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olga55 [171]
2 years ago
13

A particle on a spring moves in simple harmonic motion along the x axis between turning points at x1 = 95 cm and x2 = 135 cm. (i

) At which of the following positions does the particle have maximum speed? 95 cm 105 cm 115 cm at none of those positions (ii) At which position does it have maximum acceleration? 95 cm 105 cm 115 cm at none of those positions (iii) At which position is the greatest net force exerted on the particle? 95 cm 105 cm 115 cm at none of those positions
Physics
1 answer:
uranmaximum [27]2 years ago
8 0

Answer:

(i) x = 115\,cm, (ii) x = 95\,cm, (iii) x = 95\,cm

Explanation:

(i) x_{1} and x_{2} represent the points where particle has a velocity of zero and spring reach maximum deformation, Given the absence of non-conservative force and by the Principle of Energy Conservation, the position where particle is at maximum speed is average of both extreme positions:

x = 115\,cm

(ii) Maximum accelerations is reached at x_{1} and x_{2}.

x = 95\,cm

(iii) Greatest net forces exerted on the particle are reached at  x_{1} and x_{2}.

x = 95\,cm

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telo118 [61]

Answer:

Answer: Given m = 10 kg and . F = 20 N. Thus, the force required to accelerate the object upward direction is 20 N.

Explanation:

Answer: Given m = 10 kg and . F = 20 N. Thus, the force required to accelerate the object upward direction is 20 N.

6 0
2 years ago
Explain why, under some circumstances, it is not advisable to weld a structure that is fabricated with a 3003 aluminum alloy. Hi
Scorpion4ik [409]

The 3003 aluminum alloy is made up of 1.25% Magnesium and 0.1% Copper. This combination is designed to increase the strength of the material over other types of alloys such as those of the 1000 series. This alloy provides a medium strength and can be educated by cold work.

The alloy is not heat treatable and generally has good formability, corrosion resistance and weldability.

However, being a material that hardens by cold work, welding a 3003 Aluminum structure will cause the body to undergo recrystallization which will generate a loss in the 'resistance' of the material and the force capable of withstanding. If this aluminum will be used for structural purposes, it should not be welded. It would be better to perform the structure with a 6061 aluminum, which has similar characteristics and is not so affected by welding.

7 0
2 years ago
Imagine you can take all the atoms in a single drop of water and put them on a single line as closely packed as they can be. How
tester [92]

Answer:

option a

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Size of an atom (diameter) = 10⁻¹⁰ m

There are approximately 10²² atoms in a single drop of water. If they are put in  a straight line, the length would be

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l = 10²²×  10⁻¹⁰ m = 10¹² m

Distance between the Sun and the Earth is 1.47 × 10¹¹ m. The calculated length is greater than the distance between the Sun and the Earth.

Thus, option a is correct.

7 0
3 years ago
The temperature of the cooling water as it leaves the hot engine of an automobile is 240 °F. After it passes through the radiato
djverab [1.8K]

Answer:571.09 kJ

Explanation:

Given

Temperature of cooling water from engine exit=240^{\circ} F\approx 115.55^{\circ}C

After Passing through the radiator its temperature decreases to 175^{\circ}F\approx 79.44^{\circ}F

specific heat of water=4.184 J/g^{\circ}C

Volume of water = 1 gallon\approx 3.78 L

density of water \rho =1 gm/mL

Thus mass of water=\rho \times V=3.78\times 1=3.78 kg

Heat transferred to the surrounding is equal to heat absorbed by cooling water

Q=m\cdot c\cdot \Delta T

Q=3.78\times 4.184\times 1000\times (115.55-79.44)

Q=3.78\times 4.184\times 1000\times (36.11)

Q=571.09 kJ

4 0
3 years ago
A beam of microwaves with λ = 0.9 mm is incident upon a 9 cm slit. At a distance of 1.5 m from the slit, what is the approximate
liq [111]

Answer:

3 cm

Explanation:

According to the question,

D=1.5 m.

d=9 cm.

\lambda =0.9 mm.

Now the approximate slit's image width is equal to width of central maxima.

And width of central maxima is twice the width from center to first maxima

So,

y=2\frac{\lambda (D)}{d}.

Substitute all the variable in above equation.

y=\frac{(2)0.9\times 10^{-2} m(1.5 m) }{0.09 m}.

y=3 cm.

5 0
3 years ago
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