Answer:
the average force exerted on the ball by the bat is 11,613.27 N
Explanation:
Given;
mass of the baseball, m = 151 g = 0.151 kg
initial velocity of the baseball, u = 39.5 m/s
final velocity of the baseball, v = 45.1 m/s
time of action, t = 1.10 ms = 1.10 x 10⁻³ s
The average force exerted on the ball by the bat is calculate as;
[tex]F = ma = \frac{m(v-u)}{t} \\\\F = \frac{0.151(45.1-(-39.5))}{1.10\times 10^{-3}} \\\\F = \frac{0.151(45.1\ +\ 39.5)}{1.10\times 10^{-3}} \\\\F = 11,613.27 \ N[/tex]
Therefore, the average force exerted on the ball by the bat is 11,613.27 N
Force excerted is defined as the product of mass and rate of change of velocity .The average force exerted on the ball by the bat will be 11,613.27 N.
What is force?Force is defined as the push or pull applied to the body. Sometimes it is used to change the shape, size, and direction of the body. Force is defined as the product of mass and acceleration. Its unit is Newton.
m is the mass of the baseball = 151 g = 0.151 kg
u is the initial velocity of the baseball =39.5 m/s
v is the final velocity of the baseball = 45.1 m/s
t is time of contact = 1.10 ms = 1.10 x 10⁻³ s
The average force mathematically given as
[tex]\rm F=\frac{m(v-u)}{t} \\\\\rm F=\frac{0.151(45.1-39.85)}{1.10\times10^-3} \\\\\rm F=11,613.27 N[/tex]
Hence the average force exerted on the ball by the bat will be 11,613.27 N.
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Lolliguncula brevis squid use a form of jet propulsion to swim—they eject water out of jets that can point in different directions, allowing them to change direction quickly. When swimming at a speed of 0.15m/s0.15m/s or greater, they can accelerate at 1.2m/s21.2m/s 2 .
(a) Determine the time interval needed for a squid to increase its speed from 0.15m/s0.15m/s to 0.45m/s0.45m/s.
(b) What other questions can you answer using the data?
Answer:
a) t = 0.25 s, b) x = 0.075 m
Explanation:
a) For this exercise we will use kinematic relationships in one dimension
v = v₀ + a t
in the problem they indicate the initial velocity v₀ = 0.15 m / s, the final velocity v = 0.45 m / s and the acceleration of the squid a = 1.2 m / s²
t = [tex]\frac{v -v_o}{a}[/tex]
we calculate
t = [tex]\frac{0.45 - 0.15}{1.2}[/tex]
t = 0.25 s
b) We can also find the distance traveled during this acceleration
v² = v₀² + 2a x
x = [tex]\frac{v^2 -v_o^2 }{2a}[/tex]
let's calculate
x = [tex]\frac{0.45^2 - 0.15^2 }{2 \ 1.2}[/tex]
x = 0.075 m
Calculate the magnitude and direction of the resultant of the following forces
Answer:
add them
100+150 = 250N same direction
that's the resultant
same direction
First to answer gets brainliest
Answer:
Sodium (K)
Explanation:
Two 800 cm^3 containers hold identical amounts of a monatomic gas at 20°C. Container A is rigid. Container B has a 100 cm^2 piston with a mass of 10 kg that can slide up and down vertically without friction. Both containers are placed on identical heaters and heated for equal amounts of time.
Required:
a. Will the final temperature of the gas in A be greater, less than, or equal to the temperature in B?
b. Show both processes on a single PV diagram.
c. What are the initial pressures in containers A and B?
d. Suppose the heaters have 25 W of power and are turned on for 15s. What is the final volume of container B?
Answer:
1) Final Temperature of the gas in A will be GREATER than the temperature in B
2) Diagram of both processes on a single PV has been uploaded below
3) The Initial pressures in containers A and B is 3039.87 J/liters
4) the final volume of container B is 923.36 cm³
Explanation:
Given that;
Temperature = 20°C = 293 K
mass of piston = 10 kg
Area = 100cm³
Volume V = 800 cm³ = 0.8 L
ideal gas constant R = 8.3 J/K·mol
1)
Final Temperature of the gas in A will b GREATER than the temperature in B
2)
Diagram of both processes on a single PV has been uploaded below,
3)
Initial pressures in containers A and B
PV = nRT
P = RT/V
we substitute
P = (8.3 × 293) / 0.8
P = 2431.9 / 0.8
P = 3039.87 J/liters
Therefore, The Initial pressures in containers A and B is 3039.87 J/liters
4)
Given that;
power = 25 W
time t = 15s
the final volume of container B = ?
we know that;
work done = power × time
work done = 25 × 15 = 375
Also work done = P( V₂ - V₁ )
so we substitute
375 = 3039.87 ( V₂ - 0.8 )
( V₂ - 0.8 ) = 375 / 3039.87
V₂ - 0.8 = 0.12336
V₂ = 0.12336 + 0.8
V₂ = 0.92336 Litres
V₂ = 923.36 cm³
Therefore, the final volume of container B is 923.36 cm³
Students want to investigate the inverse relationship between the pressure and temperature of an ideal gas as predicted by the ideal gas law. Their plan is to use a gas filled cylinder with a movable piston on one end and a heater inside that can be turned on and off. The students will the measure the pressure and temperature of the gas. Which of the following refinements to this procedure will allow the students to observe the predicted relationship between pressure and temperature? Select two answers
A. Start with enough gas to have a pressure near atmospheric pressure, and repeat the experiment, removing gas from the cylinder each time.
B. Fix the piston in place so the volume of the pas remains constant.
C. Ensure the piston and cylinder walls are insulated to the gas can reach equilibrium for each set of measurements
D. Conduct the investigation under conditions of very high pressure to ensure ideal gas behavior
Answer:
Option B, Fix the piston in place so the volume of the pas remains constant
Explanation:
As we know
[tex]\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}[/tex]
The effect on variable due to another variable can be studied by keeping the third variable constant.
Hence, in order the study the variation of temperature with pressure or vice versa, the volume needs to fixed at a certain value.
Hence, option B is correct
The cart is given an initial push up the ramp. After this push, as the car moves up the ramp, the direction of the acceleration of the cart is ________ the ramp. After the reaches its highest point, turns around, and begins moving down the ramp, the direction of the acceleration of the cart is ________ the ramp. At the highest point the cart reaches on the ramp, when the cart momentarily comes to rest, the magnitude of the acceleration of the cart is _______.
Answer:
Explanation:
The only force acting on the cart is a component of its weight parallel to ramp downwards . No other force acts parallel to the ramp .
Even when the cart is moving up after the initial push , its weight is acting downwards so acceleration is acting downwards .
When the cart is stationary at the top position , its weight is acting downwards so acceleration is downwards at that moment also . When the cart is going downwards , still its weight is acting down so acceleration is acting downwards .
After this push, as the car moves up the ramp, the direction of the acceleration of the cart is _down _______ the ramp. After the reaches its highest point, turns around, and begins moving down the ramp, the direction of the acceleration of the cart is _down ________ the ramp. At the highest point the cart reaches on the ramp, when the cart momentarily comes to rest, the magnitude of the acceleration of the cart is _downwards ______.
Two blocks of the same mass but made of different material slide across a horizontal, rough surface and eventually come to rest. A graph of the kinetic energy of each block as a function of position along the surface . Which of the following is a true statement about the frictional force Ff that is exerted on the two blocks?
a. Fr=2F8, since the force of friction is represented as the slope for each of the two curves.
b. Fr.-12Fri, since the force of friction is represented as the inverse slope for each of the two curves.
c. Ff:=2Ffi, since the force of friction is represented as the inverse of the area bound by each curve and th horizontal axis.
d. Fe=1/2Fr., since the force of friction is represented as the area bound by each curve and the horizontal axis.
Answer:
a. [tex]\mathbf{F_{f_2} = 2 F_{f1}}[/tex], [tex]\mathbf { since \ the \ force \ o f \ friction \ is \ represented \ as \ the \ slope \ for \ each \ of \ the \ two \ curves.}[/tex]
Explanation:
From the information given;
Using the work-energy theorem
ΔKE = W = [tex]\mathbf{ F_f \times r}[/tex]
K = [tex]\mathbf{ F_f \times r}[/tex]
∴
[tex]\dfrac{K_1}{K_2} = \dfrac{F_{f1}}{F_{f2}} (\dfrac{r_1}{r_2})[/tex]
Since [tex]K_1 = K_2[/tex] and r_1 = 4, and r_2 = 2 (from the missing diagram which is attached below)
Then;
[tex]1 = \dfrac{F_{f1}}{F_{f2}} (\dfrac{4 \ m}{2 \ m})[/tex]
[tex]\mathbf{F_{f_2} = 2 F_{f1}}[/tex]
At what height does a 3500-kg truck have a potential energy of 90,000 J gravitational potential energy relative to the ground?
Answer:
MGH=energy
3500*9.8*h=90000
h=90000/34300
h=2.62m
Microbes such as bacteria have small positive charges when in solution. Public health agencies are exploring a new way to measure the presence of small numbers of microbes in drinking water by using electric forces to concentrate the microbes. Water is sent between the two oppositely charged electrodes of a parallel-plate capacitor. Any microbes in the water will collect on one of the electrodes.
Required:
a. On which electrode will the microbes collect?
b. How could the microbes be easily removed from the electrodes for analysis?
Answer:
The answer is below
Explanation:
a) Coulomb's law of electric force for charges at rest states that Like charges repel each other while unlike charges attract one another.
Therefore since the microbes has small positive charges, the microbe would be repelled by the positively charged electrodes and attracted by the negative charged electrodes.
Hence, the microbes would collect on the negatively charged electrodes.
b) The microbes can easily removed from the negative electrode for analysis by discharging the electrode from the source. Thereby making the electrode to be incapable of attracting the microbe.
An object undergoes constant acceleration after starting from rest and then travels 5m in the first second. Determine how far it will go in the next second
The speed will be 10 m/s after the 1st and 20 m/s after the 2nd for an average of 15 m/s. So it will travel 15 m during that 2nd second
Mark as brainlist
The object, which undergoes constant acceleration after starting from rest, will go in the next second 15 m.
What is acceleration?Acceleration is rate of change of velocity with time. Due to having both direction and magnitude, it is a vector quantity. Si unit of acceleration is meter/second² (m/s²).
Given parameters:
initial velocity of object: u = 0.
time = 1 second.
distance travelled: d= 5 m.
So, acceleration of the object: a = 2d/t² = (2×5)/1² m/s² = 10 m/s².
Hence, it will go in the next second = 1/2×a(2²-1²) m
= 1/2×10×3 m.
= 15 m.
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The stars, Rigel and Betelgeuse, are both found in the constellation Orion. Rigel is a blue supergiant, and Betelgeuse is a red supergiant. Which of the following correctly compares the temperatures of Rigel and Betelgeuse?
Answer:
batrix
Explanation:
Suppose Group A runs their experiment 3 times and calculates that the best estimate of the distance slid by red blocks is 15 + 3 feet. Group B runs a similar experiment 3 times and calculates that the best estimate of the distance slid by green blocks is 25 + 4 feet. Using what you learned in the above video, find the t' parameter for the comparison of the results of Groups A and B.
In comparing of result of experiment A and B, the result of A is by 2 factors differ from experiment B.
The given parameters;
in group A, distance traveled by the red block, s = 15 ± 3 feet
in group B, distance traveled by the green block, s = 25 ± 4 feet
To find:
the t' parameter for comparison of the results of Groups A and B.Since both groups performed the experiment at equal times, we assume the time for both motion = t
Also, assume the initial velocity of both blocks = 0
For group A, we set-up the equation of motion as follows;
[tex]s = v_0t + \frac{1}{2} at^2 \\\\15+ 3 = 0 + 0.5\times a_1t^2\\\\18 = 0.5a_1t^2\\\\t^2 = \frac{18}{0.5a_1} \\\\t^2 = \frac{36}{a_1}[/tex]
For group B, we set-up the equation of motion as follows;
[tex]25 + 4 = v_0t + \frac{1}{2} a_2t^2\\\\29 = 0.5\times a_2t^2\\\\t^2 = \frac{29}{0.5a_2} = \frac{58}{a_2}[/tex]
Solve the first equation and the second equation together;
[tex]\frac{36}{a_1} = \frac{58}{a_2} \\\\\frac{a_2}{a_1} = \frac{58}{36} \\\\\frac{a_2}{a_1} = 1.61[/tex]
The ratio of error margin of both experiments;
[tex]\frac{4}{3} = 1.33[/tex]
The resulting parameter for comparison;
[tex]parameter, t' = 1.33 \times 1.61 = 2.14 \approx 2.0[/tex]
Thus, in comparing of result of experiment A and B, the result of A is by 2 factors differ from experiment B.
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A woman on a bridge 84.5 m high sees a raft floating at a constant speed on the river below. She drops a stone from rest in an attempt to hit the raft. The stone is released when the raft has 6.00 m more to travel before passing under the bridge. The stone hits the water 4.00 m in front of the raft. Find the speed of the raft.
Answer:
V = 0.48 m/s
Explanation:
In this case, we need to analyze the given data by parts.
At first, we know that the woman is on a height of 84.5 m of a river. She drops a stone thinking that she may hit the raft that is traveling with a constant speed. When the raft is 6 m near the bridge, the woman drops the stone, and the stone hits the water when the raft is still 4 m far of the bridge.
With this given data, we can calculate the distance covered by the raft, because is traveling at a constant speed:
X = 6 - 4 = 2 m
And as it's traveling at constant speed then:
X = V.t
We have the distance of the raft, but not the time it took to cover that distance. This time will be the same time that the stone took to hit the water, therefore, if we can determine the time of the rock, well be determining the time of the raft to cover the distance, and then, we can determine it speed.
To determine the time of the rock, as the stone is going on a free fall, with an innitial speed of 0, the flight time of the rock will be:
y = gt²/2 ---> solving for t
2y/g = t²
t = √2y/g
If g = 9.8 m/s, and replacing the data we have that the flight time of the rock is:
t = √2*84.5 / 9.8
t = 4.15 s
So the rock took 4.15 s to hit the water, and it's also the time that the raft took to cover the distance of 2 m, then, it's speed:
V = X/t
V = 2 / 4.15
V = 0.48 m/sHope this helps
If a motorbike accelerates from 15m/s to 25m/s in 15 seconds how far does it travel in that time
Answer:
distance = 0.2330142 miles
= 0.2330142 mi
Explanation:
not sure if its right though....
A pitching machine is programmed to pitch baseballs horizontally at a speed of 134 km/h. The machine is mounted on a truck and aimed forward. As the truck drives toward you at a speed of 85 km/h, the machine shoots a ball toward you. A pickup truck moves to the left at a constant velocity. A pitching machine sits in the bed of the pickup truck. The pitching machine launches a baseball to the right with a different constant velocity. A man with a baseball mitt stands at rest some distance to the right of the truck. For each of the object pairings listed, determine the correct relative speed. The speed of the pitching machine relative to the truck The speed of the pitched ball relative to the truck The speed of the pitching machine relative to you The speed of the pitched ball relative to you
Answer: 134 = 143 = 151 = 166 = 176
Hope this helps!!
Sorry if it's incorrect!!
:'(
Which statement applies only to magnetic force instead of both electric and magnetic forces? O A. It acts between a north pole and a south pole. O B. It can push objects apart. O C. It can pull objects together. D. It acts between objects that do not touch.
Answer:
the answer would be A. electricity don't specify the direction of any cardinal points the flow of charges moves.
Answer:
A
Explanation:
I did the test on ap3x
Ang larong Latin at Sisiw ay________________________.
larong pinoy
Explanation:
ito ay larong Pinoy
An airplane used to drop water on brushfires is flying horizontally in a straight line at 180 mi/h at an altitude of 450 ft. Determine the distance d at which the pilot should release the water so that it will hit the fire at B. The distance at which the pilot should release the water so that it will hit the fire at B is ft.
Answer:
1399.2 ft
Explanation:
The initial velocity = 180 mph = [(180 * 5280) / (1 * 3600)] ft/s = 264 ft/s
[tex]In\ the \ horizontal\ direction(x)\\\\Initial\ velocity = v_{ox}=264\ ft/s\\\\distance\ travelled\ in\ x \ direction(x) =v_{ox}t\\\\\\For\ the\ vertical\ direction:\\\\initial\ velocity(y_{oy})=0\\\\vertical\ distance(y)=y_{oy}t+0.5gt^2\\\\but\ g\ =-32\ ft/s^2. Hence:\\\\y=0t+0.5(-32)t^2\\\\y=-16t^2\\\\At\ point\ B, y=-450, therefore:\\\\-450=-16t^2\\\\t^2=28.125\\\\t=5.3\ s\\\\The\ distance\ at\ which\ the\ pilot\ should\ release\ the\ water=x=v_{ox}t=264*5.3\\\\x=1399.2\ ft[/tex]
An important diagnostic tool for heart disease is the pressure difference between blood pressure in the heart and in the aorta leading away from the heart. Since blood within the heart is essentially stationary, this pressure difference can be inferred from a measurement of the speed of blood flow in the aorta. Take the speed of sound in stationary blood to be c.
a. Sound sent by a transmitter placed directly inline with the aorta will be reflected back to a receiver and show a frequency shift with each heartbeat. If the maximum speed of blood in the aorta is v, what frequency will the receiver detect? Note that you cannot simply use the textbook Doppler Shift formula because the detector is the same device as the source, receiving sound after reflection.
b. Show that in the limit of low blood velocity (v <
f= 2fo v/c
Answer:
a) f ’’ = f₀ [tex]\frac{1 + \frac{v}{c} }{1- \frac{v}{c} }[/tex] , b) Δf = 2 f₀ [tex]\frac{v}{c}[/tex]
Explanation:
a) This is a Doppler effect exercise, which we must solve in two parts in the first the emitter is fixed and in the second when the sound is reflected the emitter is mobile.
Let's look for the frequency (f ’) that the mobile aorta receives, the blood is leaving the aorta or is moving towards the source
f ’= fo[tex]\frac{c+v}{c}[/tex]
This sound wave is reflected by the blood that becomes the emitter, mobile and the receiver is fixed.
f ’’ = f’ [tex]\frac{c}{ c-v}[/tex]
where c represents the sound velocity in stationary blood
therefore the received frequency is
f ’’ = f₀ [tex]\frac{c}{c-v}[/tex]
let's simplify the expression
f ’’ = f₀ \frac{c+v}{c-v}
f ’’ = f₀ [tex]\frac{1 + \frac{v}{c} }{1- \frac{v}{c} }[/tex]
b) At the low speed limit v <c, we can expand the quantity
(1 -x)ⁿ = 1 - x + n (n-1) x² + ...
[tex]( 1- \frac{v}{c} ) ^{-1} = 1 + \frac{v}{c}[/tex]
f ’’ = fo [tex]( 1+ \frac{v}{c}) ( 1 + \frac{v}{c} )[/tex]
f ’’ = fo [tex]( 1 + 2 \frac{v}{c} + \frac{v^2}{ c^2} )[/tex]
leave the linear term
f ’’ = f₀ + f₀ 2[tex]\frac{v}{c}[/tex]
the sound difference
f ’’ -f₀ = 2f₀ v/c
Δf = 2 f₀ [tex]\frac{v}{c}[/tex]
where is a neutron located within an atom
Answer:
the neutron is located in the nucleus of an atom
Answer:
nucleus
Explanation:
Atoms are made up of protons and neutrons located within the nucleus, with electrons in orbitals surrounding the nucleus.
1. Hydrogen: For an electron in the lowest energy it can orbit around proton, they have a separation of 5.3 *10-11 m. If you have a 4.5*10-19J photon (bit of light) hit the electron, it will transfer all of its energy to the proton electron interaction and the electron will start orbiting at a larger radius. Assuming all the energy went into the potential energy, what is the new distance between the electron and proton.
Answer:
rₙ = 1,325 10⁻⁹ m
Explanation:
To solve this problem we use the bohr atomic model
Eₙ = -13.606 /n² [eV]
the brackets indicate that the units are in electron volts.
let's reduce the photon energy to eV
E = 4.5 10-19J (1 eV / 1.6 10⁻¹⁹ eV) = 2.8125 eV
This energy is in the visible range, so the transition must occur in this range, this is for the Balmer series whose initial number is n₀ = 2
for an atomic transition on two levels
ΔE = Eₙ - E₀ = [tex]\frac{-13.606}{n^2} + \frac{13.606}{2^2}[/tex]
2.8125 = [tex]\frac{-13.606}{n^2} + 3.4015[/tex]
[tex]\frac{13.606}{n^2}[/tex] = 3.4015 - 2.8125 = 0.589
n² = 13.606 / 0.589
n² = 23.1
n = 4.8
as n must be an integer
n = 5
taking the quantum number as far as the electron goes, we substitute in the equation for the radius
rn = n² a₀
where ao is the radius of the lowest level a₀ = 5.3 10⁻¹¹ m
rₙ = 5 2 5.3 10⁻¹¹
rₙ = 132.5 10⁻¹¹ m
rₙ = 1,325 10⁻⁹ m
an object has a mass of 2000kg what is its weight on earth
Answer:
19600 N
Explanation:
weight = mass x gravity
We know that gravity = 9.8 m/s^2 and mass = 2000 kg.
w = m x g
w = 2000 kg x 9.8 m/s^2
w = 19600 N
The weight of the object is 19600 N (newtons).
Answer:
the answer i 2000kg
Explanation:
There are 5 planets visible to the naked eye in the sky.
True
False
When a space shuttle takes off, the chemical reactions of the fuel give the shuttle the kinetic energy to leave Earth's atmosphere as shown in the figure below. The kinetic energy of the space shuttle is less than the potential energy of the fuel used. Which statement best explains this idea?
A.) The potential energy is used to overcome Earth’s gravity.
B.) The potential energy is also converted to light, thermal energy, and sound energy.
C.) The potential energy must be consumed to make the fuel burn.
D.) The potential energy is destroyed by the warmth of the reaction.
Answer:a
Explanation:
Because its has to use tihs potential energy to overcome the atmosphere so the shuttle will not go back down
How does energy move in relation to the medium in a transverse wave?
Answer:
Only the energy of the wave travels through the medium. In a transverse wave, particles of the medium vibrate up and down perpendicular to the direction of the wave. ... In a surface wave, particles of the medium vibrate both up and down and back and forth, so they end up moving in a circle.
can an object have kinetic energy if there is no motion.
No. The object has to have motion for it to have kinetic energy.
someone help me with this exercise ?
1. if a body with a mass of 350kg is subjected to a fare of 90n what will be its mass
?
Mass remains mass no matter what you do to it.
compare and contrast speed and velocity.
Speed is the time rate of an object moving from one place to another, while velocity is the rate and direction of the object's movement. They are very similar but they don't mean the same thing.
1)A rocket expels gas at high speed for a short period of time. We are going to treat the rocket as being far away from any gravitational objects.a)Draw a momentum chart for the rocket expelling gas in space.Take the initial time before expelling gas and the final time after the rocket has finished expelling gas. The rocket has an initial constant speed.Put the rocket and the expelled gas on separate rows.b)Use your chart to explainhow the speed of the rocket changes. c)Does the rocket have to keep expelling gas to stay at a constant speed
We have that
a) We Draw a graph to follow the equation
[tex]v=\frac{m_0u}{m_0-dt}[/tex]
b) The speed of the rocket changes because Momentum is conserved because with the law of conservation in mind mass is reduced thereby causing increase in speed
c) The rocket does NOT have to keep expelling gas to stay at a constant speed because On the account of maintaining speed the Rocket need not expel gas as that will only increase speed and not maintain it
a)
Let
Mass =m
Time t
Generally, the equation for mass ejection constant is mathematically given by
[tex]\phi=\frac{d_m}{d_t}[/tex]
Therefore
[tex]m=m_0-\phi t[/tex]
where
[tex]m_0=initial\ mass[/tex]
Apply the law of conservation of momentum which states that
Conservation of momentum, states that momentum can neither be lost nor gained in an isolated system
[tex]m_o\mu=mv[/tex]
[tex]v=\frac{m_0u}{m_0-dt}[/tex]
b)
Momentum is conserved because with the law of conservation in mind mass is reduced thereby causing increase in speed
c)
On the account of maintaining speed the Rocket need not expel gas as that will only increase speed and not maintain it.Therefore the answer is NO
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A 1325 kg car and a 2050 kg pickup truck approach a curve on a highway that has a radius of 255 m. At what angle should the highway engineer bank this curve so that vehicles traveling at 75.0 mi/h can safely round it regardless of the condition of their tires
Answer:
the banking angle of the road is 24.2⁰
Explanation:
Given;
speed of the vehicles considered, v = 75 mi/h
Speed in m/s ⇒ 1 mi/h --------> 0.44704 m/s
75 mi/h --------> ?
= 75 x 0.44704 m/s = 33.528 m/s
radius of the curve, r = 255 m
The banking angle of the road is calculated as;
[tex]\theta = tan^{-1} (\frac{v^2}{rg} )\\\\\theta = tan^{-1} (\frac{33.528^2}{255\times 9.8} )\\\\\theta = tan^{-1}(0.44983)\\\\\theta =24.2^0[/tex]
Therefore, the banking angle of the road is 24.2⁰
The angle of banking is 24 degrees.
What is the angle of banking?As a driver approaches a bend two equal and opposite forces act on him which are the centripetal force and the centrifugal force. The driver will have to ben through a certain angle called the angle of banking to avoid falling off.
The angle of banking depends on the speed of the vehicle and the radius of the curve.
θ = v^2/rg
speed = 75.0 mi/h or 33.5 m/s
r = 255 m
g = 9.8 ms-1
θ = tan-1 (33.5 m/s)^2/ 255 m × 9.8 ms-1
θ = tan-1(1122.3/2499)
θ = 24 degrees
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