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a_sh-v [17]
3 years ago
12

you happen to visit the moon when some people on earth see a total solar eclipse. who has a better experience of this event, you

or the friends you left behind back on earth
Physics
1 answer:
Vinil7 [7]3 years ago
5 0
The friends left on earth because they can see the total eclipse, where as you are on the moon witnessing sections get dark rather than the whole picture
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A) A 12 kg object has a velocity of 37.5 m/s. What is its momentum?
Mkey [24]

Answer:

The answer would be 450 m kg/s

Explanation/ Explanation / Example:

Provided an object traveled 500 meters in 3 minutes , to calculate the average velocity you should take the following steps: Change minutes into seconds (so that the final result would be in meters per second). 3 minutes = 3 * 60 = 180 seconds , Divide the distance by time: velocity = 500 / 180 = 2.77 m/s .

If this doesn't help let me know!

4 0
3 years ago
a jet plane traveling 1890 km/h pulls out of a dive by moving in an arc of radius 5.20km. what is planes acceleration in g's
Andrew [12]

Answer:

53 m/s^2

Explanation:

Unit conversions:

1890 km/h = 1890 km/h * 1000m/km * 1/3600 h/s = 525 m/s

5.2 km = 5200 m

Assume that the jets is traveling in perfect circular motion, we can calculate the centripetal acceleration of the motion:

a = \frac{v^2}{r}

where v = 525m/s is the velocity of the jet and r = 5200 is the radius of the arc

a = \frac{525^2}{5200} = 53m/s^2

7 0
3 years ago
The concentration of an acid or base refers to how completely it dissociates in
Novosadov [1.4K]

Answer:

False

Explanation:

The Concentration of Acid or Base is the ph of the solution.

6 0
3 years ago
Two stones are launched from the top of a tall building. One stoneis thrown in a direction 30.0^\circ above the horizontal with
Butoxors [25]

Answer:

Part A)

t(1) > t(2), the stone thrown 30 above the horizontal spends more time in the air.

Part B)

x(f1) > x(f2), the first stone will land farther away from the building.

Explanation:

<u>Part A)</u>

Let's use the parabolic motion equation to solve it. Let's define the variables:

  • y(i) is the initial height, it is a constant.
  • y(f) is the final height, in our case is 0
  • v(i) is the initial velocity (v(i)=16 m/s)
  • θ1 is the first angle, 30°
  • θ2 is the first angle, -30°

For the first stone

y_{f1}=y_{i1}+v*sin(\theta_{1})t_{1}-0.5gt_{1}^{2}              

0=y_{i1}+16*sin(30)t_{1}-0.5*9.81*t_{1}^{2}

0=y_{i1}+8t_{1}-4.905*t_{1}^{2} (1)  

For the second stone  

0=y_{i2}+16*sin(-30)t_{2}-4.905t_{2}^{2}    

0=y_{i2}-8t_{2}-4.905t_{2}^{2} (2)            

 

If we solve the equation (1) we will have:

t_{1}=\frac{-8\pm \sqrt{64+19.62*y_{i}}}{-9.81}  

We can do the same procedure for the equation (2)

t_{1}=\frac{8\pm \sqrt{64+19.62*y_{i}}}{-9.81}

We can analyze each solution to see which one spends more time in the air.

It is easy to see that the value inside the square root of each equation is always greater than 8, assuming that the height of the building is > 0. Now, to get positive values of t(1) and t(2) we need to take the negative option of the square root.

Therefore, t(1) > t(2), it means that the stone thrown 30 above the horizontal spends more time in the air.

<u>Part B)</u>

We can use the equation of the horizontal position here.

<u>First stone</u>

x_{f1}=x_{i1}+vcos(30)t_{1}

x_{f1}=0+13.86*t_{1}

x_{f1}=13.86*t_{1}

<u>Second stone</u>

x_{2}=x_{i2}+vcos(-30)t_{2}

x_{1}=0+13.86*t_{1}

x_{1}=13.86*t_{2}

Knowing that t(1) > t(2) then x(f1) > x(f2)

Therefore, the first stone will land farther away from the building.

They land at different points at different times.

I hope it helps you!

3 0
3 years ago
If you wanted to know the location of a vehicle that ran out of gas after taking a zigzag route through the city, which quantity
Papessa [141]

Explanation:

the vehicles displacement, since displacement deals with position

8 0
3 years ago
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