Answer:
s = 23.68 m
Explanation:
given,
acceleration of the sperm whale = 0.09 m/s²
initial speed of the whale, u = 0.7 m/s
final speed of the whale, v = 2.18 m/s
Distance traveled by the sperm whale = ?
Using equation of motion
v² = u² + 2 a s
2.18² = 0.7² + 2 x 0.09 x s
0.18 s = 4.2624
s = 23.68 m
Distance traveled by the whale is equal to s = 23.68 m
In what ways do the mirrors in X-ray telescopes differ from those found in optical instruments?
Explanation:
X-ray telescopes have a different design than that of optical telescopes, since x-rays can reflect off from mirror if they are struck at a particular of angle of the grazing. The X-rays are concentrated to a point in two reflections. By nesting the mirrors inside a one another , the X-ray telescope area can be enhanced.
Final answer:
X-ray telescope mirrors are designed to reflect high-energy X-rays at small angles, using precision-coated and aligned mirrors, while optical telescope mirrors only require front surface polishing and are used for visible light reflection, allowing them to be larger and more cost-effective.
Explanation:
The mirrors in X-ray telescopes, such as those used in the Chandra X-ray Observatory, are designed to reflect high-energy X-rays that are absorbed when incident perpendicular to the medium. Unlike mirrors in optical instruments, which reflect visible light usually by incident at direct angles, X-ray telescope mirrors reflect X-rays at small grazing angles, similar to a rock skipping across a lake. The design of these mirrors involves a long barrelled pathway with multiple pairs of mirrors, often coated with metals like iridium, to focus the rays at a specific point. They're precision-engineered to be extremely smooth for the most effective reflection.
In contrast, optical telescopes like reflectors use mirrors that only need the front surface polished accurately, avoiding issues like flaws and bubbles within the glass. These mirrors can be larger and more cost-effective than lenses, making them suitable for studying dimmer or more distant objects. Today's largest optical telescopes are reflectors for this reason.
A car comes to a bridge during a storm and finds the bridge washed out. The driver must get to the other side, so he decides to try leaping it with his car. The side the car is on is 20.0 m above the river, whereas the opposite side is a mere 2.1 m above the river. The river itself is a raging torrent 61.0 m wide.
A) How fast should the car be traveling just as it leaves the cliff in order to just clear the river and land safely on the opposite side?
B) What is the speed of the car just before it lands safely on the other side?
Answer:
A) The car should be traveling at 31.9 m/s.
B) The speed of the car just before it lands on the other side is 37.0 m/s.
Explanation:
Hi there!
A) Please see the attached figure for a better description of the problem. When the car reaches the other side of the river, its position vector will be r1 in the figure. The components of this vector are r1x and r1y.
If we place the origin of the frame of reference at the edge of the cliff, the components of the vector r1 will be:
r1x = 61.0 m
r1y = -20.0 m + 2.1 m = -17.9 m
The equations for the x and y-components of the position vector of the car are the following:
x = x0 + v0 · t
y = y0 + 1/2 · g · t²
Where:
x = horizontal position at a time t.
x0 = initial horizontal position.
v0 = initial velocity.
t = time.
y = vertical position at a time t.
y0 = initial vertical position.
g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).
Using the equation of the y-component of r1, we can find the time it takes the car to reach the other side of the river. We have to find the time at which the vector r1y is -17.9 m:
y = y0 + 1/2 · g · t² (y0 = 0 because the origin of the frame of reference is located at the edge of the cliff).
y = 1/2 · g · t²
-17.9 m = -1/2 · 9.8 m/s² · t²
-17.9 m / -4.9 m/s² = t²
t = 1.91 s
Now, using the equation of the x-component, we can find the initial velocity. We know that at t = 1.91 s, the horizontal component of the vector r1 is 61.0 m:
x = x0 + v0 · t (x0 = 0 because the origin of the frame of reference is located at the edge of the cliff).
x = v0 · t
61.0 m = v0 · 1.91 s
v0 = 61.0 m / 1.91 s = 31.9 m/s
The car should be traveling at 31.9 m/s.
B) The equation of the velocity vector of the car is the following:
v = (v0, g · t)
The horizontal component of the velocity vector is v0, 31.9 m/s.
Let's calculate the value of the vertical component:
vy = g · t
vy = -9.8 m/s² · 1.91 s
vy = -18.7 m/s
Then, the velocity vector of the car just before it lands on the other side is the following:
v = (31.9, -18.7) m/s
The magnitude of this vector is calculated as follows:
|v| = √[(31.9 m/s)² + (-18.7 m/s)²]
|v| = 37.0 m/s
The speed of the car just before it lands on the other side is 37.0 m/s.
Final answer:
To safely clear the river, the car must travel at 23.4 m/s at the cliff's edge, and it will land with a speed of 22.6 m/s on the other side.
Explanation:
A) To clear the river and land safely on the other side, the car should be traveling at a speed of 23.4 m/s as it leaves the cliff. This is calculated using the principles of projectile motion and conservation of energy.
B) The speed of the car just before it lands safely on the opposite side would be 22.6 m/s. This speed is also determined by energy considerations, such as the conversion of potential energy to kinetic energy.
You and your best friend are trying to pull one another toward your respective dorm rooms. You're the stronger of the two and, with a mighty tug, you drag your friend into your room.
As you are pulling your friend toward your room, the force you exert on your friend is:
A) equal in amount to the force your friend exerts on you.
B) definitely equal to three times the weight of Spongebob Squarepants.
C) less in amount than the force your friend exerts on you.
D) greater in amount than the force your friend exerts on you.
Answer:
A) equal in amount to the force your friend exerts on you.
Explanation:
According to the Newton's third law of motion every action has equal and opposite reaction. Here both the bodies exert equal and opposite forces on each other but it is just that I am able to stop myself from getting pulled by a greater force of friction by applying more normal force on the ground.
As we know that the force of friction is given as:
[tex]f=\mu.N[/tex]
where:
[tex]\rm f=\ force\ of\ friction\\N=\ applied\ normal\ force\ to\ the\ ground\\ \mu=\ coefficient\ of\ friction[/tex]
An entertainer juggles balls while doing other activities. In one act, she throws a ball vertically upward, and while it is in the air, she runs to and from a table 5.50 m away at an average speed of 3.00 m/s, returning just in time to catch the falling ball. (a) With what minimum initial speed must she throw the ball upward to accomplish this feat? (b) How high above its initial position is the ball just as she reaches the table?
To accomplish this feat, the entertainer must throw the ball upward with a minimum initial speed and reach a certain height above its initial position.
Explanation:(a) While the ball is in the air, it rises and then falls to a final position 10.0 m higher than its starting altitude. We can find the time for this by using Equation 4.22:
y = yo + voyt - (1/2)gt².
If we take the initial position yo to be zero, then the final position is y = 10 m. The initial vertical velocity is the vertical component of the initial velocity:
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Final answer:
The minimum initial speed the entertainer must throw the ball upward is 17.96 m/s, and the height of the ball just as she reaches the table is approximately 15.56 m.
Explanation:
Calculating the Minimum Initial Speed and Height of the Juggling Ball
To answer the student's question regarding the juggling entertainer's act, we need to apply concepts from kinematics, a subfield of classical mechanics in physics. First, let's find out the total time the entertainer has to throw the ball and return to the starting position. We can find the time taken to reach the table and come back with the given average speed and distance: Time = (2 × Distance) / Speed. The entertainer runs to and from a table which is 5.50 m away at an average speed of 3.00 m/s. Hence, the total time for the round trip is (2 × 5.50 m) / 3.00 m/s = 3.67 s.
Now, we use the equation of motion to calculate the initial vertical speed needed for the ball to be in the air for this duration: s = ut + 0.5 × a × t², where s is the displacement (which is 0 because the ball returns to the same position), u is the initial vertical speed, and a is the acceleration due to gravity (-9.81 m/s²). After rearranging the formulas, the initial speed u turns out to be 17.96 m/s.
To find the height of the ball as she reaches the table at 5.50 m away, we consider half the total time, which is 1.835 s. Putting this in the kinematic equation s = ut + 0.5 × a × t², we find the height to be approximately 15.56 m at that instant.
Calculate the magnitude of the gravitational force exerted by the Moon on a 79 kg human standing on the surface of the Moon. (The mass of the Moon is 7.4 × 1022 kg and its radius is 1.7 × 106 m.)
Answer:
F= 134.92 N
Explanation:
Given that
The mass of the moon ,M = 7.4 x 10²² kg
The mass of the man ,m = 79 kg
The radius ,R= 1.7 x 10⁶ m
The force exerted by moon is given as
[tex]F=G\dfrac{Mm}{R^2}[/tex]
Now by putting the values in the above equation we get
[tex]F=6.67\times 10^{-11}\times \dfrac{79\times 7.4\times 10^{22}}{(1.7\times 10^6)^2}\ N\\F=134.92 N[/tex]
Therefore the force will be 134.92 N.
F= 134.92 N
a third resistor is added in parallel with the first two.
(picture shows 2 resistors connected in parallel to battery)
What happens to the current in the battery?
A. remains the same
B. increases
C. decreases
What happens to the terminal voltage of the battery?
A. remains the same
B. increases
C. decreases
Answer:
C.
A
Explanation:
Question 1
When a third resistor is added in parallel to the first two. The effective resistance of the circuit increases R_eq. When the resistance is increased in a circuit while the battery provides a constant voltage the current decreases as per Ohm's Law:
I = V / R_eq
Current and effective resistance are inversely proportional. Hence, the current in the battery decreases.
Question 2
The terminal voltage remains the same because the amount of push required to move electrons in a path remains same. Adding more paths would not require more push, unless resistance is added in the same path i.e series. Hence, terminal voltage of the battery remains the same.
Final answer:
Adding a third resistor in parallel decreases the total resistance and increases the current drawn from the battery, but the terminal voltage of the battery ideally remains the same.
Explanation:
When a third resistor is added in parallel with the first two resistors across a battery, the total resistance of the circuit is reduced. According to Ohm's law, the total current ('I') drawn from the battery is equal to the voltage ('V') divided by the total resistance ('R'), as in the formula I = V/R. When the total resistance decreases due to the addition of another resistor in parallel, the total current provided by the battery increases. Therefore, the current in the battery increases.
As for the terminal voltage of the battery, it remains the same assuming the battery is ideal. In an ideal circuit, adding additional resistors in parallel does not change the voltage across the resistors; they all share the same potential difference as that of the battery. Real-world batteries, however, may experience a slight drop in terminal voltage due to internal resistance, but this effect is typically ignored in basic circuit analysis.
What fraction of the copper's electrons has been removed? Each copper atom has 29 protons, and copper has an atomic mass of 63.5.
Answer:
9.09*10^-13
Explanation:
Certain values of the problem where omitted,however the omitted values were captured in the solution.
Step1:
Avogadro's number (NA) I = 6.02*10^23 atoms/mole.
Step2:
To determine the number of moles of copper that are present, thus: Using the mass and atomic mass :
n = m/A
n = 50.0g/63.5g/mol
Therefore, since the are 29 protons per atom, I the number of protons can be determined as follows :
Np = nNA*29 protons /atom
Np=(50.0gm/63.5g/mol)(6.02*10^23 atoms/mol) * (29 protons / C u atom)
Np= 1.375*10^25 protons
Note that there are same number of electrons as protons in a neutral atom, I therefore the removal of electrons to give the copper a net change, hence the result is 1.375*10^25
Step3:
To determine the number electrons , removed to leave a net charge of 2.00Uc, then remove -2 .00Uc of charge, so that the number of electrons to be removed are as follows :
Ne(removed)=
Q/qe= -2.00*10^-6c/-1.60* 10^-19c
Ne(removed)=1.25*10^13 electrons removed
Step4:
To calculate the fraction of copper's electron by taking the ratio of the number of electrons initially present:
Ne,removed/Ne,initially=1.25*10^13/ 1.37*10^25 = 9.09*10^-13
A battery having an emf of 9.63 V delivers 118 mA when connected to a 60.0 Ω load. Determine the internal resistance of the battery.
Answer:
21.6 ohm
Explanation:
We are given that
EMF=E=9.63 V
Current=I=118 mA=[tex]118\times 10^{-3} A[/tex]
[tex]1 mA=10^{-3} A[/tex]
Resistance=[tex]R=60\Omega[/tex]
We have to find the internal resistance of the battery.
We know that
[tex]V=E-Ir[/tex]
We know that V=IR
[tex]IR=E-Ir[/tex]
[tex]IR+Ir=E[/tex]
[tex]I(R+r)=E[/tex]
[tex]R+r=\frac{E}{I}[/tex]
Substitute the values
[tex]60+r=\frac{9.63}{118\times 10^{-3}}[/tex]
[tex]60+r=81.6[/tex]
[tex]r=81.6-60[/tex]
[tex]r=21.6\Omega[/tex]
Hence, the internal resistance of the battery=21.6 ohm
Answer:
21.6 ohm
Explanation:
EMF of the battery, E = 9.63 V
Current, i = 118 mA = 0.118 A
Resistance, R = 60 ohm
Let the internal resistance of the cell is r.
[tex]i = \frac{E}{R + r}[/tex]
R + r = 9.63 / 0.118
60 + r = 81.6
r = 21.6 ohm
A particle moves so that its position (in meters) as a function of time (in seconds) is r = i ^ + 4t2 j ^ + tk ^. Write expressions for (a) its velocity and (b) its acceleration as functions of time.
Answer:
a.V=8tj+k
b.a=8j
Explanation:
Given:
Position r= i+4t^2j +tk
Nb r is position in metre and time in seconds
a.velocity is change in position/ change in time
v= ∆r/∆t =dr/dt
V=d ( i+ 4t^2j+tk)/dr
Differenting with respect to (t)
V=8tj+K
b.acceleration = change in velocity/change in time
a= ∆v/∆r =dv/dt
a=d (8tj+k)/dt
a= 8j
Answer:
(a) velocity, v = 8t j + k
(b) acceleration, a = 8 j
Explanation:
The position of the particle as a function of time is given as;
r = i + 4t² j + t k --------------------(i)
(a) To get the expression of its velocity, v, find the derivative of its position with respect to time by differentiating equation (i) with respect to t as follows;
v = dr / dt = 0 + 8t j + k
v = dr / dt = 8t j + k
v = 8t j + k ----------------------(ii)
Therefore, the equation/expression for the particle's velocity (v) is
v = 8t j + k
(b) To get the expression of its acceleration, a, find the derivative of its velocity with respect to time by differentiating equation (ii) with respect to t as follows;
a = dv / dt = t j + 0
a = dv / dt = t j
a = 8 j
Therefore, the expression for the particle's acceleration, a, is a = 8 j
At standard temperature and pressure (0 ∘C∘C and 1.00 atmatm ), 1.00 molmol of an ideal gas occupies a volume of 22.4 LL. What volume would the same amount of gas occupy at the same pressure and 25 ∘C∘C ?
Answer:
Final volume will be 24.45 L
Explanation:
We have given initial temperature [tex]T_1=0^{\circ}C=0+273=273K[/tex]
Pressure is [tex]P_1=1atm[/tex]
Volume occupied [tex]V_1=22.4lL[/tex]
From ideal gas equation [tex]PV=nRT[/tex]
[tex]\frac{PV}{T}=constant[/tex]
So [tex]\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}[/tex]
Final temperature [tex]T_2=25+273=298K[/tex]
Pressure is remain constant so [tex]P_2=1atm[/tex]
We have to fond the volume [tex]V_2[/tex]
So [tex]\frac{1\times 22.4}{273}=\frac{1\times V_2}{298}[/tex]
[tex]V_2=24.45L[/tex]
So final volume will be 24.45 L
A rock is thrown with a velocity v0, at an angle of α0 from the horizontal, from the roof of a building of height h. Ignore air resistance. Calculate the speed of the rock just before it strikes the ground, and show that this speed is independent of α0.
Answer:
[tex]V=\sqrt{V_{0}^{2}+2gy}[/tex]
Explanation:
Data given,
[tex]velocity,v =v_{0}\\ angle =\alpha _^{0}[/tex]
since the motion part is describe by a projectile motion, the acceleration along the horizontal axis is zero
Hence using the equation v=u+at we have the following equation ,
the velocity along the horizontal axis to be
[tex]V_{x}=V_{0}cos\alpha _{0} \\[/tex]
the velocity along the vertical axis to be
[tex]V_{y}=V_{0}sin\alpha _{0}-gt \\[/tex]
the magnitude of this velocity can be determine using Pythagoras theorem
[tex]V^{2}=V_{x}^{2} +V_{y} ^{2}[/tex]
if we substitute the expressions we have
[tex]V^{2}=V_{0}^{2}cos\alpha _{0}^{2} +(V_{0}sin\alpha _{0}-gt)\\expanding \\V^{2}=V_{0}^{2}(cos\alpha _{0}^{2}+sin\alpha _{0}^{2})-2gtsin\alpha _{0}+(gt)^{2}\\(cos\alpha _{0}^{2}+sin\alpha _{0}^{2})=1\\V^{2}=V_{0}^{2}-2gtsin\alpha _{0}+(gt)^{2}\\[/tex]
[tex]V^{2}=V_{0}^{2}-2gtV_{0}sin\alpha _{0}+(gt)^{2}\\V^{2}=V_{0}^{2}-2g(V_{0}sin\alpha _{0}-\frac{1}{2}gt^{2} )\\V_{0}sin\alpha _{0}-\frac{1}{2}gt^{2}=distance=y\\V^{2}=V_{0}^{2}-2gy\\ for upward \\V^{2}=V_{0}^{2}+2gy\\V=\sqrt{V_{0}^{2}+2gy}[/tex]
The speed of the rock, just before it strikes the ground (maximum speed) can be given as,
[tex]v=\sqrt{v_0^2-2gh} \\[/tex]
Witch is independent to angle [tex]a_0[/tex].
What is speed of the object just before hitting the ground?The speed of the object falling from a height achieved the maximum speed, just before hitting the ground.
Given information-
The rock is thrown with a velocity [tex]v_0[/tex].
The rock is thrown with a angle of [tex]a_0[/tex].
The height of the building is [tex]h[/tex].
The height of the building can be given as,
[tex]h=v_0\times\sin(a_0)-\dfrac{1}{2}gt[/tex] .........1
Let the above equation as equation 1,
The horizontal velocity of the rock can be given as,
[tex]v_h=v_0\times\cos (a_0)[/tex].
The vertical velocity of the rock can be given as,
[tex]v_v=v_0\times\sin(a_0)-gt[/tex]
Here, [tex]g[/tex] is the gravitational force and [tex]t[/tex] is time.
Now the magnitude of the velocity can be given as,
[tex]v=\sqrt{v_h^2+v_v^2} \\v=\sqrt{(v_0\times\cos(a_0))^2+(v_0\times\sin(a_0)-gt)^2} \\v=\sqrt{v_0^2\times\cos^2(a_0)+v_0^2\times\sin(a_0)^2+(gt)^2-2gtv_0\times\sin(a_0)} \\v=\sqrt{v_0^2(\cos^2(a_0)+\times\sin(a_0)^2)-2g(v_0\times\sin(a_0)-\dfrac{1}{2}gt^2} \\v=\sqrt{v_0^2(1)-2g(v_0\times\sin(a_0)-\dfrac{1}{2}gt^2} \\[/tex]
Put the values of [tex]h[/tex] from equation 1 to the above equation as,
[tex]v=\sqrt{v_0^2-2gh} \\[/tex]
Hence the speed of the rock, just before it strikes the ground (maximum speed) can be given as,
[tex]v=\sqrt{v_0^2-2gh} \\[/tex]
Witch is independent to angle [tex]a_0[/tex].
Learn more about the velocity of free falling body here;
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A simple elevator ride can teach you quite a bit about the normal force as this rider below can (hopefully) tell you. There are three different scenarios given, detailing the rider\'s experience in an unnamed hotel. For each scenario, calculate the normal force, FN,1-3, acting on the rider if his mass is m = 76.6 kg and the acceleration due to gravity g = 9.81 m/s2. In scenario 1, the elevator has constant velocity. In scenario 2 the elevator is moving with upward acceleration a2 = 4.84 m/s2. Finally, in scenario 3, unfortunately for the rider, the cable breaks and the elevator accelerates downward at a3 = 9.81 m/s2.
FN1= ___________
FN2= ___________
FN3= ___________-
Answer:
FN1 = 751.5 N
FN2 = 1122.2 N
FN3 = 0
Explanation:
Scenario 1 :
The elevator has constant velocity.The normal force, can adopt any value, as needed by Newton's 2nd Law, in order to fit this general expression:
Fnet = m*a
In the first scenario, as the elevator is moving at a constant speed, this means that no external net force is present.
The two forces that act on the rider, are gravity (always present, downward) and the normal force, as follows:
Fnet = Fn - m*g = m*a
For scenario 1:
Fnet = 0 ⇒ Fn = m*g = 76.6 kg * 9.81 m/s² = 751. 5 N
Scenario 2In this scenario, the elevator has an upward acceleration of 4.84 m/s², so the Newton's 2nd Law is as follows:
Fnet = FN - m*g = m*a
⇒ FN = m* ( g+ a) = 76.6 kg* (9.81 m/s² + 4.84 m/s²) = 1,122.2 N
Scenario 3As the elevator is in free fall, this means that a = -g, so, in this condition, the normal force is just zero, as it can be seen from the following equation:
FN-mg = m*a
If a = -g,
⇒ FN -mg = -mg ⇒ FN=0
Final answer:
The normal forces for a person in an elevator are as follows: in scenario 1 with constant velocity, it is 751.686 N; in scenario 2 with upward acceleration, it is 1122.59 N; and in scenario 3 during free fall, it is 0 N.
Explanation:
The question requires us to calculate the normal force acting on a person in an elevator under different scenarios using Newton's second law.
Scenario 1: Constant Velocity
In scenario 1, since the elevator is moving with a constant velocity, the acceleration is 0 [tex]m/s^2[/tex], so the normal force ([tex]FN_1[/tex]) will be equal to the weight of the person. That is:
[tex]FN_1[/tex] = m * g = 76.6 kg * 9.81 [tex]m/s^2[/tex] = 751.686 N
Scenario 2: Upward Acceleration
In scenario 2, the elevator is accelerating upward, so the normal force ([tex]FN_2[/tex]) will be more than the weight of the person. The equation will be:
[tex]FN_2[/tex] = m * (g + a2) = 76.6 kg * (9.81 [tex]m/s^2[/tex] + 4.84 [tex]m/s^2[/tex]) = 76.6 kg * 14.65 [tex]m/s^2[/tex] = 1122.59 N
Scenario 3: Downward Acceleration (Free Fall)
In scenario 3, since the cable breaks, the elevator and the person inside will be in free fall, thus experiencing the same acceleration downward as the acceleration due to gravity. This means there will be no normal force acting on the person ([tex]FN_3[/tex] = 0 N) because they are in free fall.
If a certain silver wire has a resistance of 8.60 Ω at 29.0°C, what resistance will it have at 42.0°C?
Answer:
9.027 Ω
Explanation:
Using,
R = R₀(1+αΔt)
R = R₀(1+α[t₂-t₁]).......................... Equation 1
Where R = the value of the resistance at the final temperature, R₀ = the value of the resistance at the initial temperature, α = Temperature coefficient of resistance, t₂ = Final temperature, t₁ = Initial temperature.
Given: R₀ = 8.6 Ω, t₂ = 42 °C, t₁ = 29 °C
Constant : 0.003819/°C
Substitute into equation 1
R = 8.6(1+0.003819[42-29])
R = 9.027 Ω.
Hence the resistance = 9.027 Ω
9.02Ω
Explanation:The resistivity of a conductor increases as temperature increases. In this case, silver, which is a great conductor, will increase its resistivity linearly over a range of increasing temperatures . The relationship between resistance(R) and temperature(T) is given as;
R = R₀ (1 + α(T - T₀)) --------------(i)
where;
R and R₀ are the final and initial resistances of the material (silver in this case)
α = temperature coefficient of resistivity of the material (silver) = 0.0038/°C
T and T₀ are the final and initial temperatures.
From the question;
R₀ = 8.60Ω
T₀ = 29.0°C
T = 42.0°C
Substitute these values into equation (i);
R = 8.60 (1 + 0.0038(42.0 - 29.0))
R = 8.60(1 + 0.0038(13))
R = 8.60(1 + 0.0494)
R = 8.60(1.0494)
R = 9.02Ω
Therefore, the resistance at 42.0°C is 9.02Ω
A helium-filled weather balloon has a 0.90 m radius at liftoff where air pressure is 1.0 atm and the temperature is 298 K. When airborne, the temperature is 210 K, and its radius expands to 3.0 m. What is the pressure at the airborne location
Answer:
0.019 atm
Explanation:
Assume ideal gas, so PV/T is constant where P is pressure, V is volume and a product of radius R cubed and a constant C, T is the temperature
[tex]\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}[/tex]
[tex]P_2 = P_1\frac{V_1}{V_2}\frac{T_2}{T_1}[/tex]
[tex]P_2 = P_1\frac{CR_1^3}{CR_2^3}\frac{T_2}{T_1}[/tex]
[tex]P_2 = P_1\left(\frac{R_1^3}{R_2^3}\right)^3\frac{T_2}{T_1}[/tex]
[tex]P_2 = 1\left(\frac{0.9}{3}\right)^3\frac{210}{298}[/tex]
[tex]P_2 = 0.3^3*0.7 = 0.019 atm[/tex]
The student's question about the change in pressure of a helium-filled weather balloon as it expands and cools at altitude can be answered using the ideal gas law. By comparing initial and final conditions of pressure, volume, and temperature, and using the equation P2 = P1V1T2 / (T1V2), the new pressure can be determined.
Explanation:The student is asking how to determine the pressure inside a helium-filled weather balloon when it rises to an altitude where the external conditions change. We are given the initial temperature, pressure, and radius of the balloon at liftoff, and the radius at its airborne location where the external temperature has decreased. To solve this problem, the ideal gas law is used in combination with the assumption that the balloon expands isotropically (uniformly in all directions). Since the internal pressure of the balloon must balance the external air pressure plus the pressure due to the tension in the balloon's material, we need to use a modified version of the ideal gas law that accounts for changes in temperature and volume to find the new pressure.
Given that the temperature and volume of the balloon change upon reaching altitude, if the volume and temperature of a gas are changed and the amount of gas (number of moles) remains constant, the ideal gas law (PV = nRT) can be rearranged to show the relationship between initial and final states:
P1V1/T1 = P2V2/T2
Where P1, V1, and T1 are the initial pressure, volume, and temperature and P2, V2, and T2 are the final pressure, volume, and temperature, respectively. Since we know all variables except for P2, we can solve for P2 by rearranging the equation to:
P2 = P1V1T2 / (T1V2)
However, it is important to note that we must convert the volumes from radius measurements to actual volumes using the formula for the volume of a sphere, V = (4/3)πr3, and we must use absolute temperatures in Kelvin.
Using this equation with the provided values (making sure to convert units where necessary), the student will be able to determine the pressure at the airborne location for the weather balloon.
Technician A says a change in circuit resistance will change the amount of current in the circuit. Technician B says a change in circuit voltage will change the amount of current in the circuit. Who is right?
Answer:
Both technician A and B are right
Explanation:
According to ohm's law current flowing in a circuit is equal to [tex]i=\frac{V}{R}[/tex], here i is current V is voltage and R is resistance of the circuit
From the relation we can see that current in the circuit is dependent on both voltage and resistance
So if we change the resistance then current also changes and if we change the resistance then also current changes
So both Technician A and B are right
A 79.7 kg base runner begins his slide into second base while moving at a speed of 4.77 m/s. The coefficient of friction between his clothes and Earth is 0.635. He slides so that his speed is zero just as he reaches the base. The acceleration of gravity is 9.8 m/s 2 . What is the magnitude of the mechanical energy lost due to friction acting on the runner? Answer in units of J.
To solve this problem we will apply the concept related to the kinetic energy theorem. Said theorem states that the work done by the net force (sum of all forces) applied to a particle is equal to the change experienced by the kinetic energy of that particle. This is:
[tex]\Delta W = \Delta KE[/tex]
[tex]\Delta W = \frac{1}{2} mv^2[/tex]
Here,
m = mass
v = Velocity
Our values are given as,
[tex]m = 79.7kg[/tex]
[tex]v = 4.77m/s[/tex]
Replacing,
[tex]\Delta W = \frac{1}{2} (79.7kg)(4.77m/s)^2[/tex]
[tex]\Delta W = 907J[/tex]
Therefore the mechanical energy lost due to friction acting on the runner is 907J
In terms of the variables in the problem, determine the time, t, after the launch it takes the balloon to reach the target. Your answer should not include h.
Answer:
[tex]t=\dfrac{d}{v_0cos(\theta )}[/tex]
Explanation:
The background information:
A student throws a water balloon with speed v0 from a height h = 1.8m at an angle θ = 29° above the horizontal toward a target on the ground. The target is located a horizontal distance d = 9.5 m from the student’s feet. Assume that the balloon moves without air resistance. Use a Cartesian coordinate system with the origin at the balloon's initial position.
The time it takes for the balloon to reach the target is equal to the target distance [tex]d[/tex] divided by the horizontal component of the velocity [tex]v_0[/tex]:
[tex]t=\dfrac{d}{v_x}[/tex]
where [tex]v_x[/tex] is the horizontal component of the velocity [tex]v_0[/tex], and it is given by
[tex]v_x=v_0cos(\theta)[/tex];
Therefore, we have
[tex]\boxed{t=\dfrac{d}{v_0cos(\theta)} }[/tex]
The time it takes for the balloon to reach the target can be found by solving the equation x = xo + vot + at². By rearranging the equation and substituting the given values, the time can be determined. For example, if the balloon has an initial vertical velocity of 21.2 m/s and lands 10.0 m below its starting altitude, it will spend 3.79 s in the air.
The time it takes for the balloon to reach the target, denoted by t, can be determined by the vertical motion of the balloon. The equation x = xo + vot + at² can be used to solve for t, since the only unknown in the equation is t.
By rearranging the equation and substituting the given values, we can solve for t.
For example, if the balloon has an initial vertical velocity of 21.2 m/s and lands 10.0 m below its starting altitude, it will spend 3.79 s in the air.
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Speedy Sue, driving at 32.0 m/s, enters a one-lane tunnel. She then observes a slow-moving van 170 m ahead traveling with velocity 5.50 m/s. Sue applies her brakes but can accelerate only at ?2.00 m/s2 because the road is wet. Will there be a collision?If yes, determine how far into the tunnel and at what time the collision occurs.
If no, determine the distance of closest approach between Sue's car and the van, and enter zero for the time.
Distance in meters?Speed in seconds?
Answer:
10.89 seconds
229.895 m
Explanation:
Distance van travels
[tex]x_v=170+5.5t+\dfrac{1}{2}0t^2\\\Rightarrow x_v=170+5.5t[/tex]
Position of car
[tex]x_c=0+32t+\dfrac{1}{2}-2t^2\\\Rightarrow x_c=32t-t^2[/tex]
They are equal
[tex]170+5.5t=32t-t^2\\\Rightarrow 17+5.5t-32t+t^2\\\Rightarrow t^2-26.5t+170=0\\\Rightarrow 10t^2-265t+1700=0[/tex]
[tex]t=\frac{-\left(-265\right)+\sqrt{\left(-265\right)^2-4\cdot \:10\cdot \:1700}}{2\cdot \:10}, \frac{-\left(-265\right)-\sqrt{\left(-265\right)^2-4\cdot \:10\cdot \:1700}}{2\cdot \:10}\\\Rightarrow t=15.6, 10.89\ s[/tex]
The collision occurs at 10.89 seconds
[tex]x_v=170+5.5\times 10.89\\\Rightarrow x_v=229.895\ m[/tex]
Collision occurs at 229.895 m from the starting point
Two steamrollers begin 115 mm apart and head toward each other, each at a constant speed of 1.10 m/sm/s . At the same instant, a fly that travels at a constant speed of 2.20 m/sm/s starts from the front roller of the southbound steamroller and flies to the front roller of the northbound one, then turns around and flies to the front roller of the southbound once again, and continues in this way until it is crushed between the steamrollers in a collision. What distance does the fly travel?
When we Using formula of relative speed
Then v = v1+ v2After that Put the value into the formulav = 1.10 + 1.10v = 2.20 m/sThen We need to calculate the time for the two steamrollers to meet each otherthen we Used the formula of timethen t = d/uWhen we put the value into the formula
Then t = 115/2.20Then t = 52.3 secAfter that, We need to calculate the distance of fly travelthen Using the formula of distanceThen d = vtWhen we put the value into the formula
Then d = 2.20 multiply by 52.3Then d = 115.06 mThus, Hence proof The distance of fly travel is 115.06 m.Find out more information about distance here:
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Final answer:
The fly travels a distance of approximately 115 mm before the steamrollers collide when both steamrollers and the fly start at the same time and approach each other at given constant speeds.
Explanation:
The question involves calculating the distance a fly travels, given that it is flying back and forth between two objects moving towards each other. We know the steamrollers start 115 mm apart and each moves at 1.10 m/s towards the other, while the fly travels at a constant speed of 2.20 m/s. To find the distance the fly travels before the steamrollers meet, we first need to determine how long it takes for the steamrollers to collide.
The steamrollers are moving towards each other, so their relative speed is the sum of their individual speeds, which is 1.10 m/s + 1.10 m/s = 2.20 m/s. Since they start 115 mm apart, which is 0.115 meters, the time it takes for them to meet is the distance divided by their relative speed, 0.115 m / 2.20 m/s = 0.05227 seconds. In this time, since the fly is traveling at 2.20 m/s, the distance it covers is the fly's speed multiplied by the time, which gives us 2.20 m/s * 0.05227 seconds = 0.11499 meters, or approximately 115 mm.
2. A rocket blasts off vertically from rest on the launch pad with a constant upward acceleration of 2.5 . At 30.0 s after blast off, the engines suddenly fail, and the rocket begins free fall. a. What is the highest point reached by the rocket? b. How long after it is launched does the rocket crash?
Answer:
a)The highest point reached by the rocket is 1412 m
b)The rocket crashes after 54.7 s
Explanation:
Hi there!
The equations of height and velocity of the rocket are the following:
h = h0 + v0 · t + 1/2 · a · t² (while the engines work).
h = h0 + v0 · t + 1/2 · g · t² (when the rocket is in free fall).
v = v0 + a · t (while the engines work).
v = v0 + g · t (when the rocket is in free fall).
Where:
h = height of the rocket at a time t.
h0 = initial height of the rocket.
v0 = initial velocity.
t = time.
a = acceleration due to the engines.
g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).
v = velocity of the rocket at a time t.
First, let's find the velocity and height reached by the rocket until the engines fail:
h = h0 + v0 · t + 1/2 · a · t²
Let's set the origin of the frame of reference at the launching point so that h0 = 0. Since the rocket starts from rest, v0 = 0. So after 30.0 s the height of the rocket will be:
h = 1/2 · a · t²
h = 1/2 · 2.5 m/s² · (30.0 s)²
h = 1125 m
Now let's find the velocity of the rocket at t = 30.0 s:
v = v0 + a · t (v0 = 0)
v = 2.5 m/s² · 30.0 s
v = 75 m/s
After 30.0s the rocket will continue to ascend with a velocity of 75 m/s. This velocity will be gradually reduced due to the acceleration of gravity. When the velocity is zero, the rocket will start to fall. At that time, the rocket is at its maximum height. So, let's find the time at which the velocity of the rocket is zero:
v = v0 + g · t
0 = 75 m/s - 9.8 m/s² · t (v0 = 75 m/s because the rocket begins its free-fall motion with that velocity).
-75 m/s / -9.8 m/s² = t
t = 7.7 s
Now, let's find the height of the rocket 7.7 s after the engines fail:
h = h0 + v0 · t + 1/2 · g · t²
The rocket begins its free fall at a height of 1125 m and with a velocity 75 m/s, then, h0 = 1125 m and v0 = 75 m/s:
h = 1125 m + 75 m/s · 7.7 s - 1/2 · 9.8 m/s² · (7.7 s)²
h = 1412 m
The highest point reached by the rocket is 1412 m
b) Now, let's calculate how much time it takes the rocket to reach a height of zero (i.e. to crash) from a height of 1412 m.
h = h0 + v0 · t + 1/2 · g · t² (v0 = 0 because at the maximum height the velocity is zero)
0 = 1412 m - 1/2 · 9.8 m/s² · t²
-1412 m / -4.9 m/s² = t²
t = 17 s
The rocket goes up for 30.0 s with an acceleration of 2.5 m/s².
Then, it goes up for 7.7 s with an acceleration of -9.8 m/s².
Finally, the rocket falls for 17 s with an acceleration of -9.8 m/s²
The rocket crashes after (30.0 s + 7.7 s + 17 s) 54.7 s
(a)The highest point reached by the rocket is 1412 m
(b)The rocket crashes after 54.7 s
Equation of motions:The height reached by the rocket until the engines fail is:
[tex]h = h_0+ v_0 t + \frac{1}{2} a t^2[/tex]
here, h = height of the rocket at a time t.
h₀ = initial height of the rocket.
v₀ = initial velocity.
t = time.
a = acceleration due to the engines.
At the time of the launch h₀= 0 and v₀ = 0
[tex]h = \frac{1}{2} a t^2\\\\h = \frac{1}{2} \times2.5\times(30)^2\\\\h = 1125 m[/tex]
The velocity of the rocket at t = 30.0 s:
[tex]v = a t\\\\v = 2.5 \times30.0\\\\v = 75 m/s[/tex]
The rocket will continue to ascend with a velocity of 75 m/s until it is finally zero due to gravitational force. Now the time for which the rocket continues to ascend is given by:
[tex]0 = v - g t\\\\0 = 75 - 9.8 t \\\\\frac{-75}{ -9.8} = t\\\\t = 7.7 s[/tex]
The height gained by the rocket 7.7 s after the engines fail:
[tex]h' = h + v t + \frac{1}{2} g t^2\\\\h' = 1125+ 75\times 7.7 - .5\times9.8 \times (7.7 s)^2\\\\h' = 1412 m[/tex]
So, the highest point reached by the rocket is 1412 m
(b) Now,the time talen to crash is the time takn to fall down from the height of 1412m
[tex]h' =0 \times t +\frac{1}{2} g t^2\\\\1412 = 0.5\times9.8\times t^2\\\\t^2=\frac{1412}{9.8}\\\\ t = 17 s[/tex]
Total time taken to crash is 30.0s + 7.7s + 17s = 54.7s
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Astronomers have no theoretical explanation for the ""hot Jupiters"" observed orbiting some other stars. (T/F)
Answer:
Astronomers have no theoretical explanation for the ""hot Jupiters"" observed orbiting some other stars.
False
Explanation:
The “hot Jupiters” joint word startes to be used to be able to describe planets like 51 Pegasi b, a planet with a 10-day-or-less orbit and a mass 25% or greater than Jupitere, circling a sun-like star planet in 1995, which was found by astronomers Michel Mayor and Didier Queloz, who were awarded the 2019 Nobel Prize for Physics along with the cosmologist James Peebles for their “contributions to our understanding of the evolution of the universe and Earth’s place in the cosmos.”
Now we know a total of 4,000-plus exoplanets, but only a few more than 400 meet the definition of the enigmatic hot Jupiters which, tell us a lot about how planetary systems form, and what kinds of conditions cause extreme results.
In a 2018 paper in the Annual Review of Astronomy and Astrophysics, astronomers Rebekah Dawson of the Pennsylvania State University and John Asher Johnson of Harvard University reviewed on how hot Jupiters might have formed, and would be the meaning for the rest of the planets in the galaxy.
An airplane is flying with a velocity of v0 at an angle of α above the horizontal. When the plane is a distance h directly above a dog that is standing on level ground, a suitcase drops out of the luggage compartment.
Part A
How far from the dog will the suitcase land? You can ignore air resistance.
Take the free fall acceleration to be g.
Final answer:
To find how far the suitcase lands from the dog, one must calculate the time it takes for the suitcase to hit the ground using its vertical motion and multiply that time by the horizontal component of its initial velocity, with no need to consider air resistance.
Explanation:
To determine how far from the dog the suitcase will land, we need to break down the initial velocity of the suitcase (v0) into its horizontal (vx) and vertical components (vy). The flight time of the suitcase is primarily determined by its vertical motion, governed by the equation y = vyt + 0.5gt2, where y is the vertical displacement (in this case, equal to -h since the suitcase is falling down), t is the time, and g is the acceleration due to gravity. Since we ignore air resistance, the horizontal velocity remains constant throughout the flight. Therefore, the horizontal distance d from the dog can be found using d = vxt.
To extract the horizontal (vx) and vertical (vy) components of the initial velocity v0, we use the equations: vx = v0cos(α) and vy = v0sin(α). Finally, solving for t using the vertical motion equation and substituting it into the horizontal distance equation gives us the required distance d.
A hot-water bottle contains 715 g of water at 51∘C. If the liquid water cools to body temperature (37 ∘C), how many kilojoules of heat could be transferred to sore muscles?
Answer:
[tex]Q=41.90kJ[/tex]
Explanation:
The heat lost by the water in the cooling process is transferred to the muscles. Therefore, we must calculate this water lost heat, which is defined as:
[tex]Q=mc\Delta T[/tex]
Where m is the water's mass, c is the specific heat capacity of the water and [tex]\Delta T=T_f-T_0[/tex] is the change in temperature. Replacing the given values:
[tex]Q=715g(4186\frac{J}{g^\circC}})(51^\circ C-37^\circ C)\\Q=41901.86J\\Q=41.90kJ[/tex]
The description for a certain brand of house paint claims a coverage of 475 ft²/gal.
(a) Express this quantity in square meters per liter.
(b) Express this quantity in an SI unit.
(c) What is the inverse of the original quantity?
Answer:
(a) 11.66 square meters per liter
(b) 11657.8 per meters
(c) 0.00211 gal per square feet
Explanation:
(a) 475ft^2/gal = 475ft^2/gal × (1m/3.2808ft)^2 × 1gal/3.7854L = 11.66m^2/L
(b) 475ft^2/gal = 475ft^2/gal × (1m/3.2808ft)^2 × 264.17gal/1m^3 = 11657.8/m
(c) Inverse of 475ft^2/gal = 1/475ft^3/gal = 0.00211gal/ft^3
A photographer uses his camera, whose lens has a 50mm focal length, to focus on an object 5.0m away. He then wants to take a picture of an object that is 60cm away.
How far must the lens move to focus on this second object?
Answer:
f₂ = 0.019 m
Explanation:
Let's analyze this exercise a bit, when taking a picture the image should always be in the same place, position of the CCD, let's use the builder's equation to find this distance from the image (i)
1 / f = 1 / o + 1 / i
Where f is the focal length and "o, i" are the distances to the object and image, respectively
1 / i = 1 / f - 1 / o
Let's reduce the magnitudes to the SI system
f = 50 mm = 0.050 m
o = 5.0 m
Let's calculate
1 / i = 1 / 0.050 - 1 / 5.0 = 20- 0.2 = 19.8
i = 0.020 m
Now the object is 60 cm, rotates the lens and has a new focal length
o₂ = 60 cm = 0.60 m
1 / f = 1 / 0.60 + 1 / 0.020 = 1.66 + 50 = 51.66
f₂ = 0.019 m
A series of parallel linear water wave fronts are traveling directly toward the shore at 15.5 cm/s on an otherwise placid lake. A long concrete barrier that runs parallel to the shore at a distance of 3.10 m away has a hole in it. You count the wave crests and observe that 75.0 of them pass by each minute, and you also observe that no waves reach the shore at ±62.3cm from the point directly opposite the hole, but waves do reach the shore everywhere within this distance.A) How wide is the hole in the barrier?B) At what other angles do you find no waves hitting the shore?
Answer: (a) 62.9cm
Explanation: see attachment below
You are red and your friend is green. You stand 2 meters from the mirror. Your friend stands 1 meter from the mirror. Would your friend appear to be in a different position to anyone else, in a different position?
Answer:
Explanation:
It is given that red is 2 m from the mirror and green is 1 m from the mirror so the image of green and red will be formed 1 and 2 m behind the mirror respectively.
Green will be seen at the same distance from the mirror when seen from different position to anyone else.
The above can be explained by the given diagram
A cylindrical shell of radius 7.1 cm and length 251 cm has its charge density uniformly distributed on its surface. The electric field intensity at a point 25.2 cm radially outward from its axis (measured from the midpoint of the shell ) is 37400 N/C.(a) What is the net charge on the shell?
(b) What is the electric field at a point 4.07 cm from the axis? The value of Coulomb’s constant is 8.99 × 10^9 N • m^2/C^2.
Answer:
0.00000131569788654 C
0
Explanation:
R = Radius = 7.1 cm
L = Length of shell = 251 cm
r = 25.2 cm
E = Electric field = 37400 N/C
Electric field is given by
[tex]E=\dfrac{2kq}{rL}\\\Rightarrow q=\dfrac{ErL}{2k}\\\Rightarrow q=\dfrac{37400\times 0.252\times 2.51}{2\times 8.99\times 10^{9}}\\\Rightarrow q=0.00000131569788654\ C[/tex]
The net charge on the shell is 0.00000131569788654 C
Here, 4.07<7.1 cm which means r'<R
From Gauss law the electric at that point is 0
Determine the Mach number at the exit of the nozzle. The gas constant of carbon dioxide is R = 0.1889 kJ/kg·K. Take its constant pressure specific heat and specific heat ratio at room temperature to be cp = 0.8439 kJ/kg·K and k = 1.288.
Answer:
[tex] MA_1 = \frac{50 m/s}{\sqtr{1.288*188.9 J/Kg K * 1200 K}}=0.093[/tex]
[tex] MA_2 =\frac{1163.074 m/s}{\sqrt{1.288 *188.9 J/Kg K * 400 K}}=3.73[/tex]
Explanation:
Assuming this problem: "Carbon dioxide enters an adiabatic nozzle at 1200 K with a velocity of 50 m/s and leaves at 400 K. Assuming constant specific heats at room temperature, determine the Mach number (a) at the inlet and (b) at the exit of the nozzle. Assess the accuracy of the constant specific heat assumption."
Part a
For this case we can assume at the inlet we have the following properties:
[tex] T_1 = 1200 K, v_1 = 50 m/s [/tex]
We can calculate the Mach number with the following formula:
[tex] MA_1 = \frac{v_1}{c_1} = \frac{v_1}{\sqrt{kRT}}[/tex]
Where k represent the specific ratio given k =1.288 and R would be the universal gas constant for the carbon diaxide given by: [tex] R= 188.9 J/ Kg K[/tex]
And if we replace we got:
[tex] MA_1 = \frac{50 m/s}{\sqtr{1.288*188.9 J/Kg K * 1200 K}}=0.093[/tex]
Part b
For this case we can use the same formula:
[tex] MA_2 = \frac{v_2}{c_2} [/tex]
And we can obtain the value of v2 from the total energy of adiabatic flow process, given by this equation:
[tex] c_p T_1 + \frac{v^2_1}{2}=c_p T_2 + \frac{v^2_2}{2}[/tex]
The value of [tex] C_p = 0.8439 K /Kg K = 843.9 /Kg K[/tex] and the value fo T2 = 400 K so we can solve for [tex] v_2[/tex] and we got:
[tex] v_2= \sqrt{2c_p (T_1 -T_2) +v^2_1}=1163.074 m/s[/tex]
And now we can replace on this equation:
[tex] MA_2 = \frac{v_2}{c_2} [/tex]
And we got:
[tex] MA_2 =\frac{1163.074 m/s}{\sqrt{1.288 *188.9 J/Kg K * 400 K}}=3.73[/tex]
Final answer:
To determine the Mach number at the exit of the nozzle, we can use the isentropic flow equations.
Explanation:
To determine the Mach number at the exit of the nozzle, we need to use the isentropic flow equations. The Mach number (M) at the exit of the nozzle can be calculated using the equation:
M = sqrt( 2/(k-1) * ( (P/Pref) ^ ((k-1)/k) - 1) )
Where:
k is the specific heat ratio (given as 1.288)P is the pressure at the exit of the nozzlePref is the reference pressure (1 atm)Given the information provided in the question, we can substitute the values into the equation to calculate the Mach number.
During a braking test, a car is brought to rest beginning from an initial speed of 60 mi/hr in a distance of 120 ft. With the same constant deceleration, what would be the stopping distance s from an initial speed of 80 mi/hr?
Using physics principles and the kinematic equation, the problem calculates the stopping distance of a car decelerating from 80 mi/hr, based on known stopping distance at 60 mi/hr. It involves converting speeds and applying algebraic manipulation to solve for the new distance.
Explanation:To solve the problem of finding the stopping distance from an initial speed of 80 mi/hr, given the stopping distance from 60 mi/hr is 120 ft, we use the principle of physics that relates deceleration, distance, and speed. This approach requires converting speeds from miles per hour to feet per second, applying the kinematic equation v² = u² + 2as, and solving for the unknown stopping distance.
First, speeds are converted from miles per hour to feet per second. Given the initial situation: a car decelerates from 60 mi/hr to rest over 120 feet. Converting 60 mi/hr to feet per second gives 88 feet per second (using 1 mi = 5280 feet and 1 hour = 3600 seconds). Applying v² = u² + 2as (where v is final speed, u is initial speed, a is acceleration, and s is stopping distance) allows us to calculate the deceleration using the initial conditions. Next, using the same deceleration, we calculate the stopping distance from 80 mi/hr (converted to feet per second).
The detailed calculation involves algebraic manipulation of the kinematic equation to solve for the new stopping distance using the derived constant deceleration from the 60 mi/hr case. It demonstrates how a car's stopping distance increases with a square of the speed, illustrating the critical relationship between speed, stopping distance, and safety.
The stopping distance from an initial speed of 80 mi/hr would be approximately 195.555 ft, assuming the same constant deceleration as the first scenario.
To find the stopping distance from an initial speed of 80 mi/hr, we first need to determine the acceleration of the car during braking. Using the initial velocity of [tex]60 mi/hr[/tex] and the stopping distance of 120 ft, we can calculate the acceleration using the kinematic equation:
[tex]\[ v_f^2 = v_i^2 + 2as \][/tex]
Given that [tex]\( v_f = 0 \)[/tex] (the car comes to rest), [tex]\( v_i = 60 \) mi/hr, and \( s = 120 \)[/tex]ft, we rearrange the equation to solve for [tex]\( a \)[/tex]:
[tex]\[ a = \frac{{v_f^2 - v_i^2}}{{2s}} \][/tex]
Substituting the values:
[tex]\[ a = \frac{{0 - (60 \, \text{mi/hr})^2}}{{2 \times 120 \, \text{ft}}} \]\[ a \approx -33.333 \, \text{ft/s}^2 \][/tex]
Now, using this acceleration value, we can find the stopping distance[tex](\( s \))[/tex] from an initial speed of 80 mi/hr. With [tex]\( v_i = 80 \) mi/hr[/tex], we convert it to feet per second and use it in the same kinematic equation:
[tex]\[ v_i = 80 \, \text{mi/hr} \times 1.46667 \, \text{ft/s/mi/hr} = 117.333 \, \text{ft/s} \]\[ s = \frac{{-(117.333 \, \text{ft/s})^2}}{{2 \times (-33.333 \, \text{ft/s}^2)}} \]\[ s \approx 195.555 \, \text{ft} \][/tex]
Therefore, the stopping distance from an initial speed of [tex]80 mi/hr[/tex]would be approximately 195.555 ft, assuming the same constant deceleration as the first scenario.