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3 years ago

When play stops in the game of hockey, to restart the action what is that called? (Hint: there are several points on the rink)

Physics

1 answer:

QveST [7]3 years ago

3 0

Answer:

Face - off

Explanation:

Face - off is a term used in ice hockey to start periods of play and also to restart the play/action after a previous stoppage.

This face - off involves two opposing players from the 2 opposing teams standing opposite each other at an approximate distance of one sticks blade with the game official dropping the puck between them. Thereafter, two players will then to try to gain possession of the puck.

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A ball is dropped from rest from the top of a building of height h. At the same instant, a second ball is projected vertically u

uranmaximum [27]

Answer:

a) $t = \sqrt{\frac{h}{2g}}$

b) Ball 1 has a greater speed than ball 2 when they are passing.

c) The height is the same for both balls = 3h/4.

Explanation:

a) We can find the time when the two balls meet by equating the distances as follows:

$y = y_{0_{1}} + v_{0_{1}}t - \frac{1}{2}gt^{2}$

Where:

$y_{0_{1}}$ : is the initial height = h

$v_{0_{1}}$ : is the initial speed of ball 1 = 0 (it is dropped from rest)

$y = h - \frac{1}{2}gt^{2}$ (1)

Now, for ball 2 we have:

$y = y_{0_{2}} + v_{0_{2}}t - \frac{1}{2}gt^{2}$

Where:

$y_{0_{2}}$ : is the initial height of ball 2 = 0

$y = v_{0_{2}}t - \frac{1}{2}gt^{2}$ (2)

By equating equation (1) and (2) we have:

$h - \frac{1}{2}gt^{2} = v_{0_{2}}t - \frac{1}{2}gt^{2}$

$t=\frac{h}{v_{0_{2}}}$

Where the initial velocity of the ball 2 is:

$v_{f_{2}}^{2} = v_{0_{2}}^{2} - 2gh$

Since $v_{f_{2}}^{2}$ = 0 at the maximum height (h):

$v_{0_{2}} = \sqrt{2gh}$

Hence, the time when they pass each other is:

$t = \frac{h}{\sqrt{2gh}} = \sqrt{\frac{h}{2g}}$

b) When they are passing the speed of each one is:

For ball 1:

$v_{f_{1}} = - gt = -g*\sqrt{\frac{h}{2g}} = - 0.71\sqrt{gh}$

The minus sign is because ball 1 is going down.

For ball 2:

$v_{f_{2}} = v_{0_{2}} - gt = \sqrt{2gh} - g*\sqrt{\frac{h}{2g}} = (\sqrt{1} - \frac{1}{\sqrt{2}})*\sqrt{gh} = 0.41\sqrt{gh}$

Therefore, taking the magnitude of ball 1 we can see that it has a greater speed than ball 2 when they are passing.

c) The height of the ball is:

For ball 1:

$y_{1} = h - \frac{1}{2}gt^{2} = h - \frac{1}{2}g(\sqrt{\frac{h}{2g}})^{2} = \frac{3}{4}h$

For ball 2:

$y_{2} = v_{0_{2}}t - \frac{1}{2}gt^{2} = \sqrt{2gh}*\sqrt{\frac{h}{2g}} - \frac{1}{2}g(\sqrt{\frac{h}{2g}})^{2} = \frac{3}{4}h$

Then, when they are passing the height is the same for both = 3h/4.

I hope it helps you!

7 0

3 years ago

Which subatomic particle identifies the element?

grin007 [14]

A. protons.ekgfsbdbsi

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3 years ago

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X-rays are used instead of visible light because X-rays have shorter ______________ than visible light, allowing them to produce

pashok25 [27]

The answer is wavelengths.

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3 years ago

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Two particles having charges of 0.440 nC and 11.0 nC are separated by a distance of 1.80 m . 1) At what point along the line con

Harlamova29_29 [7]

Answer:

a) 0.3 m

b) r = 0.45 m

Explanation:

given,

q₁ = 0.44 n C and q₂ = 11.0 n C

assume the distance be r from q₁ where the electric field is zero.

distance of point from q₂ be equal to 1.8 -r

now,

E₁ = E₂

$\dfrac{K q_2}{(1.8-r)^2} = \dfrac{K q_1}{r^2}$

$(\dfrac{1.8-r}{r})^2= \dfrac{q_2}{q_1}$

$\dfrac{1.8-r}{r}= \sqrt{\dfrac{11}{0.44}}$

1.8 = 6 r

r = 0.3 m

<h3>b) zero when one charge is negative.</h3>

let us assume q₁ be negative so, distance from q₁ be r

from charge q₂ the distance of the point be 1.8 +r

now,

E₁ = E₂

$\dfrac{K q_2}{(1.8+r)^2} = \dfrac{K q_1}{r^2}$

$(\dfrac{1.8+r}{r})^2= \dfrac{q_2}{q_1}$

$\dfrac{1.8+r}{r}= \sqrt{\dfrac{11}{0.44}}$

1.8 =4 r

r = 0.45 m

4 0

4 years ago

A 24-cm circumference loop of wire has a resistance of 0.14 Ω. The loop is placed between the poles of an electromagnet, and a f

Triss [41]

Answer:

The induced current in the loop is 1.2 A

Explanation:

Given;

length of the wire, L = 24 cm = 0.24 m

resistance of the wire, R = 0.14 Ω.

magnetic field strength, B = 0.55 T

time, t = 15 ms = 15 x 10⁻³ s

Circumference of a circle is given as;

L = 2πr

0.24 = 2πr

r = 0.24 / 2π

r = 0.0382 m

Area of a loop is given as;

A = πr²

A = π (0.0382)²

A = 0.004585 m²

Induced emf is given as;

$emf = -\frac{\delta \phi}{dt}$

Ф = ΔB x A

Ф = ( 0 - 0.55 T) x 0.004585 m²

Ф = -0.002522 T.m²

$emf = - (\frac{-0.002522}{15*10^{-3}} )\\\\emf = 0.168 \ V$

According to ohm's law;

V = IR

Where;

I is current

The induced current in the loop is calculated as;

I = V / R

I = 0.168 / 0.14

I = 1.2 A

Therefore, the induced current in the loop is 1.2 A

4 0

3 years ago