All categories

3 years ago

How long does a player sit out of a game of handball for committing a second or third foul?

Physics

1 answer:

yulyashka [42]3 years ago

5 0

Answer:

Walking’ - If a handball player takes more than three steps without dribbling (bouncing the ball) or holds the ball for more than 3 seconds without bouncing it, shooting or passing, then that is deemed ‘walking' and possession is lost.

'Double dribble’ - Handball players cannot receive the ball and bounce it, then hold the ball, and bounce it again. This is termed ‘double dribble’ and is against the rules.

Askmeanything2♡

You might be interested in

How much force is applied if a 130kg mass is accelerated at 5 m/s^2?

STALIN [3.7K]

Answer:

650N

Explanation:

f= ma

130kg x 5 m/s^2 = 650N

5 0

3 years ago

A well-insulated bucket of negligible heat capacity contains 129 g of ice at 0°C.

Luba_88 [7]

Answer:

The final equilibrium temperature of the system is $T = 12.48^oC$

For the ice it would melt completely the mass that would remain is Zero

Explanation:

In the following question we are provided with

Mass of the ice $M_{i}$ = 129 g = 0.129 kg

Mass of the steam $M_s$ = 19 g = 0.019 kg

Initial temperature is $T_i$ = 0°C

Temperature of steam $T_s$ = 100°C

Following the change of state of water in the question

The energy required by ice to change to water is mathematically given as

$Q_A = M_iL_f$

Where $L_f$ is a constant known as heat of fusion and the value is $334*10^3 J/kg$

$Q_A = 0.129 *334 *10^3 = 43086 J$

The energy been released when the steam changes to water is mathematically given as

$Q_B = M_s * L_v$

Where $L_v$ is a constant known as heat of vaporization and the value is $2256*10^3J/kg$

$Q_B = 0.019 * 2256*10^3 = 42864J$

The energy released when the temperature of water decrease from 100°C to 0°C is

$Q_C = M_s *C_water ($ 100°C)

Where $C_{water}$ is the specific heat of water which has a value $4186J/kg \cdot K$

$Q_C = 0.019 *4186*100 = 7953.4$

Looking at the values we obtained we noticed that ]

$Q_B + Q_C > Q_A$

What this means is that the ice will melt

bearing in mind the conservation of energy

looking the way at which water at different temperature were mixed according to the question

Heat lossed by the vapor = heat gained by ice

$Q_B + M_s *C_{water}(100-T) = Q_A + M_i C_{water} T$

$T = \frac{Q_B+M_s *C_{water}(100^oC)-Q_A}{(M_s *C_{water})+(M_i*C_{water})}$

$T = \frac{42864+7953.4-43086}{(0.019+0.129)(4186)}$

$T = 12.48^oC$

3 0

3 years ago

3. A sprinter leaves the starting blocks with an acceleration of 4.5 m/s2. What is the

UkoKoshka [18]

Hi there! :)

$\large\boxed{v_{f} = 18 m/s}$

Use the following kinematic equation to solve for the final velocity:

$v_{f} = v_{i} + at$

In this instance, the runner started from rest, so the initial velocity is 0 m/s. We can rewrite the equation as:

$v_{f} = at$

Plug in the given acceleration and time:

$v_{f} = 4.5 * 4 = 18 m/s$

5 0

3 years ago

A dinner plate falls vertically to the floor and breaks up into three pieces, which slide horizontally along the floor. Immediat

Olegator [25]

Answer:

Explanation:

There will be conservation of momentum along horizontal plane because no force acts along horizontal plane.

momentum of first piece = .320 kg x 2 m/s

= 0.64 kg m/s along x -axis.

momentum of second piece = .355 kg x 1.5 m/s

= 0.5325 kg m/s along y- axis .

Let the velocity of third piece be v and it is making angle of θ with x -axis .

Horizontal component of its velocity = .100 kg x v cosθ = .1 v cosθ

vertical component of its velocity = .100 kg x v sinθ = .1 v sinθ

For making total momentum in the plane zero

.1 v cosθ = 0.64 kg m/s

.1 v sinθ = 0.5325 kg m/s

Dividing

Tanθ = .5325 / .64 = .83

θ = 40⁰.

The angle will be actually 180 + 40 = 220 ⁰ from positive x -axis.

6 0

2 years ago

Read 2 more answers

A comparison of a machine’s work output with its work input is called __??__.

Aloiza [94]

Ohhhhh its called a input machine

7 0

3 years ago