16.2 *10⁻⁶C

a= 40 cm = 40*10⁻²m = 0.4m

Problem development

(a)Magnitude of the electric field at 0 cm :

We replace r=0 in the formula (2) , then, E=0

(b) Magnitude of the electric field at 10.0 cm from the center of the sphere

r<a , We apply the Formula (2):

$E=9*10^{9} *\frac{16.2*10^{-6} }{0.4^{3} } *0.1$

E= 227.8*10³ N/C

r=a, We apply the Formula (1) :

$E=\frac{9*10^{9}*16.2*10^{-6} }{0.4^{2} }$

E= 911.25*10³ N/C

(d) Magnitude of the electric field at 59.5 cm from the center of the sphere

r>a , We apply the Formula (1) :

$E=\frac{9*10^{9}*16.2*10^{-6} }{0.595^{2} }$

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