# Add me on i m v u at Chloe0989

## Answers

what

Step-by-step explanation:

i would but my school blocks it

Step-by-step explanation:

1- staff

2- a white key on the keyboard

3- last option (D)

4- treble and bass

5- it tells you where g is on the staff

6- abcdefg

7- I believe it’s legerdemain lines

8- bass

9- 2 staves of 5 lines

10- middle c

1. staff

2. a white key

3. tell musicians how to figure the names

4. treble and bass

5. it tells you where g is

6. a b c d e f g

7. leger lines

8. bass

9.5

not sure of 10

1, 4

Step-by-step explanation:

1. [tex]tan \theta = \frac{v^2}{Rg}[/tex]

For the first part, we just need to write the equation of the forces along two perpendicular directions.

We have actually only two forces acting on the car, if we want it to go around the track without friction:

- The weight of the car, mg, downward

- The normal reaction of the track on the car, N, which is perpendicular to the track itself (see free-body diagram attached)

By resolving the normal reaction along the horizontal and vertical direction, we find the following equations:

[tex]N cos \theta = mg[/tex] (1)

[tex]N sin \theta = m \frac{v^2}{R}[/tex] (2)

where in the second equation, the term [tex]m\frac{v^2}{R}[/tex] represents the centripetal force, with v being the speed of the car and R the radius of the track.

Dividing eq.(2) by eq.(1), we get the following expression:

[tex]tan \theta = \frac{v^2}{Rg}[/tex]

2. [tex]F=\frac{m}{R}(w^2-v^2)[/tex]

In this second situation, the cars moves around the track at a speed

[tex]wv[/tex]

This means that the centripetal force term

[tex]m\frac{v^2}{R}[/tex]

is now larger than before, and therefore, the horizontal component of the normal reaction, [tex]N sin \theta[/tex], is no longer enough to keep the car in circular motion.

This means, therefore, that an additional radial force F is required to keep the car round the track in circular motion, and therefore the equation becomes

[tex]N sin \theta + F = m\frac{w^2}{R}[/tex]

And re-arranging for F,

[tex]F=m\frac{w^2}{R}-N sin \theta[/tex] (3)

But from eq.(2) in the previous part we know that

[tex]N sin \theta = m \frac{v^2}{R}[/tex]

So, susbtituting into eq.(3),

[tex]F=m\frac{w^2}{R}-m\frac{v^2}{R}=\frac{m}{R}(w^2-v^2)[/tex]

UV = 14 ft and m∠TUV = 45°

ST = 20 ft, UV = 14 ft, and m∠UST = 98°

Step-by-step explanation:

step-by-step explanation:

amount of oatypop that contains the prize = 40%

amount of oatypop which doesn't contain the prize = (100-40)=60%

probability that hannah will win a prize

= s +f s +f f s+ff f s+ff f f s+ infinity.where f= failure,s= success

= or 60%

probability that hannah only has to buy 3 or less boxes before getting a prize= (s) and ( f s) and (ff s) and (f f f s)

= = 87.04 %

step-by-step explanation: