Ask question

Ask question

All categories

liberstina [14]

4 years ago

7

Sportsmanship means following the rules of the game and respecting the judgment of referees and officials. What is another examp

le of sportsmanship?

2 answers:

Ludmilka [50]4 years ago

6 0

Another example of sportsmanship is playing equally with other opponents and treated appropriately.

Alika [10]4 years ago

4 0

Answer:

Following are the examples of sportsmanship:

Respecting opponent and teammates,
Helping opponent and teammates if they fall or get injured.
Shake hands with opponent before start and at the end of game.

Explanation:

A true sportsman always respect his teammates and opponents for their good performance. He take a sports like a sports not a war and a true sportsman does not take his opponent as rival but as a a sportsman. If opponent falls in the playground a true sportsman does not hesitate to help him.

Following are the examples of sportsmanship:

Respecting opponent and teammates,
Helping opponent and teammates if they fall or get injured.
Shake hands with opponent before start and at the end of game.

You might be interested in

A luggage handler pulls a suitcase of mass 19.6 kg up a ramp inclined at an angle 24.0 ∘ above the horizontal by a force F⃗ of m

Dvinal [7]

(a) 638.4 J

The work done by a force is given by

$W=Fd cos \theta$

where

F is the magnitude of the force

d is the displacement of the object

$\theta$ is the angle between the direction of the force and the displacement

Here we want to calculate the work done by the force F, of magnitude

F = 152 N

The displacement of the suitcase is

d = 4.20 m along the ramp

And the force is parallel to the displacement, so $\theta=0^{\circ}$ . Therefore, the work done by this force is

$W_F=(152)(4.2)(cos 0)=638.4 J$

b) -328.2 J

The magnitude of the gravitational force is

W = mg

where

m = 19.6 kg is the mass of the suitcase

$g=9.8 m/s^2$ is the acceleration of gravity

Substituting,

$W=(19.6)(9.8)=192.1 N$

Again, the displacement is

d = 4.20 m

The gravitational force acts vertically downward, so the angle between the displacement and the force is

$\theta= 90^{\circ} - \alpha = 90+24=114^{\circ}$

Where $\alpha = 24^{\circ}$ is the angle between the incline and the horizontal.

Therefore, the work done by gravity is

$W_g=(192.1)(4.20)(cos 114^{\circ})=-328.2 J$

c) 0

The magnitude of the normal force is equal to the component of the weight perpendicular to the ramp, therefore:

$R=mg cos \alpha$

And substituting

m = 19.6 kg

g = 9.8 m/s^2

$\alpha=24^{\circ}$

We find

$R=(19.6)(9.8)(cos 24)=175.5 N$

Now: the angle between the direction of the normal force and the displacement of the suitcase is 90 degrees:

$\theta=90^{\circ}$

Therefore, the work done by the normal force is

$W_R=R d cos \theta =(175.4)(4.20)(cos 90)=0$

d) -194.5 J

The magnitude of the force of friction is

$F_f = \mu R$

where

$\mu = 0.264$ is the coefficient of kinetic friction

R = 175.5 N is the normal force

Substituting,

$F_f = (0.264)(175.5)=46.3 N$

The displacement is still

d = 4.20 m

And the friction force points down along the slope, so the angle between the friction and the displacement is

$\theta=180^{\circ}$

Therefore, the work done by friction is

$W_f = F_f d cos \theta =(46.3)(4.20)(cos 180)=-194.5 J$

e) 115.7 J

The total work done on the suitcase is simply equal to the sum of the work done by each force,therefore:

$W=W_F + W_g + W_R +W_f = 638.4 +(-328.2)+0+(-194.5)=115.7 J$

f) 3.3 m/s

First of all, we have to find the work done by each force on the suitcase while it has travelled a distance of

d = 3.80 m

Using the same procedure as in part a-d, we find:

$W_F=(152)(3.80)(cos 0)=577.6 J$

$W_g=(192.1)(3.80)(cos 114^{\circ})=-296.9 J$

$W_R=(175.4)(3.80)(cos 90)=0$

$W_f =(46.3)(3.80)(cos 180)=-175.9 J$

So the total work done is

$W=577.6+(-296.9)+0+(-175.9)=104.8 J$

Now we can use the work-energy theorem to find the final speed of the suitcase: in fact, the total work done is equal to the gain in kinetic energy of the suitcase, therefore

$W=\Delta K = K_f - K_i\\W=\frac{1}{2}mv^2\\v=\sqrt{\frac{2W}{m}}=\sqrt{\frac{2(104.8)}{19.6}}=3.3 m/s$

6 0

3 years ago

Consider an ideal monatomic gas of N particles with mass m in thermal equilibrium at a temperature T. The gas is contained in a

harina [27]

Answer:

$K.E.=\dfrac{3}{2}KT$

Explanation:

Given that

Number of particle =N

Equilibrium temperature= T

Side of cube = L

Gravitational acceleration =g

The kinetic energy of an atom given as

$K.E.=\dfrac{3}{2}KT$

Where

Equilibrium temperature= T

Boltzmann constant =K

K =1.380649×10−23 J/K

3 0

3 years ago

Which state of matter is most likely represented in the image?

dedylja [7]

It’s a liquid state !

4 0

3 years ago

Read 2 more answers

A 3.80-m-long, 500 kg steel beam extends horizontally from the point where it has been bolted to the framework of a new building

Virty [35]

Answer:

Explanation:

The question is ;

3.80 m-long, 500. kg steel beam extends horizontally from the point

where it has been bolted to the framework of a new building under

construction. A 67.0 kg construction worker stands at the far end of

the beam. What is the magnitude of the torque about the point where the beam is bolted into place?

Solution:

Here two torques are in action

a) torque T1 due to weight of worker at the edge of the beam - at center of the beam

b) torque T2 due to weight of the uniform beam - at the point where the beam is bolted

So we first calculate the torque produced due to weight of the worker;

We can see that;

Distance of worker from the center of the beam = 1.9m

Mass of the worker = 67 kg

Value of g= 9.8 m/sec2

T1=force × distance from point of rotation

Here force is weight of the worker which is = mass × g=67×9.8=656.6 N

So the torque is

T1= 656.6×1.9=12225.892 Nm

or

T1 = 12225.892 Nm

Now torque by the beam itself ;

Length of the beam from its center point to the bolt point = 1.9 m

Weight force of beam acting at center point= mass× g= 500×9.8=4900 N

Torque T2 at bolt point by the beam weight = weight of beam × length of beam from its center to the bolt point = 4900×1.9=9310 Nm

T2=9310Nm

So total torque = T1+T2= 12225.892+9310=21565.892 Nm

8 0

3 years ago

The energy transferred to the water in 100 seconds was 155 000 J. specific heat capacity of water = 4200 J/kg °C

skelet666 [1.2K]

Answer:

0.37 kg

Explanation:

I'm not a professor myself, but this is how I worked it out:

using the graph, after 100 seconds, the temperature is 100 degrees Celsius.

If we now substitute everything into the specific heat capacity equation, making the mass "m", we would come up with:

4200 = 155000/(m x 100)

If we rearrange and solve for m, we get 0.37 kg.

I'm not sure if I have done this correctly, feel free to correct me.

Hope this helps!

4 0

3 years ago

Read 2 more answers