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Elena-2011 [213]
4 years ago
14

A crate slides down a ramp at a constant velocity. the ramp has an incline of 28 degrees above the horizontal the crate weighs 2

00 Newton’s
Physics
1 answer:
Sophie [7]4 years ago
5 0

First is the coefficient of static friction between the map

and the crate.

And the second one is at what angle does the crate begin to

slide if the mass is doubled.

1.    To look for the coefficient of static friction:

ΣF = 0

μ x m x g x cos θ = m * g * sin θ

μ = sin θ/ cos θ = tan θ

μ = tan 24 = 0.45

2.   At what angle… the solution is:

The static friction or μ does not depend

on mass, so the angle is still 24.

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Answer:

B. Nueral Receptors

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4 years ago
A 1.00 -kg object slides to the right on a surface having a coefficient of kinetic friction 0.250 (Fig. P8.62a). The object has
LekaFEV [45]

The distance D where the object comes to rest is 1.08.m.

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(c) the distance D where the object comes to rest.

W_{total} =ΔKE ⇒ -0.25*1*9.8*D = 0-1/2*1*2.3^{2}

⇒D=\frac{0.5*2.3^{2} }{2.45}

⇒1.08.m

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5 0
2 years ago
Which of the following is a definition of glass transition temperature?
aleksandr82 [10.1K]

Answer:

c. The temperature at which a glass transforms from a solid to liquid.

Explanation:

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4 years ago
Calculate the mass (in SI units) of (a) a 160 lb human being; (b) a 1.9 lb cockatoo. Calculate the weight (in English units) of
kondaur [170]

Explanation:

1. Force applied on an object is given by :

F = W = mg

(a) A 160 lb human being, F = 160 lb

g = acceleration due to gravity, g = 32 ft/s²

m=\dfrac{F}{g}

m=\dfrac{160\ lb}{32\ ft/s^2}

m = 5 kg

(b) A 1.9 lb cockatoo, F = 1.9 lb

m=\dfrac{F}{g}

m=\dfrac{1.9\ lb}{32\ ft/s^2}

m = 0.059 kg

2. (a) A 2300 kg rhinoceros, m = 2300 kg

W=2300\ kg\times 32\ ft/s^2=73600\ lb

(b) A 22 g song sparrow, m = 22 g = 0.022 kg

W=0.022\ kg\times 32\ ft/s^2=0.704\ lb

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3 years ago
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