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Vsevolod [243]
3 years ago
8

You attach a 1.10 kg block to a horizontal spring that is fixed at one end. You pull the block until the spring is stretched by

0.200 m and release it from rest. Assume the block slides on a horizontal surface with negligible friction. The block reaches a speed of zero again 0.100 s after release (for the first time after release). What is the maximum speed of the block (in m/s)?
Physics
1 answer:
dangina [55]3 years ago
6 0

Answer:

6.3 m/s

Explanation:

m = mass of the block = 1.10 kg

k = spring constant of the spring

x = stretch in the spring = 0.2 m

t = time taken by block to come to zero speed first time = 0.100 s

T = Time period of oscillation

Time period of oscillation is given as

T = 2t

T = 2 (0.1)

T = 0.2 s

Time period is also given as

T = 2\pi \sqrt{\frac{m}{k}}

0.2 = 2(3.14) \sqrt{\frac{1.10}{k}}

k = 1084.6 N/m

v = maximum speed of the block

using conservation of energy

Maximum kinetic energy = Maximum spring potential energy

(0.5) m v² = (0.5) k x²

m v² = k x²

(1.10) v² = (1084.6) (0.2)²

v = 6.3 m/s

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Answer: true

Explanation:

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3 years ago
the resistance of a wire of length 80cm and of uniform area of cross-section 0.025cmsq., is found to be 1.50 ohm. Calculate spec
vivado [14]
Specific\ resistance\ =resistivity\\
From\ formula\ on \ resistance:\ R= \frac{pL}{A}\ p-resistivity,\ L-length,\\ A-area\ of\ cross\ section\\
p= \frac{R*A}{L}= \frac{1,5Ohm*0,025*10^{-4}m^2 }{80* 10^{-2}m }=0,0003515625 Ohm*m
3 0
3 years ago
Consider two points in an electric field. The potential at point 1, V1, is 33 V. The potential at point 2, V2, is 175 V. An elec
Mnenie [13.5K]

Answer:

ΔU  = e(V₂ - V₁) and its value ΔU = -2.275 × 10⁻²¹ J

Explanation:

Since the electric potential at point 1 is V₁ = 33 V and the electric potential at point 2 is V₂ = 175 V, when the electron is accelerated from point 1 to point 2, there is a change in electric potential ΔV which is given by ΔV = V₂ - V₁.

Substituting the values of the variables into the equation, we have

ΔV = V₂ - V₁.

ΔV = 175 V - 33 V.

ΔV = 142 V

The change in electric potential energy ΔU = eΔV = e(V₂ - V₁) where e = electron charge = -1.602 × 10⁻¹⁹ C and ΔV = electric potential change from point 1 to point 2 = 142 V.

So, substituting the values of the variables into the equation, we have

ΔU = eΔV

ΔU = eΔV

ΔU = -1.602 × 10⁻¹⁹ C × 142 V

ΔU = -227.484 × 10⁻¹⁹ J

ΔU = -2.27484 × 10⁻²¹ J

ΔU ≅ -2.275 × 10⁻²¹ J

So, the required equation for the electric potential energy change is

ΔU  = e(V₂ - V₁) and its value ΔU = -2.275 × 10⁻²¹ J

5 0
3 years ago
A coffee filter of mass 1.5 grams dropped from a height of 3 m reaches the ground with a speed of 0.7 m/s. How much kinetic ener
Mademuasel [1]

The kinetic energy gained by the air molecules is 0.0437 J                

<h3 />

Given:

Mass of a coffee filter, m = 1.5 g

Height from which it is dropped, h = 3 m

Speed at ground, v = 0.7 m/s

Initially, the coffee filter has potential energy. It is given by :

P =mgh

P = 1.5 × 10⁻³ kg × 9.8 m/s² × 3m

P = 0.0441 J

Finally, it will have kinetic energy. It is given by :

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E= \frac{1}{2}×1.4 × 10⁻³ × (0.7)²

E = 0.000343 J

The  kinetic energy Kair did the air molecules gain from the falling coffee filter is :

E = 0.000343 -  0.0441

  = 0.0437 J

So, the kinetic energy Kair did the air molecules gain from the falling coffee filter is 0.0437 J

Learn more about kinetic energy here:

brainly.com/question/8101588

#SPJ4

8 0
2 years ago
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Novosadov [1.4K]

Answer:

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where f is the focal length of the cornea-crystalline system, p and q are the distances to the object and the image.

In this case, the distance to the image on the retina is constant, about 3 cm. Therefore depending on the distance to the object) p = the focal length must change

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let's apply this expression to our case

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correct answer is C

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3 years ago
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