Answer:
the net electric charge of the iron atom is 2.24 × 10⁻¹⁸ C
Explanation:
Given the data in the question;
we know that neural iron atoms should have same number of protons and electrons i.e 26 and 26 but in the sun's atmosphere, many of their electrons stripped away.
so; we have 26 protons and 12 electrons, meaning ( 26-12=14) 14 electrons have been stripped away.
when we loss electron, we gain positive charge
so, the Net charge of the iron atom of with 26 protons and 12 electrons will be;
q = 26 ( proton charge) + 12( electron charge )
we know that; a singular proton has a charge of 1.6×10⁻¹⁹ C
protons carry negative charge while electrons carry negative
we substitute
q = 26 ( 1.6×10⁻¹⁹ ) + 12( -1.6×10⁻¹⁹ )
q = 2.24 × 10⁻¹⁸ C
Therefore, the net electric charge of the iron atom is 2.24 × 10⁻¹⁸ C
If there is an outlier on your graph, you should ignore it.
O True
O False
Answer:
false
Explanation:
21. Which substance is a gas at 45 degrees C?* A. Ethanol O B. Hexane C. Methane O D. Methanol
Answer:
C. Methane
Explanation:
A chemical equation can be defined as a reaction between two chemical elements.
For a chemical reaction (equation) to be balanced, the condition which must be met is that the number of atoms contained in the reactants must be equal to the number of atoms in the products.
This ultimately implies that, the mass and charge of the chemical equation are both balanced properly.
Methane is a hydrocarbon compound formed by the reaction between four (4) hydrogen atoms with a carbon atom. It belongs to the alkane group of hydrocarbons and has the following physical and chemical properties; colorless, highly flammable, non-toxic and odorless gas. The empirical formula for Methane is CH4.
Methane is a gas at 45 degrees celsius. This ultimately implies that, at 45°C methane is a gas because it is still within the range of standard temperature and pressure.
An RLC series circuit has an applied voltage of 277 volts. The current through the resistor is 3.5 A. What is the apparent power of the circuit
Answer:
the apparent power of the circuit is 969.5 VA.
Explanation:
Given;
Applied voltage, E = 277 V
current flowing in the circuit, I = 3.5 A
Apparent occurs when the applied voltage and current are out of phase because of reactance. This apparent power of the circuit is calculated as;
[tex]P = EI\\\\P= 277 \ \times \ 3.5 \\\\P = 969.5 \ VA[/tex]
Therefore, the apparent power of the circuit is 969.5 VA.
Consider a runner in lane 1 with a speed v1 and a runner in lane 4 with a speed v4 , where the ratio of the speeds v4/v1 is 2 . The radius of lane 4 is b times the radius of lane 1, and both runners have the same mass. Which of the following claims is correct about the net forces F1 and F4 exerted on the runners in lanes 1 and 4, respectively?
a. F4 is equal to 2/b F1.
b. F4 is equal to 4/b F1.
c. F4 is equal to 2/b^2 F1.
d. F4 is equal to 4/b^2 A F1.
Answer:
[tex]F_4=\frac{4}{b}F_1[/tex]
Explanation:
Centripetal force is the net force acting on a body which makes it move along a curved path. This force is always towards the center of curvature.
The centripetal force is given by:
F = mv² / r
where m is the mass of the body, v is the velocity of the body, r is the radius, F is the centripetal force and v²/r is the centripetal acceleration.
Given that:
[tex]\frac{v_4}{v_1}=2\\\\v_4=2v_1\\\\ Also, radius\ of\ lane\ 4(r_4)=2*radius\ of\ lane\ 1\\\\Let\ r_1\ be\ radius\ of\ lane\ 1,m=mass\ of\ runner\ in\ lane\ 1=mass\ of\ runner\ in\ lane\ 4\ \\and\ v_4=velocity\ of\ runner\ in\ lane\ 4.\ Hence:\\\\r_4=b*r_1=br_1\\\\The\ centripetal\ force\ for\ lane\ 4(F_4)\ is:\\\\F_4=\frac{mv_4^2}{r_4}\\\\ F_4=\frac{m(2v_1)^2}{br_1}\\\\F_4= \frac{4mv_1^2}{br_1} \\\\But\ F_1=\frac{mv_1^2}{r_1}\\\\Hence\ F_4=\frac{4}{b}F_1[/tex]
F4 is equal to 4/b F1. A further explanation is provided below.
According to the question,
[tex]\frac{v_4}{v_1} =2[/tex][tex]\frac{r_4}{r_1} = b[/tex]The net force of the object in uniform circular motion will be:
→ [tex]F = \frac{mv^2}{r}[/tex]
∴ [tex]F_1 = \frac{m_1 v_1^2}{r_1}[/tex]
[tex]F_4 = \frac{m_4 v_4^2}{r_4}[/tex]
Now,
→ [tex]\frac{F_1}{F_4} = \frac{m_1}{m_4}\times (\frac{v_1}{v_4} )^2\times \frac{r_4}{r_1}[/tex]
[tex]\frac{F_1}{F_4} = \frac{1}{4}\times b[/tex]
[tex]F_1 = \frac{b}{4} F_4[/tex]
[tex]F_4 = \frac{4}{b} F_1[/tex]
Thus the above response i.e., "option b" is correct.
Learn more:
https://brainly.com/question/4527006
Identify the characteristics of an oppressor.
Oppressors have a(n)
attitude toward others. They have a need for
.
Answer:
Oppressors have a repressive attitude toward others. They have a need for power and dominance.
Explanation:
Oppression is the painful experience of deliberate arbitrariness, violence and abuse of power inflicted on an individual, society or group of people. The term repression is often used synonymously for this.
Oppression arises from the general, also unconscious, assumption that a certain group of people is inferior or without rights. Oppression is seldom limited to governmental activities. Individuals can also be victims of oppression if they lack the solidarity of a social group.
Oppressive individuals, in turn, are those who exert repression on others. They are generally narcissistic and megalomaniacal people, who have a wrong self-perception of their virtues and negatively weigh the virtues of other people.
PLEASE HELPPPPPP!!!!
Answer:
b
Explanation:
i just did that question
Suppose you have a total charge qtot that you can split in any manner. Once split, the separation distance is fixed. How do you split the charge to achieve the greatest force
Answer:
Answer is explained in the explanation section below.
Explanation:
Solution:
We know from the Coulomb's Law that, Coulomb's force is directly proportional to the product of two charges q1 and q2 and inversely proportional to the square of the radius between them.
So,
F = [tex]\frac{Kq1q2}{r^{2} }[/tex]
Now, we are asked to get the greatest force. So, in order to do that, product of the charges must be greatest because the force and product of charges are directly proportional.
Let's suppose, q1 = q
So,
if q1 = q
then
q2 = Q-q
Product of Charges = q1 x q2
Now, it is:
Product of Charges = q x (Q-q)
So,
Product of Charges = qQ - [tex]q^{2}[/tex]
And the expression qQ - [tex]q^{2}[/tex] is clearly a quadratic expression. And clearly its roots are 0 and Q.
So, the highest value of the quadratic equation will be surely at mid-point between the two roots 0 and Q.
So, the midpoint is:
q = [tex]\frac{Q + 0}{2}[/tex]
q = Q/2 and it is the highest value of each charge in order to get the greatest force.
Which gives off more heat, a potato heated to 375°F or a 20-lb turkey heated to the same temperature? Explain your answer.
Answer:
the heat emitted by potatoes is much less than the heat emitted by turkey
Explanation:
The heat transfer is given by
Q = m [tex]c_{e}[/tex] ΔT
where m is the mass of the body, c_e is the specific heat and ΔT is the change in temperature of the body
in this case they ask us to appear the heat emitted by the potatoes and the turkey.
From the tables found the specious colors of both
potato c_e = 0.82 Kcal / kg ºC
turkey c_e = 0.72 Kcal / kg ºC
Since the two bodies have the same initial temperature and we assume that they reach the same final (ambient) temperature, the ratio of the heat emitted is
Q₁ / Q₂ = [tex]\frac{ m_1 \ c_{e1} }{m_2 \ c_{e2} }[/tex]
where we use the rise 1 for the potatoes and the su index 2 for the turkey
the mass of the potatoes can be estimated at approximately m = 1 ka
we substitute
Q₁ / Q₂ = [tex]\frac{1 \ 0.82}{ 20 0.72}[/tex]
Q₁ /Q₂ = 0.0569
Q₁ = 0.0569 Q₂
therefore the heat emitted by potatoes is much less than the heat emitted by turkey
At t = 0, a car registers at 30 miles/hr. Forty seconds later, the car’s velocity is now at 50 miles/hr. Assuming constant acceleration, find the value of this acceleration.
Select one:
a. 20 miles/hr2
b. 0.5 miles/hr2
c. 0.5 miles/s2
d. 1.38 x 10-4 miles/s2
e. None of the given choices.
A 3.0-kilogram object is placed on a frictionless track at point A and released from rest.
(Assume the gravitational potential energy of the system to be zero at point C.)
- 4.0 m
m-3.0 kg
3.0 m
2.0 m
1.0 m
0.0 m
b. Calculate the kinetic energy of the object at point B. [Show all work, including the equation
and substitution with units.] [2]
Answer:
Explanation:
From the given information:
The gravitational potential energy at A is:
[tex]PE_A = mgh[/tex]
GIven that:
mass(m) = 3
height (h) = 3
[tex]PE_A =3\times 9.81 \times 3[/tex]
[tex]\mathbf {PE_A =88.29 \ J }[/tex]
b)
Using the conservation of mechanical energy:
[tex]PE_A+KE_A = PE_B + KE_B[/tex]
[tex]88.29 +0 = 3(9.8)(1) + KE_B[/tex]
[tex]88.29 = 29.4 + KE_B[/tex]
[tex]KE_B = 88.29 -29.4[/tex]
[tex]\mathbf{ KE_B= 58.89 \ J}[/tex]
From the data provided and the calculations done, the potential energy at A is 88.3 J while the kinetic energy at B is 58.9 J.
What is the Potential energy at A?The potential energy of a body is given as:
PE = mghwhere:
m = mass of the object
g = acceleration due to gravity
h = h
At A:
m= 3.0 kg
g = 9.8 m/s^2
h = 3.0 m
PE = 3 × 9.8 × 3
PE = 88.3 J
What is the kinetic energy at B?Kinetic energy, KE of a body is given as:
KE = 1/2 × mv^2where:
m is mass of the bodyv is velocity of the bodyFrom the law of conservation of energy, the sum of potential energy and Kinetic energy at every point is equal.
PE + KE at A = PE + KE at B
KE at B = PE + KE at A - PE at B
At A, PE = 88.3 J and KE = 0
At B; h = 1.0 m
PE = 3 × 9.8 × 1
PE = 29.4 J
Therefore, KE at B = 88.3 + 0 - 29.4
KE at B = 58.9 J
Therefore, the potential energy at A is 88.3 J while the kinetic energy at B is 58.9 J.
Learn more about potential energy and Kinetic energy at: https://brainly.com/question/14427111
For both circuits: Determine the potential difference on and the current through each resistor. Show your work and then check your answer with PhET
Answer:
Answer is explained in the explanation section below.
Explanation:
Note: This question is incomplete and lacks the necessary diagram of the circuits to solve for this question. However, I have found similar question on the internet and dragged the diagrams out of it. I have attached the diagrams of the circuits for your ease. It has two circuits to be solved. Attachment 1 refers to first circuit and Attachment 2 refers to second circuit.
Solution:
We are asked to find the potential difference and current through each resistor.
So,
Calculations for Circuit 1: Please refer to Attachment 1
First we need to find the Resistance:
Req = [tex]\frac{9 . 6}{9+6} +3[/tex]
R = 6.6 Ohm
Now, we know that:
V = IR
So,
I = V/R
V = 12 V
I = 12 V/ 6.6 ohm
I = 1.82 Amperes.
Now,
The Potential Difference through 3 ohm resistor:
P.D = [tex]V_{3}[/tex] = 1.82 x 3
[tex]V_{3}[/tex] = 5.45 V
Now,
The Potential Difference through 6 ohm and 9 ohm resistor:
[tex]V_{6/9}[/tex] = 12V - 5.45V
[tex]V_{6/9}[/tex] = 6.545V
Now,
The current through 6 ohm is:
[tex]I_{6}[/tex] = [tex]\frac{6.545}{6}[/tex]
[tex]I_{6}[/tex] = 1.09 A
Now,
The Current through 9 ohm is:
[tex]I_{9}[/tex] = [tex]\frac{6.545}{9}[/tex]
[tex]I_{9}[/tex] = 0.72 A
Similarly,
Calculations for Circuit 2: Please refer to attachment 2:
Req = [tex]\frac{(30+50).90}{90+(30+50)} +20[/tex]
Req = 62.35 ohm
I = V/R
I = 12 V/ 62.35ohm
I = 0.192 Amperes.
Now,
The The Potential Difference through 20 ohm resistor:
[tex]V_{20}[/tex] = 20 x 0.192
[tex]V_{20}[/tex] = 3.849 V
Now,
The The Potential Difference through 90 ohm resistor:
[tex]V_{90/(30+50)}[/tex] = 12 - 3.849
[tex]V_{90/(30+50)}[/tex] = 8.150
Now,
The Current through 90 ohm is:
[tex]I_{90}[/tex] = 8.1509/90
[tex]I_{90}[/tex] = 0.0905 Amperes
Now,
The Current through 30 and 50 ohm is:
[tex]I_{30/50}[/tex] = 0.192 - 0.0905
[tex]I_{30/50}[/tex] = 0.101 A
Now,
The Potential Difference through 50 ohm resistor:
[tex]V_{50}[/tex] = 50 x 0.101
[tex]V_{50}[/tex] = 5.094 V
The Potential Difference through 30 ohm resistor:
[tex]V_{30}[/tex] = 30 x 0.101
[tex]V_{30}[/tex] = 3.0566 V
How does a tire of a car interact with the road
Answer:
friction
Explanation:
the road is the pathway for the vehicle so the tures create friction to stay sturdie
If the coefficient of static friction is 0.35 and the normal force is 80 newtons, what is the maximum frictional force of the surface acting on the object? A. 9.8 newtons B. 28 newtons C. 80 newtons D. 23 newtons E. 35 newtons
Answer:
Option B. 28 N
Explanation:
From the question given above, the following data were obtained:
Coefficient of static friction (μ) = 0.35
Normal force (N) = 80 N
Frictional force (F) =?
The coefficient of static friction, frictional force and the normal force are related according to the following equation:
Coefficient of friction = frictional force / normal force
μ = F/N
With the above formula, we can obtain the frictional force as follow:
Coefficient of static friction (μ) = 0.35
Normal force (N) = 80 N
Frictional force (F) =?
μ = F/N
0.35 = F / 80
Cross multiply
F = 0.35 × 80
F = 28 N
Thus the frictional force is 28 N
Answer:
Option B. 28 N
Explanation:
A square plate of copper with 50.0-cm sides has no net charge and is placed in a region of a uniform electric field of 80.0 kN/C directed perpendicularly to the plate. Find (a) the charge density of each face of the plate and (b) the total charge on each face.
Answer:
σ = ±708 nC/m²
Q = ±177 nC
Explanation:
given data
Side of copper plate L = 50 cm
Electric field, E = 80 kN/C
solution
we get here Charge density,σ that is express as
σ = E x ε₀ ....................1
here ε₀ is Permittivity of free space that is 8.85 x 10⁻¹² C²/Nm²
so put value in eq1we get
σ = 80 x 10³ x 8.85 x 10⁻¹²
σ = 708 x 10⁻⁹ C/m²
σ = 708 nC/m²
and
now we get here total change on each faces
Q = σ A ...............2
Q = 708 x 10⁻⁹ x (0.50)²
Q = 177 nC
How much power would it take to move a chair 5 meters in 15 seconds using a force of 51 newtons?
Answer:
the power used by the person is 17 W.
Explanation:
Given;
applied force, F = 51 N
distance through which the chair is used, d = 5 m
time of motion, t = 15 s
The power used by the person is calculated as;
[tex]P = FV\\\\P = F \ \times \ \frac{d}{t} \\\\P = 51 \ \times \ \frac{5}{15} \\\\P = 17 \ W[/tex]
Therefore, the power used by the person is 17 W.
The existence of the dwarf planet Pluto was proposed based on irregularities in Neptune's orbit. Pluto was subsequently discovered near its predicted position. But it now appears that the discovery was fortuitous, because Pluto is small and the irregularities in Neptune's orbit were not well known. To illustrate that Pluto has a minor effect on the orbit of Neptune compared with the closest planet to Neptune:
Required:
a. Calculate the acceleration due to gravity at Neptune due to Pluto when they are 4.50x10^12 m apart, as they are at present. The mass of Pluto is 1.4x10^22 kg.
b. Calculate the acceleration due to gravity at Neptune due to Uranus, presently about 2.50x10^12 m apart, and compare it with that due to Pluto. The mass of Uranus is 8.62x10^25 kg.
Answer:
4.61 * 10^-14 m/s ; 9.20 * 10^-10 m/s
Explanation:
To acceleration due to gravity, a
a = GM / r²
Mass of pluto = 1.4x10^22 kg
Distance apart, r = 4.50x10^12 m
G = Gravitational constant = 6.67 * 10^- 11
a = (6.67*10^-11 * 1.4x10^22) / (4.50x10^12)²
a = 9.338 * 10^(-11 + 22) / 20.25x10^24
a = 0.461 * 10^(11 - 24)
a = 0.461 * 10^-13
a = 4.61 * 10^-14 m/s
B.)
Mass of Uranus = 8.62x10^25 kg
Distance apart, r = 2.50x10^12 m
a = (6.67*10^-11 * 8.62x10^25 ) / (2.50x10^12 )²
a = 57.4954 * 10^(-11 + 25) / 6.25x10^24
a = 9.199264 * 10^(14 - 24)
a = 9.199264 * 10^-10
a = 9.20 * 10^-10 m/s
An object initially at rest accelerates at 5 meters per second2 until it attains a speed of 30 meters per second. What distance does the object move while accelerating?
A) 30 m
B) 3 m
C) 90 m
D) 600 m
Answer:
Distance s = 90 meter
Explanation:
Given:
Initial velocity u = 0 m/s
Final velocity v = 30 m/s
Acceleration a = 5 m/s²
Find:
Distance s
Computation:
v² = u² + 2as
30² = 0² + 2(5)(s)
Distance s = 90 meter
You kick a soccer ball with a speed of 12 m/s at an angle of 21. How long does it take the ball to reach the top of its trajectory?
Answer:
Time, t = 0.439 seconds
Explanation:
Given the following data;
Angle, x = 21°
Initial velocity, u = usinx = 12Sin21° = 4.3004m/s
We know that acceleration due to gravity is equal to 9.8m/s
Final velocity = 0m/s since it reached its highest trajectory.
To find the time, we would use the first equation of motion;
V = U + gt
0 = 4.3004 + (-9.8)t
0 = 4.3004 - 9.8t
9.8t = 4.3004
t = 4.3004/9.8
t = 0.439 seconds
Therefore, it will take the ball 0.439 seconds to reach the top of its trajectory.
Answer: 0.44 s
Explanation:
that’s what I got correct on my test in physics ;)
When light bulbs are connected in parallel and you unscrew one light bulb, the other light bulbs will
a. All go out
b. All get dimmer
c. All get brighter
d. All stay on
a. All go out
b. All get dimmer
c. All get brighter
d. All stay on
A box of textbooks is on the middle shelf in the bookroom 1.3 m from the floor. If the janitor relocates the box to a shelf that is 2.6 m from the floor, how much work does he do
Answer:
Explanation:
Work done by janitor = increase in potential energy of book
Potential energy = mgh , m is mass , h is displacement in height and g is acceleration due to gravity .
here displacement in height of book = 2.6 - 1.3 = 1.3 m
so h = 1.3 m
m = mass of the box
increase in potential energy of box = m x 9.8 x 1.3
= 12.74 m
Work done by janitor = 12.74 m Joule .
An air conditioner removes heat steadily from a house at a rate of 750 kJ/min while drawing electric power at a rate of 5.25 kW. Determine (a) the COP of this air conditioner and (b) the rate of heat transfer to the outside air.
Answer:
[tex]Q'_H=1065kJ/min[/tex]
Explanation:
From the question we are told that
Heat Remove rate Q_o= 750 kJ/min
Power Draw P=5.25
a)Generally the coefficient of performance (COP) of an air conditioner is mathematically given as
[tex]COP=\frac{Q_o}{P}[/tex]
[tex]COP=\frac{750kJ/min}{5.25kw}[/tex]
Where
[tex]750kJ/min=12.5kJ/sec[/tex]
[tex]COP=\frac{12.5kJ/sec}{5.25kw}[/tex]
[tex]COP=2.38[/tex]
b)Generally the equation for power is mathematically represented as
[tex]P=Q_H-Q_o[/tex]
[tex]P+Q_o=Q_H[/tex]
[tex]Q_H=P+Q_o[/tex]
[tex]Q_H=5.25+12.5[/tex]
[tex]Q_H=17.75kW[/tex]
Generally rate of heat transfer to the outside air is mathematically given as
[tex]Q'_H=17.75*60[/tex]
[tex]Q'_H=1065kJ/min[/tex]
How do you think the sock and underwear became charged?
Is Obesity primarily due to Genetics?
what is work??
ASAP who pls
A force of 22.04 N is applied tangentially to a wheel of radius 0.340 m and gives rise to an angular acceleration of 1.20 rad/s2. What is the moment of inertia of the wheel
Answer:
the moment of inertia of the wheel is is 6.245 kg.m²
Explanation:
Given;
applied force, f = 22.04 N
radius of the wheel, r = 0.34 m
angular acceleration, a = 1.2 rad/s²
The moment of inertia of the wheel is calculated as;
[tex]I = \frac{\tau}{\alpha } \\\\I = \frac{Fr}{\alpha } \\\\I = \frac{22.04 \ \times \ 0.34 }{1.2} \\\\I = 6.245 \ kg.m^2[/tex]
Therefore, the moment of inertia of the wheel is is 6.245 kg.m²
what
must always be
included on the
graph
I NEED HELP WITH THE LAST QUESTION PLS HELP!! (The one below 8)
Answer:
I. 6 cells .
II. Series connection.
Explanation:
I. Determination of the number of cells needed.
From the question given above,
Total voltage (V) = 9 V
1.5 V = 1 cell
Number of cells needed =?
The number of cells needed to make the 9V battery can be obtained as follow:
1.5 V = 1 cell
Therefore,
9 V = 9 V × 1 cell / 1.5 V
9 V = 6 cells
Thus, 6 cells of 1.5 V each is needed
II. Determination of the connection line
Total voltage (Vₜ) = 9 V
Cell 1 (V₁) = 1.5 V
Cell 2 (V₂) = 1.5 V
Cell 3 (V₃) = 1.5 V
Cell 4 (V₄) = 1.5 V
Cell 5 (V₅) = 1.5 V
Cell 6(V₆ ) = 1.5 V
For parrall connection:
Vₜ = V₁ = V₂ = V₃ = V₄ = V₅ = V₆
9 V = 1.5 V =... = 1.5 V
For series connection:
Vₜ = V₁ + V₂ + V₃ + V₄ + V₅ + V₆
9 = 1.5 + 1.5 + 1.5 + 1.5 + 1.5 + 1.5
9 V = 9 V
From the illustration above, we can see that series connection of each cells will give a total volt of 9 V unlike the parallel connection which resulted to 1.5 V.
Therfore, the cells should be arranged in series connection
⚠️Pls help me this is due soon!⚠️
professional is the answer
3rd
A sports car moving at constant velocity travels 130 m in 4.9 s . If it then brakes and comes to a stop in 3.8 s , what is the magnitude of its acceleration (assumed constant) in m/s2
Answer:
the magnitude of the constant acceleration is 6.98 m/s²
Explanation:
Given;
distance traveled at a constant velocity, d = 130 m
time of motion, t = 4.9 s
The constant velocity is calculated as;
[tex]v = \frac{d}{t} \\\\v = \frac{130}{4.9} \\\\v = 26.53 \ m/s[/tex]
The car comes to a stop in 3.8 s.
When the car comes to a complete stop, the final velocity of the car will be zero.
the constant acceleration is calculated as;
v = u + at
0 = u + at
at = -u
a = -u/t
a = (-26.53) / (3.8)
a = -6.98 m/s²
Therefore, the magnitude of the constant acceleration is 6.98 m/s²
Could someone please help me out with this problem? Thank you so much!! :)
Answer:
67.17 m
Explanation:
From the question given above, the following data were obtained:
Initial velocity (u) = 32 m/s
Angle of projection (θ) = 20°
Acceleration due to gravity (g) = 9.8 m/s²
Range (R) =?
The range can be obtained as follow:
R = u²Sine 2θ / g
R = 32² × Sine (2×20) / 9.8
R = 1024 × Sine 40 / 9.8
R = (1024 × 0.6428) / 9.8
R = 67.17 m
Thus, the range of the ball is 67.17 m
