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3 years ago

13

Consider a bicycle that has wheels with a circumference of 2 m. Solve for the linear speed of the bicycle when it's wheels rotat

e at 1 revolution per second

1 answer:

Yuliya22 [10]3 years ago

4 0

Answer:0.3183m/s

Explanation: V=wr

V=linear velocity

W= angular velocity

P=2πr

r=p/2π

r=2/2π

r= 1/3.142

r=0.3183

V= 0.3183*1

V=0.3183m/s

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Two gliders on an air track collide in a perfectly elastic collision. Glider A has a mass of 1.1 kg and is initially travelling

Eva8 [605]

m1= mass 1 = 1.1 kg

Vi1 = initial velocity 1 = 2.7 m/s

m2= 2.4 kg

V2i = -1.9 m/s

We assume east as positive and west as negative.

Apply the formulas:

Vf1 = ?

$vf1=(\frac{m1-m2}{m1+m2})Vi1+(\frac{2m2}{m1+m2})Vi2$

Replacing:

$Vf1=\frac{(1.1-2.4)}{(1.1+2.4)}2.7+\frac{(2\times2.4)}{(1.1+2.4)}-1.9$ $Vf1=(\frac{-1.3}{3.5})2.7+(\frac{4.8}{3.5})-1.9$ $Vf1=-1-2.6=-3.6\text{ m/s}$

Answer: 3.6 m/s west

6 0

1 year ago

: A 70 kg man and a 12 kg sled are on the frictionless ice of a frozen lake, 25 m apart but connected by a rope of negligible ma

e-lub [12.9K]

Answer:

$x_1 = 3.74m$

Explanation:

given,

mass of man = 70 kg

mass of sled = 12 kg

F = m a_s

$a_s = \dfrac{F}{m}$

$a_s = \dfrac{8.2}{12}$

$a_s = 0.68\ m/s^2$

$F = m a_m$

$a_m = \dfrac{F}{m}$

$a_m = \dfrac{8.2}{70}$

$a_m = 0.12\ m/s^2$

$x_1+x_2 = 25$

$\dfrac{1}{2}a_ct^2+ \dfrac{1}{2}a_mt^2 = 25$

$(a_c+a_m)t^2=50$

$(0.12+0.68)t^2=50$

$t = \sqrt{\dfrac{50}{0.8}}$

t = 7.90 s

$x_1 = \dfrac{1}{2}a_ct^2$

$x_1 = 0.5\times 0.12 \times 7.90^2$

$x_1 = 3.74m$

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3 years ago

Suppose an event is measured to be at a = (0,-2, 3, 5) in one reference frame. Find the components of this event in another refe

LenaWriter [7]

Answer:

The components of the moving frame is (8.07c, -2, 3, 9.493)

Solution:

As per the question:

Velocity of moving frame w.r.t original frame $v_{m}$ 0.85c

Point 'a' of an event in one reference frame corresponds to the (x, y, z, t) coordinates of the plane

a = (0, - 2, 3, 5)

Now, according the the question, the coordinates of moving frame, say (X, Y, Z, t'):

New coordinates are given by:

X = $\frac{x - v_{m}t}{\sqrt{1 - \frac{v_{m}^{2}}{c^{2}}}}$

X = $\frac{0 - 0.85c\times 5}{\sqrt{1 - \frac{(0.85c)^{2}}{c^{2}}}}$

X = $8.07 c$

Now,

Y = y = - 2

Z = z = 3

Now,

$t' = \frac{t - \frac{vx}{c}^{2}}{\sqrt{1 - (\frac{v}{c})^{2}}}$

$t' = \frac{5 - 0}{\sqrt{1 - (\frac{0.85c}{c})^{2}}} = 9.493 s$

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3 years ago

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Basile [38]

Saying english so we can help u

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