What is the momentum of a 20.0 kg scooter traveling at 5.00 m/s?

Answers

Answer 1

Answer:

The answer is 100 kgm/s

Explanation:

To find the momentum of an object given it's mass and velocity we use the formula

momentum = mass × velocity

From the question

mass = 20 kg

velocity = 5 m/s

We have

momentum = 20 × 5

We have the final answer as

100 kgm/s

Hope this helps you


Related Questions

What is the product of 0.0322 cm x 6.5 cm rounded to the correct number of significant figures?

Answers

Answer:

2.093 × 10⁻³ cm²

Explanation:

Given:

0.0322 cm × 6.5 cm

Find:

Product

Computation:

⇒ 0.0322 cm × 6.5 cm

⇒ 0.2093 cm²

2.093 × 10⁻³ cm²

I need help with this answer

Answers

Answer:

Single Replacement

Explanation:

why bananas are curved

Answers

Because they built:different

they are bananas and bananas are curved because they grow curved

An airplane flies with a constant speed of 540 miles per hour. How far can it travel in 1 1/4 hours?

Answers

Answer:

Hey!

Your answer is 675 MILES!

Explanation:

Using the S = D x T formula...

We first need to convert any values... 1 1/4 hours--75 mins (1.25hrs)

So now we substitute into the formula!

D (distance)= 540 x 1.25 = 675     (Distance = Speed x Time)

Distance Travelled= 675 MILES!

I HOPE THIS HELPED YOU!

A 20-kg bicycle carrying a 50-kg girl is traveling with a kinetic energy of 2,290 J.
What is the velocity of the girl and bicycle?
8 m/s
16 m/s
64 m/s
4 m/s

Answers

Answer:

64 m/s

Explanation:

64m/s is the answer of question

64/m/s the answer

12) If a man weighs 900 N-on the Earth, what would he weigh on Jupiter, where the acceleration due to gravity is 25.9 m/s?

Answers

Answer:

Force=Mass*acceleration

on earth, acceleration=9.81 m/s^2

900 N=Mass*9.81 m/s^2

Mass=91.74 Kg

F=Mass*acceleration(Jupiter)

F=91.74Kg*25.9m/s

F=2376.066 N on Jupiter

Plz mark me as brainliest if u found it helpful

4. A tankful of liquid has a volume
of 0.2m3. What is the volume in (a)
lities (b) cm3 (c)ml​

Answers

Explanation:

sjaaqqjajkslxxjxn vbnvisowpsjsndncnmcnvngieoowpwisnxxnnccnnckfkdkdjcncjcmdmcmcmdmcmdmdmcmcmcdjqowieurisosksjsjsosoapqpskdj

A boy kicked off a cliff and lands 151m away 45s later. What was the initial velocity? How tall is the cliff?

Answers

s = 1/2 g t^2 = 1/2 * 9.8 * 45^2 = 9922 m
That’s the height of the cliff

The original sideways velocity = 151/45 = 3.35 m/s

NO ONE WILL HELP, PLEASE PLEASE HELP, I HAVE AN HOUR TO GET 4 PAGES DONE! Find the average speed of a marble that takes 6 seconds to roll 30 m across a gymnasium floor.

Answers


distance
speed = — ————
time
s= 30m
———
6s
the average speed of the marble is 5s/m

NEED HELP DUE AT 11:59!! A ball is thrown horizontally from the top of
a building 130 m high. The ball strikes the
ground 53 m horizontally from the point of
release.
What is the speed of the ball just before it
strikes the ground?
Answer in units of m/s.

Answers

Answer:

Since the ball was thrown horizontally, there was no vertical component in that force. and hence, the initial vertical velocity of the ball is 0 m/s and the initial horizontal velocity is r.

We are given:

initial velocity  (u) = 0 m/s     [vertical]

final velocity (v) = v m/s  [vertical]

time taken to reach the ground (t) = t seconds

acceleration (a) = 10 m/s/s   [vertical , due to gravity]

height from the ground (h) = 130 m

displacement (s) = 53 m [horizontal]

Solving for time taken:

From the third equation of motion:

s = ut + 1/2 at²

130 = (0)(t) + 1/2 * (10) * t²

130 = 5t²

t² = 26

t = √26 seconds  or   5.1 seconds

Final Horizontal velocity of the ball

Since the horizontal velocity of the ball will remain constant:

the ball covered 53 m in 5.1 seconds [horizontally]

horizontal velocity of the ball = horizontal distance covered / time taken

Velocity of the ball = 53 / 5.1

Velocity of the ball = 10.4 m/s

Answer:

51.51519 m/s

Explanation:

Given: [tex]a_{x} =0[/tex] [tex]a_{y} -g[/tex] [tex]v_{yo} =0[/tex] [tex]x_{o} =0[/tex] [tex]x=53[/tex][tex]y_{o} =130[/tex]

X-direction                           | Y-direction

[tex]x=x_{o} +v_{xo}t[/tex]                         | [tex]y=y_{o} +v_{yo}t+\frac{1}{2}a_{y}t^2[/tex]

[tex]53=0v_{xo}(5.15078)[/tex]                 | [tex]0=130+\frac{1}{2}(-9.8)t^2[/tex]

[tex]53=v_{xo} (5.15078)[/tex]                    | [tex]-130=-4.9t^2[/tex]

[tex]\frac{53}{5.15078} =v_{xo}[/tex]                             |  [tex]\sqrt{\frac{-130}{-4.9} }=\sqrt{t^2}[/tex]

[tex]10.2897=v_{xo}[/tex]                            | [tex]5.15078=t[/tex]

[tex]v=\sqrt{v_{y}^2+ v_{x}^2}[/tex]                            | [tex]v_{y}^2 =v_{yo}+2a_{y} d[/tex]

[tex]v=\sqrt{(50.27771)^2+(10.2897)^2}[/tex] | [tex]\sqrt{v_{y}^2} =\sqrt{2(-9.8)(0-130)}[/tex]

[tex]v=51.51519 m/s[/tex]                        | [tex]v_{y}=50.47771[/tex]

A 2.60-kg box is sliding across the horizontal floor of an elevator. The coefficient of kinetic friction between the box and the floor is 0.380. Determine the kinetic frictional force that acts on the box for each of the following cases. (a) The elevator is stationary. N (b) The elevator is accelerating upward with an acceleration whose magnitude is 1.20 m/s2. N (c) The elevator is accelerating downward with an acceleration whose magnitude is 1.20 m/s2. N Additional Materials

Answers

Answer:

a) F = 9.69 N

b) F = 10.88 N

c) F = 8.51 N

Explanation:

a) The kinetic frictional force when the elevator is stationary is the following:

[tex] F_{k} = \mu_{k}N = \mu(mg) [/tex]    

Where:

F(k) is the kinetic frictional force

N is the normal force = mg

m: is the mass = 2.60 kg

g: is the gravity = 9.81 m/s²

μ(k) is the coefficient of kinetic friction = 0.380

[tex] F_{k} = \mu(mg) = 0.380*2.60 kg*9.81 m/s^{2} = 9.69 N [/tex]    

b) When the elevator is accelerating upward with acceleration "a" equal to 1.20 m/s²

[tex] F_{k} = \mu_{k}N = \mu[m(g + a)] [/tex]    

The normal is equal to mg plus ma because the elevator is accelerating upward

[tex] F_{k} = \mu m(g + a) = 0.380*2.60 kg(9.81 m/s^{2} + 1.20 m/s^{2}) = 10.88 N [/tex]    

c) When the elevator is accelerating downward with a =1.20 m/s² we can find the kinetic frictional force similar to the previous case:

[tex] F_{k} = \mu m(g - a) = 0.380*2.60 kg(9.81 m/s^{2} - 1.20 m/s^{2}) = 8.51 N [/tex]    

I hope it helps you!

A +5.00 pC charge is located on a sheet of paper.
(a) Draw to scale the curves where the equipotential surfaces due to these charges intersect the paper. Show only the surfaces that have a potential (relative to infinity) of 1.00 V, 2.00 V, 3.00 V, 4.00 V, and 5.00 V.
(b) The surfaces are separated equally in potential. Are they also separated equally in distance?
(c) In words, describe the shape and orientation of the surfaces you just found.

Answers

Answer:

a)    V = - x ( σ / 2ε₀)

c)  parallel to the flat sheet of paper

Explanation:

a) For this exercise we use the relationship between the electric field and the electric potential

          V = - ∫ E . dx        (1)

for which we need the electric field of the sheet of paper, for this we use Gauss's law. Let us use as a Gaussian surface a cylinder with faces parallel to the sheet

       Ф = ∫ E . dA = [tex]q_{int}[/tex] /ε₀

the electric field lines are perpendicular to the sheet, therefore they are parallel to the normal of the area, which reduces the scalar product to the algebraic product

          E A = q_{int} /ε₀

area let's use the concept of density

        σ = q_{int}/ A

       q_{int} = σ A

          E = σ /ε₀

as the leaf emits bonnet towards both sides, for only one side the field must be

          E = σ / 2ε₀

         we substitute in equation 1 and integrate

      V = - σ x / 2ε₀  

       V = - x ( σ / 2ε₀)

if the area of ​​the sheeta is 100 cm² = 10⁻² m²

      V = - x  (10⁻²/(2 8.85 10⁻¹²) = - x  ( 5.6 10⁻¹⁰)

       

      x = 1 cm     V = -1   V

      x = 2cm     V = -2   V

This value is relative to the loaded sheet if we combine our reference system the values ​​are inverted

       V ’= V (inf) - V

       x = 1 V = 5

       x = 2 V = 4

       x = 3 V = 3

   

These surfaces are perpendicular to the electric field lines, so they are parallel to the sheet.

 

In the attachment we can see a schematic representation of the equipotential surfaces

b) From the equation we can see that the equipotential surfaces are parallel to the sheet and equally spaced

c) parallel to the flat sheet of paper

If a bird applies a 5 N upward force on a branch to lift the branch of ground to a
height of 24 m, how much work did the bird do?

Answers

Work equals Force times Distance. 5*24= 120

please help me. i have 2 hours

Answers

the missing word is clockwise moment. I hope this helps good luck

A car traveling 21 m/s is accelerated uniformly at the rate of 2.2 m/s ^2 for 6.9 s. What is the car’s final speed?

Answers

Answer:

36m/s

Explanation:

v=u+at

v=21+(2.2×6.9)

v=21+15.18

v=36.18

v=36m/s

How many sets of planets would you need to create the mass of the Sun?

Answers

Answer:

volume about 1.3 million Earths could fit inside the Sun the mass of the Sun is 1.989 X 10 to the 30 kg about 333000 times the mass of the Earth

Explanation:

step-by-step explanation hope this answer your question

A pile of bricks of mass M is being raised to the tenth floor of a building of height H = 4y above the ground by a crane that is on top of the building. During the first part of the lift, the crane lifts the bricks a vertical distance h1=3y in a time t1=4T. During the second part of the lift, the crane lifts the bricks a vertical distance h2=y in t2=T. Which of the following correctly relates the power P1 generated by the crane during the first part of the lift to the power P2 generated by the crane during the second part of the lift?
A. P2=4P1
B. P2=43P1
C. P2=P1
D. P2=34P1
E. P2=13P1

Answers

Complete Question

A pile of bricks of mass M is being raised to the tenth floor of a building of height H = 4y above the ground by a crane that is on top of the building. During the first part of the lift, the crane lifts the bricks a vertical distance h1=3y in a time t1=4T. During the second part of the lift, the crane lifts the bricks a vertical distance h2=y in t2=T. Which of the following correctly relates the power P1 generated by the crane during the first part of the lift to the power P2 generated by the crane during the second part of the lift?

[tex]A.\ \ P_2=4P_1[/tex]

[tex]B.\ \  P_2=\frac{4}{3} P1[/tex]

[tex]C.\ \  P_2=P_1[/tex]

[tex]D. \ \ P_2=\frac{3}{4} P_1[/tex]

[tex]E. \ \ \ P_2=\frac{1}{3} P_1[/tex]

Answer:

The correct option is  B

Explanation:

From the question we are told that  

    The mass of the brick is  M

    The  height height of the 10th floor is  H =  4y

     The height attained during the first part of the lift is  [tex]h_1 =  3y[/tex]

     The time taken is [tex]t_1 =  4T[/tex]

    The height attained during the second part of the lift is  [tex]h_2  = y[/tex]

    The time taken is  [tex]t_2  =  T[/tex]

 

Generally the velocity of the crane during the first lift is mathematically represented as

         [tex]v_1  =  \frac{h_1}{t_1}[/tex]

=>      [tex]v_1  =  \frac{3y}{4T}[/tex]

Generally the velocity of the crane during the first lift is mathematically represented as

         [tex]v_1  =  \frac{h_2}{t_2}[/tex]

=>      [tex]v_1  =  \frac{y}{T}[/tex]

Generally the power generated during the first lift is  

     [tex]P_1 =  F_1 *  v_1[/tex]

Here [tex]F_1[/tex] force applied during the first lift which is mathematically represented as

       [tex]F_1  =  M  *  g[/tex] here g is the acceleration  due to gravity

So

       [tex]P_1 =  Mg * \frac{3y}{4T}[/tex]

Generally the power generated during the second lift is  

     [tex]P_2 =  F_2 *  v_2[/tex]

Here [tex]F_2[/tex] force applied during the second lift which is mathematically represented as

       [tex]F_2  =  M  *  g[/tex] here g is the acceleration  due to gravity

So

       [tex]P_2 =  Mg * \frac{y}{T}[/tex]

So the ratio  of the first power to the second power is  

      [tex]\frac{P_1}{P_2}  =  \frac{Mg * \frac{3y}{4T}[}{Mg * \frac{y}{T}}[/tex]

=>   [tex]\frac{P_1}{P_2}  = \frac{3}{4}[/tex]

=>    [tex]P_2 = \frac{4}{3} P_1[/tex]

A compass needle moves when it is near a wire with electric current. which of the following explains this phenomenon?

Answers

Answer:

D

Explanation:

Your chosen answer is correct .

Hope this helps :)

The electric current creates a magnetic field that explains this phenomenon. So, the correct option is D.

What is Magnetic field?

A magnetic field is defined as the vector field that describes the magnetic effect on electric charges, electric currents and magnetic materials where a charge moving in a magnetic field experiences a force perpendicular to its velocity and magnetic field.

The magnetic field is explained as the region around a magnet that has a magnetic force where all magnets have north and south poles. Opposite poles attract each other, while like poles repel each other. The electric current creates a magnetic field that explains a compass needle moves when it is near a wire with electric current phenomenon.  

Therefore, the correct option is D.

Learn more about Magnetic field, here:

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3 Points
The half-value layer for lead for g rays from 60Co is 12.8 mm. An electrometer gave a count rate of 35 pulses per second when at a
certain distance from a source, with no lead between source and detector. If 6 mm of lead were then inserted between source and
detector, the count rate (in pulses per second) would be nearest to:
A. 16
OB. 8
OC. 20
D. 25
OE. 75

Answers

Answer:

  D.  25

Explanation:

The number of pulses per second (p) for x mm of lead can be modeled by the exponential equation ...

  p = 35(0.5^(x/12.8))

Then for x=6, the value of p is ...

  p = 35·0.5^(6/12.8) ≈ 25.29

The expected count rate would be nearest 25 pps.

Researchers have measured the acceleration of racing greyhounds as a function of their speed; a simplified version of their results is shown in (Figure 1). The acceleration at low speeds is constant and is limited by the fact that any greater acceleration would result in the dog pitching forward because of the force acting on its hind legs during its power stroke. At higher speeds, the dog's acceleration is limited by the maximum power its muscles can provide.

How far does the dog run until its speed reaches 4.0 m/s?

Answers

[tex]\Huge\boxed{\boxed{\dfrac{4}{5}\ \text{meters}}}[/tex]

Let's start by finding the time it takes for the dog to reach a velocity of [tex]4[/tex] m/s.

We can use the following equation, where [tex]v_i[/tex] is initial velocity, [tex]v_f[/tex] is final velocity, [tex]t[/tex] is time, and [tex]a[/tex] is acceleration.

[tex]v_f-v_i=at[/tex]

We're trying to solve for [tex]t[/tex] first, so divide both sides by [tex]a[/tex].

[tex]\dfrac{v_f-v_i}{a}=t[/tex]

Substitute in the known values.

[tex]\dfrac{4-0}{10}=t[/tex]

[tex]\dfrac{4}{10}=t[/tex]

[tex]\dfrac{2}{5}=t[/tex]

Now, we can use the following formula to find the distance.

[tex]s=v_it+\dfrac{1}{2}at^2[/tex]

Substitute in the known values.

[tex]s=0*\dfrac{2}{5}+\dfrac{1}{2}*10*(\frac{2}{5})^2[/tex]

Anything multiplied by [tex]0[/tex] is

[tex]s=\dfrac{1}{2}*10*(\frac{2}{5})^2[/tex]

Just simplify from there.

[tex]s=\dfrac{1}{2}*10*\dfrac{4}{25}[/tex]

[tex]s=5*\dfrac{4}{25}[/tex]

[tex]s=\dfrac{20}{25}[/tex]

[tex]s=\boxed{\dfrac{4}{5}}[/tex]

Hey there!

The hind legs of the dog create it to accelerate.

We know the mass of the dog is 36km (m).

We know that the acceleration of the dog it 10m/s² (a).

Find the average force with the formula [ f = ma ] where m = mass and a = acceleration.

f = 36*10

f = 360 newtons

We don't know the traveled distance but we do know that the starting speed of the dog was 0m/s and the ending speed was 4m/s.

We can use the formula [ vf² = vo² + 2ad ] where vf = final speed, vo = starting speed, a = acceleration, and d = distance.

We know all the variables except the distance, so we are going to solve for d.

(4)² = (0)² + 2(10)(d)

16 = 0 + 20d

16 = 20d

16/20 = 20d/20

0.8 = d

Therefore, the dog runs 0.8 meters until it reaches 4m/s.

Best of Luck!

What type of climate would be near cold ocean currents? PLEASE HELP QUICK!!

Answers

Answer:

hot humid with lots of rain.

Explanation:

ocean currents act as conveyer belts of warm and cold water sending heat to the polar regions and helping the tropical areas cool off, thus influencing both weather and climate. the tropics are particularly rainy because heat absorption , and thus ocean evaporation, is highest.

Mild with cooler weather than similar latitudes type of climate would be near cold ocean currents.

What is cold ocean currents?

On the eastern side of ocean basins, lower latitudes or the tropics are where cold currents migrate in a direction toward the equator. They carry cold water into regions of the continent that have warm water, which are usually found on the west coastlines.

The Labrador Ocean Current and East Greenland Current flows are two noteworthy examples of cold currents.

From the perspectives of ecology and climate, ocean currents are also advantageous. For example, they are responsible for changing global temperatures.

Learn more about cold ocean currents here:

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You have purchased an inexpensive USB oscilloscope (which measures and displays voltage waveforms). You wish to determine if the oscilloscope has an error bias; in other words, you wish to determine if the errors made by the oscilloscope have a population mean that is not equal to zero. So you use a very accurate voltmeter to find the measurement errors for 13 different measurements made by your USB oscilloscope. A data file containing these measurements is HTMean1.csvPreview the document . Do a statistical analysis on this data to determine if the oscilloscope has an error bias.

Answers

Complete Question

The complete question is shown on the first uploaded image

Answer:

1

  A

2

A

Explanation:

From the question the data given for Error (mV) is -15

-15.17

8.67

-13.74

-20.69

-6.96

-1.36

-2.96

-9.26

3.11

-14.12

6.39

-14.77

Generally

The null hypothesis is [tex]H_o : \mu = 0[/tex]

The alternative hypothesis is [tex]H_a : \mu \ne 0[/tex]

The sample size is n = 13

Here [tex]\mu[/tex] represents the true error bias (i.e population error bias)

Generally the sample error bias is mathematically represented as

[tex]\= x = \frac{ \sum x_i}{n}[/tex]

=> [tex]\= x = \frac{ -15.17 + 8.67 + (-13.74) + \cdots + (-14.77) }{13}[/tex]

[tex]\= x = -7.37[/tex]

Generally the standard deviation is mathematically represented as

[tex]\sigma = \sqrt{\frac{\sum (x_i - \= x )^2}{n} }[/tex]

=> [tex]\sigma = \sqrt{\frac{ (-15.17-( -7.37) )^2 + (8.67 -( -7.37) )^2 + \cdots + (-14.77 -( -7.37) )^2 }{13} }[/tex]

=> [tex]\sigma = \sqrt{ 119.385}[/tex]

=> [tex]\sigma = 10.926[/tex]

Generally the test statistics is mathematically represented as

[tex]t = \frac{\= x - \mu }{\frac{\sigma }{\sqrt{n} } }[/tex]

=> [tex]t = \frac{ -7.37 - 0 }{\frac{10.926}{\sqrt{13} } }[/tex]

=> [tex]t = -2.838[/tex]

Generally the p-value is mathematically represented as

[tex]p-value = 2 P(t < -2.432)[/tex]

From the z-table  [tex]P(t < -2.432) =  0.0075 [/tex]

So  [tex]p-value  =  2* 0.0075 [/tex]

=>  [tex]p-value  = 0.015 [/tex]

So given that  p-value is  less than the [tex] \alpha = 0.05[/tex] then we reject the null hypothesis and conclude that the oscilloscope has an error bias

   

Anuja hit a golf ball on a level field at 70 degrees and 40 degrees with the same total speed as shown below.
70°
40°
Which launch angle causes the ball to be in the air for the longest time?
o not enough information
40 degrees
70 degress
times are the same

Answers

Answer:

At 40 deg  Vy = V sin 40

at 70 deg Vy = V sin 70

So the ball launched at 70 deg has the greatest vertical velocity and will remain in the air the longest:

Since t = Vy / g    time for to reach zero vertical velocity and also the time for the ball to reach velocity Vy on the downward path

Mathis kicked a ball on a level surface at 30∘ and 60∘ with the same total speed as shown below.

Which launch angle results in the greater maximum height for the ball?

Answer: CORRECT (SELECTED)

60

please help me with this physics problem​

Answers

Answer:

It says the answer. (5 marks.)

Explanation:

The FitnessGram Pacer Test is a multistage aerobic capacity test that progressively gets more difficult as it continues. The 20 meter pacer test will begin in 30 seconds. Line up at the start. The running speed starts slowly, but gets faster each minute after you hear this signal. A single lap should be completed each time you hear this sound. Remember to run in a straight line, and run as long as possible.

Reset Answers
Practice Energy transformations (Ouestion 5 of 10)
Q5
A penny is dropped from the top of the Statue of Liberty. The Statue of Liberty is 93 m tall If a penny has about 3 Joules of gravitational potential energy at the top, how much
potential energy will have transformed into kinetic energy at the half way point (46.5m) ?

Answers

Explanation:

At a height of 93 m, the gravitational potential energy is given by :

P = mgh

Where, m is the mass of a penny

We can find its mass.

[tex]m=\dfrac{P}{gh}\\\\m=\dfrac{3}{9.8\times 93}\\\\m=0.00329\ kg[/tex]

We need to find how much potential energy will have transformed into kinetic energy at the half way point (46.5m). It can be calculated as :

[tex]E=mgh'\\\\E=0.00329\times 9.8\times 46.5\\\\E=1.499\ J[/tex]

Hence, a penny will have transferred 1.499 J of potential energy into kinetic energy.

25° 20' 42" S 131° 2' E
1. What is the place?

Answers

That place is the "Uluru- Kara That National Park"

If dx denotes the change in position of an object and dt denotes the corresponding time interval, then instantaneous velocity is given by:

Answers

Answer:

[tex]Velocity=\frac{dx}{dt}[/tex]

Explanation:

Remember that instantaneous velocity is just a measure to know the velocity that any object has at any point given in time, so we just need to know the distance it has travel, which would be the change in position, and the time it took that change in position to occurr, this means distance by time, so we just divide dx by dt and we have the solution for instantaneous velocity.

If a change in position as denoted by [tex]dx[/tex] and [tex]dt[/tex] change in time, the instantaneous velocity will be given by,

[tex]v = \dfrac {dx}{dt}[/tex]

What is Velocity?It can be defined by the change in position of the object over time. This is a vector quantity. Vector quantity is a quantity that has both magnitude and direction.

Instantaneous velocity:The velocity of the object at a point of time is known as instantaneous velocity. Instantaneous velocity can be calculated by the ratio of change in position to the elapsed point of time.

[tex]v = \dfrac {dx}{dt}[/tex]

Where,

[tex]v[/tex] - instantaneous velocity

[tex]dt[/tex]   - change in distance (position)

[tex]dt[/tex]- change in time

Therefore, if a change in position as denoted by [tex]dx[/tex] and [tex]dt[/tex]change in time, the instantaneous velocity will be given by,

[tex]v = \dfrac {dx}{dt}[/tex]

Learn more about  Instantaneous velocity.

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f F= {mango, apple, banana, orange)​

Answers

Answer:

n(F) = 4

Explanation:

Cardinality of a set is the number of elements in that set. Given the set.

F= {mango, apple, banana, orange)​, we are to determine the cardinality of the set i.e the amount of fruit present in the set. Cardinality of the set F is represented as n(F).

Since there are 4 different fruit in the given set F, hence the cardinality of the set F is n(F) = 4

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How do two interacting objects exert equal and opposite forces on each other when they collide, even though they have different masses?

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Answer:

Did a little research.

Explanation:

The Law of Action-Reaction:

Newton's third law of motion is naturally applied to collisions between two objects. In a collision between two objects, both objects experience forces that are equal in magnitude and opposite in direction. Such forces often cause one object to speed up (gain momentum) and the other object to slow down (lose momentum). According to Newton's third law, the forces on the two objects are equal in magnitude. While the forces are equal in magnitude and opposite in direction, the accelerations of the objects are not necessarily equal in magnitude. In accord with Newton's second law of motion, the acceleration of an object is dependent upon both force and mass. Thus, if the colliding objects have unequal mass, they will have unequal accelerations as a result of the contact force that results during the collision.Consider the collision between the club head and the golf ball in the sport of golf. When the club head of a moving golf club collides with a golf ball at rest upon a tee, the force experienced by the club head is equal to the force experienced by the golf ball. Most observers of this collision have difficulty with this concept because they perceive the high speed given to the ball as the result of the collision. They are not observing unequal forces upon the ball and club head, but rather unequal accelerations. Both club head and ball experience equal forces, yet the ball experiences a greater acceleration due to its smaller mass. In a collision, there is a force on both objects that causes an acceleration of both objects. The forces are equal in magnitude and opposite in direction, yet the least massive object receives the greatest acceleration.

Consider the collision between a moving seven ball and an eight ball that is at rest in the sport of table pool. When the seven ball collides with the eight ball, each ball experiences an equal force directed in opposite directions. The rightward moving seven ball experiences a leftward force that causes it to slow down; the eight ball experiences a rightward force that causes it to speed up. Since the two balls have equal masses, they will also experience equal accelerations. In a collision, there is a force on both objects that causes an acceleration of both objects; the forces are equal in magnitude and opposite in direction. For collisions between equal-mass objects, each object experiences the same acceleration.

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