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Deffense [45]
3 years ago
9

A ball on a frictionless plane is swung around in a circle at constant speed. The acceleration points in the same direction as t

he velocity vector.
a. True
b. False
Physics
2 answers:
emmainna [20.7K]3 years ago
8 0

Answer:

False

Explanation:

You have a circle so think back to circular motion. Theres 2 directions, centripetal and tangential. The problem tells you there's a constant tangential speed so tangential acceleration is 0. However there is a centripetal acceleration acting on the ball that holds it in its circular motion (i.e. tension, or gravity). Since centripetal is perpendicular to the tangential direction, acceleration and velocity are in different directions.

Kitty [74]3 years ago
4 0
I believe it is False, only because the plane is Frictionless. Hope this helps, good luck.
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The dragster has a mass of 1.3 Mg and a center of mass at G. A parachute is attached at C provides a horizontal braking force of
adell [148]

Answer:

The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is  16.33 m/s²

Explanation:

The additional information to the question is embedded in the diagram attached below:

The height between the dragster and ground is considered to be 0.35 m since is not given ; thus in addition win 0.75 m between the dragster and the parachute; we have: (0.75 + 0.35) m = 1.1 m

Balancing the equilibrium about point A;

F(1.1) - mg (1.25) = ma_a (0.35)

1.8v^2(1.1) - 1200(9.8)(1.25) = 1200a(0.35)

1.8v^2(1.1) - 14700 = 420 a   ------- equation (1)

F_x = ma_x \\ \\ = 1.8v^2 = 1200 \ a             --------- equation (2)

Replacing equation 2 into equation 1 ; we have :

{1.1 * 1200 \ a} - 14700 = 420 a

1320 a - 14700 = 420 a

1320 a -  420 a =14700

900 a = 14700

a = 14700/900

a = 16.33 m/s²

The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is  16.33 m/s²

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The mathematical and proportional relationship between mL and cm ^ 3 said us that 1cm ^ 3 is equivalent to 1mL.

If the density is considered as the amount of mass per unit volume we will have to

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here,

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Replacing we have that

\rho = \frac{40g}{23cm^3}

\rho = 1.739g/cm^3

As 1mL = 1cm^3 we have that the density in g/mL is,

\rho = 1.739g/mL

6 0
3 years ago
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