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3 years ago

Two passengers are on a moving sidewalk at the airport. Passenger A is

Physics

1 answer:

Oksanka [162]3 years ago

8 0

Answer:

V = - 2[m/s]

Explanation:

This is a classic problem of relative velocities, in order to solve it we must understand each of the velocities, given in the initial data of the problem.

VA = velocity of passenger A, = 1.5[km/h], it is the same velocity of the sidewalk

VB = 2 + 1.5 = 3.5 [km/h], that is the velocity observed by a person outside from the sidewalk.

And from the perspective of passenger B the speed of passanger A is:

VA-VB = - 2 [m/s]

It means that the passenger B is seen how passenger A is getting close to him and then passed.

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A short term affect of anorexia might include wight loss

Serggg [28]

Is there a question....

3 0

3 years ago

A disk rotates at constant angular acceleration, from angular position θ1 = 16.0 rad to angular position θ2 = 76.0 rad in 5.30 s

Oliga [24]

Answer:

(a) the angular velocity at θ1 is 11.64 rad/s

(b) the angular acceleration is 0.12 rad/ $s^{2}$

(c) the angular position was the disk initially at rest is - 428.27 rad

Explanation:

Given information :

θ1 = 16 rad

θ2 = 76 rad

ω2 = 11 rad/s

t = 5.3 s

(a) The angular velocity at θ1

First, we use the angular motion equation for constant acceleration

Δθ = (ω1+ω2)t/2

θ2 - θ1 = (ω1+ω2)t/2

ω1 + ω2 = 2 (θ2 - θ1) / t

ω1 = (2 (θ2 - θ1) / t ) - ω2

= (2 (76-16) / 5.3) - 11

= 11.64 rad/s

(b) the angular acceleration

ω2 = ω1 + α t

α t = ω2 - ω1

α = (ω2 - ω1)/t

= (11.64 - 11) / 5.3

= 0.12 rad/ $s^{2}$

(c) the angular position was the disk initially at rest, θ0

at rest ω0 = 0

ω2^2 = ω01 t + 2 α Δθ

2 α Δθ = ω2^2

θ2 - θ0 = ω2^2 / 2 α

θ0 = θ2 - (ω2^2) / 2 α

= 76 - ( $11^{2}$ / 2 x 0.12

= 76 - 504.16

= - 428.27 rad

4 0

3 years ago

If the cold temperature reservoir of a Carnot engine is held at a constant 306 K, what temperature should the hot reservoir be k

Paraphin [41]

Efficiency η of a Carnot engine is defined to be:
η = 1 - Tc / Th = (Th - Tc) / Th 
where 
Tc is the absolute temperature of the cold reservoir, and 
Th is the absolute temperature of the hot reservoir. 

In this case, given is η=22% and Th - Tc = 75K 
Notice that although temperature difference is given in °C it has same numerical value in Kelvins because magnitude of the degree Celsius is exactly equal to that of the Kelvin (the difference between two scales is only in their starting points). 

Th = (Th - Tc) / η 
Th = 75 / 0.22 = 341 K (rounded to closest number) 
Tc = Th - 75 = 266 K 

Lower temperature is Tc = 266 K 
Higher temperature is Th = 341 K

6 0

3 years ago

6. A football is kicked with an initial speed of 10.2 m/s at an angle of 40.00 above the horizontal. It lands on the ground 2.12

igor_vitrenko [27]

Answer:

7.81 m/s

Explanation:

Given,

initial speed, u = 10.2 m/s

angle of inclination, θ = 40°

time, t = 2.12 s

Horizontal component of the velocity: