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Ber [7]
3 years ago
10

Convert 1.5 radian into milliradians.

Physics
1 answer:
Alex3 years ago
4 0

Answer:

1500 milliradians

Explanation:

Data provided in the question:

1.5 radians

Now,

1 radians consists of  1000 milliradians

1 milli = 1000

thus for the 1.5 radians, we have

1.5 radians = 1.5 multiplied by 1000 milliradians

or

1.5 radians = 1500 milliradians

Hence, after the conversion

1.5 radians equals to the value 1500 milliradians

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Two converging lenses are placed 30 cm apart. The focal length of the lens on the right is 20 cm while the focal length of the l
Masja [62]

Answer:

a)   I2 = 3 (o-10) / (o- 30) , b)   h ’/h=  3 (o-10) / o (o-30)

Explanation:

The builder's equation is

          1 / f = 1 / o + 1 / i

Where f is the focal length, or e i are the distance to the object and image, respectively

As the separation between the lenses is greater than the focal distances, we must work them individually and separately. Let's start with the leftmost lens with focal length f = 15 cm

Let's calculate the position of the image of this lens

         1 / i1 = 1 / f - 1 / o

         1 / i1 = 1/15 - 1 / o

         i1 = o 15 / (o-15)

Let's calculate the distance to the image of the second lens, for this the image of the first is the distance to the object of the second

        o2 = d-i1

We write the builder equation

       1 / f2 = 1 / o2 + 1 / i2

       1 / i2 = 1 / f2 -1 / o2

       1 / i2 = 1 / f2 - 1 / (d-i1)

       1 / i2 = 1/20 - 1 / (d-i1)            (1)

Let's evaluate the last term

      d-i1 = d - 15 o / (o-15)

      d-i1 = (d (o-15) - 15 o) / (o-15)

      d- i1 = (30 or -30 15 -15 o) / (o-15)

      d-i1 = (15 or - 450) / (o- 15)

      d-i1 = = (15 or -450) / (o-15)

replace in 1

      1 / i2 = 1/20 - (or - 15) / (15 or -450)

      1 / i2 = [(15 o-450) - (o-15) 20] / (15 or -150)

      1 / i2 = (15 or - 450 - 20 or + 300) / (15 or - 150)

      1 / i2 = (-5 or -150) / (15 or -150)

      1 / i2 = (or -30) / (3 or - 30)

      I2 = 3 (o-10) / (o- 30)

Part B

The height of the image, we use the magnification equation

     m = h ’/ h = - i / o

     h ’= - h i / o

In our case

     h ’= h i2 / o

     h ’= h 3 (o-10) / o (o-30)

If they give the distance to the object it is easier

5 0
3 years ago
A mass weighting 16 lbs stretches a spring 3 inches. The mass is in a medium that exerts a viscous resistance of 20 lbs when the
const2013 [10]

Answer:

The equation for the object's displacement is u(t)=0.583cos11.35t

Explanation:

Given:

m = 16 lb

δ = 3 in

The stiffness is:

k=\frac{m}{\delta } =\frac{16}{3} =5.33lb/in

The angular speed is:

w=\sqrt{\frac{k}{m} } =\sqrt{\frac{5.33*386.4}{16} } =11.35rad/s

The damping force is:

F_{D} =cu

Where

FD = 20 lb

u = 4 ft/s = 48 in/s

Replacing:

c=\frac{F_{D} }{u} =\frac{20}{48} =0.42lbs/in

The critical damping is equal:

c_{c} =\frac{2k}{w} =\frac{2*5.33}{11.35} =0.94lbs/in

Like cc>c the system is undamped

The equilibrium expression is:

u(t)=u(o)coswt+u'(o)sinwt\\u(o)=7=0.583\\u'(o)=0\\u(t)=0.583coswt\\u(t)=0.583cos11.35t

3 0
3 years ago
The pictures below show a bottle made of plastic and a block made of concrete.
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5 0
3 years ago
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A 10 m long uniform beam weighing 100 N is supported by two ropes at the ends. If a 400 N person sits at 2.0 m from one end of t
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Answer:

T1 = 130N, T2 = 370N

Explanation:

In order for the system to be at rest, the sum of all forces must be zero and the torque around a point on the beam must be zero.

1. forces:

Let tension in rope 1 be T1 and in rope 2 be T2:

ma = T1 + T2 - 100N - 400N = 0

(1) T1 + T2 = 500N

2. torque around the center point of the beam:

τ = r x F = 5*T1 + 3*400N - 5*T2 = 0

(2) T1 - T2 = -240N

Solving both equations:

T1 = 130N

T2 = 370N

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3 years ago
A boxed 12.0 kg computer monitor is dragged by friction 5.50 m up along the moving surface of a conveyor belt inclined at an ang
satela [25.4K]

Nah a a yuloure

Explanation:

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