You stand 17.5 m from a wall holding a tennis ball. You throw the tennis ball at the wall at an angle of 22.5∘ from the ground with an initial speed of 24.5 m/s. At what height above its initial position does the tennis ball hit the wall? Ignore any effects of air resistance.

A charged particle is injected into a uniform magnetic field such that its velocity vector is perpendicular to the magnetic field vector. Ignoring the particle's weight, the particle will follow a circular path.

Option D

Explanation:

Magnetic force causes charged particles to move in spiral paths. The Particle accelerators keep the protons to follow circular paths when it is in the magnetic field. Velocity has a change in direction but magnitude remains the same when this condition exists.

The magnetic force exerted on the charged particle is given by the formula:

         [tex]F=q v B \sin \theta[/tex]

where

q is the charge

v is the velocity of the particle

B is the magnetic field

[tex]\theta[/tex] is the angle

In this problem, the velocity is perpendicular to the magnetic field vector, hence

[tex]\theta[/tex] = [tex]90^{\circ}[/tex] and sin[tex]\boldsymbol{\theta}[/tex] =sin 90 degree = 1.

So applying the formula,

the force is simply [tex]F=q v B[/tex]

Also, the force is perpendicular to both B and v and so according to the right-hand rule, we have:

This means that the force acts as a centripetal force, so it will keep the charged particle in a uniform circular motion.

Answer:

[tex]8.64283\times 10^{-14}\ N[/tex]

Explanation:

q = Charge of proton = [tex]1.6\times 10^{-19}\ C[/tex]

v = Velocity of proton = [tex]0.267\times c[/tex]

c = Speed of light = [tex]3\times 10^8\ m/s[/tex]

B = Magnetic field = 0.00687 T

[tex]\theta[/tex] = Angle = [tex]101^{\circ}[/tex]

Magnetic force is given by

[tex]F=qvBsin\theta\\\Rightarrow F=1.6\times 10^{-19}\times (0.267\times 3\times 10^8)\times 0.00687\times sin101\\\Rightarrow F=8.64283\times 10^{-14}\ N[/tex]

The magnetic force acting on the proton is [tex]8.64283\times 10^{-14}\ N[/tex]

Answer:

[tex]Q_{2}=1200cm^{3}/s[/tex]

Explanation:

Given data

Q₁=200cm³/s

We know that:

[tex]F=n\frac{vA}{l}\\[/tex]

can be written as:

ΔP=F/A=n×v/L

And

Q=ΔP/R

As

n₂=6.0n₁

So

Q=ΔP/R

[tex]Q=\frac{nv}{lR}\\ \frac{Q_{2}}{n_{2}}= \frac{Q_{1}}{n_{1}}\\ Q_{2}=\frac{Q_{1}}{n_{1}}*(n_{2})\\Q_{2}=\frac{200}{n_{1}}*6.0n_{1}\\ Q_{2}=1200cm^{3}/s[/tex]

Answer:

The length of the lake is 340 meters.

Explanation:

It is given that, a hiker determines the length of a lake by listening for the echo of her shout reflected by a cliff at the far end of the lake. She hears the echo 2 s after shouting. We need to find the length of the lake.

The distance covered by the person in 2 s is :

[tex]d=vt[/tex]

v is the speed of sound

[tex]d=340\ m/s\times 2\ s[/tex]

[tex]d=680\ m[/tex]

The length of the lake is given by :

[tex]l=\dfrac{d}{2}[/tex]

[tex]l=\dfrac{680\ m}{2}[/tex]

l = 340 meters

So, the length of the lake is 340 meters. Hence, this is the required solution.

Answer:

The statements which are true are:

1) A person's power output limits the amount of work that he or she can do in a given time span.

2) The SI unit of power is the watt.

4) Power can be considered the rate at which work is done.

Explanation:

Power can be defined as the the rate at which the work is done. Mathematically,

[tex]\large{P = \dfrac{W}{t}}[/tex]

where '[tex]W[/tex]' is the amount of work done in time '[tex]t[/tex]'.

Also one can perform some work in the expense of the energy that he/she consumes. So alternatively power can also be defined as the rate at which the energy is expended.

The SI unit of work done is Joule. So the unit of power in SI system is

[tex]Js^{-1}[/tex] or Watt.

Answer:

T = 2 T₀

Explanation:

To answer this question let's write the expression for electrical conductivity

    σ = n e2 τ / m*

The relationship with resistivity is

       ρ = 1 /σ

Whereby the resistance

        R = ρ L / A = 1 /σ  L / A

We see that there is no explicit relationship between time and resistance, there is only a dependence on the life time (τ) that depends on the properties of the material, not on its diameter or length.

As also the average velocity or electron velocity of electrons is constant, the time to cross 2 mm in length is twice as long as the time to cross a mm in length

 T = 2 T₀

Answer:

You could use Malus's Law. Malus's Law tells us that if you have a polarized wave (of intensity I 0 0 ) passing through a polarizer the emerging intensity ( Y OR ) will be proportional to the square cosine of the angle between the polarization direction of the incoming wave and the axis of the polarizer.

Explanation:

OR: I = I 00 ⋅ cos two ( e )

The magnifying power of an astronomical telescope will be:

"0.095".

According to the question,

Radius of curvature, R = 5.5 mm

Focal length of eyepiece, [tex]F_e[/tex] = 2.9 cm

We know that,

→ Focal length of mirror,

F₀ = [tex]\frac{Radius \ of \ curvature}{2}[/tex]

By substituting the values,

    = [tex]\frac{5.5}{2}[/tex]

    = 2.75 mm or,

    = 0.278 cm

hence,

The telescope's magnification be:

= [tex]\frac{F_0}{F_e}[/tex]

= [tex]\frac{0.275}{2.9}[/tex]

= 0.095

Thus the above approach is correct.

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Final answer:

The magnifying power of the astronomical telescope using the given values is approximately 0.19.

Explanation:

In order to find the magnifying power of an astronomical telescope, we need to use the formula:

Magnifying Power = Angular Magnification = (focal length of objective) / (focal length of eyepiece)

Given that the radius of curvature of the reflecting mirror is 5.5 mm (which is equal to 0.55 cm) and the focal length of the eyepiece is 2.9 cm, we can substitute these values into the formula to find the magnifying power.

Magnifying Power = (radius of curvature of mirror) / (focal length of eyepiece)

Magnifying Power = 0.55 cm / 2.9 cm

Magnifying Power = 0.19

Therefore, the magnifying power of the astronomical telescope is approximately 0.19.

Answer:

(a) 0.3 C

(b) 0.9 J

Explanation:

(a)

Given:

Current drawn (I) = 0.50 mA = 0.50 × 10⁻³ A

Terminal voltage (V) = 3.0 V

Time of operation (t) = 10 min = 10 × 60 = 600 s

Charge flowing through the toy car 'Q' is given as:

[tex]Q=It[/tex]

Plug in the given values and solve for 'Q'. This gives,

[tex]Q=(0.50\times 10^{-3}\ A)(600\ s)\\\\Q=0.3\ C[/tex]

Therefore, 0.3 C charge flows the toy car.

(b)

Energy lost by the battery is equal to the product of power consumed by the battery and time of operation.

Power consumed by the battery is given as:

[tex]P=VI[/tex]

Plug in the given values and solve for 'P'. This gives,

[tex]P=(3.0\ V)(0.50\times 10^{-3}\ A)\\\\P=1.5\times 10^{-3}\ W[/tex]

Therefore, the energy lost by the battery is given as:

[tex]E=P\times t\\\\E=1.5\times 10^{-3}\ W\times 600\ s\\\\E=0.9\ J[/tex]

Therefore, the energy lost by the battery is 0.9 J

The given question is incomplete. The complete question is as follows.

For your senior project, you would like to build a cyclotron that will accelerate protons to 10% of the speed of light. The largest vacuum chamber you can find is 60 cm in diameter.

What magnetic field strength will you need?

Explanation:

Formula for the strength of magnetic field is as follows.

      B = [tex]\frac{mv}{qr}[/tex]

Here,    m = mass of proton = [tex]1.67 \times 10^{-27}[/tex] kg

      v = velocity = 10% of [tex]3 \times 10^{8}[/tex] = [tex]3 \times 10^{7}[/tex] m/s

      q = charge of proton = [tex]1.6 \times 10^{-19} C[/tex]

      r = radius = [tex]\frac{60}{2}[/tex] = 30 cm = 0.30 m   (as 1 m = 100 cm)

Therefore, magnetic field will be calculated as follows.

            B = [tex]\frac{mv}{qr}[/tex]

               = [tex]\frac{1.67 \times 10^{-27} \times 3 \times 10^{7}}{1.6 \times 10^{-19} C \times 0.30 m}[/tex]

               = [tex]\frac{5.01 \times 10^{-20}}{0.48 \times 10^{-19}}[/tex]

               = 1.0437 T

Thus, we can conclude that magnetic field strength is 1.0437 T.

The student queries about the construction of a cyclotron to accelerate protons to a specific velocity. The radius and rotational period of protons within the cyclotron are calculated using the magnetic field strength and the desired velocity.

The student is interested in building a cyclotron that can accelerate protons to 10% of the speed of light, with specific constraints on the vacuum chamber dimensions. In physics, particularly in the field of particle accelerators, a cyclotron is a type of particle accelerator that uses a combination of an electric field and a constant magnetic field to increase the kinetic energy of charged particles. The radius of the cyclotron, which determines the maximum orbit size for the particles being accelerated, is a critical design parameter and can be calculated based on the desired kinetic energy of the particles and the strength of the magnetic field.

In the context of a cyclotron, the student might need to calculate the rotational period and maximum radius of proton orbits within given specifications such as the strength of the magnetic field and desired velocity. Understanding the principles behind cyclotrons and particle acceleration is essential for this project, which falls under the umbrella of advanced physics topics.

Answer:

T = 600K

Explanation:

See attachment below.

Answer:

speed

Explanation:

The slope of a  line on any distance-time graph represents the speed of the object.

Velocity  only comes in when there is speed of the object in a particular direction.

Answer:

GRadient= 0.192 mb / km ,   the correct answer is a

Explanation:

The pressure gradient would be considered linear so we can use a proportional rule to find the gradient

          Gradient = Dp / Dd

          Gradient = (1045 -997) / 250

          Gradient = 48/250

          GRadient= 0.192 mb / km

The correct answer is a

Answer: 4.08kg/J

Explanation: Please find the attached file for the solution

Answer:

Entropy = 4.08 kj/k

Explanation:

From energy balance in first law of thermodynamics, we have;

Δv(i)+ ΔU(h2o) = 0

Thus;

[MCp(T2 - T1)]iron + [MCp(T2 - T1)]water = 0

Where Cp is specific heat capacity

For iron, Cp = 0.45 Kj/kg°C and for water, Cp = 4.18 Kj/kg°C

From question, Mass of iron =25kg while mass of water = 100kg

And Initial temperature of iron (T1) = 350°C while initial temperature of water(T1) = 18°C

Thus,

[25 x 0.45(T2 - 350)] + [100 x 4.18(T2 - 18)] = 0

11.25T2 - 3937.5 + 418T2 - 7524 = 0

So,

429.25T2 = 11461.5

T2 = 26.7 °C

Now for entropy, we have convert the temperature from degree celsius to kelvins.

Thus, for iron T1 = 350 + 273 = 623K and for water, T1 = 18 + 273 = 291 K. Also, T2 = 26.7 + 273 = 299.7K.

The entropy changes will be;

For iron ;

Δs(i) = MCp(In(T2/T1)) = 25 x 0.45(In(299.7/623)) = -8.23 Kj/k

Now, for water;

Δs(water) = MCp(In(T2/T1)) = 100 x 4.18(In(299.7/291)) = 12.31 kj/k

Thus, total entropy will be the sum of that of iron and water.

Δs(total) = 12.31 kj/k - 8.23 Kj/k = 4.08 kj/k

Answer:

Better Equilibrium Maintenance for better accuracy...

Explanation:

In the Galileo's experiment, there is no utilization of two equal masses at a time. However, as we can see in a Atwood Machine, there are two equal masses involved that make the whole system to be in a state of equilibrium and ultimately the better measurements of acceleration due to gravity.

The induced electromotive force (EMF) in a generator coil is directly proportional to the area of the coil, frequency of rotation, and the magnetic field strength, but not to the coil's internal resistance. Changing any of these factors as described, except resistance, results in a proportional change in the induced EMF.

The behavior of induced electromotive force (EMF) in a generator is explained by Faraday's law of electromagnetic induction, which states that the induced EMF in a coil is proportional to the rate of change of magnetic flux through the coil.

These principles help us to understand the operation of an electric generator, where mechanical work is converted into electrical energy.

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The work done pumping the liquid out of the tank relies on the weight of the mass being lifted, the gravity, and the height it is being lifted to. Calculations are given for lifting all the fluid and two-thirds of it to two different heights.

To calculate the work done pumping the liquid out of the tank, we first need to find the volume of the tank, which is 4 meters * 3 meters * 6 meters, giving a total of 72 cubic meters of rubbing alcohol. Multiplying this by the density of the alcohol (786 kg/m^3) gives us the total mass of the alcohol, 56,592 kg. The work done by gravity when an object is lifted is equal to the weight of the object (mass*gravity) multiplied by the distance it is lifted (height). Therefore, we can calculate the work done pumping the liquid out of the tank:

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The work done to pump all of the liquid out over the top of the tank is calculated as 1665568.8 J, and 4441516.8 J through a spout 2 meters above the tank. To pump out two-thirds of the liquid over the top, the work is 1109203.2 J, and through the spout, it is 2957875.2 J. Each part uses the weight of the liquid and the distance it needs to be moved to calculate the work done.

Part a :

To find the work done, we need to calculate the force required to move the liquid to the top of the tank and the distance it needs to be moved.

Calculate the volume of the tank:

Volume = length × width × depth = 4 m × 3 m × 6 m = 72 m³.

Calculate the mass of the liquid:

Mass = density × volume = 786 kg/m³ × 72 m³ = 56652 kg.

Calculate the weight of the liquid:

Weight = mass × gravity = 56652 kg × 9.8 m/s² = 555189.6 N.

Calculate the center of mass:

The center of mass for a full tank is at half the depth:

3 meters.

Calculate the work done:

Work = weight × height = 555189.6 N × 3 m = 1665568.8 J (Joules).

Part b :

Here, the height the liquid needs to be moved is 6 m (tank height) + 2 m (spout height):

6 + 2 = 8 meters.

Work = weight × height

Work = 555189.6 N × 8 m = 4441516.8 J.

Part c :

Calculate the volume of two-thirds of the tank:

Volume = (2/3) × 72 m³ = 48 m³.

Calculate the mass of two-thirds of the liquid:

Mass = 786 kg/m³ × 48 m³ = 37728 kg.

Calculate the weight of two-thirds of the liquid:

Weight = 37728 kg × 9.8 m/s² = 369734.4 N.

Since the tank is still of uniform depth, the center of mass for the remaining liquid will be halfway up the tank's current depth:

1.5 meters (half of the 3 meters left).

Work = weight × height

Work = 369734.4 N × 3 m = 1109203.2 J.

Part d :

Height the liquid needs to be moved = 6 m (tank height) + 2 m (spout height) = 8 meters.

Work = weight × height

Work = 369734.4 N × 8 m = 2957875.2 J.

Answer:

12 volts.

Explanation:

Equal to the emf of battery. internal resistance won't count because the internal resistance is only apparent when a current passes through the battery.

The voltmeter reading is the terminal voltage, which is slightly less than the EMF due to the internal resistance of the battery and the small current drawn by the voltmeter. The exact value in volts is not provided due to the unknown internal resistance.

Answer:

Explanation:

Minimum intensity occurs due to destructive interference of sound. For it to take place ,

path difference = odd multiple of half wave length

When moved by .025 m ,

path difference created = ( x + .25 ) - ( x - .25 )

= 2 x .25

= .5 m

So .5 = half wave length

wave length = 2 x .5

= 1 m

frequency = velocity / wave length

= 343 / 1

= 343 Hz

B )

Now when frequency is increased , wave length will be decreased . For maximum intensity , constructive interference will have to take place .

For that

path difference = integral multiple of wave length

0.5 = 1 x wave length

wave length = .5

frequency = 343 / .5

= 686 Hz

Answer:

Zero work done,since the body isn't acting against  or by gravity.

Explanation:

Gravitational force is usually  considered as work done against gravity (-ve) and work by gravity ( +ve ) and also When work isn't done by or against gravity work done in this case is zero.

Gravitational force can be define as that force that attracts a body to any other phyical body or system that have mass.

The planet been considered as our system in this case is assumed to have mass, and ought to demonstrate such properties associated with gravitational force in such system. Such properties include the return of every object been thrown up as a result of gravity acting downwards. The orbiting nature of object along an elliptical part when gravitational force isn't acting on the body and it is assumed to be zero.

Answer:

Negative

Explanation:

First law of thermodynamic also known as the  law of conservation of energy states that the total energy of an isolated system is constant; energy can be transformed from one form to another, but can be neither created nor destroyed.

The first law relates  relates changes in internal energy to heat added to a system and the work done by a system by the conservation of energy.

The first law is mathematically given as ΔU  = [tex]U_{F}[/tex] - [tex]U_{0}[/tex] = Q - W

Where Q  = Quantity of heat

            W = Work done

From the first law The internal energy has the symbol U. Q is positive if heat is added to the system, and negative if heat is removed; W is positive if work is done by the system, and negative if work is done on the system.

Analyzing the pistol when it raises in isothermal and when it falls in  isobaric state.The following can be said:

In the Isothermal compression of a gas there is work done on the system to decrease the volume and increase the pressure. For work to be done on the system it is a negative work done then.

In the Isobaric State An isobaric process occurs at constant pressure. Since the pressure is constant, the force exerted is constant and the work done is given as PΔV.If a gas is to expand at a constant pressure, heat should be transferred into the system at a certain rate.Isobaric is a fuction of heat which is Isothermal Provided the pressure is kept constant.

In Isobaric definition above it can be seen that " Heat should be transferred into the system ata certain rate. For heat to be transferred into the system work is deinitely been done on the system thereby favouring the negative work done.

Answer: radius r = 42360.7km

Height above ground = 35950.7km

The height of the satellite above the ground is about 100 times the height of the ISS above the ground.

This is a high orbit.

Explanation: a synchronous satellite of mass m, revolving around earth with angular speed w, having a radius of travel r will experience centripetal force F = m*r*w^2*

But w = 2¶/T

F = m*r*(2¶/T)^2

F = (4*m*r*¶^2)/T^2

For the same body on the surface of the earth of radius R, the force F will be F =mg

According to newton's law,

(4*m*r*¶^2)/T^2 is proportional to 1/r^2

also mg is proportional to 1/R^2

Therefore,

(4*m*r*¶^2)/T^2 = K/r^2,

mg = K/R^2

Equating the two we get

K = gR^2 = (4*r^3*¶^2)/T^2 (where K is a constant equal to the product of mass of earth M and gravitational constant.)

r^3 = (g*R^2*T^2)/(4x3.142^2)

Substituting values of g=9.81m/s2

R = 6400000m (radius of earth)

T = 60x60x24 = 86400s (synchronous orbit has period equal one day)

r = 42350775.04m = 42350.7km

Height above ground H = r - R

H = 42350.7 - 6400 = 35950.7km

Please verify with calculator. Thanks

Answer with Explanation:

Mass of pendulum bob, m=0.453 kg

Speed, [tex]v_1=[/tex]2.58 m/s

a.r=75.1 cm=[tex]75.1\times 10^{-2}m[/tex]=0.751 m

[tex] 1cm=10^{-2} m[/tex]

Tension in the pendulum cable is given  by

Tension=Centripetal force+force due to gravity

[tex]T=\frac{mv^2}{r}+mg[/tex]

Where [tex]g=9.8 m/s^2[/tex]

Substitute the values

[tex]T=\frac{0.453(2.58)^2}{75.1\times 10^{-2}}+0.453\times 9.8[/tex]

[tex]T=8.45 N[/tex]

b.When the pendulum reaches its highest point,then

Final velocity, [tex]v_2=0[/tex]

According to law of conservation of energy

[tex]mgh_1+\frac{1}{2}mv^2_1=mgh_2+\frac{1}{2}mv^2_2[/tex]

[tex]gh_1+\frac{1}{2}v^2_1=gh_2+\frac{1}{2}v^2_2[/tex]

[tex]h_1=0[/tex]

Substitute the values

[tex]9.8\times 0+\frac{1}{2}(2.58)^2=9.8\times h_2+\frac{1}{2}(0)^2[/tex]

[tex]3.3282=9.8h_2[/tex]

[tex]h_2=\frac{3.3282}{9.8}=0.34 m[/tex]

The angle mad  by cable with the vertical=[tex]cos\theta=\frac{0.751-0.34}{0.751}=0.55[/tex]

[tex]\theta=cos^{-1}(0.55)=56.6^{\circ}[/tex]

c.When the pendulum reaches at highest point then

Acceleration, a=0

Therefore, the tension  in the pendulum cable

[tex]T=mgcos\theta[/tex]

Substitute the values

[tex]T=0.453\times 9.8cos56.6[/tex]

[tex]T=2.4 N[/tex]

Answer:

B. 0.224 m/s²

Explanation:

Given:

Mass of the object (m) = 20 kg

The forces acting on the object are:

[tex]\vec{F_1}=3\vec{i}\ N\\\\\vec{F_2}=5\vec{j}\ N\\\\\vec{F_3}=(\vec{i}-3\vec{j})\ N[/tex]

Now, the net force acting on the object is equal to the vector sum of the forces acting on it. Therefore,

[tex]\vec{F_{net}}=\vec{F_1}+\vec{F_2}+\vec{F_3}\\\\\vec{F_{net}}=3\vec{i}+5\vec{j}+\vec{i}-3\vec{j}\\\\\vec{F_{net}}=(3+1)\vec{i}+(5-3)\vec{j}\\\\\vec{F_{net}}=(4\vec{i}+2\vec{j})\ N[/tex]

Now, the magnitude of the net force is equal to the square root of the sum of the squares of its components and is given as:

[tex]|\vec{F_{net}}|=\sqrt{4^2+2^2}\\\\|\vec{F_{net}}|=\sqrt{20}\ N[/tex]

Now, from Newton's second law, the magnitude of acceleration is equal to the ratio of the magnitude of net force and mass. So,

Magnitude of acceleration is given as:

[tex]|\vec{a}|=\dfrac{|\vec{F_{net}}|}{m}\\\\|\vec{a}|=\frac{\sqrt{20}\ N}{20\ kg}\\\\|\vec{a}|=0.224\ m/s^2[/tex]

Therefore, option (B) is correct.

Answer: e

Explanation:

Argument against the person, circumstantial.

Answer:

C) equal to zero

Explanation:

Electric potential is calculated by multiplying constant and charge, then dividing it by distance. The location that we want to measure is equidistant from two particles, mean that the distance from both particles is the same(r2=r1). The charges of the particle have equal strength of magnitude but the opposite sign(q2=-q1). The resultant will be:V = kq/r

ΔV= V1 + V2= kq1/r1 + kq2/r2

ΔV= V1 + V2= kq1/r1 + k(-q1)/(r)1

ΔV= kq1/r1 - kq1/r1

ΔV=0

The electric potential equal to zero

Answer:

The required angular velocity (ω) will be [tex]0.313~rads^{-1}[/tex].

Explanation:

Due to the rotation of the space station the astronauts experience a centripetal acceleration towards the centre of the space station. If '[tex]\large{a_{c}}[/tex]', 'ω' and 'R' represent the centripetal acceleration, angular velocity of the space station and the radius of the space station respectively, then

[tex]a_{c} = \omega^{2}.R[/tex]

As according to the problem the space station has to rotate in such an angular velocity that it produces the same "artificial gravity" as Earth's surface, we can write

[tex]a_{c} = g = 9.8 ms^{-2}[/tex]

Also given [tex]R = \dfrac{diameter~of~the~space~station}{2} = \dfrac{200 m}{2} = 100 m[/tex]

Therefore we can write,

[tex]&& a_{c} = g = \omega^{2}.R\\&or,& \omega = \sqrt{\dfrac{g}{R}} = \sqrt{\dfrac{9.8 ms^{-1}}{100 m}} = 0.313~rads^{-1}[/tex]

Answer:

180 N

Explanation:

We know that acceleration is the rate of change of speed per unit time hence

[tex]a=\frac {v_f-v_i}{t}[/tex] where v and t are velocity and time respectively, f and i represent final and initial.

Also, from Newton's law of motion, F=ma and replacing a with the above then

[tex]F=m\frac {v_f-v_i}{t}[/tex]

Substituting 1.5 Kg for mass, m -14 m/s for i and 10 m/s for for v then

[tex]F=1.5\times \frac {10--14}{0.2}=180 N[/tex]

Therefore, the force is 180 N

The average force exerted by a 1.5 kg ball on the floor, falling with a speed of 14 m/s and rebounding with a speed of 10 m/s over 0.20 seconds, is 180 N.

The subject of this question is Physics and from the concept of impulse and momentum. The change in momentum equals the product of force and the time over which the force is applied. So, we can calculate the force using this formula: Force = Change in momentum / Time.

In this case, the ball's momentum changes by the difference in velocity multiplied by the mass of the ball, which is (14 m/s + 10 m/s) * 1.5 kg. The reason the velocities are added is because the direction of velocity changes, making the speed of ball before and after striking the floor of equal magnitude but opposite in direction. So, the change in momentum becomes 36 kg*m/s. Given that the time is 0.20 seconds, the force would be 36 kg*m/s divided by 0.20 s, or 180 N.

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Explanation:

Beloware attachments containing the complete question and solution.

Answer:

The answer to the question is

The ladybug begins to slide

Explanation:

To solve the question we assume that the frictional force of the ladybug and the gentleman bug are the same

Where the  frictional force equals [tex]F_{Friction}[/tex] = μ×N = m×g×μ

and the centripetal force is given by m·ω²·r

If we denote the properties of the ladybug as 1 and that of the gentleman bug as 2, we have

m₁×g×μ = m₁·ω²·r₁ ⇒ g×μ = ω²·r₁

and for the gentleman bug we have

m₂×g×μ = m₂·ω²·r₂ ⇒ g×μ = ω²·r₂

But r₁ = 2×r₂

Therefore substituting the values of r₁ =2×r₂ we have

g×μ = ω²·r₁ = g×μ = ω²·2·r₂

Therefore   ω²·r₂ = 0.5×g×μ for the ladybug. That is the ladybug has to overcome half the frictional force experienced by the gentleman bug before it start to slide

The ladybug begins to slide