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4 years ago

5

A large centrifuge is used to expose aspiring astronauts to accelerations similar to those experienced in rocket launches and at

mospheric reentries. At what angular velocity is the centripetal acceleration 10 g if the rider is 15.0 m from the center of rotation

1 answer:

melomori [17]4 years ago

6 0

Answer:

ω = 2.55 rad/sec

Explanation:

Assuming no other external forces acting in the horizontal plane, the only force keeping the rider in a circular path of a radius equal to his distance to the center of rotation, is the centripetal force.

According to Newton's 2nd law, in the horizontal direction, we have:

F = Fc = m*a = m*ω²*r

We know that a = ac = 10*g = 98.0 m/s², and that r = 15.0 m.

Replacing these values in (1), and solving for ω, we get:

ω = √98.0m/s²/ 15.0 m = 2.55 rad/sec

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Noah is loading the ark and the last animal on board is a stubborn 1500-kg elephant who refuses to budge. Noah and his family pu

Oxana [17]

The coefficient of sliding friction is 0.514

Explanation:

We start by writing the equations of motion of the elephant along the two directions, parallel and perpendicular, to the incline.

Along the parallel direction we have:

$F- mg sin \theta - \mu_k R = ma$ (1)

where :

F = 10,000 N is the force applied by Noah

$mg sin \theta$ is the component of the weight parallel to the incline, where:

m is the mass

g = 9.8 m/s^2 the acceleration of gravity

$\theta=10^{\circ}$ is the angle of incline

$\mu_k R$ is the force of friction, where:

$\mu_k$ is the coefficient of friction

R is the normal reaction

and a is the acceleration

Perpendicular direction:

$R-mg cos \theta =0$ (2)

where $mg cos \theta$ is the component of the weight perpendicular to the incline

From (2) we find

$R=mg cos \theta$

And substituting into (1)

$F-mg sin \theta - \mu_k mg cos \theta = ma$

We know that the elephant moves at constant speed, so the acceleration is zero:

a = 0

So the equation becomes

$F-mg sin \theta - \mu_k mg cos \theta=0$

And we can re-arrange it to find the coefficient of friction:

$F-mg sin \theta - \mu_k mg cos \theta=0\\\mu_k = \frac{F-m g sin \theta}{mg cos \theta}=\frac{10000-(1500)(9.8)(sin 10)}{(1500)(9.8)(cos 10)}=0.514$

Learn more about friction and inclined planes:

brainly.com/question/5884009

#LearnwithBrainly

4 0

3 years ago

Francesca, who likes physics experiments, dangles her watch from a thin piece of string while the jetliner she is in takes off f

Tom [10]

Answer:

$v_f = 82.3 m/s$

Explanation:

When a suspended string makes and angle of 25 degree while aircraft is accelerating on runway then we can use force equations to find the acceleration of the aircraft

So we will have

$T sin\theta = ma$

$T cos\theta = mg$

now from above two equations we have

$a = g tan\theta$

$a = (9.81) tan25$

$a = 4.57 m/s^2$

now if it took 18 s to take off we can have final speed of the aircraft given by

$v_f - v_i = at$

$v_f - 0 = (4.57)(18)$

$v_f = 82.3 m/s$

6 0

3 years ago

Consider a pipe of length L that is open at both ends. What are the wavelengths of the three lowest-pitch tones produced by this

lina2011 [118]

Answer:

2 L, L, 2 L/3

Explanation:

For a pipe of length l open at both ends, the length of the pipe is given by :

$l=\dfrac{n\lambda}{2}$

or

$\lambda=\dfrac{2l}{n}$

For first pitch tone, n = 1

$\lambda_1=2l$

For first pitch tone, n = 2

$\lambda_2=\dfrac{2l}{2}$

$\lambda_2=l$

For first pitch tone, n = 3

$\lambda_3=\dfrac{2l}{3}$

$\lambda_3=\dfrac{2l}{3}$

So, the wavelengths of the three lowest-pitch tones produced by this pipe are 2 L, L, 2 L/3 respectively. Hence, the correct option is (d)

8 0

3 years ago

If the coefficient of static friction is 0.357, and the same ladder makes a 58.0° angle with respect to the horizontal, how far

zavuch27 [327]

Answer: d= 0.57* l

Explanation:

We need to check that before ladder slips the length of ladder the painter can climb.

So we need to satisfy the equilibrium conditions.

So for ∑Fx=0, ∑Fy=0 and ∑M=0

We have,

At the base of ladder, two components N₁ acting vertical and f₁ acting horizontal

At the top of ladder, N₂ acting horizontal

And Between somewhere we have the weight of painter acting downward equal to= mg

So, we have N₁=mg

and also mg*d*cosФ= N₂*l*sin∅

So,

d= $\frac{N2}{mg}*l$ * tan∅

Also, we have f₁=N₂

As f₁= чN₁

So f₁= 0.357 * 69.1 * 9.8

f₁= 241.75

Putting in d equation, we have

d= $\frac{241.75}{69.1*9.8} *l$ * tan 58

d= 0.57* l

So painter can be along the 57% of length before the ladder begins to slip

3 0

3 years ago

Read 2 more answers

Which is a diatomic molecule? <br> A. Ar <br> B. CO <br> C. CO2 <br> D. NaCl

snow_tiger [21]

Answer: B. CO

Explanation:

Diatomic molecules are those that are formed by two atoms of the same chemical element (homonuclear diatomic molecule) or different chemical element (heteronuclear diatomic molecule).

In this sense, oxygen is a homonuclear diatomic molecule because it is formed by two atoms of the same element ( $O_{2}$ ) and Carbon monoxide ( $CO$ ) is heteronuclear diatomic molecule.

Sodium Chloride $NaCl$ is not a diatomic molecule because is a product of ionization, but it can be diatomic in its gas phase with a polar covalent bond.

3 0

3 years ago