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Margaret [11]
2 years ago
13

From a penalty kick, the ball rebounds off the goalkeeper back to the player who took the kick. That player then kicks the ball

into the goal. What is the correct restart
Physics
1 answer:
Alexxx [7]2 years ago
7 0

The correct restart is for there to be another kick off taken in this type of scenario.

<h3>What is Kick-off in Soccer?</h3>

This is a method of restarting play in which the ball is put on the center circle and passed by a player.

Kick offs occur at the beginning of any halves or when a goal is scored during the match which is why it's the most appropriate choice.

Read more about Soccer here brainly.com/question/12597997

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What are the uses of rockets ​
lora16 [44]

They are used to be able to go out of Earth and into space to Be able to explore and toBe able to prove and tell about Earth and space

8 0
3 years ago
Two ice skaters stand in the middle of an ice rink. Drew has a mass of 75 kg, and Lily has a mass of 55 kg. Drew holds Lily, and
ss7ja [257]

PART a)

Before Drew throw Lily in forwards direction they both stays at rest

So initial speed of both of them is zero

So here we can say that initial momentum of both of them is zero

So total momentum of the system initially = ZERO

PART b)

Since there is no external force on the system of two

so there will be no change in the momentum of this system and it will remain same as initial momentum

So final momentum of both of them will be ZERO

PART c)

As we know that momentum of both will be zero always

so we have

P_1 + P_2 = 0

75(v) + 55(2) = 0

v = 1.47 m/s in opposite direction

7 0
3 years ago
Read 2 more answers
If it takes a ball dropped from rest 2.261 s to fall to the ground, from what height H was it released? Express your answer in m
algol [13]

Answer:

Height, H = 25.04 meters

Explanation:

Initially the ball is at rest, u = 0

Time taken to fall to the ground, t = 2.261 s

Let H is the height from which the ball is released. It can be calculated using the second equation of motion as :

H=ut+\dfrac{1}{2}at^2

Here, a = g

H=\dfrac{1}{2}gt^2            

H=\dfrac{1}{2}\times 9.8\times (2.261)^2

H = 25.04 meters

So, the ball is released form a height of 25.04 meters. Hence, this is the required solution.

7 0
3 years ago
The parking brake on a 1000 kg Cadillac has failed, and it is rolling slowly, at 1 mph , toward a group of small children. Seein
iogann1982 [59]

Answer:

0.5 mph in the opposite direction

Explanation:

m_1 = Mass of Cadillac = 1000 kg

v_1 = Velocity of Cadillac = 1 mph

m_2 = Mass of Volkswagen = 2000 kg

v_2 = Velocity of Volkswagen

In order to know the speed the system must have the momentum exchange

As the linear momentum of the system is conserved

m_1v_1+m_2v_2=0\\\Rightarrow v_2=-\dfrac{m_1v_1}{m_2}\\\Rightarrow v_2=\dfrac{1000\times 1}{2000}\\\Rightarrow v_2=-0.5\ mph

The speed of the impact is given by 0.5 mph in the opposite direction

8 0
3 years ago
To tugboats pull a disabled supertanker. Each tug exerts a constant force of 1.50•10^6 N, one at an angle 19.0° west of north, a
irina1246 [14]

Answer:

Work done by a tug boat, W = 1.735 x 10⁸ J

Explanation:

Given,

The of each tugboat, F = 1.5 x 10⁶ N

The angle of each tugboat forms with the resultant force, θ = 19°

The displacement of the supertanker, s = 710 m

The individual tugboat will be responsible for the displacement, d = 710/2

                                                                                                               = 355 m

The displacement component in each tugboat direction = 355 · sin θ meter

Therefore, the work done by each tugboat is

                                           W = F x S    joules

Substituting the values in the above equation

                                            W = 1.5 x 10⁶  x  355 · sin θ

                                                = 1.735 x 10⁸ J

Hence, the work done by each tugboat is, W = 1.735 x 10⁸ J                        

6 0
3 years ago
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