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nirvana33 [79]
4 years ago
11

A block of mass m = 1.5 kg is released from rest at a height of H = 0.81 m on a curved frictionless ramp. At the foot of the ram

p is a spring whose spring constant is k = 250.0 N/m. What is the maximum compression of the spring, x?
Physics
1 answer:
kogti [31]4 years ago
7 0

Answer:

0.31 m

Explanation:

m = mass of the block = 1.5 kg

H = height from which the block is released on ramp = 0.81 m

k = spring constant of the spring = 250 N/m

x = maximum compression of the spring

using conservation of energy

Spring potential energy gained by spring = Potential energy lost by block

(0.5) k x² = mgH

(0.5) (250) x² = (1.5) (9.8) (0.81)

x = 0.31 m

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A submarine is 2.99 ✕ 102 m horizontally from shore and 1.00 ✕ 102 m beneath the surface of the water. A laser beam is sent from
Vsevolod [243]

Answer:

a The diagram of the situation is shown on the first uploaded image

b the angle of  incidence  beam striking the water is \theta = 49.63^o

c  the angle of  refraction  beam striking the water is r = 59.7^o

d The angle the refracted beam make with respect to the horizontal is = 30.3^o

e The height of the target above sea level is  

                   h= 125.05m

Explanation:

From the diagram we see that the angle of the beam striking the water is

                   tan \theta = \frac{100}{85}

                        \theta = tan^{-1}(\frac{100}{85} )

                           = 49.63^o

According to Snell's law

                   \mu_{water} *sin(i) = \mu_{air}  *sin(r)

Where \mu_{water } is the refractive index of water =  1.333

           i is the angle of incidence

          \mu_{air} is the refractive index of air  = 1

            r is the angle of refraction

 Substituting values accordingly

          1.33 * sin (40.37) = 1 * sin(r)

    Making r the subject of the formula

                       r = sin^{-1}(\frac{1.333 *sin(40.37)}{1})

                          = 59.7^o

looking at the diagram we can see that to  obtain the angle the refraction beam makes with the horizontal   by subtracting the angle refraction from 90°

                 i.e  90 -59.7 = 30.3 °

From the diagram we see that the height  target above sea level can be obtained by this relation

                   tan \theta = \frac{h}{214}\\

Where h is the is the height

                   tan(30.3) = \frac{h}{214}

                         h = 214 * tan (30.3)

                            =125.05m

           

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Answer:

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What type of volcano forms when there is a series of violent and gentle eruptions resulting in layering of lava and ash?
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3 years ago
What is the average speed for the entire graph?
Maru [420]

Answer:

Explanation:

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7 0
3 years ago
The aircraft wing from problem 6 experiences temperature extremes that span 210 degrees Celsius. The component for the wing will
ExtremeBDS [4]

Answer:

α = 2,857 10⁻⁵ ºC⁻¹

Explanation:

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         ΔL = L_f - L₀

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3 years ago
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