The diagram shows two balls before they collide.

2 balls with grey arrows pointing to them from the outside. The left ball has below it m subscript 1 = 0.6 kilograms v subscript 1 = 0.5 meters per second. The right ball has below it m subscript 2 = 0.5 kilograms v subscript 2 = negative 0.2 meters per second.
What is the momentum of the system after the collision?
1. 0.0 kg • m/s
2. 0.2 kg • m/s
3. 0.3 kg • m/s
4. 0.4 kg • m/s

Answers

Answer 1

Answer:

The Answer is B)0.2 kg • m/s

Explanation:

I made a 100 on my test. Sorry if I'm late but hope I helped.

Answer 2

Answer:

B. 0.2 kg x m/s

Explanation:


Related Questions

Please help will mark as brainiest.

12. An object with a mass of 1250 g has a speed of 17 m/s. What is the object's momentum? Round
answer to the nearest hundredth.

13. How fast is Team USA's Oracle sailing during the America's Cup? The boat has a mass of 1320 kg and
travels with a momentum of 32023.2 kg x m/s Round answer to the nearest hundredth.

Answers

Answer:

Explanation:

Momentum is mass multiplied by velocity, p = mv

Velocity is just speed but with a direction. So they can essentially be considered the same thing as long as you acknowledge this fact and understand where each is applied.

12)

First convert the grams to kilograms, those are the standard units us physicists use.

p = (1.25 kg)(17 m/s)

=21.25 kg*m/s

13)

p = mv

32,023.2 kg*m/s = (1,320 kg)(x m/s)

x = 24.26 m/s

If an ice cube with the mass of 5.0 grams melts in a closed system such as a closed glass jar what is the mass of the liquid water after the ice cube completely melts

Answers

If an ice cube with the mass of 5.0 grams melts in a closed system such as a closed glass jar, what is the mass of the liquid water after the ice cube completely melts? 5.0 grams. Which type of energy transformation does NOT describe potential

A circus tightrope walker weighing 800 N is standing in the middle of a 15 meter long cable stretched between two posts. The cable was originally horizontal. The lowest point of the cable is now at his feet and is 30 cm below the horizontal. Assume the cable is massless. What is the tension in the cable

Answers

Answer:

T = 10010 N

Explanation:

To solve this problem we must use the translational equilibrium relation, let's set a reference frame

X axis

       Fₓ-Fₓ = 0

       Fₓ = Fₓ

whereby the horizontal components of the tension in the cable cancel

Y Axis  

        [tex]F_{y} + F_{y} - W =0[/tex]

        2[tex]F_{y}[/tex] = W

let's use trigonometry to find the angles

        tan θ = y / x

        θ = tan⁻¹ (0.30 / 0.50 L)

        θ = tan⁻¹ (0.30 / 0.50 15)

        θ = 2.29º

the components of stress are

         F_{y} = T sin θ

we substitute

       2 T sin θ = W

       T = W / 2sin θ

        T = [tex]\frac{ 800}{ 2sin 2.29}[/tex]

        T = 10010 N

Two trains, each having a speed of 22 km/h, are headed at each other on the same straight track. A bird that can fly 60 km/h flies off the front of one train when they are 51 km apart and heads directly for the other train. On reaching the other train it flies directly back to the first train, and so forth. (We have no idea why a bird would behave in this way.) What is the total distance the bird travels before the trains collide

Answers

Answer:

the total distance the bird travels before the trains collide is 69.54 km

Explanation:

Given the data in the question,

dA = 22km/h × t

dB = 51km - 22km/h × t

now we find the time, when the two trains collide

dA = Db

22km/h × t =  51km - 22km/h × t

44km/h × t =  51km

t = 51km / 44km/h

t = 1.159 hrs

so the bird can fly back and forth for 1.159 hrs  before the train collide.

hence, distance travelled by the bird in total will be;

d = v × t

we substitute

d = 60 km/h × 1.159 h

d = 69.54 km

Therefore, the total distance the bird travels before the trains collide is 69.54 km

A fisherman notices that his boat is moving up and down periodically without any horizontal motion, owing to waves on the surface of the water. It takes a time of 2.30 s for the boat to travel from its highest point to its lowest, a total distance of 0.660 m . The fisherman sees that the wave crests are spaced a horizontal distance of 5.50 m apart. How fast are the waves traveling

Answers

Answer:

v = 1.2 m/s

Explanation:

The wavelength of the waves is given as the horizontal distance between the crests:

λ = wavelength = 5.5 m

Now, the time period is given as the time taken by boat to move from the highest point again to the highest point. So it will be equal to twice the time taken by the boat to travel from highest to the lowest point:

T = Time Period = 2(2.3 s) = 4.6 s

Now, the speed of the wave is given as:

[tex]v = f\lambda[/tex]

where,

v= speed of wave = ?

f = frequency of wave = [tex]\frac{1}{T} = \frac{1}{4.6\ s} = 0.217\ Hz[/tex]

Therefore,

[tex]v = (0.217\ Hz)(5.5\ m)\\[/tex]

v = 1.2 m/s

A level test track has a coefficient of road adhesion of 0.80, and a car being tested has a coefficient of rolling friction that is approximated as 0.018 for all speeds. The vehicle is tested unloaded and achieves the theoretical minimum stop in 180 ft (from brake application). The initial speed was 60 mi/h. Ignoring aerodynamic resistance, what is the unloaded braking efficiency

Answers

Answer:

the unloaded braking efficiency is 84.6 %

Explanation:

Given the data in the question;

by Ignoring aerodynamic resistance; we can find the theoretical stopping distance using the following formula

S = (Y[tex]_{b}[/tex]( V₁² - V₂²)) / ( 2g( ηbμ + [tex]f_{rl}[/tex] ± sin∅[tex]_{g}[/tex]))

now given that the tracked is levelled, ∅[tex]_{g}[/tex] = 0, also Y[tex]_{b}[/tex] = 1.04 for level or flat road

Speed V₁ = 60mil/hr = (60×5280)/(1×60×60) = 316800ft/3600s = 88ft/s

now, we substitute in our values to get the braking efficiency;

180ft = (1.04( (88ft/s)² - 0²)) / ( 2(32.2( (ηb/100)(0.80) + (0.018) ± sin(0°)))

180ft = 8053.76 / ( 64.4)(0.008ηb + 0.018)

180ft = 8053.76 / ( 0.5152ηb + 1.1592)

180( 0.5152ηb + 1.1592)  = 8053.76

( 0.5152ηb + 1.1592) = 8053.76 /180

0.5152ηb + 1.1592 = 44.7431

0.5152ηb = 44.7431 - 1.1592

0.5152ηb = 43.5839

ηb = 43.5839 / 0.5152

ηb = 84.596 ≈ 84.6 %

Therefore,  the unloaded braking efficiency is 84.6 %

why should we not use nuclear energy

Answers

Answer:

Barriers to and risks associated with an increasing use of nuclear energy include operational risks and the associated safety concerns, uranium mining risks, financial and regulatory risks, unresolved waste management issues, nuclear weapons proliferation concerns, and adverse public opinion.

Explanation:

In this system, ______kinetic_______ energy from the moving water of the faucet turns the turbine. The Mechanical energy of the spinning turbine generates Electical energy which is transformed into Thermal energy that causes the temperature of the water to increase. The water then becomes steam and gives off more Thermal energy into the atmosphere.

Answers

Answer:

The MECHANICAL energy of the water is transformed into KINETIC ENERGY

Explanation:

In this exercise you are asked to complete the sentence so that it makes sense.

When the tap is opened, the water has a mechanical energy given by the potential energy due to the height and the energy accumulated by the pressure of the system inside the pipe.

Therefore the sentence must be completed with

The MECHANICAL energy of the water is transformed into KINETIC ENERGY from the moving tap that turns the turbine ...

Answer:

dawg this helped me

Explanation:

i had to do the exact same question thanks homie

Match each letter to the description


A woman walks in a

straight line with the sun

to her right at six o'clock in

the morning. Toward

which pole does she walk?


A line of latitude.


A man walks in a straight

line with the sun to his

right at six o'clock in the

evening. Toward which

pole does he walk?


A line of longitude.

Answers

A woman walks in a  straight line with the sun to her right at six o'clock in the morning.

The sun rises East of her, so the woman is walking toward the North pole.

A man walks in a straight line with the sun to his right at six o'clock in the  evening.

The sun sets West of him, so the man is walking toward the South pole.

The woman and the man are both walking along lines of constant longitude.

When researchers replicate a study, they are seeking to __________.
A.
prove that the hypothesis upon which the study was founded is untestable
B.
develop a new hypothesis
C.
change the study to provide new results
D.
support or reject the hypothesis upon which the study was founded



Please select the best answer from the choices provided


A
B
C
D

Answers

The best answer I think is D) it’s the best one

Answer:

D

Explanation:

right edge 2022

Cathode ray tubes (CRTs) used in old-style televisions have been replaced by modern LCD and LED screens. Part of the CRT included a set of accelerating plates separated by a distance of about 1.54 cm. If the potential difference across the plates was 26.5 kV, find the magnitude of the electric field (in V/m) in the region between the plates.

Answers

Answer:

E = 1,720,779.221 or 1.720779221 * 10^ 6V/m

Explanation:

The electric field between the parallel conducting plates is given by

E =V / d

where V is the potential difference and d is the distance between the plates.

E = 26.5 kV/ 1.54 cm

Now we have to convert into proper units

26.5 kv= 26.5 * 1000 v=  26500 volts

1  kv= 1000 volts

1.54 cm = 1.54/ 100 m= 0.0154m

1m = 100cm

Now putting the values

E= 26500/0.0154 = 1,720,779.221 V/m

The Electric field is equal to E= 1,720,799.221 or 1.7220799221 * 10 ^6 Volts per meter.

In scientific notation this can be written as 1.7220799221 *10^6 V/m

"45 meters north" is an example of

Answers

Answer:

Displacement

Explanation:

The quantity 45m north is a typical example of displacement.

Displacement is the distance traveled by a body in a specific direction. Displacement is a vector quantity with both magnitude and direction.

When we are specifying the displacement of a body, the direction must be indicated accurately. Therefore, the quantity given is displacement


Does changing the height of point C affect the speed of the coaster car at point D?​

Answers

Without friction, NO.

The speed at D depends only on the difference in height between A and D. Whatever happens between them doesn't matter.

The speed of the coaster car at point D will be affected if  the height of point C is changed.

Potencial Energy:

It is the enrgy in a body due to the position of differnt part of the object or system.

As we increase the the hight of the car the potetial enrgy increase, the gravitational acceleration on car will be more due to the high of the point C.

Therefore, the speed of the coaster car at point D will be affected if  the height of point C is changed.

To know more about  speed of the coaster car,

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Lenz’s Law allows us to find _______.

the direction of the induced current.
the magnitude of the induced emf.
the direction of the induced emf.
the magnitude of the induced current.

Answers

Answer:

Explanation:

a

The moons phases are caused by

A. Eclipse of sun

B. Planets moving across the face of the moon

C. The alignment of the Earth, moon, and sun

D. The alignment of the planets.

Please help me!!!!

Answers

C. The alignment of the Earth, moon, and sun.

A 35.8 kg box initially at rest is pushed 2.38 m along a rough, horizontal floor with a constant applied horizontal force of 108.915 N. If the coefficient of friction between box and floor is 0.256, find the work done by the applied force. The acceleration of gravity is 9.8 m/s 2 . Answer in units of J.

Answers

Answer:

The work done by the applied force is 259.22 J.

Explanation:

The work done by the applied force is given by:

[tex] W = F*d [/tex]

Where:

F: is the applied horizontal force = 108.915 N

d: is the distance = 2.38 m  

Hence, the work is:

[tex] W = F*d = 108.915 N*2.38 m = 259.22 J [/tex]

Therefore, the work done by the applied force is 259.22 J.

I hope it helps you!                                                

a 250.0 g snowball of radius 4.00 cm starts from rest at the top of the peak of a roof and rolls down a section angled at 30.0 degrees

Answers

Answer:

The response to this question is as follows:

Explanation:

The whole question and answer can be identified in the file attached, please find it.

The force diagram of all the forces acting on the snowball include the normal force acting upwards, the weight of the snowball acting downwards and the frictional force acting horizontal.

The given parameters;

mass of the snow ball, m = 250 gradius of the snow ball, r = 4 cmangle of inclination of the plane, θ = 30 ⁰

The force diagram of all the forces acting on the snowball is calculated as follows;

                                     ↑ N

                                     ⊕  → F

                                      ↓ W

Where;

N is the normal force on the snowballF is the frictional force on the snowballW is the weight of the ball

Thus, the force diagram of all the forces acting on the snowball include the normal force acting upwards, the weight of the snowball acting downwards and the frictional force acting horizontal.

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A magnifying glass uses a convex lens of focal length 6.25 cm. When it is held 5.20 cm in front of an object, what is the image distance?
(Mind your minus signs)
(Unit=cm)

Answers

Answer:

The answer is -30.95.

Explanation:

Use the lens equation: 1/focal length = 1/object distance + 1/image distance. The answer comes out to -30.95. This is correct on Acellus.

When it is held 5.20 cm in front of an object, the image distance will be "-30.95 cm". To understand the calcultaion, check below.

Convex lens

According to the question,

Object distance, u = -5.20 cm

Focal length, f = 6.25 cm

By using the Lens formula, we get

→ [tex]\frac{1}{f} = \frac{1}{v} - \frac{1}{u}[/tex]

or,

→ [tex]\frac{1}{v} = \frac{1}{f} + \frac{1}{u}[/tex]

By substituting the values, we get

     [tex]= \frac{1}{6.25} - \frac{1}{5.20}[/tex]

  [tex]\frac{1}{v} = -\frac{21}{650}[/tex]

By applying cross-multiplication,

   v = -30.95 cm

Thus the above answer is correct.  

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A stone is dropped from the top of a high cliff with zero initial velocity. In which system is the net momentum zero as the stone falls freely

Answers

Answer:

A system that includes the stone and the earth.

Explanation:

If the system of being dropped from the height of the cliff consists of just the stone alone, then it means that its momentum will certainly undergo changes as it falls freely. However, If the system is now expanded to include not only the stone but also the Earth, then it implies that the momentum of the stone which is in the downward direction will be equal and opposite to the momentum of the Earth in the upwards direction towards the stone. Therefore, the momentum will cancel out and net momentum will be zero.

A system of stone and earth can result to a net zero momentum.

Conservation of linear momentum

The principle of conservation of linear momentum states that the sum of the initial momentum is equal to the sum of final momentum.

[tex]m_1u_1 + m_2 u_2 = m_1v_1 + m_2 v_2[/tex]

A system that consists a linear system of stone and earth can result to a net zero momentum.

Thus, a system of stone and earth can result to a net zero momentum.

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A physics student spends part of her day walking between classes or for recreation, during which time she expends energy at an average rate of 280 W. The remainder of the day she is sitting in class, studying or resting; during these activities, she expends energy at a rate of 100 W. If she expends a total of 1.1 x 10^7 J of energy in a 24 hour day, how much of the day did she spend walking

Answers

RESULT: Twalk= 3.64 hr

The time of the day she spent walking is equal to 3.70 hrs.

What is power?

Power can be explained as the rate of doing work in unit time. The SI unit of measurement of power is J/s or Watt (W). Power can be described as a time based quantity. The mathematical expression for power can be represented as mentioned below.

Power = work/time

P = W/t

Given, the energy spends part of her day walking, Ew = 280 W

The energy is spent by sitting in the class, Es = 100 W

The total energy spends, Et = 1.1 × 10⁷J

[tex]E_w \times t + E_s(24\times 60\times 60-t)= 1.1 \times 10^7J[/tex]

[tex]280 \times t + 100(24\times 60\times 60-t)= 1.1 \times 10^7[/tex]

280 t + 0.86 × 10⁷ - 100 t = 1.1 × 10⁷

180 t = 0.24 × 10⁷

t =  0.24 × 10⁷/180 × 3600

t = 3.70 hr

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Under what conditions will a moving 0.030 kg marble and a moving 2.43 kg rock have the same kinetic energy

Answers

Answer:

To have the same kinetic energy the speed of the marble must be 9 times the speed of rock.

Explanation:

The general formula of kinetic energy is given as follows:

[tex]K.E = \frac{1}{2}mv^{2}[/tex]

where,

K.E = Kinetic Energy

m = mass of the object

v = speed of the object

So, for the marble and rock to have same kinetic energy, we can write:

[tex]K.E_{marble} = K.E_{rock}\\\\\frac{1}{2}m_{marble}v_{marble}^{2} = \frac{1}{2}m_{rock}v_{rock}^{2}\\\\(0.03\ kg)v_{marble}^{2} = (2.43\ kg)v_{rock}^{2}\\\\taking\ square\ root\ on\ both\ sides:\\v_{marble} = \sqrt{\frac{2.43\ kg}{0.03\ kg}}v_{rock}\\\\v_{marble} = 9\ v_{rock}[/tex]

Hence, to have the same kinetic energy the speed of the marble must be 9 times the speed of rock.

Which of the following is a mixture?
a air
biron
Chydrogen
d nickel

Answers

The answer is Chydrogen

A beaker is filled with water to the rim. Gently you place a toy duck in the beaker and some of the water spills out. The duck floats. The weight of the beaker with the duck floating in it is Group of answer choices less than its weight before adding the duck. the same as its weight before adding the duck. greater or less than its weight before adding the duck, depending on the weight of the duck. greater than its weight before adding the duck.

Answers

Answer:

Explanation:

The same as its weight before adding the duck

When a moving object collides with an object that isn't moving, what happens to the kinetic energy of each object?

Answers

The kinetic energy of the object being hit will increase and the the potential energy will decrease. The kinetic energy of the object that collided with the other said object will increase and there will be not potential energy. This is my personal knowledge on the matter.

Although the internet states otherwise:

In the extreme case, multiple objects collide, stick together, and remain motionless after the collision. Since the objects are all motionless after the collision, the final kinetic energy is also zero; the loss of kinetic energy is a maximum.

All the objects are motionless, so kinetic energy of each object is zero after the collision.

What is Kinetic Energy?

The kinetic energy of an object is defined as the energy which is  possesses due to its motion. It is the work required to accelerate a body of a given mass from rest to its stated velocity. This energy is gained during its acceleration, the body maintains the kinetic energy as long as its momentum does not change.

Kinetic Energy can be expressed as

[tex]K.E.=[/tex] [tex]1/2 mv^2[/tex]

Where, m is the mass of the object

v is the velocity.

It is expressed in joules (J).

After the collision all the objects are at rest, therefore, the final kinetic energy is also zero which shows maximum loss of kinetic energy. Such collisions are called perfectly inelastic.

Thus, all the objects are motionless, so kinetic energy of each object is zero after the collision.

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How heavier elements formed during stellar nucleosynthesis and evolution?

Answers

Answer:

i honestly think its 21

Explanation:

da memes

10 + 10 =21

If a person weighs 140 lb'on Earth, their mass in kilograms is

Answers

Answer:

70 kg

Explanation:

divide it by 2

Hope this helped!

Answer:

63.502932 Kilograms

Explanation:

A jet aircraft is traveling at 233 m/s in horizontal flight. The engine takes in air at a rate of 122 kg/s and burns fuel at a rate of 3.36 kg/s. The exhaust gases are ejected at 487 m/s relative to the aircraft. Find the thrust of the jet engine. Answer in units of N.

Answers

Answer:

33624.32 N

Explanation:

From second law of motion, thrust of engine can be computed as

F= Δp/ Δt

= m1 * Δv + m2 v(o)

F= m1 *[v(o) - v] + m2 *v(o).............eqn(1)

Where

v(o) = initial velocity of the gas when ejected= 487 m/s

v= final velocity=233 m/s

m1= initial rate The engine takes in air at =122 kg/s

m2= rate of burning fuel= 3.36 kg/s.

If we input the given values into eqn (1) we have

F= 122×[ 487-233] + [3.36×487]

F=30988+1636.33

=33624.32 N

Hence, the thrust of the jet engine is 33624.32 N

Traumatic brain injury such as concussion results when the head undergoes a very large acceleration. Generally, an acceleration less than 800 m/s2 lasting for any length of time will not cause injury, whereas an acceleration greater than 1,000 m/s2 lasting for at least 1 ms will cause injury. Suppose a small child rolls off a bed that is 0.55 m above the floor. If the floor is hardwood, the child's head is brought to rest in approximately 1.7 mm. If the floor is carpeted, this stopping distance is increased to about 1.4 cm. Calculate the magnitude and duration of the deceleration in both cases, to determine the risk of injury. Assume the child remains horizontal during the fall to the floor. Note that a more complicated fall could result in a head velocity greater or less than the speed you calculate.

Answers

Answer:

a) wooden floor  

      a = 3170.6 m / s²,    t = 1.03 10⁻³ s

the child's traumatic injury to the brain

b) the floor is carpeted

      a = 385 m / s²

no injuries are created in the child

Explanation:

To solve this exercise we can use the energy conservation relations, let's start by looking for the speed of the child when he is just reaching the ground

starting point. When you get out of bed

         Em₀ = U = m g h

final point. Just when it hits the floor

          [tex]Em_{f}[/tex] = K = ½ m v²

as there is no friction, energy is conserved

           Em₀ = Em_{f}

           mgh = ½ m v²

           v² = 2 gh

let's calculate

          v² = 2 9.8 0.55

          v² = 10.78

           v = 3.28 m / s

Now let's use the concepts of kinematics to find the deceleration. The case of the wooden floor, where the distance for the deceleration is

x = 1.7 mm = 0.0017 m

      v² = v₀² - 2 a y

as the child stops the final velocity is zero

      0 = v₀² - 2a y

      a = v₀² / 2y

let's calculate

      a = [tex]\frac{10.78}{2 \ 0.0017}[/tex]

      a = 3170.6 m / s²

Let's find the time that braking lasts

        v = v₀ - a t

         0 = v₀ - at

          t = v₀ / a

         t = 3.28 / 3170.6

         t = 1.03 10⁻³ s

hence the child's traumatic injury to the brain

second case the floor is carpeted, in this case the stopping distance is

x = 1.4 cm = 0.014 m

we look for acceleration

         a = v₀² / 2y

         a = [tex]\frac{10.78}{2 \ 0.014}[/tex]

         a = 385 m / s²

therefore no injuries are created in the child

In conclusion we see that with the wooden floor there is silence and with the carpeted floor there is no

Which of the physical variables listed below will change when you change the area of the capacitor plates (while keeping the battery connected).

a. Capacitance
b. Charge on the plates
c. Voltage across the plates
d. Net electric field between the plates
e. Energy stored in the capacitor

Answers

Answer:

a. Capacitance

b. Charge on the plates  

e. Energy stored in the capacitor

Explanation:

Let A be the area of the capacitor plate

The capacitance of a capacitor is given as;

[tex]C = \frac{Q}{V} = \frac{\epsilon _0 A}{d} \\\\[/tex]

where;

V is the potential difference between the plates

The charge on the plates is given as;

[tex]Q = \frac{V\epsilon _0 A}{d}[/tex]

The energy stored in the capacitor is given as;

[tex]E = \frac{1}{2} CV^2\\\\E = \frac{1}{2} (\frac{\epsilon _0 A}{d} )V^2[/tex]

Thus, the physical variables listed that will change include;

a. Capacitance

b. Charge on the plates  

e. Energy stored in the capacitor

A 56-kg skater initially at rest throws a 5.0-kg medicine ball horizontally to the left. Suppose the ball is accelerated through a distance of 1.0 m before leaving the skater's hand at a speed of 7.0 m/s. Assume the skater and the ball to be point-like and the surface to be frictionless and ignore air resistance. Use a vertical y-axis with the positive direction pointing up and a horizontal x-axis with the positive direction pointing to the right. What will happen to the skater and the ball after the ball is thrown

Answers

Answer:

[tex]a_2=24.5\ \text{m/s}^2[/tex] towards right

[tex]a_1=2.1875\ \text{m/s}^2[/tex] towards left

Explanation:

[tex]m_1[/tex] = Mass of skater = 56 kg

[tex]m_2[/tex] = Mass of ball = 5 kg

v = Final velocity of ball = 7 m/s

u = Initial velocity of ball = 0

s = Distance the ball moved in the hand of the skater = 1 m

Moving left is considered and moving right is considered positive.

From kinematic equations of motion we have

[tex]v^2-u^2=2as\\\Rightarrow a=\dfrac{v^2-u^2}{2s}\\\Rightarrow a=\dfrac{7^2-0^2}{2\times 1}\\\Rightarrow a=24.5\ \text{m/s}^2[/tex]

So, the ball will move towards right with a magnitude of acceleration [tex]a_2=24.5\ \text{m/s}^2[/tex].

The force on the ball will be

[tex]F_2=m_2a_2\\\Rightarrow F_2=5\times 24.5\\\Rightarrow F_2=122.5\ \text{N}[/tex]

The force on the ball is [tex]122.5\ \text{N}[/tex]

The reaction force on the skater will be equal to the force on the ball but will have opposite direction.

[tex]-F_1=F_2\\\Rightarrow -m_1a_1=F_2\\\Rightarrow a_1=-\dfrac{122.5}{56}\\\Rightarrow a_1=-2.1875\ \text{m/s}^2[/tex]

So, the skater will move towards left with a magnitude of acceleration [tex]2.1875\ \text{m/s}^2[/tex]

After the ball is thrown, the skater will accelerate to the left at 2.19 m/s² while the ball will accelerate to the right at 24.5 m/s².

The given parameters

Mass of the skater, m1 = 56 kgMass of the ball, m2 = 5 kgDistance traveled by the ball, d = 1 mSpeed of the ball, v = 7 m/s

The acceleration of the ball as it leaves the skaters hand is calculated as follows;

[tex]v^2 = u^2 + 2as\\\\ v^2 = 0 + 2as\\\\ a = \frac{v^2}{2s} \\\\ a = \frac{7^2}{2(1)} \\\\ a = 24.5 \ m/s^2[/tex]

The force experienced by the ball is calculated as follows;

[tex]F = ma\\\\ F_b = 5 \times 24.5\\\\ F_b = 122.5 \ N[/tex]

The force experienced by the skater is equal in magnitude to that of the ball but opposite in direction.

[tex]F_s = -F_b\\\\ F_s = -122.5 \ N[/tex]

The acceleration of the skater after the ball was thrown is calculated as follows;

[tex]F = ma \\\\ a = \frac{F}{m} \\\\ a = \frac{-122.5 }{56} \\\\ a =-2.19 \ m/s^2[/tex]

Thus, after the ball is thrown, the skater will accelerate to the left at 2.19 m/s² while the ball will accelerate to the right at 24.5 m/s².

Learn more about Newton's third law of motion here: https://brainly.com/question/25998091

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