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stellarik [79]
3 years ago
14

An object is attached to a vertical spring and bobs up and down between points A and B. Where is the object located when its kin

etic energy is a minimum?
a. at either A or B
b. midway between A and B
c. one-third of the way between A and B
d. one-fourth of the way between A and B
Physics
1 answer:
gtnhenbr [62]3 years ago
4 0

Answer:

a. at either A or B

Explanation:

Kinetic energy may be defined as the energy of the system or an object which is due to its velocity of the object it possess.

In the context, an object having mass  is attached to spring which is vertical and the object moves up and down due to spring effect between points A and B. Now these points A and B are the extreme points after which the object bounces back.

At point A and B, the velocity of the object becomes zero and hence the kinetic energy of a body varies directly proportional to its velocity.

                  i.e. Kinetic energy $= \frac{1}{2} \text{mass} \times (\text{ velocity})^2$  

You might be interested in
What is 16.558 m/s rounded to three significant figures?
In-s [12.5K]

Answer:

Option C. 16.6 m/s

Explanation:

To round this 16.558 m/s to 3sf, we need to count the number beginning from 1. When we get to the 3rd number( ie 5), we'll examine the fourth number(i.e 5)to see if it less than five or greater. If it less than five, then we'll discard it. But if it five or greater, we'll approximate it and add it to the 3rd number.

So.

16.558 m/s = 16.6m/s to 3sf

3 0
4 years ago
A merry-go-round of radius R, shown in the figure, is rotating at constant angular speed. The friction in its bearings is so sma
mel-nik [20]

The angular speed of the merry-go-round reduced more as the sandbag is

placed further from the axis than increasing the mass of the sandbag.

The rank from largest to smallest angular speed is presented as follows;

[m = 10 kg, r = 0.25·R]

              {} ⇩

[m = 20 kg, r = 0.25·R]

              {} ⇩

[m = 10 kg, r = 0.5·R]

              {} ⇩

[m = 10 kg, r = 0.5·R] = [m = 40 kg, r = 0.25·R]

              {} ⇩

[m = 10 kg, r = 1.0·R]

Reasons:

The given combination in the question as obtained from a similar question online are;

<em>1: m = 20 kg, r = 0.25·R</em>

<em>2: m = 10 kg, r = 1.0·R</em>

<em>3: m = 10 kg, r = 0.25·R</em>

<em>4: m = 15 kg, r = 0.75·R</em>

<em>5: m = 10 kg, r = 0.5·R</em>

<em>6: m = 40 kg, r = 0.25·R</em>

According to the principle of conservation of angular momentum, we have;

I_i \cdot \omega _i = I_f \cdot \omega _f

The moment of inertia of the merry-go-round, I_m = 0.5·M·R²

Moment of inertia of the sandbag = m·r²

Therefore;

0.5·M·R²·\omega _i = (0.5·M·R² + m·r²)·\omega _f

Given that 0.5·M·R²·\omega _i is constant, as the value of  m·r² increases, the value of \omega _f decreases.

The values of m·r² for each combination are;

Combination 1: m = 20 kg, r = 0.25·R; m·r² = 1.25·R²

Combination 2: m = 10 kg, r = 1.0·R; m·r² = 10·R²

Combination 3: m = 10 kg, r = 0.25·R; m·r² = 0.625·R²

Combination 4: m = 15 kg, r = 0.75·R; m·r² = 8.4375·R²

Combination 5: m = 10 kg, r = 0.5·R; m·r² = 2.5·R²

Combination 6: m = 40 kg, r = 0.25·R; m·r² = 2.5·R²

Therefore, the rank from largest to smallest angular speed is as follows;

Combination 3 > Combination 1 > Combination 5 = Combination 6 >

Combination 2

Which gives;

[<u>m = 10 kg, r = 0.25·R</u>] > [<u>m = 20 kg, r = 0.25·R</u>] > [<u>m = 10 kg, r = 0.5·R</u>] > [<u>m = </u>

<u>10 kg, r = 0.5·R</u>] = [<u>m = 40 kg, r = 0.25·R</u>] > [<u>m = 10 kg, r = 1.0·R</u>].

Learn more here:

brainly.com/question/15188750

6 0
2 years ago
In asexual reproduction the new plant is identical to its parent.<br><br><br> true<br><br><br> false
Step2247 [10]

Answer:

YES

Explanation:

Asexual means the plant is identical to the parent

5 0
3 years ago
Read 2 more answers
A 34-kg child runs with a speed of 2.8 m/s tangential to the rim of a stationary merrygo-round. The merry-go-round has a momentu
tekilochka [14]

Answer: 0.43\ rad/s

Explanation:

Given

Mass of child m=34\ kg

speed of child is v=2.8\ m/s

Moment of inertia of merry go round is I=510\ kg.m^2

radius r=2.31\ m

Conserving the angular momentum

\Rightarrow mvr=I\omega \\\Rightarrow 34\times 2.8\times 2.31=510\times \omega\\\\\Rightarrow \omega=\dfrac{219.912}{510}\\\Rightarrow \omega=0.43\ rad/s

7 0
3 years ago
If a cable insulation is described as RHW, you would be able to use the cable for what maximum temperature?
BartSMP [9]
167•F which is c


Tell me my wrong from right
7 0
3 years ago
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