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maw [93]
3 years ago
11

Deneb and the Sun are two stars. Deneb has a diameter that is 109-200 times greater than that of the sun. When seen from earth,

Physics
1 answer:
saw5 [17]3 years ago
7 0
I believe it is 
B)The Sun is closer than Deneb


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During spring tide, the sun, earth, and moon are in a straight line. This causes ...............
enyata [817]

Answer:

This causes higher average tidal ranges. The gravitational pull of the Sun and moon on Earth combined cause high tides that will be higher and low tides that will be lower than average.

4 0
3 years ago
Which of the following statements about colloids is FALSE?
ycow [4]
I believe that the answer to this would be B




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3 0
3 years ago
Select the correct answer. Rita is a registered dietician. What does her work entail? A. prescribing medication for clients B. c
yarga [219]

Answer:

D

Explanation:

The answer is D because:

A) Only doctors are allowed to prescribe medications.

B) Rita is not a chef/cook.

C) Rita is not a personal trainer

D) The job of a dietican is to provide reccomendations to their clients in order for them to implement a healthy lifestyle via consuming what is best for them.

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4 0
3 years ago
Water is not used as thermometric liquid.why?​
Dima020 [189]

Answer:

Water cannot be used in thermometer because of its higher freezing point and lower boiling point than other liquids .  If water is used in a thermometer , it will start phase change at 0degree\\C and 100degreeC and will not measure temperature , out of this range . This range is very small as compared  to other liquids as mercury , having freezing point about −39degreeC and boiling point 356degreeC.

Explanation:

8 0
3 years ago
A small object moves along the x-axis with acceleration ax(t) = −(0.0320m/s3)(15.0s−t). At t = 0 the object is at x = -14.0 m an
Ber [7]

Complete Question

A small object moves along the x-axis with acceleration ax(t) = −(0.0320m/s3)(15.0s−t)−(0.0320m/s3)(15.0s−t). At t = 0 the object is at x = -14.0 m and has velocity v0x = 7.10 m/s. What is the x-coordinate of the object when t = 10.0 s?

Answer:

The position of the object at t = 10s is  X  =  38.3 \  m

Explanation:

From the question we are told that

The acceleration along the x axis is  a_{x}t  =  -(0.0320\ m/s^3)(15.0 s- t)- (0.0320\ m/s^3)

  The position of the object at t = 0 is  x = -14.0 m

  The velocity at t = 0 s is  v_{0}x = 7.10 m/s

Generally from the equation for acceleration along x axis we have that

     a_x = \frac{dV_{x}}{dt}  = -0.032 (15- t)

=>   \int\limits  {dV_{x}} \, = \int\limits  {-0.032(15- t)} \, dt

=>   V_{x} = -0.032 [15t - \frac{t^2 }{2} ]+ K_1

At  t =0  s   and  v_{0}x = 7.10 m/s

=>   7.10  = -0.032 [15(0) - \frac{(0)^2 }{2} ]+ K_1

=>   K_1 = 7.10      

So  

      \frac{dX}{dt}  = -0.032 [15t - \frac{t^2 }{2} ]+ K_1

=>  \int\limits dX  = \int\limits [-0.032 [15t - \frac{t^2 }{2} ]+ K_1] }{dt}

=>  X  =  -0.032 [ 15\frac{t^2}{2}  - \frac{t^3 }{6} ]+ K_1t +K_2

At  t =0  s   and   x = -14.0 m

  -14  =  -0.032 [ 15\frac{0^2}{2}  - \frac{0^3 }{6} ]+ K_1(0) +K_2

=>   K_2 = -14

So

     X  =  -0.032 [ 15\frac{t^2}{2}  - \frac{t^3 }{6} ]+ 7.10 t -14

At  t = 10.0 s

      X  =  -0.032 [ 15\frac{10^2}{2}  - \frac{10^3 }{6} ]+ 7.10 (10) -14

=>   X  =  38.3 \  m

             

     

5 0
3 years ago
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