Answer:
Distance from starting point= 242.249871 miles
Explanation:
Distance covered on bearing of 40 degree=a= [tex]1.4*110\ mph[/tex]
Distance covered on bearing of 40 degree=a=154 miles
Distance covered on bearing of 130 degree=b=[tex]1.7*110\ mph[/tex]
Distance covered on bearing of 40 degree=b=187 miles
Angle between bearing=[tex]130-40[/tex]
Angle between bearing=90 degree
With the angle of 90 degree, Distance from starting point can be calculated from Pythagoras theorem.
Distance from starting point=[tex]\sqrt{a^2+b^2}[/tex]
Distance from starting point=[tex]\sqrt{154^2+187^2}[/tex]
Distance from starting point= 242.249871 miles
Answer:
Distance from starting point= 242.249871 miles
Distance covered on bearing of 40 degree=a=
Distance covered on bearing of 40 degree=a=154 miles
Distance covered on bearing of 130 degree=b=
Distance covered on bearing of 40 degree=b=187 miles
Angle between bearing=
Angle between bearing=90 degree
With the angle of 90 degree, Distance from starting point can be calculated from Pythagoras theorem.
Distance from starting point=
Distance from starting point=
Distance from starting point= 242.249871 miles
I got this from the other person so mark them brainliest
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
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 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.
A large box of mass M is pulled across a horizontal, frictionless surface by a horizontal rope with tension T. A small box of mass m sits on top of the large box. The coefficients of static and kinetic friction between the two boxes are μs and μk, respectively.Find an expression for the maximum tension T max for which the small box rides on top of the large box without slipping.
Answer:
Explanation:
Given
Mass of big box is M and small box is m
Tension T will cause the boxes to accelerate
[tex]T=(M+m)a[/tex]
where a=acceleration of the boxes
Now smaller box will slip over large box if the acceleration force will exceed the static friction
i.e. for limiting value
[tex]\mu _smg=ma[/tex]
[tex]a=\mu _s\cdot g[/tex]
thus maximum tension
[tex]T=\mu _s(M+m)g[/tex]
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.
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 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.
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 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]
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 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
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.
An electric motor rotating a workshop grinding wheel at a rate of 44.2 rev/min is switched off with a constant deceleration of 2.69 rad/s 2 . How long does it take for the grinding wheel to stop?
Answer: time(t) = 1.72s
Explanation: [tex]Angular acceleration =\frac{angular velocity}{time taken}[/tex]
Let angular acceleration = α = [tex]2.69rad/s^{2}[/tex]
ω = angular velocity = 44rev/ min
the angular velocity is in rev/ min but we need to have it in rad/s , thus we do so below
recall that 1 rev = 2π and 1 min = 60s, [tex]\frac{44.2 * 2\pi }{60} \\\\ \frac{44.2 * 2*3.142}{60} \\\\= 4.63rads^{-1}[/tex]
hence 44.2rev/min = 44.2 * 2π/ 60 = 44.2 * 2 *3.142/ 60
thus angular velocity = 4.63rad/s
time taken = angular velocity/ angular acceleration
time taken = 4.63/2.69
time taken = 1.72s
A 975-kg elevator accelerates upward at 0.754 m/s2, pulled by a cable of negligible mass. Find the tension force in the cable.
To solve this problem we will apply the concepts of equilibrium and Newton's second law.
According to the description given, it is under constant ascending acceleration, and the balance of the forces corresponding to the tension of the rope and the weight of the elevator must be equal to said acceleration. So
[tex]\sum F = ma[/tex]
[tex]T-mg = ma[/tex]
Here,
T = Tension
m = Mass
g = Gravitational Acceleration
a = Acceleration (upward)
Rearranging to find T,
[tex]T = m(g+a)[/tex]
[tex]T = (975)(9.8+0.754)[/tex]
[tex]T= 10290.15N[/tex]
Therefore the tension force in the cable is 10290.15N
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
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
The plates of a parallel-plate capacitor are 3.00 mm apart, and each carries a charge of magnitude 79.0 nC. The plates are in vacuum. The electric field between the plates has a magnitude of 5.00 x 10^6 V/m.
Part A) What is the potential difference between the plates?Part B) What is the area of each plate?Part C) What is the capacitance?
To solve this problem we will apply the concepts of the potential difference such as the product between the electric field and the distance, then we will use two definitions of capacitance, the first depending on the Area and the second depending on the load to find the Area. Finally we will look for capacitance with the values already obtained in the first sections of this problem
PART A) Potential Difference is
[tex]V = Ed[/tex]
Here,
E = Electric Field
d = Distance
Replacing,
[tex]V = (5*10^6)(3.0*10^{-3})[/tex]
[tex]V = 15000V= 15kV[/tex]
PART B) Capacitance of the capacitor is
[tex]C = \frac{\epsilon_0 A}{d}[/tex]
Here,
A = Area
[tex]\epsilon_0[/tex] = Permittivity Vacuum
d = Distance
Rearranging to find the Area we have,
[tex]A = \frac{Cd}{\epsilon_0}[/tex]
We know at the same time that Capacitance is the charge per Voltage, then
[tex]C = \frac{Q}{V}[/tex]
Replacing at this equation we have that
[tex]A = \frac{Qd}{V\epsilon_0}[/tex]
[tex]A = \frac{(79*10^{-9})(3*10^{-3})}{(15000)(8.853*10^{-12})}[/tex]
[tex]A = 1.78mm^2[/tex]
PART C)
Capacitance is given by,
[tex]C = \frac{Q}{V}[/tex]
[tex]C =\frac{79*10^{-9}}{15000}[/tex]
[tex]C =5.26pF[/tex]
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;
https://brainly.com/question/24520854
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.
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.
A book rests on the shelf of a bookcase. The reaction force to the force of gravity acting on the book is 1. The force of the shelf holding the book up. 2. The force exerted by the book on the earth. 3. The weight of the book. 4. The frictional force between book and shelf. 5. None of these.
Answer: the force exerted by the book on the earth
The reaction force to the force of gravity acting on a book resting on a shelf is the force exerted by the shelf, also known as the normal force. This force counteracts the force of gravity and is equal to the weight of the book. Understanding this concept requires knowledge of Newton's Second and Third Laws of Motion.
Explanation:The reaction force to the force of gravity acting on a book resting on a shelf is the Force exerted by the shelf on the book. The shelf applies a Normal force, perpendicular to its surface, that counteracts the force of gravity pulling down on the book. This is directly tied into Newton's Third Law, which states that for every action, there is an equal and opposite reaction.
The normal force exerted by the shelf is a type of contact force and is exactly equal to the weight of the book (which is the force of gravity acting on the book). If the normal force were weaker, the book would begin to sink into the shelf. If it were stronger, the book would start to lift off of the shelf. Frictional force also plays a role here, preventing the book from sliding off the shelf, but it is not the reaction force to gravity in this case. Newton's Laws of Motion, particularly the Second and Third Laws, are key to understanding these dynamics.
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In a single wire, how much current would be required to generate 1 Tesla magnetic field at a 2 meter distance away from the wire?
Answer:
12.56 A.
Explanation:
The magnetic field of a conductor carrying current is give as
H = I/2πr ............................... Equation 1
Where H = Magnetic Field, I = current, r = distance, and π = pie
Making I the subject of the equation,
I = 2πrH............... Equation 2
Given: H = 1 T, r = 2 m.
Constant: π = 3.14
Substitute into equation 2
I = 2×3.14×2×1
I = 12.56 A.
Hence, the magnetic field = 12.56 A.
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
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
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.
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.
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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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]
An object falls a distance h from rest. If it travels 0.540h in the last 1.00 s, find (a) the time and (b) the height of its fall.
Answer
given,
distance of fall = h
initial speed = 0 m/s
it travels 0.540 h in the last 1.00 s
Speed of the fall = [tex]\dfrac{Distance}{time}[/tex]
Speed = [tex]\dfrac{0.540\ h-0}{1}[/tex]
S = 0.540h m/s
time of the fall
using equation of motion
v = u + g t
0.540 h = 0 + 9.8 t
t = 0.055h s
The time of fall is equal to 0.055h s
height of fall = ?
again using equation of motion
v² = u² + 2 g s
(0.540h)² = 0 + 2 x 9.8 x s
s = 0.0148 h²
Hence, the time of fall of the object = 0.055h s.
the distance of fall of object = 0.0148 h²
The object has been falling for a total time of 3.13 seconds and the total height from which it fell is 47.9 meters, given that it covered 0.540 of its total height in the last 1 second of free fall.
Explanation:Solution for Time and Height of a Free-Falling ObjectLet's tackle this physics problem step by step to find both the time the object has been falling and the total height from which it fell. We'll use the known acceleration due to gravity (g = 9.81 m/s²) and apply kinematic equations.
Given that the object travels 0.540h in the last 1.00 s, we can set up two equations using the kinematic formula h = ½gt², where h is the height, g is the acceleration due to gravity, and t is the time in seconds:
Total height fallen (h) after time t: h = ½g(t²)Height fallen in the last 1 second (0.540h): 0.540h = ½g(t²) - ½g((t-1)²)We solve the second equation for t, which will then be used to find the total height using the first equation.
Solving the second equation: 0.540h = ½g(t²) - ½g((t-1)²)
Expand and simplify the equation: 0.540h = ½g(t²) - ½g(t² - 2t + 1)
0.540 = t² - ½g(t² - 2t + 1)
Now, assuming g is approximately 9.8 m/s², we can plug in values and solve for t:
0.540 = t² - 0.5(9.81)(t² - 2t + 1)
After solving for t, we find:
(a) Total time of fall, t = 3.13 s(b) Total height fallen, h = 0.5(9.81)(3.13²) = 47.9 mThis calculation assumes that air resistance is negligible and the acceleration due to gravity is constant.