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Irina-Kira [14]
3 years ago
5

A box is sitting stationary on a ramp that is 42° to the horizontal. The box has a gravitational force of 112.1 N. What is the m

inimum amount of force that can be applied to the box to get the box to move?
Physics
2 answers:
ale4655 [162]3 years ago
5 0

Answer:

Answer is 80

Explanation:

I just took the tets

nordsb [41]3 years ago
4 0

Force of gravity along the inclined plane

F_x = F_g sin\theta

here we have

F_g = 112.1 N

\theta = 42 degree

now we have

F_x = (112.1 N)sin42

F_x = 75 N

so here we require 75 N force along the incline to move the box upwards

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A ball bearing of radius of 1.5 mm made of iron of density
Serjik [45]

Answer:

\boxed{\sf Viscosity \ of \ glycerine \ (\eta) = 14.382 \ poise}

Given:

Radius of ball bearing (r) = 1.5 mm = 0.15 cm

Density of iron (ρ) = 7.85 g/cm³

Density of glycerine (σ) = 1.25 g/cm³

Terminal velocity (v) = 2.25 cm/s

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To Find:

Viscosity of glycerine (\sf \eta)

Explanation:

\boxed{ \bold{v =  \frac{2}{9}  \frac{( {r}^{2} ( \rho -  \sigma)g)}{ \eta} }}

\sf \implies \eta =  \frac{2}{9}  \frac{( {r}^{2}( \rho -  \sigma)g )}{v}

Substituting values of r, ρ, σ, v & g in the equation:

\sf \implies \eta =  \frac{2}{9}  \frac{( {(0.15)}^{2}  \times  (7.85 - 1.25) \times 980.6)}{2.25}

\sf \implies \eta =  \frac{2}{9}  \frac{(0.0225 \times 6.6 \times 980.6)}{2.25}

\sf \implies \eta =  \frac{2}{9}  \times  \frac{145.6191}{2.25}

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6 0
3 years ago
You tie a cord to a pail of water and swing the pail in a vertical circle of radius 0.710 mm . What minumum speed must the pail
Blababa [14]

Answer:

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