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const2013 [10]
2 years ago
12

A14 is running toward the goal. her teammate, a12, kicks the ball toward a14. a14 did not see the ball coming and the ball strik

es her in the back of the hand as her hand is at her side. the proper call would be to:
Physics
1 answer:
harkovskaia [24]2 years ago
3 0

The Available options are:

A. Award an indirect free kick to Team B for incidental contact.

B.  Award a direct free kick to Team B.

C.  Allow play to continue as the handling was incidental but issue a caution to the player at the next stoppage of play.  

D.  Allow play to continue.

Answer:

Allow play to continue.

Explanation:

IFAB (International Football Association Board) has approved various means to clarify the issue of handball in foodtball.

Hence, some of the cases include:

1. Should the the ball touches a player's hand/arm immediately from their own head/body/foot or the head/body/foot of another player.

2. the ball touches a player's hand/arm close to their body and has not made their silhouette unnaturally bigger.

3. a player is falling and the ball touches their hand/arm when it is between their body and the ground (but not extended to make the body bigger).

4. Additionally, should the goalkeeper attempt to clear a ball from a teammate but fails, the goalkeeper is allowed to handle the ball.

Hemce, Since a14 who did not see the ball coming and the ball strikes her in the back of the hand as her hand is at her side. the proper call would be to: ALLOW PLAY CONTINUE

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A proton of mass mp is released from rest just above the lower plate and reaches the top plate with speed vp. An electron of mas
vodka [1.7K]

Answer:

v_e=\sqrt{\frac{m_pv_p^2}{m_e}}

Explanation:

You can consider that the force that acts over the proton is the same to the force over the electron. This is because the electric force is given by:

F=qE

F_p=F_e

where E is the constant electric field between the parallel plates, and is the same for both electron and proton. Also, the charge is the same.

by using the Newton second law for the proton, and by using kinematic equation for the calculation of the acceleration you can obtain:

m_pa_p=qE\\\\a_p=\frac{v_p^2}{2d}\\\\\frac{m_pv_p^2}{2d}=qE

(it has been used that vp^2 = v_o^2+2ad) where d is the separation of the plates, ap the acceleration of the proton, vp its velocity and mp its mass.

By doing the same for the electron you obtain:

\frac{m_ev_e^2}{2d}=qE

we can equals these expressions for both proton and electron, because the forces qE are the same:

\frac{m_pv_p^2}{2d}=\frac{m_ev_e^2}{2d}\\\\v_e=\sqrt{\frac{m_pv_p^2}{m_e}}

4 0
2 years ago
Cassy shoots a large marble (Marble A, mass: 0.06 kg) at a smaller marble (Marble B, mass: 0.03 kg) that is sitting still. Marbl
Lostsunrise [7]
Conservation of linear momentum:

m*v inital = m*v final

0.06*0.7 + 0.03*0 = 0.06*(-0.2) + 0.03*v

(my algebra, or use ur calculator: 0.06*.07=0.042, etc ... or ur teacher may think you got some help)

0.06*(0.7+0.2)=0.03*v, v = 0.06*0.9/0.03=1.8 m/s

Answer 1.8 m/s (positive, to the right).

 

4 0
2 years ago
Ricardo and Jane are standing under a tree in the middle of a pasture. An argument ensues, and they walk away in different direc
Advocard [28]

Answer:

the direction that should be walked by Ricardo to go directly to Jane is 23.52 m, 24° east of south

Explanation:

given information:

Ricardo walks 27.0 m in a direction 60.0 ∘ west of north, thus

A= 27

Ax =  27 sin 60 = - 23.4

Ay = 27 cos 60 = 13.5

Jane walks 16.0 m in a direction 30.0 ∘ south of west, so

B = 16

Bx = 16 cos 30 = -13.9

By = 16  sin 30 = -8

the direction that should be walked by Ricardo to go directly to Jane

R = √A²+B² - (2ABcos60)

   = √27²+16² - (2(27)(16)(cos 60))

   = 23.52 m

now we can use the sines law to find the angle

tan θ = \frac{R_{y} }{R_{x} }

         = By - Ay/Bx -Ax

         = (-8 - 13.5)/(-13.9 - (-23.4))

     θ  = 90 - (-8 - 13.5)/(-13.9 - (-23.4))

         = 24° east of south

4 0
2 years ago
A 20.0 kg curling stone travels 30.0 m along the ice surface. If the frictional force is 10.0 N, the thermal energy produced is
evablogger [386]
The thermal energy is where the work of friction comes from.  That is what stops it eventually.  In this case a counter force of 10N is applied over the distance of 30.0m.  The energy is given by Force*Distance.  Here this is 300J.  This friction work is the thermal energy.
4 0
2 years ago
Read 2 more answers
In this problem, you will analyze a system composed of two blocks, 1 and 2, of respective masses m1 and m2. To simplify the anal
Dafna1 [17]

Answer:

a) p = m1 v1 + m2 v2 , b) dp / dt = m1 a1 + m2 a2 , c) It is equivalent to force

dp / dt = 0

Explanation:

In this problem we have two blocks and the system is formed by the two bodies.

Part A. Initially they ask us to find the moment of the whole system

    p = m1 v1 + m2 v2

Part B.

Find the derivative

     dp / dt = m1 dv1dt + m2 dv2 / dt

     dp / dt = m1 a1 + m2 a2

Part C.

Let's analyze the dimensions

     m a = [kg] [m / s2] = [N]

It is equivalent to force

Part d

Acceleration is due to a net force applied

Part e

The acceleration of block 1 is due to the force exerted by block 2 during the moment change

Part f

Force of block 1 on block 2

True f12 = m1a1        f21 = m2a2

Part g

By the law of action and reaction are equal magnitude F12 = f21

Part H

     dp / dt = 0

Isolated system F12 = F21 and the masses are constant. The total moment is only redistributed

7 0
2 years ago
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