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pashok25 [27]
2 years ago
12

A factory robot drops a 10 kg computer onto a conveyor belt running at 3.1 m/s. The materials are such that μs = 0.50 and μk = 0

.30 between the belt and the computer. How far is the computer dragged before it is riding smoothly on the belt?
Physics
1 answer:
frutty [35]2 years ago
8 0

Answer:

x = 1.63 m

Explanation:

mass (m) = 10 kg

μk = 0.3

velocity (v) = 3.1 m/s

Assuming that most of the computers weight is applied on the belt instantaneously, we can apply the constant acceleration equation below

x = v^{2}/2a

where a = μk.g , therefore

x = v^{2}/2μk.g

x = (3.1 x 3.1)/(2 x 0.3 x 9.8)

x = 1.63 m

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Which particles in an atom are acted upon by the strong force? A. Electrons and neutrons B. Neutrons and protons O C. Protons an
BARSIC [14]

\large \sf \pmb{B) \: Neutrons \:  and  \: Protons}

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2 years ago
Which has the most kinetic energy from its motion?
victus00 [196]

Answer:

option c

Explanation:

K.E=1/2mv(squared)

K.E=1/2x900x80x80

K.E=450x6400=2880000J

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6 0
3 years ago
A projectile is launched from ground level at an angle of 50.0 degrees above the horizontal with a speed of 30.0 m/s. If the pro
Allushta [10]

Answer:

4.69 s

Explanation:

u = Initial velocity = 30 m/s

\theta = Angle of launch = 50^{\circ}

g = Acceleration due to gravity = 9.81\ \text{m/s}^2

Time taken by the projectile to land is given by

t=\dfrac{2u\sin\theta}{g}\\\Rightarrow t=\dfrac{2\times 30\times \sin50^{\circ}}{9.81}\\\Rightarrow t=4.69\ \text{s}

The time the projectile is in the air before it contacts the ground again is 4.69 s.

6 0
3 years ago
A light beam strikes a piece of glass at a 65.00 ∘ incident angle. The beam contains two wavelengths, 450.0 nm and 700.0 nm, for
Leokris [45]

Answer:

Explanation:

Lower the refractive index ,  higher the wave length

Refractive index becoming high reduces the velocity and hence wavelength as frequency remains unchanged.

So refractive index  1.4831 will correspond to wavelength of 450 nm.

For  refractive index is 1.4831 , angle of incidence i , angle of refraction r .

Sin i / Sinr = 1.4831

sin65 / sinr = 1.4831

sir r = sin65 / 1.4831

= .9063 / 1.4831

= .6111

r = 37.67 degree

For  refractive index is 1.4754 , angle of incidence i

Sin i / Sinr = 1.4754

sin65 / sinr = 1.4754

sir r = sin65 / 1.4754

= .9063 / 1.4754

= .61427

r = 37.9 degree

angle between two refracted ray

37.9 -37.67

= 0. 23 degree

6 0
3 years ago
Read 2 more answers
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