Which shows the correct order of forces, from weakest to strongest?

You are riding on a carousel that is rotating at a constant 24 rpm. It has an inside radius of 4 ftand outside radius of 12 ft.

Nookie1986 [14]

Answer:

magnitude of the Coriolis acceleration is 44.235 ft/s² and the direction of the acceleration is along the axis of transmission

Explanation:

Given the data in the question;

Speed of carousel N = 24 rpm

From the diagram below, selected path direction defines the Axis of slip.

Hence, The Coriolis is acting along the axis of transmission

Now, we determine the angular speed ω of the carousel.

ω = 2πN / 60

we substitute in the value of N

ω = (2π × 24) / 60

ω = 2.5133 rad/s

Next, we convert the given velocity from mph to ft/s

we know that; 1 mph = 1.4667 ft/s

so

$V_{slip$ = 6 mph = ( 6 × 1.4667 ) = 8.8002 ft/s

Now, we determine the magnitude of the Coriolis acceleration

$a_c$ = 2( $V_{slip$ × ω )

we substitute

$a_c$ = 2( 8.8002 ft/s × 2.5133 rad/s )

$a_c$ = 44.235 ft/s²

Hence, magnitude of the Coriolis acceleration is 44.235 ft/s² and the direction of the acceleration is along the axis of transmission

7 0

3 years ago

I'm very confused. Thanks for whoever helps me :)

sergij07 [2.7K]

(C). Remember gravity provides an acceleration of 9.81m/s^2, so the y component of velocity initial is zero because it isn’t already falling, and we have the height, so basically we use the kinematic equation vf^2=vi^2+2ad, substitute given values and you get vf^2=2(9.81)(65) which is 1275, when you take the square root you get 35.7m/s for final velocity
(B). Then you use vf=vi+at to get the equation 35.7=(9.81)t, when you divide out you get 3.64s for time t
(A). Finally, since we assume that there is no acceleration or deceleration horizonatally, we just multiply the time taken for it to hit the ground and the initial speed ((3.64)(35.7)) to get 129.96, with significant figures I would round that to 130 metres.
**this is in the order that I felt was easiest to answer**

6 0

3 years ago

A particle's position along the x-axis is described by. x(t)= At+Bt^2where t is In seconds: x is in meters: and the constants A

DanielleElmas [232]

Answer

given,

position of particle

x(t)= A t + B t²

A = -3.5 m/s

B = 3.9 m/s²

t = 3 s

a) x(t)= -3.5 t + 3.9 t²

velocity of the particle is equal to the differentiation of position w.r.t. time.

$\dfrac{dx}{dt}=\dfrac{d}{dt}(-3.5t + 3.9t^2)$

$v= -3.5 + 7.8 t$ ------(1)

velocity of the particle at t = 3 s

v = -3.5 + 7.8 x 3

v = 19.9 m/s

b) velocity of the particle at origin

time at which particle is at origin

x(t)= -3.5 t + 3.9 t²

0 = t (-3.5 + 3.9 t )

t = 0, $t=\dfrac{3.5}{3.9}$

t = 0 , 0.897 s

speed of the particle at t = 0.897 s

from equation (1)

v = -3.9 + 7.8 t

v = -3.9 + 7.8 x 0.897

v = 3.1 m/s

6 0

4 years ago

A(n) 71.1 kg astronaut becomes separated from the shuttle, while on a space walk. She finds herself 70.2 m away from the shuttle

denpristay [2]

Answer:

10.347 minutes.

Explanation:

According to F = ma, she exerts force on camera of the magnitude

F = 0.67Kg*12m/ $s^{2}$ = 8.04N, assuming it took her one second to accelerate camera to 12m/s, then by newtons third law, which says every action has equal and opposite reaction , the camera exerts the same amount of force on the astronaut which gives her acceleration of a = $\frac{8.04N}{70.2Kg} = 0.1130801680m/s^2$ .

and velocity of V = 0.1130801680m/s.

at this velocity , the astronaut has to cover the distance of 70.2 meters, it will take her 620.7985075s = 10.347 min to get to the shuttle (using S = vt).

4 0

3 years ago

Calculate the mass of an object that has a momentum of 100kg x m/sec and velocity of 4 m/sec

lidiya [134]

P= mv

where p is momentum
m is mass
v is velocity

so it's given p= 100kgm/sec
v= 4m/s

so putting in the formula

100= m × 4

m = 25kg

3 0

3 years ago