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Nat2105 [25]
3 years ago
9

The 25-lb slender rod has a length of 6 ft. using a collar of negligible mass, its end a is confined to move along the smooth ci

rcular bar of radius 3√2 ft. end b rests on the floor, for which the coefficient of kinetic friction is b = 0.4. if the bar is released from rest when  = 30, determine the angular acceleration of the bar at this instant.
Physics
1 answer:
sp2606 [1]3 years ago
3 0
Which of the following is NOT a factor in efficiency?<span><span>A.metabolism</span><span>B.type of movement</span><span>C.muscle efficiency</span><span>D.digestion</span></span>
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If 710- nm and 660- nm light passes through two slits 0.65 mm apart, how far apart are the second-order fringes for these two wa
alexira [117]

0.23 mm far apart are the second-order fringes for these two wavelengths on a screen 1.5 m away.

<h3>Given wavelengths 710nm and 660nm,0.65mm apart two slits, and a screen 1.5m away.</h3>

Position of n the order fringe = n λ D / d

for n = 2

position = 2 λ D / d

λ = 710 nm , D = 1.5m

d = .65 x 10⁻³

position 1 = 2 x 710 x 10⁻⁹ x 1.5 / .65 x 10⁻³

= 3276.92 x 10⁻⁶ m

= 3.276x 10⁻³ m

= 3.276mm .

For λ = 660 nm

position = 2 λ D / d

λ = 660 nm , D = 1.5 m

d = .65 x 10⁻³

position 2 = 2 x 660 x 10⁻⁹ x 1.5 / .65 x 10⁻³

= 3046.15 x 10⁻⁶ m

= 3.046 x 10⁻³ m

= 3.046 mm .

Difference between their position

= 3.276mm ₋ 3.046 mm

= 0.23 mm .

To know more about Fringes refer to:  brainly.com/question/15649748

#SPJ4

7 0
1 year ago
A man has 887.5 J of kinetic energy while running with a velocity of 5 m/s. What is his mass?
monitta

Answer:

The mass of the man is 71 kg

Explanation:

Given;

kinetic energy of the man, K.E = 887.5 J

velocity of the man, v = 5 m/s

The mass of the man is calculated as follows;

K.E = ¹/₂mv²

where;

m is the mass of the man

2K.E = mv²

m = 2K.E / v²

m = (2 x 887.5) / (5)²

m = 71 kg

Therefore, the mass of the man is 71 kg

7 0
3 years ago
Calculate the force that the 4kg block exerts on the 10kg block
Kryger [21]
Acceleration=force/mass=28/(10+4)=2m/s^2

force10kg=ma=10*2
force4kg=ma=(10*2)=20
the4 kg is pushing against the 10kg block

vf=vi+at
-10=20*28/14 * t
t=30/2=15sec

i hope this can help you.
8 0
3 years ago
Use the information from the graph to answer the question.
galina1969 [7]

The displacement of the object as determined from the velocity-time graph is 562.5 m.

<h3>What is a velocity-time graph?</h3>

A velocity-time graph is a graph of the velocity of an object plotted in the vertical or y-axis of the graph against the time taken on the horizontal or x-axis.

The displacement of an object can be obtained from its velocity-time graph by calculating the total area under the graph.

The total area under the graph = area of triangle + area of rectangle

Area of triangle = b*h/2 =

Area of triangle = 25 * (35 - 10)/2 = 312.5 m

Area of rectangle = l * b

Area of rectangle = 10 * 25 = 250 m

Total area = (312.5 + 250) m

Total area = 562.5 m

Therefore, the displacement of the object is 562.5 m

In conclusion, the total area of a velocity-time graph gives the displacement.

Learn more about velocity-time graph at: brainly.com/question/28064297

#SPJ1

5 0
1 year ago
the jet plane travels along the vertical parabolic path. when it is at point a it has speed of 200 m/s, which is increasing at t
givi [52]

Explanation:

Here is the complete question i guess. The jet plane travels along the vertical parabolic path defined by y = 0.4x². when it is at point A it has speed of 200 m/s, which is increasing at the rate .8 m/s^2. Determine the magnitude of acceleration of the plane when it is at point A.

→ The tangential component of acceleration is rate of increase in the speed of plane so,

a_{t} = v = 0.8 m/s^{2}

→ Now we have to find out the radius of curvature at point A which is 5 Km (from the figure).

dy/dx = d(0.4x²)/dx

         = 0.8x

Take the derivative again,

d²y/dx² = d(0.8x)/dx

          = 0.8

at x= 5 Km

dy/dx = 0.8(5)

         = 4

p = \frac{[1+ (\frac{dy}{dx})^{2}]^{\frac{3}{2} }   }{\frac{d^{2y} }{dx^{2} } }

now insert the values,

p = \frac{[1+(4)^{2}]^{\frac{3}{2} }  }{0.8}  = 87.62 km

→ Now the normal component of acceleration is given by

a_{n} = \frac{v^{2} }{p}

    = (200)²/(87.6×10³)

aₙ = 0.457 m/s²

→ Now the total acceleration is,

a = [(a_{t})^{2} +(a_{n} )^{2} ]^{0.5}

a = [(0.8)^{2} + (0.457)^{2}]^{0.5}

a = 0.921 m/s²

4 0
3 years ago
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