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lyudmila [28]
3 years ago
11

The tangential acceleration of a cart moving at a constant speed in a horizontal circle is:

Physics
1 answer:
qaws [65]3 years ago
7 0

Answer:

a=0                                  ∵ \alpha=0\ rad.s^{-1}

Explanation:

The tangential acceleration of a cart moving at a constant speed in a circle is:

The angular velocity is constant when the circular speed is constant.

We know that the (instantaneous) tangential velocity of such object is given by:

v=r.\omega

Now for angular acceleration we have a constant angular speed:

\alpha=0\ rad.s^{-1}

And angular acceleration is related to tangential acceleration as:

a=r.\alpha

\Rightarrow a=0

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Answer:

11.98 N

Explanation:

Normal force =   mg =  2.03 * 9.81

coeff of static friction must be overcome for the book to begin moving

       .602 = F / (2.03 * 9.81)   = 11.98  N

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An electron is accelerated from rest through a potential difference. After acceleration the electron has a de Broglie wavelength
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Answer:

3x10⁴v

Explanation:

Using

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880nm = 6.6E-34/√ 2.9.1E-31 x me

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4 0
4 years ago
a ball is thrown inclined in to the air with a initial velocity of u. if it reaches the maximum height in 6 seconds , find the r
olga55 [171]

Answer:

11:1

Explanation:

At constant acceleration, an object's position is:

y = y₀ + v₀ t + ½ at²

Given y₀ = 0, v₀ = u, and a = -g:

y = u t − ½g t²

After 6 seconds, the ball reaches the maximum height (v = 0).

v = at + v₀

0 = (-g)(6) + u

u = 6g

Substituting:

y = 6g t − ½g t²

The displacement between t=0 and t=1 is:

Δy = [ 6g (1) − ½g (1)² ] − [ 6g (0) − ½g (0)² ]

Δy = 6g − ½g

Δy = 5½g

The displacement between t=6 and t=7 is:

Δy = [ 6g (7) − ½g (7)² ] − [ 6g (6) − ½g (6)² ]

Δy = (42g − 24½g) − (36g − 18g)

Δy = 17½g − 18g

Δy = -½g

So the ratio of the distances traveled is:

(5½g) / (½g)

11 / 1

The ratio is 11:1.

8 0
3 years ago
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