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AleksandrR [38]
3 years ago
15

1. Dixie has a bag of coins. The bag contains 10

Mathematics
1 answer:
d1i1m1o1n [39]3 years ago
6 0

Answer:

Its A 11/15

Step-by-step explanation:

There are 15 probabilities and 4 of them are pennies which you cant choose from that elimenates 4 possible outomes so subtract the 4 you get 11.

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anastassius [24]

Answer:

please make it a bit clear can see properly

Step-by-step explanation:

4 0
3 years ago
An object launched straight up at a speed of 29.4 meters per second has a height, h, in meters of h , t seconds after the object
Nostrana [21]

Answer:

h = 44.06 meters (maximum height)

the time the object takes to complete this whole path is 6 seconds, this is why the time at which the object reaches its maximum height will between 0 and 6 seconds

Step-by-step explanation:

To solve this question, we need to first recognize that this is a constant acceleration problem, specifically, it can be thought of as a projectile motion problem.

Recall, the equations of motion:

1) v^2 - v_0^2 = 2a(s - s_0)\\2) s = v_0^2  + \frac{1}{2} at^2\\3) v = v_0 + at

What do we already know?

  • v_0 = 29.4 ms^-1
  • The launch is straight up
  • a = -9.81 ms^-2 this is the gravitational acceleration g
  • s_0 = 0 m, since our reference point is at s = 0, (the ground)

We can use use the Eq(1):

we know that when any object is launched up, at maximum height its velocity is going to be zero, v = 0 ms^-2

v^2 - v_0^2 = 2a(s - s_0)\\0^2 - (29.4)^2 = 2(-9.81)(s- 0)\\s = 44.06 m

this is the maximum height!

Why does t have to between zero and six?

We can answer this using a bit visualization, if you think about the second equation

s = v_0 t - \frac{1}{2}at^2\\ s = 29.4t - 4.905t^2

this is the equation of the whole trajectory that object makes.

and if you solve this by making s = 0, you will get the times at which the object was at the ground. the times will be 0s and 5.99s.

so the amount of time the object takes to go through this whole path is 6 seconds and this why the object will only reach its maximum height in between this time interval.

hope this helps :)

5 0
3 years ago
Please help with this
xenn [34]

Answer:

Step-by-step explanation:

a

5 0
3 years ago
Read 2 more answers
Braving blizzard conditions on the planet Hoth, Luke Skywalker sets out at top speed in his snowspeeder for a rebel base 4800 mi
____ [38]

Answer:

v_top = 2400 mi/hr

v_w = 400 mi/h

Step-by-step explanation:

Given:

- Total distance D = 4800 mi

- Headwind journey time taken t_up= 3 hr

- Tailwind journey time taken t_down = 2 hr

Find:

Find the top speed of Luke's snow speeder and the speed of the wind.

Solution:

- The speed of Luke v_l is in stationary frame is given by:

                           v_l = v_w + v_l/w

Where,

           v_w: Wind speed

           v_l/w: Luke speed relative to wind.

- The top speed is attained on his returned journey with tail wind. We will use distance time relationship to calculate as follows:

                         v_top = D / t_down

                         v_top = 4800 / 2

                        v_top = v_down = 2400 mi/hr

- Similarly his speed on his journey up with head wind was v_up:

                         v_up = D / t_up

                         v_up = 4800 / 3

                        v_up = 1600 mi/hr    

- Now use the frame relations to find the wind speed v_w:

                         v_down = v_w + v_l/w

                         v_up = -v_w + v_l/w    

- Solve equations simultaneously:

                         2400 =  v_w + v_l/w

                         1600 =  -v_w + v_l/w

                         4000 = 2*v_l/w

                         v_l/w = 2000 mi/h

                         v_w = 400 mi/h

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