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NASA scientists suggest using rotating cylindrical spacecraft to replicate gravity while in a weightless environment. Consider such a spacecraft that has a diameter of d = 148 m. What is the speed v, in meters per second, the spacecraft must rotate at its outer edge to replicate the force of gravity on earth?

Answers

Answer 1

Answer:

The speed of the spacecraft at its outer edge is 26.93 m/s.

The given parameters;

diameter of the spacecraft, d = 148 m
radius of the spacecraft, r = 74 m

The speed of the spacecraft at its outer edge is calculated as follows;

$F_g = F_c\n\nmg = (mv^2)/(r) \n\nv^2 = rg\n\nv = √(rg) \n\nv = √((74)(9.8)) \n\nv = 26.93 \ m/s$

Thus, the speed of the spacecraft at its outer edge is 26.93 m/s.

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Answer 2

Answer:

Answer:

Explanation:

Given

diameter of spacecraft $d=148\ m$

radius $r=74\ m$

Force of gravity $F_g$ =mg

where m =mass of object

g=acceleration due to gravity on earth

Suppose v is the speed at which spacecraft is rotating so a net centripetal acceleration is acting on spacecraft which is given by

$F_c=(mv^2)/(r)$

$F_c=F_g$

$(mv^2)/(r)=mg$

$(v^2)/(r)=g$

$v=√(gr)$

$v=√(1450.4)$

$v=38.08\ m/s$

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The radius of a typical human eardrum is about 4.15 mm. Calculate the energy per second received by an eardrum when it listens to sound that is at the threshold of hearing, assumed to be 1.20E-12 W/m2

Answers

The energy per second received by an eardrum is $6.4884 * 10^(-17) watt$

Calculation of the energy per second;

The area should be

$= \pi r^2\n\n= 3.14 * 0.00415m\n\n= 5.407 * 10^(-5)m^2$

Now

The power should be

$= 1.2 * 10^(-12) * 5.407 * 10^(-5)\n\n= 6.4884 * 10^(-17) watt$

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Answer:

Power energy per second will be equal to $6.4884* 10^(-17)watt$

Explanation:

We have given radius of human eardrum r = 4.15 mm = 0.00415 m

Intensity at threshold of hearing $I=1.2* 10^(-12)w/m^2$

Area is given by $A=\pi r^2=3.14* 0.00415^2=5.407* 10^(-5)m^2$

We know that power is given by $P=I* A=1.2* 10^(-12)* 5.407* 10^(-5)=6.4884* 10^(-17)watt$

So power energy per second will be equal to $6.4884* 10^(-17)watt$

A golf pro swings a golf club, striking a golf ball that has a mass of 55.0 g. The club is in contact with the ball for only 0.00340 s. After the collision, the ball leaves the club at a speed of 46.0 m/s. What is the magnitude of the average force (in N) exerted on the ball by the club?

Answers

To solve this problem it is necessary to apply the concepts related to Newton's second law and the equations of motion description for acceleration.

From the perspective of acceleration we have to describe it as

$a = (\Delta v)/(\Delta t)$

Where,

$\Delta v$ = Velocity

$\Delta t$ = time

At the same time by the Newton's second law we have that

F = ma

Where,

m = mass

a = Acceleration

Replacing the value of acceleration we have

$F = m ((\Delta v)/(\Delta t))$

Our values are given as,

$m = 55*10^(-3)Kg$

$v = 46m/s$

$t = 0.00340s$

Replacing we have,

$F = m ((\Delta v)/(\Delta t))$

$F = (55*10^(-3))((46)/(0.00340))$

$F = 744.11N$

Therefore the magnitude of the average force exerted on the ball by the club is 744.11N

While the block hovers in place, is the density of the block (top left) or the density of the liquid (bottom center) greater?

Answers

Answer:

for the body to float, the density of the body must be less than or equal to the density of the liquid.

Explanation:

For a block to float in a liquid, the thrust of the liquid must be greater than or equal to the weight of the block.

Weight is

W = mg

let's use the concept of density

ρ_body = m / V

m = ρ_body V

W = ρ_body V g

The thrust of the body is given by Archimedes' law

B = ρ_liquid g V_liquid

as the body floats the submerged volume of the liquid is less than or equal to the volume of the block

ρ_body V g = ρ_liquid g V_liquid

ρ_body = ρ liquid Vliquido / V_body

As we can see, for the body to float, the density of the body must be less than or equal to the density of the liquid.

A forest fire sends carbon monoxide and ash into the air. This is an example of the__1__ affecting the__2___ .1.
A. Hydrosphere
B Lithosphere
C. Atmosphere
2
A Lithosphere
B Atmosphere
C Biosphere

Answers

1. B Lithosphere
2. B Atmosphere

lithosphere and biosphere

srry ik im super late but to help other people theres the answer

An object is projected with speed of 4ms at an angle of 60° to horizontal. Calculate the time of flight of the object. (g=10ms2)

Answers

Given :

∅ = 60⁰

u = 4 m/s

g = 10m/s²

to find :

T = ?

Solution :

as per formula,

$t = (2u \: sin \theta)/(g)$

now put the value : $t \: = (2 * 4 * sin \: 60)/(10)$

as we know $sin60 \: = ( √(3) )/(2)$

therefore,

$t \: = (8 * ( √(3) )/(2) )/(10)$

as we solve this we get,

$t \: = ( 2√(3) )/(5)$

that's t = 0.69 sec

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0.8 seconds

Explanation:

time of flight = 2u/g

u=4m/s

g=10

= 8/10

= 0.8 sec

just a trial...not sure!!!

A rock is pulled back in a slingshot as shown in the diagram below. The elastic on the slingshot is displaced 0.2 meters from its initial position. The rock is pulled back with a force of 10 newtons.When the rock is released, what is its kinetic energy?

Answers

The rock takes 8.16s to return to its release point. Given that the elastic band provides a speed of 40m/s to the rock in 10 cm stretch.

What will be the speed of the rock?

Initial speed of the rock, u = 40m/s

Final position of the rock s = 0m taking the release point as reference. The rock takes 8.16s to return to its release point. Given that the elastic band provides a speed of 40m/s to the rock in 10 cm stretch.

Nuclear energy is a useful source of power but has disadvantages. The disadvantage of nuclear energy is it produces dangerous waste.

Initial speed of the rock, u = 40m/s

Final position of the rock s = 0m taking the release point as reference

From the second equation of motion:

solving above we get:

t = 0s or t = 8.16s, t =0 seconds is neglected since it represents the initial position which is the same as the final position at t = 8.16s

So, the rock takes 8.16 seconds to return to the release point.

Therefore, The rock takes 8.16s to return to its release point. Given that the elastic band provides a speed of 40m/s to the rock in 10 cm stretch.

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Answer:

Explanation:

i don't know

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