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goldfiish [28.3K]

2 years ago

6

During a soccer game, a player grabs and holds an opponent's shirt outside of the penalty box. After the foul is called, what ki

ck is awarded to put the ball back into play?
a
Penalty Kick
b
Indirect Free Kick
c
Kickoff
d
Direct Free Kick

2 answers:

Mila [183]2 years ago

8 0

D direct free kick
Hope this helps

Alex17521 [72]2 years ago

3 0

D. Direct Free Kick

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You pull straight up on the string of a yo-yo with a force 0.35 N, and while your hand is moving up a distance 0.16 m, the yo-yo

jarptica [38.1K]

Answer:

a) 0.138J

b) 3.58m/S

c) (1.52J)(I)

Explanation:

a) to find the increase in the translational kinetic energy you can use the relation

$\Delta E_k=W=W_g-W_p$

where Wp is the work done by the person and Wg is the work done by the gravitational force

By replacing Wp=Fh1 and Wg=mgh2, being h1 the distance of the motion of the hand and h2 the distance of the yo-yo, m is the mass of the yo-yo, then you obtain:

$Wp=(0.35N)(0.16m)=0.056J\\\\Wg=(0.062kg)(9.8\frac{m}{s^2})(0.32m)=0.19J\\\\\Delta E_k=W=0.19J-0.056J=0.138J$

the change in the translational kinetic energy is 0.138J

b) the new speed of the yo-yo is obtained by using the previous result and the formula for the kinetic energy of an object:

$\Delta E_k=\frac{1}{2}mv_f^2-\frac{1}{2}mv_o^2$

where vf is the final speed, vo is the initial speed. By doing vf the subject of the formula and replacing you get:

$v_f=\sqrt{\frac{2}{m}}\sqrt{\Delta E_k+(1/2)mv_o^2}\\\\v_f=\sqrt{\frac{2}{0.062kg}}\sqrt{0.138J+1/2(0.062kg)(2.9m/s)^2}=3.58\frac{m}{s}$

the new speed is 3.58m/s

c) in this case what you can compute is the quotient between the initial rotational energy and the final rotational energy

$\frac{E_{fr}}{E_{fr}}=\frac{1/2I\omega_f^2}{1/2I\omega_o^2}=\frac{\omega_f^2}{\omega_o^2}\\\\\omega_f=\frac{v_f}{r}\\\\\omega_o=\frac{v_o}{r}\\\\\frac{E_{fr}}{E_{fr}}=\frac{v_f^2}{v_o^2}=\frac{(3.58m/s)}{(2.9m/s)^2}=1.52J$

hence, the change in Er is about 1.52J times the initial rotational energy

5 0

3 years ago

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Which of the following is true about all electromagnetic waves in a vacuum

Helen [10]

Electro waves in a vacuum air is deals with this and electricity when the air and the electricity it makes electro magnets.

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3 years ago

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what evidence can you cite that the interstellar medium contains both gas and dust? (select all that apply.)

Basile [38]

(a) The gas of interstellar medium can be detected from the radiations of photons of wavelength 21 cm.

(b) The gas of interstellar medium can be detected from the absorption lines present in the light from distant stars, which must be caused by a medium of density and temperature other than that of the stars emitting the lights.

<h3>What is interstellar medium?</h3>

Interstellar medium is the matter and radiation that exist in the space between the star systems in a galaxy.

<h3>Evidence that interstellar medium contains both gas and dust</h3>

The gas of interstellar medium can be detected from the radiations of photons of wavelength 21 cm.
The gas of interstellar medium can be detected from the absorption lines present in the light from distant stars, which must be caused by a medium of density and temperature other than that of the stars emitting the lights.

Learn more about interstellar medium here: brainly.com/question/4173326

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2 years ago

If a 40,000g cannon ball is 35m above the Earth's surface, how much potential energy does the cannon

mihalych1998 [28]

Answer: 21

Explanation:

because 9 + 10 = 21

8 0

3 years ago

A satellite of Mars, called Phobos, has an orbital radius of 9.4 ✕ 106 m and a period of 2.8 ✕ 104 s. Assuming the orbit is circ

FromTheMoon [43]

Answer:

$6.27*10^{23}kg$

Explanation:

assume

M= mass of Mars

m=mass of phobos

r=orbital radius

T=period

we can apply F=ma to this orbital motion (considering the cricular motion laws)

where,

$F=\frac{GMm}{r^{2} }$ and a=rω^2

where ω= $\frac{2\pi }{T}$ and G is the universal gravitational constant.

G = 6.67 x 10-11 N m2 / kg2

$F=ma\\\frac{GMm}{r^{2} }=mr(\frac{2\pi }{T} )^{2}\\ M=\frac{r^{3}}{G} (\frac{2\pi }{T} )^{2}\\M=\frac{(9.4*10x^{6} )^{3}*(2\pi )^{2} }{(2.8*10^{4}) ^{2} *6.67*10^{-11} } \\M=6.27*10^{23}kg$

6 0

3 years ago