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3 years ago

5

How do Newton’s laws of motion explain why it is important to keep the ice smooth on a hockey rink so that players can pass a pu

ck as quickly as possible?

1 answer:

notka56 [123]3 years ago

6 0

Answer:

Newton's first law

Explanation:

Newton's first law states that if a body is at rest or moving at a constant speed in a straight line, it will remain at rest or keep moving in a straight line at constant speed unless it is acted upon by a force. Therefore, when the ice is smooth, friction gets lesser, and the force acted on that Puck will be decreased.

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How to prove formula for volume of a sphere ?

atroni [7]

Answer:

The volume of radius is $\frac{4}{3}$ × π × radius³ Proved

Explanation:

Given as :

We know that volume of sphere is v = $\frac{4}{3}$ × π × radius³

Or, v = $\frac{4}{3}$ × π × r³

Let prove the volume of sphere

So, From the figure of sphere

At the height of z , there is shaded disk with radius x

Let Find the area of triangle with side x , z , r

<u>From Pythagorean theorem</u>

x² + z² = r²

Or, x² = r² - z²

Or, x = $\sqrt{r^{2}-z^{2} }$

Now, Area of shaded disk = Area = π × x²

Where x is the radius of disk

Or, Area of shaded disk = π × ( $\sqrt{r^{2}-z^{2} }$ ) ²

∴ Area of shaded disk = π × (r² - z²)

Again

<u>If we calculate the area of all horizontal disk, we can get the volume of sphere</u>

So, we simply integrate the area of all disk from - r to + r

i.e volume = $\int_{-r}^{r} \Pi(r^{2}-z^{2} )dz$

Or, v = $\int_{-r}^{r} \Pi r^{2}dz$ - $\int_{-r}^{r} \Pi z^{2}dz$

Or, v = π r² (r + r) - π $\frac{r^{3} -(-r)^{3})}{3}$

Or, v = π r² (r + r) - π $\frac{2r^{3}}{3}$

Or, v = 2πr³ - π $\frac{2r^{3}}{3}$

Or, v = 2πr³ ( $\frac{3-1}{3}$ )

Or, v = 2πr³ × $\frac{2}{3}$

∴ v = $\frac{4}{3}$ × π × r³

Hence, The volume of radius is $\frac{4}{3}$ × π × radius³ Proved . Answer

5 0

3 years ago

A 20 watt light bulb is left burning inside a refrigerator operating on a reverse Carnot cycle. If the refrigerator also draws 2

Gwar [14]

Explanation:

Given that,

Power drawn by the refrigerator = 20 watt

Power of burning light inside = 20 watt

Room temperature = 27°C

We need to calculate the temperature

Using formula of coefficient of refrigerator

$COR_{R}=\dfrac{Q_{c}}{W}$

$\dfrac{Q_{c}}{W}=\dfrac{T_{c}}{T_{H}-T_{c}}$

Put the value into the formula

$\dfrac{20}{20}=\dfrac{T_{c}}{T_{H}-T_{c}}$

$\dfrac{T_{c}}{T_{H}-T_{c}}=1$

$T_{h}=2T_{c}$

$T_{c}=\dfrac{T_{h}}{2}$

$T_{c}=\dfrac{27+273}{2}$

$T_{c}=150\ K$

$T_{c}$ is less than $T_{h}$

Therefore, refrigerator cools its interior below room temperature while the bulb is on.

Hence, This is the required solution.

8 0

4 years ago

Two wooden boxes of equal mass but different density are held beneath the surface of a large container of water. Box A has a sma

Helga [31]

Answer:

<em>Which box has the greater acceleration?</em>

E. Box A

Explanation:

<em>The question is incomplete:</em>

<em>Which box has the greater acceleration?</em>

The bouyant force exerted by the water is equal in both boxes, because it depends on the volume displaced (that is the same for both boxes) and the density of the water.

But, the weight of each boxes is different, according to their density.

For the Box A the acceleration will be:

$m_aa_a=gV(\rho_w-\rho_a)\\\\\rho_aVa_a=gV(\rho_w-\rho_a)\\\\a_a=g\frac{(\rho_w-\rho_a)}{\rho_a}$

The same applies for the Box B:

$a_b=g\frac{(\rho_w-\rho_b)}{\rho_b}$

If we express the ratio of the accelerations, we have:

$a_a/a_b=\frac{(\rho_w-\rho_a)}{\rho_a}*\frac{\rho_b}{(\rho_w-\rho_b)}\\\\$

$a_a/a_b=\frac{(\rho_w-\rho_a)}{(\rho_w-\rho_b)} \frac{\rho_b}{\rho_a}$

We know that both densities are lower than water, because they accelerate upward to the surface when they are released (if they were more dense than water, they would sink more).

We will treat the densities as relative to water, so it becomes rho_w=1.

If we distribute the product, and know that the density of B is higher than the density of A, and both are higher than the product of the densities, we have:

$\rho_w=1\\\\\frac{a_a}{a_b}=\frac{(1-\rho_a)}{(1-\rho_b)} \frac{\rho_b}{\rho_a}=\frac{\rho_b-\rho_a\rho_b}{\rho_a-\rho_a\rho_b}\\\\\\\rho_b>\rho_a>\rho_a\rho_b>0\\\\\\\frac{a_a}{a_b}=\frac{\rho_b-\rho_a\rho_b}{\rho_a-\rho_a\rho_b}>1\\\\a_a>a_b$

The acceleration of A is higher than the acceleration of B.

7 0

4 years ago

. Boxes are sitting on a conveyor belt as the conveyor is turned on, moving the boxes toward the right. The belt reaches full sp

oksian1 [2.3K]

Answer:

The acceleration of the boxes is 1.5 ft/s²

The displacement of the boxes during the speed-up period is 0.1875 ft.

Explanation:

Hi there!

Let´s convert the 45 ft/min into ft/s:

45 ft/min · 1 min/ 60 s = 0.75 ft/s

It takes the belt 0.5 s to reach this speed. Then, the acceleration of the boxes will be:

a = v/t

Where:

a = acceleration.

v = velocity.

t = time.

a = 0.75 ft/s / 0.5 s

a = 1.5 ft/s²

The acceleration of the boxes is 1.5 ft/s²

The equation of displacement is the following:

x = x0 + v0 · t + 1/2 · a · t²

Where:

x = position of the boxes at time t.

x0 = initial position.

v0 = initial velocity.

t = time.

a = acceleration.

Since the origin of the frame of reference is located at the point where the boxes begin to move, x0 = 0. Since the boxes were initially at rest, v0 = 0. Then:

x = 1/2 · a · t²

x = 1/2 · 1.5 ft/s² · (0.5 s)²

x = 0. 1875 ft

The displacement of the boxes during the speed-up period is 0.1875 ft.

5 0

3 years ago

What is the equilibrium constant of pure water at 25C

Tcecarenko [31]

Answer:

1.0 x 10-14.

Explanation:

We then replace the term on the right side of this equation with a constant known as the water dissociation equilibrium constant, Kw. In pure water, at 25C, the [H3O+] and [OH-] ion concentrations are 1.0 x 10-7 M. The value of Kw at 25C is therefore

6 0

4 years ago