Answer:
C. both A and B
Explanation:
Sana nakatulong
Two resistors 5 ohm's each can be combined to give an equivalent resistance of
Answer: 10 Ω or 2.5 Ω
Explanation:
In series resistance R = 5 Ω + 5 Ω =10 Ω. If resistors are parallel, resistance is
1/R = 1/5Ω+ 1/5Ω and R = 2,5 Ω
what is the power output needed from a motor to lift in the absence of friction a mass of 1.5 ×10⁴ kg 25 m in 6.0 s at constant speed
Answer:
the power output needed is 61.25 × 10⁴
Explanation:
The computation of the power output needed is shown below;
Given that
m = 1.5 ×10⁴ kg
v = 25m ÷ 6.0 s
And g = 9.8 m/sec^2
Now based on the above information
p = f × v
= mg × v
= 1.5 ×10⁴ × 9.8 × 25 ÷ 6
= 61.25 × 10⁴
Hence, the power output needed is 61.25 × 10⁴
For each of the following scenarios, describe the force providing the centripetal force for the motion: (Be very specific, and give some reasoning to your answer.) a. a car making a turn b. a child swinging around a pole c. a person sitting on a bench facing the center of a carousel d. a rock swinging on a string e. the Earth orbiting the Sun.
Answer:
a) tires rubbing, b) the weight has a component parallel to the floor
c) he child's back support, d) The tension of the rope and weight
Explanation:
In this exercise, we are asked to indicate the origin of the forces for the centripetal movement parallel to the rope, e) gravitational force
a) When a car turns, the centripetal force has two origins
* The tires rubbing against the road
* If the road has a lean angle, the component of the weight directed towards the center of the circle also contributes to the centripetal force.
b) the child in general has some degree of inclination with respect to the post, for which the weight has a component parallel to the floor that is responsible for the centripetal movement of the system
c) The bench rotates together with the carousel, so the child's back support is the response to the centripetal force
d) The tension of the rope has two components: the component perpendicular to the movement and the component of the weight (parallel to the rope) the difference of these two forces is the centripetal force
e) The gravitational force of the sun on the earth is what creates the centripetal motion
WHEN AN OBJECT IS DROPPED WHAT HAPPENDS TO THE POTENTIAL AND KENETIC ENERGY AS IT FALLS?
Answer:
The sum of an object's potential and kinetic energies is called the object's mechanical energy. As an object falls its potential energy decreases, while its kinetic energy increases. The decrease in potential energy is exactly equal to the increase in kinetic energy.
Explanation:
Answer:
the potential decrease and kinetic increase
Explanation:
because it goes from a state of rest to a state of movement
a pool and stops at the
Which best describes his
motion?
A. Linear
B. Projectile
C. Vibrational
In Linear motion the swimmer swims
Answer: A
Explanation: Linear motion.
What is one benefit to measuring your body’s flexibility?
A.
meeting the national requirement for flexibility
B.
determining your muscular strength
C.
tracking your flexibility improvements over time
D.
increasing the length of your life
Answer:
C
Explanation:
if you measure your body's flexibility then you can keep track of how flexible you have gotten over time
Part D
How do supersonic flights create sonic boom? Please explain in detail.
Answer: A sonic boom is a sound associated with shock waves created when an object travels through the air faster than the speed of sound. Sonic booms generate enormous amounts of sound energy, sounding similar to an explosion or a thunderclap to the human ear. The crack of a supersonic bullet passing overhead or the crack of a bullwhip are examples of a sonic boom in miniature. Sonic booms due to large supersonic aircraft can be particularly loud and startling, tend to awaken people, and may cause minor damage to some structures. They led to prohibition of routine supersonic flight over land. Although they cannot be completely prevented, research suggests that with careful shaping of the vehicle, the nuisance due to the sonic booms may be reduced to the point that overland supersonic flight may become a feasible option. A sonic boom does not occur only at the moment an object crosses the speed of sound; and neither is it heard in all directions emanating from the supersonic object. Rather the boom is a continuous effect that occurs while the object is travelling at supersonic speeds. But it affects only observers that are positioned at a point that intersects a region in the shape of a geometrical cone behind the object. As the object moves, this conical region also moves behind it and when the cone passes over the observer, they will briefly experience the boom.
Explanation:
As incredible as the Concorde was, the sonic booms created by its supersonic flights were so disruptive that most countries restricted or completely prohibited the aircraft from flying over land. The sonic boom, at its worst, would be heard as a very loud thunder clap that was right overhead. The force of the boom rattled windows and loosened roof tiles. But even when the sonic boom sounded like a “softer” distant thunder clap, it was distracting to people and caused disruption of sleep and interruptions in activity. Imagine that you are driving on your way to work, and with clear skies overhead, you suddenly hear the sound of thunder. Your immediate responses are most likely surprise, shock, and an instinctive search for the source. Being caught by surprise in certain situations is rather annoying, and in others, potentially dangerous. In 1964, the FAA and NASA conducted a six-month sonic boom research project in Oklahoma City – without warning residents beforehand. The experiment consisted of eight sonic booms, every day, for six months. 15,000 complaints and a class action lawsuit were filed. The government lost on appeals. Great idea, guys, just brilliant. When the Concorde was originally designed, in the early 1960s, governments and airlines around the world lined up to place orders. The plane did an around-the-world publicity trip, and was well-received. But as awareness of the sonic boom effect grew, almost every country banned the aircraft. Only the US, Great Britain, and France allowed the Concorde to enter their airspace, and then only to cities in close proximity to the ocean – NYC, London, Paris, and Washington, DC. The Concorde was specifically designed for supersonic flight (specifically, Mach 2) and was very fuel-inefficient at subsonic speeds (less than Mach 1). Unfortunately, it was thus not feasible to fly at supersonic speed over water and then at subsonic speed over land.
What causes a sonic boom?
When any object moves, it creates waves in front of and behind it. Think of the waves that a boat creates at its bow and stern. In front, the waves are compressed together as the boat sails forward. Behind, the waves spread out away from the boat. In this case, you only see the waves on the surface of the water, and it appears two-dimensional. Similar principles are at play with aircraft. In front of the nose of a plane, air is pushed together and compressed as the aircraft flies forward. Behind the plane, the air creates waves that radiate out and away in the shape of a cone – three-dimensionally.
Answer:
A sonic boom is caused by the shock waves created when an object travels through the air faster than the speed of sound.
Explanation:
When any object moves, it creates waves in front of and behind it. Think of the waves that a boat creates at its bow and stern. In front, the waves are compressed together as the boat sails forward. Behind, the waves spread out away from the boat. Similar principles are at play with aircraft. In front of the nose of a plane, air is pushed together and compressed as the aircraft flies forward. Behind the plane, the air creates waves that radiate out and away in the shape of a cone – three-dimensionally. Things get interesting, and complicated, when you fly faster than the speed of sound – supersonic flight. The nose of a supersonic aircraft pushes ahead of its forward waves. These waves get in the way of the airplane, causing compression which results in a shock wave. Actually, this creates two shocks, one forming as the aircraft passes the front of the wave and then another as it leaves the wave. The shock wave generated stays mostly behind the aircraft, and radiates out in a cone
A magnetic field is passing through a loop of wire whose area is 0.014 m^2. The direction of the magnetic field is parallel to the normal to the loop, and the magnitude of the field is increasing at the rate of 0.19 T/s.
a. Determine the magnitude of the emf induced in the loop.
b. Suppose the area of the loop can be enlarged or shrunk. If the magnetic field is increasing as in part (a), at what rate (in m^2/s) should the area be changed at the instant when B = 1.6 T if the induced emf is to be zero? (Give the magnitude of the rate of change of the area.) (m^2/s). Explain whether the area is to be enlarged or shrunk.
Answer:
a. 0.00266v
b. -0.0016625
the area should be shrunk
Explanation:
the magnitude of the EMF induced in the loop
= area * rate
= 0.014 * 0.19
= 0.00266 V
B we are to solve for the rate at which the are has to be change with B = 1.6
δA/δt = -A/B * dB/dt
= (-0.014 * 0.19) /1.6
= -0.0016625
the sign is negative so the EMT is negative and so the area has to be shrunk.
a car is travelling at 18m/s accelerates ti 30m/s in 3seconds. what's the acceleration of the car
[tex] \Large {\underline { \sf {Required \; Solution :}}}[/tex]
We have ―
Initial velocity, u = 18 m/sFinal velocity, v = 30 m/sTime taken, t = 3 secondsWe've been asked to calculate acceleration.
[tex]\qquad \implies\boxed{\red{\sf{ a = \dfrac{v-u}{t} }}}\\[/tex]
a denotes accelerationv denotes final velocityu denotes initial velocityt denotes time[tex] \twoheadrightarrow \quad \sf {a = \dfrac{30-18}{3} \; ms^{-2} } \\ [/tex]
[tex] \twoheadrightarrow \quad \sf {a =\cancel{ \dfrac{12}{3} \; ms^{-2} }} \\ [/tex]
[tex]\twoheadrightarrow \quad \boxed{\red{\sf{ a = 4 \; ms^{-2} }}}\\[/tex]
Therefore, acceleration of the car is 4 m/s².
how far must 5N force pull a 50g toy car if 30J of energy are transferred?
Answer: 6 m
Explanation:
30 = 5 * d
d = 30/5
d = 6 m
Which of the following is a true statement for a child's toy spinning in a circle at constant speed?
a) The speed is constant, the velocity is constant, and the angular velocity is constant. b) The speed is constant, the velocity is not constant, and the angular velocity is constant. c) The speed is constant, the velocity is not constant, and the angular velocity is not constant. d) The speed is constant, the velocity is constant, and the angular velocity is not constant.
explain your answer.
Answer:
B
Explanation:
I am pretty sure it is B as the speed is obviously constant but the velocity is not constant as it defines as the rate of speed AND DIRECTION meaning that it is not constant as it always changes direction. And angular velocity is constant.
A 10-kg ball rolls at a rate of 2.98 m/s. What is the kinetic energy of the ball?
Answer:92
Explanation:
Prove the correctness of this equation s=vt + 1/2at
Answer:
The formula is dimensionally correct.
Explanation:
Given
[tex]s = ut + \frac{1}{2}at^2[/tex]
Required
Prove its correctness
Write out the dimension of each:
[tex]s = M^0LT^0[/tex] --- displacement
[tex]ut = M^0LT^{-1} * T[/tex] --- velocity * time
[tex]\frac{1}{2}at^2 = M^0LT^{-2} * T^2[/tex] --- acceleration * square of time
The expression becomes:
[tex]s = ut + \frac{1}{2}at^2[/tex]
[tex]M^0LT^0 = M^0LT^{-1} * T + M^0LT^{-2} * T^2[/tex]
Apply law of indices
[tex]M^0LT^0 = M^0LT^{-1+1} + M^0LT^{-2+2}[/tex]
[tex]M^0LT^0 = M^0LT^{0} + M^0LT^{0}[/tex]
[tex]M^0LT^0 = M^0LT^{0}[/tex]
Both sides of the equation are equal
A piano tuner stretches a steel piano wire with a tension of 765 N. The steel wire has a length of 0.800 m and a mass of 6.00 g . What is the frequency f1 of the string's fundamental mode of vibration
Answer:
the frequency of the fundamental mode of vibration is 199.6 Hz
Explanation:
Given;
tension of the piano wire, T = 765 N
length of the steel wire, L = 0.8 m
mass of the steel wire, m = 6.00 g = 6 x 10⁻³ kg
The frequency of the fundamental mode of vibration is calculated as;
[tex]f_o = \frac{1}{2l} \sqrt{\frac{T}{\mu} }[/tex]
where;
μ is the mass per unit length [tex]= \frac{6.0 \times 10^{-3}}{0.8} = 7.5 \times 10^{-3} \ kg/m[/tex]
[tex]f_o = \frac{1}{2l} \sqrt{\frac{T}{\mu} } \\\\f_o = \frac{1}{2\times 0.8} \sqrt{\frac{765}{7.5 \times 10^{-3}} } \\\\f_o = 199.6 \ Hz[/tex]
Therefore, the frequency of the fundamental mode of vibration is 199.6 Hz
A factory produces steel components for use by a number of different
companies. The density of the steel used by the factory is 8000 kg/mº,
A partoutor steel oube produced by the factory has a side length of 5 em,
5 cm
5 cm
5cm
Calculate the mass of one of these cubes. Write your answer in
kilograms
Answer:
1 kg
Explanation:
Since density ρ = m/v where m = mass of steel cube and v = volume of steel cube = L³ where L= side length of steel cube = 5 cm = 0.05 m
Now, m = ρv = ρL³
Since ρ = 8000 kg/m³,
m = ρL³
= 8000 kg/m³ × (0.05 m)³
= 8000 kg/m³ × 0.000125 m³
= 1 kg
Personally, do you think you will benefit from the discoveries of Archimedes’ and Pascal’s Principles? How?
Answer:
whats the question
Explanation:
customer complains of an aluminum foil mixed with his salad? what will you do?
Answer:
offer them a refund or a remake/ replacement of food item whilst apologizing sincerely
A barometer reads 780 mm Hg. Mercury has a density of 1.36 x 10^4 kg /m^3.
What is the pressure of the atmosphere in N / m^2?
A 1.1 x 10^4 N/m^2
B 1.1 x 10^5 N/m^2
C 1.1 x 10^7 N/m^2
D 1.1 x 10^8 N/m^2
How do we get to the answer? According to the mark scheme, it's B.
Answer:
B
Explanation:
Pressure = density × g × height
[tex]p \: = (1.36 \times {10}^{4} )(10)(780 \times {10}^{ - 3} ) \\ p = 1.1 \times {10}^{5} [/tex]
The pressure of the atmosphere, when a barometer reads 780 mm Hg. Mercury which a density of 1.36 x 10^4 kg /m^3 is B 1.1 x 10^5 N/m^2
This problem can be solved using the formula below
P = dgh................. Equation 1
Where P = Pressure of the atmosphere, d = density of the mercury, h = height of the mercury, g = acceleration due to gravity.
From the question,
Given: d = 1.36×10⁴ kg/m³, h = 780 mm = 0.78 m,
Constant: g = 10 m/s²
Substitute these values into equation 1
P = (1.36×10⁴)(10)(0.78)
P = 10.608×10⁴ N/m²
P ≈ 1.1×10⁵ N/m²
Hence the right answer is B. 1.1×10⁵ N/m²
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What is R2 in the circuit?
WILL GIVE BRAINLIEST !!!!
Answer:
1. Rₑq = 4 Ω
2. R₂ = 6 Ω
3. Vₜ = 12 V, V₁ = 12 V, V₂ = 12 V
4. Iₜ = 3 A, I₁ = 1 A, I₂ = 2 A
Explanation:
1. Determination of the equivalent resistance
Voltage (V) = 12 V
Current (I) = 3 A
Resistance (Rₑq) =?
V= IRₑq
12 = 3 × Rₑq
Divide both side by 3
Rₑq = 12 / 3
Rₑq = 4 Ω
Thus, the equivalent resistance (Rₑq) = 4 Ω
2. Determination of R₂.
Equivalent resistance (Rₑq) = 4 Ω
Resistance 1 (R₁) = 12 Ω
Resistance 2 (R₂)
Since the resistor are in parallel arrangement, the value of R₂ can be obtained as follow:
Rₑq = R₁ × R₂ / R₁ + R₂
4 = 12 × R₂ / 12 + R₂
Cross multiply
4(12 + R₂) = 12R₂
48 + 4R₂ = 12R₂
Collect like terms
48 = 12R₂ – 4R₂
48 = 8R₂
Divide both side by 8
R₂ = 48 / 8
R₂ = 6 Ω
3. Determination of the total voltage (Vₜ), V₁ and V₂.
From the question given above, the total voltage is 12 V
Since the resistors are arranged in parallel connection, the same voltage will go through them.
Thus,
Vₜ = V₁ = V₂ = 12 V
4. Determination of the total current (Iₜ), I₁ and I₂
From the question given above, the total current (Iₜ) is 3 A
Next, we shall determine I₁. Since the resistors are arranged in parallel connection, different current will pass through each resistor respective.
Vₜ = V₁ = 12 V
R₁ = 12 Ω
I₁ =?
V₁ = I₁R₁
12 = I₁ ×12
Divide both side by 12
I₁ = 12 / 12
I₁ = 1 A
Next, we shall determine I₂. This can be obtained as follow:
Iₜ = 3 A
I₁ = 1 A
I₂ =?
Iₜ = I₁ + I₂
3 = 1 + I₂
Collect like terms
I₂ = 3 – 1
I₂ = 2 A
2. Oscar walked to the milktea house for a snacks. He walk 2 kilometers
east, 2 kilometers south, and 2 kilometers west. What distance did he
cover? What was the displacement?
The distance Oscar covered was 6km while the displacement was 2km.
What is Displacement?Displacement is defined as the vector whose length is the shortest
distance from the initial to the final position of a point M.
Distance = Total length of space between points
= 2km+ 2km+ 2km = 6km.
Displacement = 2km.
This is because it is the shortest distance from the positions in this
scenario.
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Which of the following is true about scientific knowledge?
OA. It can be changed as new information becomes available.
OB. It cannot be changed under any circumstances.
OC. It can be changed only if all scientists approve the change.
OD. It can be changed without information that supports the change.
The true statement about scientific knowledge is it can be changed as new information becomes available.
option A
What is scientific knowledge?Science is a dynamic process, and scientific knowledge is continually evolving as new data and evidence are discovered.
Scientific theories and hypotheses are constantly being tested and refined, and they can be modified or even rejected if new evidence contradicts them.
This is one of the key features of the scientific method - that it is self-correcting and always open to revision based on new evidence.
Thus, the true statement about scientific knowledge is it can be changed as new information becomes available.
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A steel bullet (66.45g) located 103.15 cm above the ground is shot in the horizontal direction. It travels 360.25 cm in the horizontal direction before it hits the ground. Calculate the momentum of the steel ball in the horizontal direction. Enter your results using the measured units, i.e., grams, centimeters and seconds. Use scientific notation for this calculation.
Answer:
The answer is "[tex]4950 \frac{g\ cm }{s}[/tex]"
Explanation:
Consider vertical component of motion:
[tex]\to s_y=u_yt+ \frac{1}{2} at^2\\\\1.0315=0+\frac{1}{2} \times 9.81 \times t^2\\\\1.0315=\frac{1}{2} \times 9.81 \times t^2\\\\1.0315=4.905 \times t^2\\\\t^2= \frac{1.0315}{4.905}\\\\t=0.207059711\\\\[/tex]
Considering the horizontal components:
[tex]\to v_x=x\times t\\\\[/tex]
[tex]=03.6025 \times 0.207059711\\\\=0.745 \frac{m}{s}\\\\\to p=mv\\\\=66.45 \times0.745 \times 10^2\\\\=4950\ \frac{g\ cm }{s}[/tex]
1. A 455 g mass, hanging at rest on a spring, stretches the spring 22.4 cm beyond its relaxed
position. What is the spring constant of that spring?
Answer:
19.9 N/m
Explanation:
From the question,
Applying Hook's law
F = Ke.................. Equation 1
Where F = Force on the spring, k = spring constant, e = extension
But the force on the spring is the weight of the mass
Therefore,
mg = ke.................. Equation 2
Where m = mass. g = acceleration due to gravity
make e the subject of the equation
e = mg/e................ Equation 3
Given: m = 455 g = 0.455 kg, e = 22.4 cm = 0.224 m,
Constant: g = 9.8 m/s²
Substitute these values into equation 3
e = (0.455×9.8)/0.224
e = 19.9 N/m
The spring constant of the given spring is 20 N/m.
The given parameters:
Mass attached, m = 455 g = 0.455 kgExtension of the spring, x = 22.4 cm = 0.224 mThe spring constant is calculated by applying Hooke's law as follows;
[tex]F = kx\\\\mg = kx\\\\k = \frac{mg}{x} \\\\k = \frac{0.455 \times 9.8}{0.224} \\\\k = 20 \ N/m[/tex]
Thus, the spring constant of the spring is 20 N/m.
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In one cycle of any heat engine
A. the net work done is larger than the heat exhausted.
B. more heat flows from the engine than enters the engine.
C. The internal energy of the engine does not change.
D. the net heat flow is zero.
Answer:
D. the net heat flow is zero
Explanation:
In one cycle of any heat engine, three things happen:
Heat is added; this causes high temperature in the engine ([tex]Q_H[/tex])Some of the energy from that input heat is used to perform work (W). The rest of the heat is removed at a relatively cold temperature ([tex]Q_C[/tex]).[tex]Q_H = W + Q_C[/tex]
Conclusively, the net heat flow is zero
explain the working and performance of a centrifugal clutch with a sketch
a centrifugal clutch works, as the name suggests, through centrifugal force. ... The rotation of the hub forces the shoes or flyweights outwards until they come into contact with the clutch drum, the friction material transmits the torque from the flyweights to the drum. The drive is then connected
A small rubber wheel is used to drive a large pottery wheel. The two wheels are mounted so that their circular edges touch. The smalldrive-wheel has a radius of 2.20 cm and accelerates at the rate of 8.00 rad/s2, and it is in contact with the pottery wheel (radius 28.0 cm). Both wheels move without slipping.The rubber drive wheel rotates in the clockwise sense.
Required:
a. Find the angular acceleration (both magnitude and direction) of the large pottery wheel.
b. Calculate the tune it takes the pottery wheel to reach its required speed of 60 rpm. if both wheels start from rest.
Answer:
[tex]0.629\ \text{rad/s}^2[/tex] counterclockwise
[tex]9.98\ \text{s}[/tex]
Explanation:
[tex]r_1[/tex] = Small drive wheel radius = 2.2 cm
[tex]\alpha_1[/tex] = Angular acceleration of the small drive wheel = [tex]8\ \text{rad/s}^2[/tex]
[tex]r_2[/tex] = Radius of pottery wheel = 28 cm
[tex]\alpha_2[/tex] = Angular acceleration of pottery wheel
As the linear acceleration of the system is conserved we have
[tex]r_1\alpha_1=r_2\alpha_2\\\Rightarrow \alpha_2=\dfrac{r_1\alpha_1}{r_2}\\\Rightarrow \alpha_2=\dfrac{2.2\times 8}{28}\\\Rightarrow \alpha_2=0.629\ \text{rad/s}^2[/tex]
The angular acceleration of the pottery wheel is [tex]0.629\ \text{rad/s}^2[/tex].
The rubber drive wheel is rotating in clockwise direction so the pottery wheel will rotate counterclockwise.
[tex]\omega_i[/tex] = Initial angular velocity = 0
[tex]\omega_f[/tex] = Final angular velocity = [tex]60\ \text{rpm}\times \dfrac{2\pi}{60}=6.28\ \text{rad/s}[/tex]
t = Time taken
From the kinematic equations of linear motion we have
[tex]\omega_f=\omega_i+\alpha_2t\\\Rightarrow t=\dfrac{\omega_f-\omega_i}{\alpha_2}\\\Rightarrow t=\dfrac{6.28-0}{0.629}\\\Rightarrow t=9.98\ \text{s}[/tex]
The time it takes the pottery wheel to reach the required speed is [tex]9.98\ \text{s}[/tex]
1 ) when a ball is projected upwords its time of rising is ...............the time of falling .
a) greater than b) smaller than c) equal to d ) double
2 ) when an object falls freely under the effect of gravity , the distance moved is
a ) directly proportional to time
b ) inversely proportional to time
c ) directly proportional to square of time
d ) inversely proportional to square of time.
Answer:
correct answer is C
Explanation:
In this exercise, you are asked to complete the sentences so that the sentence makes sense.
1) in projectile launching, the only force that acts is gravity in the vertical direction, so the time of going up is EQUAL to the time of going down
correct answer C
2) when a body falls freely, the acceleration is the ratio of gravity, therefore if it starts from rest, its height is
y = v₀ t - ½ gt²
v₀ = 0
y = -1/2 g t²
so the position is not proportional to the square of the time
correct answer is C
How do particle motion and temperature change as a material absorbs heat?
A rocket sled accelerates from rest for a distance of 645 m at 16.0 m/s2. A parachute is then used to slow it down to a stop. If the parachute gives the sled an acceleration of -18.2 m/s2 and there is 500.0 m of sled track remaining after the shoot opens, will the sled stop before running off the track? Show why or why not?
Answer:
the stopping distance is greater than the free length of the track, the vehicle leaves the track before it can brake
Explanation:
This problem can be solved using the kinematics relations, let's start by finding the final velocity of the acceleration period
v² = v₀² + 2 a₁ x
indicate that the initial velocity is zero
v² = 2 a₁ x
let's calculate
v = [tex]\sqrt {2 \ 15.0 \ 645}[/tex]
v = 143.666 m / s
now for the second interval let's find the distance it takes to stop
v₂² = v² - 2 a₂ x₂
in this part the final velocity is zero (v₂ = 0)
0 = v² - 2 a₂ x₂
x₂ = v² / 2a₂
let's calculate
x₂ = [tex]\frac{ 143.666^2 }{2 \ 18.2}[/tex]
x₂ = 573 m
as the stopping distance is greater than the free length of the track, the vehicle leaves the track before it can brake
A ball rolls off an 8.0 m high building and strikes the ground 5.0 m away from the base of the building. How fast was the ball rolling?
Answer: horizontal speed is 3.9 m/s
Explanation: when ball starts to drop, its vertical speed v0 is zero.
We can calculate dropping time from s = v0t +0.5gt².
Dropping time t= √(2s/g)= √((2·8.0 m)/9.81 m/s²)= 1.277 s
Because ball travels horizontal distance s= 5.0 m
HorizontalSpeed v = s/t = 5.0 m/1.277s= 3,915 m/s