On an incline, the force causing the ball to move downwards would be gravity. Additionally, the component of gravity causing this ball to move downwards would be mgsintheta.

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7 0

3 years ago

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____ [38]

Answer:

Explanation:

D = 8.27 m ⇒ R = D / 2 = 8.27 m / 2 = 4.135 m

ω = 0.66 rev/sec = (0.66 rev/sec)*(2π rad/1 rev) = 4.1469 rad/s

We can apply the equation

Ff = W ⇒ μ*N = m*g (I)

then we have

N = Fc = m*ac = m*(ω²*R)

Returning to the equation I

μ*N = m*g ⇒ μ*m*ω²*R = m*g ⇒ μ = g / (ω²*R)

Finally

μ = (9.81 m/s²) / ((4.1469 rad/s)²*4.135 m) = 0.1379

3 0

3 years ago

ver shines light up to the surface of a flat glass-bottomed boat at an angle of 30 relative to the normal. If the index of refr

Free_Kalibri [48]

Answer:

$\beta = 41.68°$

Explanation:

according to snell's law

$\frac{n_w}{n_g} = \frac{sin\alpha}{sin30 }$

refractive index of water n_w is 1.33

refractive index of glass n_g is 1.5

$sin\alpha = \frac{n_w}{n_g}* sin30$

$sin\alpha = 0.443$

now applying snell's law between air and glass, so we have

$\frac{n_g}{n_a} = \frac{sin\alpha}{sin\beta}$

$sin\beta = \frac{n_g}{n_a} sin\alpha$

$\beta = sin^{-1} [\frac{n_g}{n_a}*sin\alpha]$

we know that $sin\alpha = 0.443$

$\beta = 41.68°$

7 0

3 years ago

A rigid tank internal energy of fluid 800kJ. Fluid loses 500kJ of heat and padle wheel does 100kJ of work. Find final internal e

Nesterboy [21]

Answer:

U₂ = 400 KJ

Explanation:

Given that

Initial energy of the tank ,U₁= 800 KJ

Heat loses by fluid ,Q= - 500 KJ

Work done on the fluid ,W= - 100 KJ

Sign -

1.Heat rejected by system - negative

2.Heat gain by system - Positive

3.Work done by system = Positive

4.Work done on the system-Negative

Lets take final internal energy =U₂

We know that

Q= U₂ - U₁ + W

-500 = U₂ - 800 - 100

U₂ = -500 +900 KJ

U₂ = 400 KJ

Therefore the final internal energy = 400 KJ

6 0

3 years ago

Define the strip attached to the magnet?​

Define the strip attached to the magnet?