A particle with mass 3×10−2 kg
k
g
and charge +3 μC
μ
C
enters a region of space where there is a magnetic field of 1 T
T
that is perpendicular to the velocity of the particle. When the particle encounters the magnetic field, it experiences an acceleration of 12 m/s2
m
/
s
2
. What is the speed of the particle when it enters the magnetic-field region?

Time taken by the ball to reach the ground is 1.8s.

When an object is released from rest in free air considering no friction, the motion is depend only on the acceleration due to gravity, g.

A ball is thrown directly downward with an initial speed of 7.30 m/s, from a height of 29.0 m

Using the second equation of motion,

s = ut + 1/2 at²

Plug the values, we get

29 = 7.3t + 1/2 x9.81t²

4.905t² + 7.3t -29 =0

t =1.79871 or −3.28699

As time can't be negative, time taken to strike the ground is  approximately 1.8 s.

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I don't think the provided solution is correct because it's not dimensionally consistent. The quantity [tex]8h-1[/tex] in particular doesn't make sense since it's mixing a distance with a dimensionless constant.

Here's how I think the proper answer should look:

The net force on the box acting perpendicular to the ramp is

[tex]\sum F_\perp = F_{\rm normal} - mg \cos(\theta) = 0[/tex]

where [tex]F_{\rm normal}[/tex] is the magnitude of the normal force due to contact with the ramp and [tex]mg\cos(\theta)[/tex] is the magnitude of the box's weight acting in this direction. The net force is zero since the box doesn't move up or down relative to the plane of motion.

The net force acting parallel the ramp is

[tex]\sum F_\| = mg\sin(\theta) - F_{\rm friction} = ma[/tex]

where [tex]mg\sin(\theta)[/tex] is the magnitude of the parallel component of the box's weight, [tex]F_{\rm friction}[/tex] is the magnitude of kinetic friction, and [tex]a[/tex] is the acceleration of the box.

From the first equation, we find

[tex]F_{\rm normal} = mg \cos(\theta)[/tex]

and since [tex]F_{\rm friction} = \mu F_{\rm normal}[/tex], we get from the second equation

[tex]mg\sin(\theta) - \mu mg\cos(\theta) = ma[/tex]

and with [tex]\mu = 0.25[/tex] and [tex]\theta=60^\circ[/tex], we get

[tex]a = g\sin(60^\circ) - 0.25g \cos(60^\circ) = \left(\dfrac{\sqrt3}2 - \dfrac18\right) g[/tex]

Let [tex]x[/tex] be the length of the ramp, i.e. the distance that the box covers as it slides down it. Then the box attains a final velocity [tex]v[/tex] such that

[tex]v^2 = 2ax[/tex]

From the diagram, we see that

[tex]\sin(\theta) = \dfrac hx \implies x = \dfrac h{\sin(60^\circ)} = \dfrac{2h}{\sqrt3}[/tex]

and so

[tex]v^2 = \dfrac{4ah}{\sqrt3} = \left(2 - \dfrac1{2\sqrt3}\right) hg = \dfrac14 \left(8 - \dfrac2{\sqrt3}\right) hg[/tex]

[tex]\implies v = \dfrac12 \sqrt{\left(8 - \dfrac2{\sqrt3}\right) hg}[/tex]

(A)The total resistance of the parallel resistors is: [tex]R_{T} = \frac{R_{1} \times R_{2} \times R_{3}}{R_{2}R_{3}+R_{1}R_{3}+R_{1}R_{2}}[/tex]

(B) The current in the overall circuit is: [tex]I_{T} = \frac{V}{R_{1}} + \frac{V}{R_{2}} + \frac{V}{R_{3}}[/tex]

(C) The current through each resistance is as follows:

For the three resistances connected in parallel, R1, R2, R3 , the total resistance, [tex]R_{T}[/tex] is calculated as follows:

[tex]R_{T} = \frac{R_{1} \times R_{2} \times R_{3}}{R_{2}R_{3}+R_{1}R_{3}+R_{1}R_{2}}[/tex]

The current in the overall circuit is calculated using the formula:

[tex]I_{T} = \frac{V}{R_{1}} + \frac{V}{R_{2}} + \frac{V}{R_{3}}[/tex]

The current through each resistance is given as follows;

Current through R1, [tex]I_{1} = \frac{V}{R_{1}}[/tex]

Current through R2; [tex]I_{2} = \frac{V}{R_{2}}[/tex]

Current through R3; [tex]I_{3} = \frac{V}{R_{3}}[/tex]

In conclusion, the voltage across resistances in parallel is the same but the current varies.

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The horizontal component of the tension in the string is a centripetal force, so by Newton's second law we have

• net horizontal force

[tex]F_{\rm tension} \sin(\theta) = \dfrac{mv^2}R[/tex]

where [tex]m=4.10\,\rm kg[/tex], [tex]v=2.85\frac{\rm m}{\rm s}[/tex], and [tex]R[/tex] is the radius of the circular path.

As shown in the diagram, we can see that

[tex]\sin(\theta) = \dfrac Rr \implies R = r\sin(\theta)[/tex]

where [tex]r=1.69\,\rm m[/tex], so that

[tex]F_{\rm tension} \sin(\theta) = \dfrac{mv^2}R \\\\ \implies F_{\rm tension} = \dfrac{mv^2}{r\sin^2(\theta)}[/tex]

The vertical component of the tension counters the weight of the mass and keeps it in the same plane, so that by Newton's second law we have

• net vertical force

[tex]F_{\rm \tension} \cos(\theta) - mg = 0 \\\\ \implies F_{\rm tension} = \dfrac{mg}{\cos(\theta)}[/tex]

Solve for [tex]\theta[/tex] :

[tex]\dfrac{mv^2}{r\sin^2(\theta)} = \dfrac{mg}{\cos(\theta)} \\\\ \implies \dfrac{\sin^2(\theta)}{\cos(\theta)} = \dfrac{v^2}{rg} \\\\ \implies \dfrac{1-\cos^2(\theta)}{\cos(\theta)} = \dfrac{v^2}{rg} \\\\ \implies \cos^2(\theta) + \dfrac{v^2}{rg} \cos(\theta) - 1 = 0[/tex]

Complete the square:

[tex]\cos^2(\theta) + \dfrac{v^2}{rg} \cos(\theta) + \dfrac{v^4}{4r^2g^2} = 1 + \dfrac{v^4}{4r^2g^2} \\\\ \implies \left(\cos(\theta) + \dfrac{v^2}{2rg}\right)^2 = 1 + \dfrac{v^4}{4r^2g^2} \\\\ \implies \cos(\theta) + \dfrac{v^2}{2rg} = \pm \sqrt{1 + \dfrac{v^4}{4r^2g^2}} \\\\ \implies \cos(\theta) = -\dfrac{v^2}{2rg} \pm \sqrt{1 + \dfrac{v^4}{4r^2g^2}}[/tex]

Plugging in the known quantities, we end up with

[tex]\cos(\theta) \approx 0.784 \text{ or } \cos(\theta) \approx -1.27[/tex]

The second case has no real solution, since [tex]-1\le\cos(\theta)\le1[/tex] for all [tex]\theta[/tex]. This leaves us with

[tex]\cos(\theta) \approx 0.784 \implies \theta \approx \cos^{-1}(0.784) \approx \boxed{38.3^\circ}[/tex]

The bus will travel a further 20 m before coming to rest.

The term acceleration has to do with a change in velocity with time. Now we have;

u = 20 m/s

v = 10 m/s

s =  60 m

Now;

v^2 = u^2 -2as

v^2 -  u^2 = -2as

(10)^2 - (20)^2 = - 2 * a * 60

a = (10)^2 - (20)^2/ - 2 * 60

a = 100 - 400/ - 2 * 60

a = 2.5 m/s^2

At that time;

v = 0 m/s

u = 20 m/s

a = 2.5 m/s^2

s = ?

Hence;

u^2 = 2as

s = u^2/2a

s = (20)^2/ 2 *  2.5

s = 400/5

s = 80 m

Hence, the bus will travel a further 20 m before coming to rest.

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The change in kinetic energy of the proton is 4.8 x 10⁻²⁰.

The change in kinetic energy of the proton is calculated as follows;

ΔK.E = W = eV

where;

V = Ed

V = 3 V/m x 0.1 m = 0.3 V

ΔK.E  = 1.6 x 10⁻¹⁹ x 0.3 = 4.8 x 10⁻²⁰ J

Thus, the change in kinetic energy of the proton is 4.8 x 10⁻²⁰ J.

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Answer:

A

Explanation:

As it begins to fall

F = ma        a = 9.81

F = .21 * 9.81 = 2.06 N

The definition of force in physics is the push or pull on a massed object that changes its velocity. An external force is an agent that has the power to alter the resting or moving condition of a body. The correct option is A.

Force is a physical factor that alters or has the potential to alter an object's state of rest or motion as well as its shape. Newton is the SI unit of force.

One of the most fundamental types of physical entities is the force. Due to the fact that force is a vector quantity, it possesses both a magnitude and a direction. Frictional force and contact force are the two categories under which all forces can be classified.

F = m × a

F = 0.21 × 9.81

F = 2.06 N

Thus the correct option is A.

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Answer:

patron

Explanation:

 The patron is something or someone who defends some cause or point of view. In the art field, for example, the patron may be considered a sponsor, that is, someone who is known for defending a particular group of people or specific situation.

  The patron is one who advocates, advises and directs. In the military, patrons are heroic figures who are chosen to defend a military unit, for example.

Explanation:

patron

Explanation:

 The patron is something or someone who defends some cause or point of view. In the art field, for example, the patron may be considered a sponsor, that is, someone who is known for defending a particular group of people or specific situation.

  The patron is one who advocates, advises and directs. In the military, patrons are heroic figures who are chosen to defend a military unit, for example.

(a) The peak voltage across the resistor is 339.41 V.

(b) The maximum current through the resistor is 0.23 A.

(c) The rms current through the resistor is 0.16 A.

(d)  The average power dissipated by the resistor is 38.4 W.

Vrms = 0.7071V₀

where;

V₀ = Vrms/0.7071

V₀ = 240/0.7071

V₀ = 339.41 V

I(rms) = V(rms)/R

I(rms) = (240)/(1,540)

I(rms) = 0.16 A

I₀ = I(rms)/0.7071

I₀ = (0.16)/0.7071

I₀ = 0.23 A

P = I(rms) x V(rms)

P = 0.16 x 240

P = 38.4 W

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(a) The angular speed of the rod, when it lands on the horizontal surface is 1.61 rad/s.

(b) The the angular acceleration of the rod, just before it touches the horizontal surface is 1.06 rad/s² .

The angular acceleration of the rod is calculated as follows;

Apply the principle of conservation of angular momentum;

τ = Iα

where;

τ = Fr = mg(L/2) cosθ

I = mL²/3

mg(L/2) cosθ =  mL²/3(α)

g(1/2) cosθ =  L/3(α)

(³/₂)g cosθ  = L(α)

(³/₂ L)g cosθ  = α

1.5Lg cosθ  = α

1.5(2.1) cos (70.4) = α

1.06 rad/s² = α

ωf² = ω₁² + 2αθ

ωf² = 0 + 2(1.06)(70.4π/180)

ωf² = 2.605

ωf = √2.605

ωf = 1.61 rad/s

Thus, the angular speed of the rod, when it lands on the horizontal surface is 1.61 rad/s.

The the angular acceleration of the rod, just before it touches the horizontal surface is 1.06 rad/s² .

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The difference in weight  of a 117 kg person as measured at sea level and at the top of the Vinson Massif, the highest peak in Antarctica is -1.847 N

To find the difference in weight, we find the change in acceleration due to gravity.

The acceleration due to gravity is the acceleration exerted by the force of gravity on an object. It is given by

g = GM/r² where

Now, to find the change in g, we differentiate g with respect to r,

So, dg/dr = d(GM/r²)

= -2GM/r³

So, the change in acceleration due to gravity, Δg is

Δg = dg/dr × Δr

= -2GM/r³ × Δr

Now, Δr = height above sea level of Vinson Massif = 5.14 × 10³ m

Substituting the values of the variables into the equation, we have

Δg = -2GM/r³ × Δr

Δg = [(-2 × 3.99 × 10¹⁴ Nm²/kg)/(6.38 × 10⁶ m)³] × 5.14 × 10³ m

Δg =[ (-7.98 × 10¹⁴ Nm²/kg)/259.69 × 10¹⁸ m³] × 5.14 × 10³ m

Δg = -0.03073 × 10⁻⁴ N/mkg × 5.14 × 10³ m

Δg = -0.1579 × 10⁻¹ N/kg

Δg = -0.01579 N/kg

Since weight W = mg where

The difference in weight is dW = mdg

Since

dW = mdg

= 117 kg × -0.01579 N/kg

= -1.847 N

So, the difference in weight  of a 117 kg person as measured at sea level and at the top of the Vinson Massif, the highest peak in Antarctica is -1.847 N

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Answer:

A. To determine jobs for people

The uniformly distributed load per meter the beam may carry will be 1.275 × 10 ⁶ N/mm.

When an object is subjected to a heavy load at a specific spot, it often experiences bending stress, which causes the object to bend and tire.

The given data in the problem is;  

Bending stress, σ = 120 N/mm2  

Moment of inertia, I = 8.5 × 106 mm⁴

Depth of beam, y = d/2 = 200/2 = 100 mm  

Length of beam, L = 8 m = 8000 mm  

Width of beam, W = 300 mm

The maximum bending moment of the beam with UDL;

[tex]\rm W = \frac{wL^2}{8}[/tex]

From the bending equation;

[tex]\rm M = \frac{\sigma I}{y_{max}} \\\\ M = \frac{120 \ N / mm^2 \times 8.5 \times 10^6 }{100 \ mm } \\\\ M = 10.2 \times 10^6 \ N - mm[/tex]

The maximum bending moment of the beam with UDL;

[tex]\rm M = \frac{wL^2}{8} \\\\ 10.2 \times 10^6 = \frac{w \times 8^2}{8} \\\\ w = 1.275 \times 10^6 \ N/mm[/tex]

Hence the uniformly distributed load per meter the beam may carry will be 1.275 × 10 ⁶ N/mm.

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The acceleration and time is mathematically given as

a=gsinθ

[tex]t=\frac{2d}{gsinθ}[/tex]

Generally, the equation for the Force of the car is mathematically given as

[tex]N = mg cos\theta[/tex]

Therefore

[tex]ma=mgsin\theta[/tex]

a=gsinθ

The acceleration of the car is mathematically given as

a=gsinθ

Generally, the equation for the final velocity of the car is mathematically given as

[tex]v^2−u^2=2as[/tex]

Therefore

[tex]v 2 −(0) 2 =2(gsin\theta)d[/tex]

[tex]v= \sqrt {2gsin\theta.d}[/tex]

​

Generally, the equation for the motion of the car is mathematically given as

v−u=at

[tex]2gsin\thetad =(gsin\theta)t[/tex]

[tex]t=\frac{2d}{gsinθ}[/tex]

In conclusion, the time it takes the front bumper to reach the bottom of the hill is

[tex]t=\frac{2d}{gsinθ}[/tex]

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The ball's height [tex]y[/tex] at time [tex]t[/tex] is given by

[tex]y = 29.2\,\mathrm m - \left(8.30\dfrac{\rm m}{\rm s}\right) t - \dfrac12 gt^2[/tex]

where [tex]g=9.80\frac{\rm m}{\mathrm s^2}[/tex].

Solve for [tex]t[/tex] when [tex]y=0[/tex]. Omitting the units, we have

[tex]29.2 - 8.30t - \dfrac g2 t^2 = 0[/tex]

I'll solve by completing the square.

[tex]29.2 - \dfrac g2 \left(t^2 + \dfrac{16.6}g t\right) = 0[/tex]

[tex]29.2 - \dfrac g2 \left(t^2 + \dfrac{16.6}g t + \dfrac{8.3^2}{g^2}\right) = -\dfracg2 \times \dfrac{8.3^2}{g^2}[/tex]

[tex]29.2 - \dfrac g2 \left(t + \dfrac{8.3}g\right)^2 = -\dfrac{8.3^2}{2g}[/tex]

[tex]\dfrac g2 \left(t + \dfrac{8.3}g\right)^2 = 29.2 + \dfrac{8.3^2}{2g}[/tex]

[tex]\left(t + \dfrac{8.3}g\right)^2 = \dfrac{58.4}g + \dfrac{8.3^2}{g^2}[/tex]

[tex]t + \dfrac{8.3}g = \pm \sqrt{\dfrac{58.4}g + \dfrac{8.3^2}{g^2}}[/tex]

[tex]t = -\dfrac{8.3}g \pm \sqrt{\dfrac{58.4}g + \dfrac{8.3^2}{g^2}}[/tex]

[tex]\implies t \approx -3.43 \text{ or } t \approx 1.74[/tex]

Ignore the negative solution; the ball hits the ground about 1.74 s after being thrown.

The acceleration due to gravity of the asteroid is 833.85 N.

F = mg

where;

Assuming the acceleration due to gravity of the asteroid = 9.81 m/s²

F = 85 x 9.81

F = 833.85 N

Thus, the acceleration due to gravity of the asteroid is 833.85 N.

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The  energy was lost to air friction is determined as 42.56 J.

The energy lost due to friction is calculated as follows;

ΔE = ΔK.E + ΔP.E

where;

ΔE = ¹/₂(0.45)(19.89² - 15.1²) + (0.45)(9.8)(12 - 30.2)

ΔE = 37.7 - 80.26

ΔE = -42.56 J

Thus, the  energy was lost to air friction is determined as 42.56 J.

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Answer:

1. Information Gathering

2. Analysis and Planning.

3. Implementation of a solution.

4. Assessment of the effectiveness of the solution.

5. Documentation of the incident.

The answer to the question is that the vertical distance the ball clear the cross bar is 1.465 metres.

We have horizontal distance as 28 m, and velocity as 18 m/s and the angle of projection 46.8 degree.

so, we can say that

In horizontal direction,

28 = (18 cos 46°) t

t = 2.2723 seconds

Now, in vertical direction,

Let be the height of the ball

h = (18 sin 46°)t - (1/2)gt²

h = (18 sin 46°)t - 4.9t²

h = ( (13.1214) × (2.2723) ) - ( (4.9) × (2.2723)² )

h = 4.5154 metre

Vertical distance the ball clear the cross bar = h - 3.05

Vertical distance the ball clear the cross bar = 4.5154 - 3.05

Vertical distance the ball clear the cross bar = 1.465

Thus, we can conclude that after solving and applying the concepts of projectile motion we find out the vertical distance the ball will clear the cross bar is 1.465 metres.

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a.

b.

Since k is the spring constant of both springs = 250 N/m. The equivalent spring constant in parallel is k' = k + k

= 2k

= 2 × 250 N/m

= 500 N/m

Now since A is the maximum distance the block is pulled from its equilibrium position, the total energy of the block is E = 1/2kA

Also, 1/2k'A² = 1/2k'x² + 1/2Mv² where

Making v subject of the formula, we have

v = √[k'(A² - x²)/M]

Substituting the values of the variables into the equation, we have

v = √[k'(A² - x²)/M]

v = √[500 N/m(A² - (0)²)/10]

v = √[50 N/m(A² - 0)]

v = [√50]A m/s

v = [5√2] A m/s

v = 7.07 A m/s

So, the speed of the block of mass when the springs are connected in parallel is 7.07 A m/s

Since velocity of spring v = ω√(A² - x²) where

Making ω subject of the formula, we have

ω = v/√(A² - x²)

Since v = 7.07 A m/s

Substituting the values of the other variables into the equation, we have

ω = v/√(A² - x²)

ω = 7.07 A m/s/√(A² - 0²)

ω = 7.07 A m/s/√(A² - 0)

ω = 7.07 A m/s/√A²

ω = 7.07 A m/s/A m

ω = 7.07 rad/s

So, the angular velocity when the two springs are in parallel is 7.07 rad/s

Since k is the spring constant of both springs = 250 N/m. The equivalent spring constant in parallel is 1/k" = 1/k + 1/k

= 2/k

⇒ k" = k/2

k" = 250 N/m ÷ 2

= 125 N/m

Now since A is the maximum distance the block is pulled from its equilibrium position, the total energy of the block is E = 1/2kA

Also, 1/2k"A² = 1/2k"x² + 1/2Mv'² where

Making v subject of the formula, we have

v = √[k"(A² - x²)/M]

Substituting the values of the variables into the equation, we have

v = √[k"(A² - x²)/M]

v = √[125 N/m(A² - (0)²)/10]

v = √[125 N/m(A² - 0)]

v = [√125]A m/s

v = [5√5] A m/s

v = 11.2 A m/s

So, the speed of the block of mass when the springs are connected in series is 11.2 A m/s

Since velocity of spring v = ω√(A² - x²) where

Making ω subject of the formula, we have

ω = v/√(A² - x²)

Since v = 11.2 A m/s

Substituting the values of the other variables into the equation, we have

ω = v/√(A² - x²)

ω = 11.2 A m/s/√(A² - 0²)

ω = 11.2 A m/s/√(A² - 0)

ω = 11.2 A m/s/√A²

ω = 11.2 A m/s/A m

ω = 11.2 rad/s

So, the angular velocity when the two springs are in series is 11.2 rad/s

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Answer:

dependent variable

Explanation:

The acceleration is 2.2 m/s^2 and the net force is  62N towards the right

The net force is the effective force that acts on a body in a give direction.

1. The net force in the y axis is; ∑Fy = 0

2. The net force along the y axis is; ∑Fx =  277 N - 215 N

3. The normal reaction is given by;  28 kg * 9.8 m/s^2 = 274.4 N

4. The net force in the x axis is  277 N - 215 N = 62N towards the right

5. The acceleration of this force is obtained from;

F = ma

62 = 28 a

a = 62/28

a = 2.2 m/s^2

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The coasts of Africa and South America were touched  125 million years ago due to continental drift.

The time can be computed from the ratio of distance and speed.

Given:

Distance = 5, 000,000

Speed = 4 cm/year = 0.04 m/year

The time is computed as:

Time = Distance / Rate

Time = 5,000,000 meters / 0.04 meters per year

Time = 125,000,000 years

Hence, the coasts of Africa and South America were touched  125 million years ago due to continental drift.

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Their velocity (in m/s) just after impact if they cling together is -0.9 m/s.

Apply the principle of conservation of linear momentum;

Pi = Pf

where;

(92.5)(5.5)  +  (116)(-6)  =  v(92.5 + 116)

508.75 - 696 = v(208.5)

-187.25 = 208.5v

v = -187.25/208.5

v = -0.9 m/s

Thus, their velocity (in m/s) just after impact if they cling together is -0.9 m/s.

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Answer:

B. why you must push harder to move a car farther.

Explanation:

acceleration (gaining speed) happens when a force acts on a mass (object). FORCE=MASSxACCELERATION

Initial speed of ball will be 2.7m/s.

When a body is rotating about an axis, then it has kinetic energy.

And this energy is called rotational kinetic energy.

It is given as -  R.K.E. = 1/2 Iω²

And if a ball is rolling without slipping.

Then the moment of inertia of the solid ball is written as -

I = 25MR²

Vi = Rω

Here it is given in the problem that-

height(h) = 0.53m

Now by the conservation of energy we can write the equation as -

1/2MVi² + 1/2Iω² = Mgh

so that -

(1/2)MVi² + (1/2)×(2/5MR²) ×(Vi/R)² = Mgh(1/2)Vi² + (1/5)Vi²

= gh(7/10)Vi² = 9.8 × 0.53

Vi = 2.7 m/s

So that the initial velocity of ball came out to be 2.7m/s after applying all concepts or rotational motion.

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(a) The maximum height reached by the projectile is 10.2 m and

(b) The horizontal distance traveled by the projectile is 40.8 m.

H = u²sin²θ/2g

H = (20²(sin(45))² / (2 x 9.8)

H = 10.2 m

T = (2u sinθ)/g

T = (2 x 20 x sin(45) )/(9.8)

T = 2.886 s

x = (v cos(45) x t

x = (20 x cos(45)) x 2.886

x = 40.8 m

Thus, the maximum height reached by the projectile is 10.2 m and the horizontal distance traveled by the projectile is 40.8 m.

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The upward force exerted on the board by the support is mathematically given as

Fu= 764.8 N

Generally, the equation for is  mathematically given as

Considering that the Net Force on the system is null

The weight of the children plus the weight of the board equals the upward force imposed on the support.

The upward force

Fu= 440 + 272 + 52.8 N

Fu= 764.8 N

In conclusion, he upward force exerted on the board by the support

Fu= 764.8 N

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