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2 years ago

15

Which of the following statements about Australian football is TRUE? A. The players tend to wear a large amount of padding. B. G

oals are scored by kicking the ball through goal posts. C. Each team has a number of set plays for both offense and defense. D. Constant movement of the ball is similar to baseball. Please select the best answer from the choices provided. A B C D

2 answers:

Naya [18.7K]2 years ago

3 0

Answer:

b. Goals are scored by kicking the ball through goal posts.

Explanation:

andreyandreev [35.5K]2 years ago

3 0

Answer:

The correct options are A , B and C

Explanation:

These are the correct options ; A. The players tend to wear a large amount of padding. B. Goals are scored by kicking the ball through goal posts. C. Each team has a number of set plays for both offense and defense.

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If different groups of scientist have access to the same data, how can they draw different conclusions?

vazorg [7]

If a group of scientists have access to one data, from the data they can draw conclusions either through mathematics or just thought experiments.

Those thought experiments is different for any scientist, no one thinks the same especially when the topic is difficult.

For example when talking about parallel universes, scientists have come up with the weirdest examples of a multiverse. Some thinking of a brane universe, while others say that its a landscape universe, quilted universe. All of their 'evidence' seems correct but they have opposite meanings.

A weird analogy is 'religion'. All the religions seem to have 'evidences' (hardly) that attract people towards it, they all make sense but that doesn't mean that their evidence is right.

----

Now if they're trying to break down the data using maths, there could be a great uncertainty and measurement error that if done enough could change the whole idea behind the data.

Interesting question, I can babble for days for this but lets keep it as that

6 0

2 years ago

Read 2 more answers

If the momentum of a system is to be conserved, which must be true of the net external force acting on the system?

kari74 [83]

Answer:

D. zero

Explanation:

For momentum of an isolated or closed system to be conserved (initial momentum must equal final momentum), the net external force acting on the system must be zero.

There is always external forces acting on a system, for this system’s momentum to remain constant, all the external forces acting on the system must cancel out, so that the net external force on the system is zero.

$F_{ext} = 0$

Therefore, the correct option is "D"

D. zero

7 0

2 years ago

Question 2 of 5 Which two statements are true for both compounds and mixtures? A. They can be separated into two or more element

Alekssandra [29.7K]

Answer:

D and A

I'm not sure about A lol

7 0

2 years ago

A skater rotates with her arms crossed at an angular speed of 8.0 rad/s and she has a moment of inertia of 100 kg/m2. She extend

solong [7]

Answer:

When her hands extends, her momen of inertia is $4.28\ kg-m^2$ .

Explanation:

Given that,

Initial angular speed, $\omega_i=8\ rad/s$

Initial moment of inertia, $I_1=100\ kg-m^2$

Final angular speed, $\omega_f=7\ rad/s$

Initially, a skater rotates with her arms crossed and finally she extends her arms. The momentum remains conserved. Using the conservation of momentum as :

$I_1\omega_1=I_2\omega_2$

$I_2$ is final moment of inertia

$I_2=\dfrac{I_1\omega_1}{\omega_2}\\\\I_2=\dfrac{100\times 8}{7}\\\\I_2=114.28\ kg-m^2$

So, when her hands extends, her momen of inertia is $4.28\ kg-m^2$ . Hence, this is the required solution.

7 0

2 years ago

A block with mass m =6.4 kg is hung from a vertical spring. When the mass hangs in equilibrium, the spring stretches x = 0.28 m.

Zanzabum

Answer

given,

mass of block (m)= 6.4 Kg

spring is stretched to distance, x = 0.28 m

initial velocity = 5.1 m/s

a) computing weight of spring

k x = m g

$k = \dfrac{mg}{x}$

$k = \dfrac{6.4 \times 9.8}{0.28}$

k = 224 N/m

b) $f = \dfrac{\omega}{2\pi}$

$\omega = \sqrt{\dfrac{k}{m}}= \sqrt{\dfrac{224}{6.4}} = 5.92 \ rad/s$

$f = \dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$

$f = \dfrac{1}{2\pi}\sqrt{\dfrac{224}{6.4}}$

$f =0.94\ Hz$

c) $v_b = -v cos \omega t$

$v_b = -5.1 \times cos (5.92 \times 0.42)$

$v_b = 4.04\ m/s$

d) $a_{max} = v \omega$

$a_{max} = 4.04 \times 5.92$

$a_{max} =23.94\ m/s^2$

e) $Y =- A sin (\omega t)$

$A = \dfrac{v}{\omega}$

$A = \dfrac{4.04}{5.92}$

A = 0.682 m

$Y =- 0.682 \times sin (5.92 \times 0.42)$

$Y =- 0.42$

Force = $m \omega^2 |Y|$

= $6.4 \times 5.92^2\times 0.42$

F = 94.20 N

4 0

3 years ago