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3 years ago

When is the average velocity of an object equal to the instantaneous velocity?

Physics

1 answer:

atroni [7]3 years ago

4 0

Answer:

Average velocity of an object is equal to the instantaneous velocity when it's acceleration is zero.

Explanation:

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Set the initial bead height to 3.00 m. Click Play. Notice that the ball makes an entire loop. What is the minimum height require

strojnjashka [21]

Answer:

h> 2R

Explanation:

For this exercise let's use the conservation of energy relations

starting point. Before releasing the ball

Em₀ = U = m g h

Final point. In the highest part of the loop

Em_f = K + U = ½ m v² + ½ I w² + m g (2R)

where R is the radius of the curl, we are considering the ball as a point body.

I = m R²

v = w R

we substitute

Em_f = ½ m v² + ½ m R² (v/R) ² + 2 m g R

em_f = m v² + 2 m g R

Energy is conserved

Emo = Em_f

mgh = m v² + 2m g R

h = v² / g + 2R

The lowest velocity that the ball can have at the top of the loop is v> 0

h> 2R

3 0

2 years ago

Middle adulthood marks the time period in which some adults face the fabled midlife crisis. What exactly is that? Do you think y

MariettaO [177]

<span>The middle life crisis occurs for men at age of 39 and for women at age of 44. At this age, people start realizing that they are getting older and experience some physical changes which seem distressful for them.</span>

<span> It affects the sense of accomplishment and progress in career. Due to a midlife crisis, people start engaging in activities to cope with the reality such as wearing clothes that are fashionable, dating someone etc. Yes, I think I will also go through one because it happens with everyone in this age. It can be avoided by accepting the fact that these changes are reality and everyone have to face it one day.</span>

6 0

3 years ago

Without effective assessment methods, teachers have no documented proof of childrens specific ____ and _____.

schepotkina [342]

Answer:

D strengths and weakneses

Explanation:

8 0

3 years ago

А A pool of water of refractive index

babymother [125]

Answer:

Apparent depth = 45 cm

Explanation:

The refractive index of water in a pool, n = 4/3

Real depth, d = 60 cm

We need to find its apparent depth when viewed vertically through air. The ratio of real depth to the apparent depth is equal to the refractive index of the material. Let the apparent depth is d'. So,

$n=\dfrac{d}{d'}\\\\d'=\dfrac{d}{n}\\\\d'=\dfrac{60}{\dfrac{4}{3}}\\\\d'=45\ cm$

So, the apparent depth is 45 cm.

3 0

2 years ago

If a hot steel tool of 1200°C was put in a bucket to cool and the bucket contained 15L of water of 15°C, and the water temperatu

Mashcka [7]

3.6 kg.

<h3>Explanation</h3>

How much heat does the hot steel tool release?

This value is the same as the amount of heat that the 15 liters of water has absorbed.

Temperature change of water:

$\Delta T = T_2 - T_1= 48\; \textdegree{\text{C}}- 15\; \textdegree{\text{C}} = 33 \; \textdegree{\text{C}}$ .

Volume of water:

$V = 15 \; \text{L} = 15 \; \text{dm}^{3} = 15 \times 10^{3} \; \text{cm}^{3}$ .

Mass of water:

$m = \rho \cdot V = 1.00 \; \text{g} \cdot \text{cm}^{-3} \times 15 \times 10^{3} \; \text{cm}^{3} = 15 \times 10^{3} \; \text{g}$ .

Amount of heat that the 15 L water absorbed:

$Q = c\cdot m \cdot \Delta T = 4.18 \; \text{J} \cdot \text{g}^{-1} \cdot \textdegree{\text{C}}^{-1} \times 15 \times 10^{3} \; \text{g} \times 33 \; \textdegree{\text{C}} = 2.06910 \times 10^{6}\; \text{J}$ .

What's the mass of the hot steel tool?

The specific heat of carbon steel is $0.49 \; \text{J} \cdot \text{g}^{-1} \cdot \textdegree{\text{C}}^{-1}$ .

The amount of heat that the tool has lost is the same as the amount of heat the 15 L of water absorbed. In other words,

$Q(\text{absorbed}) = Q(\text{released}) =2.06910 \times 10^{6}\; \text{J}$ .

$\Delta T = T_2 - T_1 = 1200\; \textdegree{\text{C}} -{\bf 48}\; \textdegree{\text{C}} = 1152\; \textdegree{\text{C}}$ .

$m = \dfrac{Q}{c\cdot \Delta T} = \dfrac{2.06910 \times 10^{6} \; \text{J}}{0.49\; \text{J} \cdot \text{g}^{-1} \cdot \textdegree{\text{C}}^{-1} \times 1152\; \textdegree{\text{C}}} = 3.6 \times 10^{3} \; \text{g} = 3.6 \; \text{kg}$ .

4 0

3 years ago