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Montano1993 [528]
2 years ago
7

How much force is needed to stretch a spring 1.2m if the spring constant is 8.5 N/m?

Physics
1 answer:
Assoli18 [71]2 years ago
6 0

Explanation:

spring constant k::8.5

and displacement x::1.2

f:? f: k×x

8.5×1.2:10.2N

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During the experiment it is determined that, as the cart rolls between two points on the track, the work done on the cart by the
natita [175]

Answer:

Explanation:

If the work done on the cart is NET work

Then the work will result in an increase in kinetic energy

KE₀ + W = KE₁

½mv₀² + W = ½mv₁²

½(0.80)(0.61²) + 0.91 = ½(0.80)v₁²

v₁ = 1.626991...

v₁ = 1.6 m/s

4 0
2 years ago
A. How long does it take light to travel through a 3.0-mm-thick piece of window glass?
hodyreva [135]

Answer:

a) 1.517\times10^{-11} s

b) 3.41 mm

Explanation:

a)

We take the speed of light, c = 3.0\times10^8 m/s and the refractive index of glass as 1.517.

Speed = distance/time

Time = distance/speed

Refractive index, n = speed of light in vacuum / speed of light in medium

n=\dfrac{c}{s}

s=\dfrac{c}{n}

t=\dfrac{d}{c/n}

t=\dfrac{dn}{c}

t=\dfrac{3\times10^{-3}\times1.517}{3.0\times10^8}

t=1.517\times10^{-11}

b)

We take the refractive index of water as 1.333.

Speed in water = speed in vacuum / refractive index of water

Distance = speed * time

d=s\times t

d=\dfrac{c}{n_w}\times \dfrac{3\times10^{-3}\times1.517}{c}

d=\dfrac{3\times10^{-3}\times 1.517}{1.333}

d = 3.41 mm

6 0
3 years ago
You use 8x binoculars were used on a warbler (14cm long) in a tree 18cm away. What angle (in degrees) does the image of the warb
mafiozo [28]

Answer:

The angle it subtend on the retina is  \theta_z = 0.44586^o    

Explanation:

From the question we are told that

     The length of the warbler is  L = 14cm = \frac{14}{100} = 0.14m

      The distance from the binoculars is    d = 18cm = \frac{18}{100} = 0.18m

        The magnification of the binoculars is  M =8

Without the 8 X binoculars the  angle made with the angular size of the object  is mathematically represented as

          \theta = \frac{L}{d}

        \theta  = \frac{0.14}{0.18}

           = 0.007778 rad

Now magnification can be represented mathematically as

         M = \frac{\theta _z}{\theta}

Where \theta_z is the angle the image of the warbler subtend on your retina when the   binoculars i.e the  binoculars zoom.

So

      \theta_z = M * \theta

=>    \theta_z =8 * 0.007778

            = 0.0622222224

Generally the conversion to degrees can be mathematically evaluated as

             \theta_z = 0.062222224 * (\frac{360 }{2 \pi rad} )

              \theta_z = 0.44586^o  

7 0
3 years ago
A wave is incident on the surface of a mirror at an angle of 41° with the normal. what can you say about its angle of reflection
Umnica [9.8K]

It's angle of reflection must be 41 degrees

we know, by the first law of reflection that angle of incidence is always equal to angle of reflection..........

7 0
3 years ago
Read 2 more answers
6/10
g100num [7]

Answer:pounds

Explanation:

7 0
2 years ago
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