Answer:
y = -2.69 m
the negative sign indicates that it is descending and the distance is less than the difference in height between the two buildings, therefore the person would be saved in the jump.
Explanation:
his problem must be solved with the missile launch equations.
Let's start by looking for the jumper's initial velocity
R = v₀² sin 2θ / g
for the long jump the angle used is tea = 45º, in the exercise they indicate that the best record is R = 7.9m
v₀² = R g / sin 2te
v₀ = [tex]\sqrt{ \frac{7.9 \ 9.8}{1 }[/tex]
v₀ = 8.80 m / s
Now suppose you jump with this speed to get to the other building, let's use trigonometry for the components of the speed
sin 45 = [tex]v_{oy}[/tex] /v₀
cos 45 = v₀ₓ / v₀
v_{oy} = v₀ sin 45
v₀ₓ = v₀ cos 45
v_{oy} = 8.8 sin 45 = 6.22 m / s
v₀ₓ = 8.8 cos 45 = 6.22 m / s
now let's calculate the sato with these speeds
x = [tex]v_{ox}[/tex] t
the minimum jump is x = 10 m
t = x / v₀ₓ
t = 10 / 6.22
t = 1.61 s
let's find the vertical distance for this time
y = v_{oy} t - ½ g t²
where zero is placed on the jump building
y = 6.22 1.61 - ½ 9.8 1.61²
y = -2.69 m
Let's analyze this result, the negative sign indicates that it is descending and the distance is less than the difference in height between the two buildings, therefore the person would be saved in the jump.
In which number are the zeros not significant?
100.0
O 0.0003
O 4.00005
O 1.0004
Answer:
0.0003
Explanation:
In the rules of Sig Figs, all zeros before with decimals are not sigificant. I.E. 0.00000000000000009. Despite how many 0's there are, only the 9 is significant. Zeros before a number is not significant. In 100, only the one is signficant in 100. with a dot at the end, the one and the two zeros are significant. hope this helps.
Answers:
the second option
Explanation:
When researchers replicate a study, they are seeking to __________.
A.
prove that the hypothesis upon which the study was founded is untestable
B.
develop a new hypothesis
C.
change the study to provide new results
D.
support or reject the hypothesis upon which the study was founded
Please select the best answer from the choices provided
A
B
C
D
Answer:
D
Explanation:
right edge 2022
A 500 kg wrecking ball is knocking down a wall. When it is pulled back to its highest point, it is at a height of 6.2 m. When it hits the wall, it is moving at 3.1 m/s. How high is the wrecking ball when it hits the wall? (Show your work and follow all of the steps of the GUESS method. Check your answer after you submit the form - it's in the feedback for this question.) |
If there is "waste" energy, does the Law of Conservation of Energy still apply?
Explanation:
Yes, the law of conservation of energy still applies even if there is waste energy.
The waste energy are the transformation products of energy from one form to another.
According to the law of conservation of energy "energy is neither created nor destroyed by transformed from one form to another in a system".
But of then times, energy is lost as heat or sound within a system.
If we take into account these waste energy, we can see that energy is indeed conserved. The sum total of the energy generated and those produced will be the same if we factor in other forms in which the energy has been transformed into.What Coulombs discovered almost 300
years ago
Answer:
ummm hehe this is my time to shine
Explanation:
MERICIA!!!!!!!!!!!!!!!!!!!!!!!
One disadvantage to experimental research is that experimental conditions do not always reflect reality.
Please select the best answer from the choices provided
T
F
Answer:
It's true I took the test on Edge.
Explanation:
Answer:
True
Explanation:
Got it right on edg
Two metal bricks are held off the edge of a balcony from the same height above the ground. The bricks are the same size but one is made of Titanium (density of 4.5 g/cm%) and one is made of Lead (density of 11.3 g/cm3) so the Lead is about twice as heavy as the Titanium. The time it takes the bricks to reach the ground will be:________.
a. less but not necessarily half as long for the heavier brick
b. about half as long for the lighter brick
c. less but not necessarily half as long for the lighter brick
d. about half as long for the heavier brick
e. about the same time for both bricks
Answer:
e.
Explanation:
Assuming that the air resistance is neglectable, both bricks are only accelerated by gravity, which produces a constant acceleration on both bricks, which is the same, according Newton's 2nd Law, as we can see below:[tex]F_{g} = m*g = m*a (1)[/tex]⇒a = g = 9.8m/s² (pointing downward)Since acceleration is constant, if both fall from the same height, we can apply the following kinematic equation:[tex]\Delta y = v_{o} * t - \frac{1}{2} *g*t^{2} (2)[/tex]
Since both bricks are held off the edge, the initial speed is zero, so (2) reduces to the following equation:[tex]h =\frac{1}{2} *g*t^{2} (3)[/tex]
Since h (the height of the balcony) is the same, we conclude that both bricks hit ground at exactly the same time.If the air resistance is not negligible, due both bricks have zero initial speed, and have the same shape, they will be affected by the drag force in similar way, so they will reach the ground at approximately the same time.what is the direction of the third force that would cause the box to remain stationary on the ramp ?
An arrow pointing from the bottom of the ramp to the top, I assume it would be friction.
Two children, each with a mass of 25.4 kg, are at fixed locations on a merry-go-round (a disk that spins about an axis perpendicular to the disk and through its center). One child is 0.78 m from the center of the merry-go-round, and the other is near the outer edge, 3.14 m from the center. With the merry-go-round rotating at a constant angular speed, the child near the edge is moving with translational speed of 11.5 m/s.
a. What is the angular speed of each child?
b. Through what angular distance does each child move in 5.0 s?
c. Through what distance in meters does each child move in 5.0 s?
d. What is the centripetal force experienced by each child as he or she holds on?
e. Which child has a more difficult time holding on?
Answer:
a) ω₁ = ω₂ = 3.7 rad/sec
b) Δθ₁ = Δθ₂ = 18.5 rad
c) d₁ = 14.5 m d₂ = 57.5 m
d) Fc1 = 273.9 N Fc2 = 1069.8 N
e) The boy near the outer edge.
Explanation:
a)
Since the merry-go-round is a rigid body, any point on it rotates at the same angular speed.However, linear speeds of points at different distances from the center, are different.Applying the definition of angular velocity, and the definition of angle, we can write the following relationship between the angular and linear speeds:[tex]v = \omega*r (1)[/tex]
Since we know the value of v for the child near the outer edge, and the value of r for this point, we can find the value of the angular speed, as follows:[tex]\omega = \frac{v_{out} }{r_{out} } = \frac{11.5m/s}{3.14m} = 3.7 rad/sec (2)[/tex]
As we have already said, ωout = ωin = 3.7 rad/secb)
Since the angular speed is the same for both childs, the angle rotated in the same time, will be the same for both also.Applying the definition of angular speed, as the rate of change of the angle rotated with respect to time, we can find the angle rotated (in radians) as follows:[tex]\Delta \theta = \omega * t = 3.7 rad/sec* 5.0 sec = 18.5 rad (3)[/tex]⇒ Δθ₁ = Δθ₂ = 18.5 rad.
c)
The linear distance traveled by each child, will be related with the linear speed of them.Knowing the value of the angular speed, and the distance from each boy to the center, we can apply (1) in order to get the linear speeds, as follows:[tex]v_{inn} = \omega * r_{inn} = 3.7 rad/sec * 0.78 m = 2.9 m/s (4)[/tex]
vout is a given of the problem ⇒ vout = 11. 5 m/s
Applying the definition of linear velocity, we can find the distance traveled by each child, as follows:[tex]d_{inn} = v_{inn} * t = 2.9m/s* 5.0 s = 14.5 m (5)[/tex]
[tex]d_{out} = v_{out} * t = 11.5 m/s* 5.0 s = 57.5 m (6)[/tex]
d)
The centripetal force experienced by each child is the force that keeps them on a circular movement, and can be written as follows:[tex]F_{c} = m*\frac{v^{2}}{r} (7)[/tex]
Replacing by the values of vin and rin, since m is a given, we can find Fcin (the force on the boy closer to the center) as follows:[tex]F_{cin} = m*\frac{v_{in}^{2}}{r_{in}} = 25.4 kg* \frac{(2.9m/s)^{2} }{0.78m} = 273.9 N (8)[/tex]
In the same way, we get Fcout (the force on the boy near the outer edge):[tex]F_{cout} = m*\frac{v_{out}^{2}}{r_{out}} = 25.4 kg* \frac{(11.5m/s)^{2} }{3.14m} = 1069.8 N (9)[/tex]
e)
The centripetal force that keeps the boys in a circular movement, is not a different type of force, and in this case, is given by the static friction force.The maximum friction force is given by the product of the coefficient of static friction times the normal force.Since the boys are not accelerated in the vertical direction, the normal force is equal and opposite to the force due to gravity, which is the weight.As both boys have the same mass, the normal force is also equal.This means that for both childs, the maximum possible static friction force, is the same, and given by the following expression:[tex]F_{frs} = \mu_{s} * m* g (10)[/tex]If this force is greater than the centripetal force, the boy will be able to hold on.So, as the centripetal force is greater for the boy close to the outer edge, he will have a more difficult time holding on.A long, straight wire carries a current of 5.20 A. An electron is traveling in the vicinity of the wire. At the instant when the electron is 4.40 cm from the wire and traveling at a speed of 6.20 * 104 m>s directly toward the wire, what are the magnitude and direction (relative to the direction of the current) of the force that the magnetic field of the current exerts on the electron
Answer:
Explanation:
Magnetic field due to current at a distance of 4.4 cm
B = 10⁻⁷ x 2 x 5.2 / 4.4 x 10⁻² [ B = 10⁻⁷ x 2i / r = ]
= 2.36 x 10⁻⁵ T.
Force on moving electron = Bqv , B is magnetic field , q is charge and v is velocity of charge .
Force = 2.36 x 10⁻⁵ x 1.6 x 10⁻¹⁹ x 6.2 x 10⁴
= 23.41 x 10⁻²⁰ N .
This force will be perpendicular to the direction of current .
Billiard ball A (0.35 kg) is struck such that it moves at 10 m/s toward a
second identical ball (Ball B). After the collision Ball A continues to move
in the same direction at 2 m/s. What is the magnitude of the velocity for
Ball B after the collision?
Before Collision:
10 m/s
A
After Collision:
2 m/s
O
Answer:
6m/s
Explanation:
Using the law of conservation of momentum which States that the sum of momentum of bodies before collision is equal to the momentum after collision.
Using the expression
m1u1 + m2u2 = (m1+m2)v
m1 and m2 are the masses
u1 and u2 are the initial velocities
v is the final velocity after collision
Substitute the given values in the formula
0.35(10)+0.35(2) = (0.35+0.35)v
3.5+0.7 = 0.7v
4.2 = 0.7v
v = 4.2/0.7
v = 6m/s
Hence the magnitude of the velocity for Ball B after the collision is 6m/s
what is momentum of a train that is 60,000 kg that is moving at velocity of 17m/s?
explain your answer
A solid sphere of radius R = 5 cm is made of non-conducting material and carries a total negative charge Q = -12 C. The charge is uniformly distributed throughout the interior of the sphere.
What is the magnitude of the electric potential V at a distance r = 30 cm from the center of the sphere, given that the potential is zero at r = [infinity] ?
Answer:
V= -3.6*10⁻¹¹ V
Explanation:
Since the charge is uniformly distributed, outside the sphere, the electric field is radial (due to symmetry), so applying Gauss' Law to a spherical surface at r= 30 cm, we can write the following expression:[tex]E* A = \frac{Q}{\epsilon_{0} } (1)[/tex]
At r= 0.3 m the spherical surface can be written as follows:[tex]A = 4*\pi *r^{2} = 4*\pi *(0.3m)^{2} (2)[/tex]
Replacing (2) in (1) and solving for E, we have:[tex]E = \frac{Q}{4*\pi *\epsilon_{0}*r^{2} } = \frac{(9e9N*m2/C2)*(-12C)}{(0.3m)^{2} y} (3)[/tex]
Since V is the work done on the charge by the field, per unit charge, in this case, V is simply:V = E. r (4)Replacing (3) in (4), we get:[tex]V =E*r = E*(0.3m) = \frac{(9e9N*m2/C2)*(-12C)}{(0.3m)} = -3.6e11 V (5)[/tex]
V = -3.6*10¹¹ Volts.The electrical potential module will be [tex]-3.6*10^-^1^1 V[/tex]
We can arrive at this answer as follows:
To answer this, we owe Gauss's law. This is because the charge is evenly distributed across the sphere. This will be done as follows:[tex]E*A=\frac{Q}{^E0} \\\\\\A=4*\pi*r^2[/tex]
Solving these equations will have:[tex]E=\frac{Q}{4*\pi*^E0*r^2} \\E= \frac{(9e9N*m2/c2)*(-12C)}{(0.3m)^2y}[/tex]
As we can see, the electric potential is carried out on the field charge. In this case, using the previous equations, we can calculate the value of V as follows:[tex]V=E*r\\V=E*0.3m= \frac{(9e9N*m^2/C2)*(-12C)}{0.3m} \\V= -3.6*10^-^1^1 V.[/tex]
More information about Gauss' law at the link:
https://brainly.com/question/14705081
is 0.8 kilograms bigger then 80 grams
Answer:
Yes
Explanation:
0.8 kilograms is equal to 800 grams
Answer:
Yes, 0.8 kilograms is greater than 80 grams
Explanation:
0.8 kilograms is equal to 800 grams and 80 grams is equal to 0.08 kilogrmas.
Sorry if I'm wrong, correct me.
The mass of 60 paper clips is 18.0 grams. What is the mass of one paper clip?
Answer:
3.333333333333333333333333333333333333333
Explanation:
3.3333333333333333333333333333333333
Which of the following is a vector quantity?
speed
distance
acceleration
When the bowling ball has fallen halfway down the building (height = 20 m), it has a speed of 19.8 m/s.
How much potential energy does the bowling ball have?
How much kinetic energy does the bowling ball have?
How much total energy (potential + kinetic) does the bowling ball have?
Of the bowling ball’s total energy, is more in the form of potential or kinetic energy?
Answer:
I think the answer is 19.8 potential energy
Explanation:
NONE.
what type of waves can only travel through a medium?
Answer:
Mechanical waves
Explanation:
Mechanical waves are the waves that can travel only through a medium. Mechanical waves are disturbance of matter and require medium to transfer the energy. There are three types of mechanical waves that include transverse wave, longitudinal wave and surface wave.
Some of the examples of mechanical waves are sound waves and seismic waves etcetera.
Hence, the correct answer is "Mechanical waves".
A stone is dropped from the top of a high cliff with zero initial velocity. In which system is the net momentum zero as the stone falls freely
Answer:
A system that includes the stone and the earth.
Explanation:
If the system of being dropped from the height of the cliff consists of just the stone alone, then it means that its momentum will certainly undergo changes as it falls freely. However, If the system is now expanded to include not only the stone but also the Earth, then it implies that the momentum of the stone which is in the downward direction will be equal and opposite to the momentum of the Earth in the upwards direction towards the stone. Therefore, the momentum will cancel out and net momentum will be zero.
A system of stone and earth can result to a net zero momentum.
Conservation of linear momentum
The principle of conservation of linear momentum states that the sum of the initial momentum is equal to the sum of final momentum.
[tex]m_1u_1 + m_2 u_2 = m_1v_1 + m_2 v_2[/tex]
A system that consists a linear system of stone and earth can result to a net zero momentum.
Thus, a system of stone and earth can result to a net zero momentum.
Learn more about conservation of momentum here: https://brainly.com/question/7538238
A vertical piston-cylinder device contains a gas at a pressure of 100 kPa. The piston has a mass of 10 kg and a diameter for 14 cm. Pressure of the gas is to be increased by placing some weights on the piston. Determine the local atmospheric pressure and the mass of the weights that will doublethe pressure of the gas inside the cylinder.
Answer:
the local atmospheric pressure is 93.63 kPa
the mass of the weights is 156.9 kg
Explanation:
Given that;
Initial pressure of gas = 100 kPa
mass of piston = 10 kg and diameter = 14 cm = 0.14 m
g = 9.81 m/s²
Now,
P_gas = P_atm + P_piston
100 = P_atm + P_piston --------- let this equation 1
P_piston = M_piston × g / A = (10 × 9.81) / π/4×d²
P_piston = 98.1 / (π/4×( 0.14 )²)
P_piston = 98.1 / 0.01539 = 6374,269 Pa = 6.37 kPa
now, from equation 1
100 = P_atm + P_piston
we substitute
100 = P_atm + 6.37
P_atm = 100 - 6.37
P_atm = 93.63 kPa
Therefore, the local atmospheric pressure is 93.63 kPa
Now for pressure of the gas in the cylinder ⇒ 2×initial pressure
Pgas_2 = 2 × 100 = 200 kPa
Pgas_2 = P_atm + P_piston + P_weight
Pgas_2 = P_gas + P_weight
we substitute
200 kPa = 100 kPa + P_weight
P_weight = 200 kPa - 100 kPa
P_weight = 100 kPa = 100,000 Pa
Also;
P_weight = M×g / A
100,000 Pa = ( M × 9.81 ) / (π/4 × (0.14)²)
100,000 × 0.01539 = M × 9.81
1539 = M × 9.81
M = 1539 / 9.81
M = 156.9 kg
Therefore, the mass of the weights is 156.9 kg
take a picture of an object in your house, describe the
energy stores and transfers that happen with it. You can be as imaginative as you wish
with the object (choose something unusual), but the stores you identify and transfers
that happen must be real.
pls give me ideas of what to take a photo of for this I'm really stuck :(
Which of the physical variables listed below will change when you change the area of the capacitor plates (while keeping the battery connected).
a. Capacitance
b. Charge on the plates
c. Voltage across the plates
d. Net electric field between the plates
e. Energy stored in the capacitor
Answer:
a. Capacitance
b. Charge on the plates
e. Energy stored in the capacitor
Explanation:
Let A be the area of the capacitor plate
The capacitance of a capacitor is given as;
[tex]C = \frac{Q}{V} = \frac{\epsilon _0 A}{d} \\\\[/tex]
where;
V is the potential difference between the plates
The charge on the plates is given as;
[tex]Q = \frac{V\epsilon _0 A}{d}[/tex]
The energy stored in the capacitor is given as;
[tex]E = \frac{1}{2} CV^2\\\\E = \frac{1}{2} (\frac{\epsilon _0 A}{d} )V^2[/tex]
Thus, the physical variables listed that will change include;
a. Capacitance
b. Charge on the plates
e. Energy stored in the capacitor
A cylindrical tank of radius R, filled to the top with a liquid, has a small hole in the side, of radius r, at distance d below the surface. Find an expression for the volume flow rate through the hole. A 4.0-mm-diameter hole is 1.0 m below the surface of a 2.0-m-diameter tank of water. What is the rate, in mm/min, at which the water level will initially drop if the water is not replenished?
Answer:
a)[tex]V=\pi *r^2 * \sqrt{2gd}[/tex]
b)[tex]dh / dt = 0.2658 mm / min[/tex]
Explanation:
From the question we are told that
Diameter of hole [tex]d_h=4mm=>0.004m[/tex]
Depth of hole [tex]D=0mm=>0.001m[/tex]
Diameter of tank [tex]d_t=2mm=>0.002m[/tex]
Generally the equation for pressure is mathematically given as
[tex]Pressure P= \rho*g*d[/tex]
[tex]P= 1/2*\rho *v^2[/tex]
Where
[tex]v = \sqrt {2gd}[/tex]
[tex]V = Area*v[/tex]
[tex]V=\pi *r^2 * \sqrt{2gd}[/tex]
Generally the level at which the water level will initially drop if the water is not replenished is mathematically given by
[tex]dh / dt = (r/R)^2 *sqrt{2gd}\\dh / dt = (2/2000)^2 *sqrt(2*9.81*1) \\dh / dt = 4.429*10^-3 mm/s \\[/tex]
Therefore the level at which the water level will initially drop if the water is not replenished
[tex]dh / dt = 0.2658 mm / min[/tex]
The rate, in mm/min, at which the water level will initially drop will be 1.0625 mm/min.
Given data:
The diameter of hole is, d = 4.0 mm = 0.004 m.
The depth of hole is, h = 1.0 m.
The diameter of tank is, d' = 2.0 m.
The given problem is based on the flow rate, which is defined as the flow of liquid through a given section per unit time.
Let us first obtain the equation of pressure as,
[tex]P=\dfrac{1}{2} \times \rho \times v^{2}[/tex]
Here, v is the velocity of efflux and its value is,
[tex]v=\sqrt{2gh} \\\\v^{2}=2gh[/tex]
And the level at which the water level will initially drop if the water is not replenished is mathematically given by,
[tex]\dfrac{dH}{dt}=(r/R)^{2} \times v[/tex]
Here,
r is the radius of hole.
R is the radius of tank.
Solving as,
[tex]\dfrac{dH}{dt}=((d/2) /(d'/2))^{2} \times \sqrt{2gh} \\\\\dfrac{dH}{dt}=((0.004/2) /(2/2))^{2} \times \sqrt{2 \times 9.8 \times 1}\\\\\dfrac{dH}{dt}=1.77 \times 10^{-5} \;\rm m/s\\\\\dfrac{dH}{dt}=1.77 \times 10^{-5} \times 6 \times 10^{4} \;\rm mm/min\\\\\dfrac{dH}{dt}=1.0625 \;\rm mm/min[/tex]
Thus, we can conclude that the rate, in mm/min, at which the water level will initially drop will be 1.0625 mm/min.
Learn more about the flow rate here:
https://brainly.com/question/11816739
If a person weighs 140 lb'on Earth, their mass in kilograms is
Answer:
70 kg
Explanation:
divide it by 2
Hope this helped!
Answer:
63.502932 Kilograms
Explanation:
If there is "waste" energy, does the Law of Conservation of Energy still apply? please don't type something random if so i'll just report it.
Explanation:
Yes, the law of conservation of energy still applies even if there is waste energy.
The waste energy are the transformation products of energy from one form to another.
According to the law of conservation of energy "energy is neither created nor destroyed by transformed from one form to another in a system".
But of then times, energy is lost as heat or sound within a system.
If we take into account these waste energy, we can see that energy is indeed conserved. The sum total of the energy generated and those produced will be the same if we factor in other forms in which the energy has been transformed into.Which is larger: 65 mph (miles per hour) or 120 kph (kilometers per hour)? As a percentage, how much faster is one than the other?
To Find :
Which is larger: 65 mph (miles per hour) or 120 kph (kilometers per hour).
Solution :
We know, 1 mph = 1.61 kph
So, 65 mph = 1.61 × 65 kph
65 mph = 104.65 kph
Since, 65 mph is 104.65 kph which is smaller than 120 kph.
Therefore, 120 kph is faster than 65 mph by ( 120 - 104.65 ) = 15.35 kph.
As waves crash into rock along the shoreline, particles of sand, shell, and other materials in the ocean water loosen tiny bits of sediment from the rock. As the waves recede, they carry the sediment away. In this scenario, which process represents weathering, and which process represents erosion?
Answer:
WEATHERING is represented by the scenario (As waves crash into rock along the shoreline, particles of sand, shell, and other materials in the ocean water loosen tiny bits of sediment from the rock).
Erosion is represented by the scenario (As the waves recede, they carry the sediment away).
Explanation:
A wave is a disturbance which travels through a medium and transfers energy from one point to another. When wind blows over a water body like the ocean, ocean waves are formed. As the generated energy from the wind is transported through the water by the waves, the can hit against rocks on the shores leading to its break down with time. WEATHERING occurs when tiny bit of sediments from rocks are loosened due to the impact of ocean waves.
Erosion can be described as the wearing away of the earth's surface due to the impact of wind, rainfall ( water) or waves. There are different types of erosion which is classified according it's cause of formation.
Wave erosion occurs when sediments such as sand, shell and other materials are carried to the shoreline by ocean waves. This erodes the shore over time as the sediments act like sandpapers.
A neutral metal bob is hanging on the bottom of a pendulum that is 15 cm long. A charged balloon is held near the metal bob and the pendulum is pulled up to a vertical angle of 20-deg. If the mass of the metal bob is 0.025kg, what is the charge on the balloon.
Answer:
Explanation:
See the figure attached
F is electrostatic force .
T cos20 = mg
T sin20 = F
Tan20 = F / mg
F = mg tan 20 = .025 x 9.8 tan20
= .09 N
Distance between bob and balloon
= 15 sin20 = 5.1 cm = .051 m
If q be the charge on balloon
F = 9 x 10⁹ x q² / .051²
= 3460 x 10⁹ q² = .09
q² = 26 x 10⁻⁶ x 10⁻⁹
q = 16.12 x 10⁻⁸ C .
A sprinter starts from rest and accelerated at a rate of 0.16 m/s over a distance of 50.0 meters. How fast is the athletes traveling at the end of the 50.0 meters?
Answer:
40m/s
Explanation:
v²=u²+2as
v²=0²+2(16)(50)
v²=160v=40m/s
A student is driving through a mountainous region where the road is at some times flat, at some times inclined upward, and at some time inclined downward. The student maintains a speed of 20 m/s on the roadway, but is required to make an emergency stop on the three sepearte occasions. On levels roadway, it takes 25 m to stop. On a downward-sloping roadway, it takes 40 m to stop. On an upward-sloping roadway, it takes 18 m to stop. Explain why the stopping distances are different. (Focus answer using work and energy, other concepts may be used as well but be sure work and energy are included.)
Answer:
Explanation:
It is frictional force of the ground that helps in bringing the vehicle to stop . In the process of stopping , negative work is done on the car by friction force to overcome its kinetic energy .
At levelled road , for stoppage
Kinetic energy of vehicle = Work done by frictional force . = friction force x displacement .
At upward slopping road , gravitational force acting downward also helps the vehicle to stop do friction has to do less work .
At upward inclined road , for stoppage
Kinetic energy of vehicle = Work done by frictional force + work done by gravitational force = (friction force + gravitational force ) x displacement .
Hence displacement is less .
At downward slopping road , friction has to do more work because friction has to do work against gravitational force acting downwards wards and kinetic energy of the vehicle also .
At downward inclined road , for stoppage
Kinetic energy of vehicle + work done by gravitational force = Work done by frictional force = friction force x displacement .
Hence displacement is more .
Hence displacement is more in the downward slopping.
What is Displacement?Displacement is defined as the change in position of an object. It is a vector quantity and has a direction and magnitude.
It is frictional force of the ground that helps in bringing the vehicle to stop . In the process of stopping , negative work is done on the car by friction force to overcome its kinetic energy .
At levelled road , for stoppage
Kinetic energy of vehicle = Work done by frictional force . = friction force x displacement .
At upward slopping road , gravitational force acting downward also helps the vehicle to stop do friction has to do less work .
At upward inclined road , for stoppage
Kinetic energy of vehicle = Work done by frictional force + work done by gravitational force = (friction force + gravitational force ) x displacement .
Hence displacement is less .
At downward slopping road , friction has to do more work because friction has to do work against gravitational force acting downwards wards and kinetic energy of the vehicle also .
At downward inclined road , for stoppage
Kinetic energy of vehicle + work done by gravitational force = Work done by frictional force = friction force x displacement .
Hence displacement is more in the downward slopping.
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