Answers
Answer:
v = √rg.
Explanation:
The Minimum speed of the stone that can have to the stone when it is rotated in a vertical circle is √rg.
Mathematical Proof ⇒
at the top point on the circle we have
T + mg = m v²/r
We know that minimum speed will be at the place when its tension will be zero.
∴ v² = rg
⇒ v = √rg.
So, the minimum speed or the critical speed is given as v = √rg.
Answer:
[tex]v=\sqrt{rg}[/tex]
Explanation:
radius of circle = R
Let T be the tension in the string.
At highest point A, the tension is equal to or more than zero, so that it completes the vertical circle. tension and weight is balanced by the centripetal force.
According to diagram,
[tex]T + mg = \frac{mv^{2}}{R}[/tex]
T ≥ 0
So, [tex]mg = \frac{mv^{2}}{R}[/tex]
Where, v be the speed at the highest point, which is called the critical speed.
[tex]v=\sqrt{rg}[/tex]
Thus, the critical speed at the highest point to complete the vertical circle is [tex]v=\sqrt{rg}[/tex].
Related Questions
A car starts from rest and accelerates at a constant rate until it reaches 70 mi/hr in a distance of 220 ft, at which time the clutch is disengaged. The car then slows down to a velocity of 40 mi/hr in an additional distance of 480 ft with a deceleration which is proportional to its velocity. Find the time t for the car to travel the 700 ft.
Answers
Answer:
[tex]T = 10.43 s[/tex]
Explanation:
During deceleration we know that the deceleration is proportional to the velocity
so we have
[tex]a = - kv[/tex]
here we know that
[tex]\frac{dv}{dt} = - kv[/tex]
so we have
[tex]\frac{dv}{v} = -k dt[/tex]
now integrate both sides
[tex]\int \frac{dv}{v} = -\int kdt[/tex]
[tex]ln(\frac{v}{v_o}) = - kt[/tex]
[tex]ln(\frac{40}{70}) = - k(t)[/tex]
[tex]kt = 0.56[/tex]
Also we know that
[tex]a = \frac{vdv}{ds}[/tex]
[tex]-kv = \frac{vdv}{ds}[/tex]
[tex]\int dv = -\int kds[/tex]
[tex](v - v_o) = -ks[/tex]
[tex](40 - 70)mph = - k (480 ft)[/tex]
[tex]-30 mph = -k(0.091 miles)[/tex]
[tex]k = 329.67[/tex]
so from above equation
[tex]t = \frac{0.56}{329.67} = 1.7 \times 10^{-3} h[/tex]
[tex]t = 6.11 s[/tex]
initially it starts from rest and uniformly accelerate to maximum speed of 70 mph and covers a distance of 220 ft
so we have
d = 220 ft = 67 m = 0.042 miles[/tex]
now we know that
[tex]d = \frac{v_f + v_i}{2} t[/tex]
[tex]0.042 = \frac{70 + 0}{2} t[/tex]
[tex]t = 4.32 s[/tex]
so total time of motion is given as
[tex]T = 4.32 + 6.11 = 10.43 s[/tex]
A fullback preparing to carry the football starts from rest and accelerates straight ahead. He is handed the ball just before he reaches the line of scrimmage. Assume that the fullback accelerates uniformly (even during the handoff), reaching the line with a velocity of 7.60 m/s. If he takes 1.07 s to reach the line, how far behind it did he start?
Answers
Answer:
x=4.06m
Explanation:
A body that moves with constant acceleration means that it moves in "a uniformly accelerated movement", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.
When performing a mathematical demonstration, it is found that the equations that define this movement are as follows.
Vf=Vo+a.t (1)\\\\
{Vf^{2}-Vo^2}/{2.a} =X(2)\\\\
X=Xo+ VoT+0.5at^{2} (3)\\
Where
Vf = final speed
Vo = Initial speed
T = time
A = acceleration
X = displacement
In conclusion to solve any problem related to a body that moves with constant acceleration we use the 3 above equations and use algebra to solve
for this problem
Vf=7.6m/s
t=1.07
Vo=0
we can use the ecuation number one to find the acceleration
a=(Vf-Vo)/t
a=(7.6-0)/1.07=7.1m/s^2
then we can use the ecuation number 2 to find the distance
{Vf^{2}-Vo^2}/{2.a} =X
(7.6^2-0^2)/(2x7.1)=4.06m
Raindrops acquire an electric charge as they fall. Suppose a 2.4-mm-diameter drop has a charge of +13 pC. In a thunderstorm, the electric field under a cloud can reach 15,000 N/C, directed upward. For a droplet exposed to this field, how do the magnitude of the electric force compare to those of the weight force and what is the dircetion of the electric force?
Answers
Answer:
567.126 x 10⁻⁶ N
[tex]5.6 x 10^{-6} N[/tex]
Explanation:
Thinking process:
[tex]13pC = 13 x 10^{-12} C[/tex]
The electric field is given by E = 15000 N/C
Electric force on the charge
= charge x electric field
= 13 x 10⁻¹² x 15000
= 195 x 10⁻⁹ N.
The force acts in upward direction as force on positive charge acts in the direction in which electric field exists.
Volume of droplet = 4/3 π R³
R = 2.4 X 10⁻³ m
Volume V = 4/3 x 3.14 x ( 2.4 x 10⁻³)³
= 57.87 x 10⁻⁹ m³
density of water = 1000 kg / m³
mass of water droplet = density x volume
= 1000 x 57. 87 x 10⁻⁹ kg
= 57.87 x 10⁻⁶ kg .
Weight = mass x g
= 57.87 x 10⁻⁶ x 9.8
= 567.126 x 10⁻⁶ N.
Therefore, the weight is more than the electric force.
The position of a ship traveling due east along a straight line is s(t) = 12t2 + 6. In this example, time t is measured in hours and position s is measured in nautical miles. We will take s = 0 to be the port of Wilmington, NC and the positive direction to be east. How far east of Wilmington is the ship and how fast is it going after one hour, that is, when t = 1?
Answers
Answer:
18 miles east; 24 mph east
Explanation:
In order to find how far east of Wilmington is the ship after 1 hour, we just need to substitute t = 1 into the formula of the position.
The equation of the position is
[tex]s(t) = 12 t^2 +6[/tex]
where t is the time. Substituting t = 1,
[tex]s(1) = 12 (1)^2 + 6 = 12+6 = 18 mi[/tex]
So, the ship is 18 miles east of Wilmington.
To find the velocity of the boat, we just need to calculate the derivative of the position, so
[tex]v(t) = s'(t) = 24 t[/tex]
And by substituting t = 1, we find the velocity after 1 hour:
[tex]v(1) = 24 (1) = 24 mph[/tex]
And the direction is east.
Initially, a 2.00-kg mass is whirling at the end of a string (in a circular path of radius 0.750 m) on a horizontal frictionless surface with a tangential speed of 5 m/s. The string has been slowly winding around a vertical rod, and a few seconds later the length of the string has shortened to 0.250 m. What is the instantaneous speed of the mass at the moment the string reaches a length of 0.250 m?
Answers
Answer:
[tex] v_f = 15 \frac{m}{s} [/tex]
Explanation:
We can solve this problem using conservation of angular momentum.
The angular momentum [tex]\vec{L}[/tex] is
[tex]\vec{L} = \vec{r} \times \vec{p}[/tex]
where [tex]\vec{r}[/tex] is the position and [tex]\vec{p}[/tex] the linear momentum.
We also know that the torque is
[tex]\vec{\tau} = \frac{d\vec{L}}{dt} = \frac{d}{dt} ( \vec{r} \times \vec{p} )[/tex]
[tex]\vec{\tau} = \frac{d}{dt} \vec{r} \times \vec{p} + \vec{r} \times \frac{d}{dt} \vec{p} [/tex]
[tex]\vec{\tau} = \vec{v} \times \vec{p} + \vec{r} \times \vec{F} [/tex]
but, as the linear momentum is [tex]\vec{p} = m \vec{v}[/tex] this means that is parallel to the velocity, and the first term must equal zero
[tex]\vec{v} \times \vec{p}=0[/tex]
so
[tex]\vec{\tau} = \vec{r} \times \vec{F} [/tex]
But, as the only horizontal force is the tension of the string, the force must be parallel to the vector position measured from the vertical rod, so
[tex]\vec{\tau}_{rod} = 0 [/tex]
this means, for the angular momentum measure from the rod:
[tex]\frac{d\vec{L}_{rod}}{dt} = 0 [/tex]
that means :
[tex]\vec{L}_{rod} = constant[/tex]
So, the magnitude of initial angular momentum is :
[tex]| \vec{L}_{rod_i} | = |\vec{r}_i||\vec{p}_i| cos(\theta)[/tex]
but the angle is 90°, so:
[tex]| \vec{L}_{rod_i} | = |\vec{r}_i||\vec{p}_i| [/tex]
[tex]| \vec{L}_{rod_i} | = r_i * m * v_i[/tex]
We know that the distance to the rod is 0.750 m, the mass 2.00 kg and the speed 5 m/s, so:
[tex]| \vec{L}_{rod_i} | = 0.750 \ m \ 2.00 \ kg \ 5 \ \frac{m}{s} [/tex]
[tex]| \vec{L}_{rod_i} | = 7.5 \frac{kg m^2}{s} [/tex]
For our final angular momentum we have:
[tex]| \vec{L}_{rod_f} | = r_f * m * v_f[/tex]
and the radius is 0.250 m and the mass is 2.00 kg
[tex]| \vec{L}_{rod_f} | = 0.250 m * 2.00 kg * v_f [/tex]
but, as the angular momentum is constant, this must be equal to the initial angular momentum
[tex] 7.5 \frac{kg m^2}{s} = 0.250 m * 2.00 kg * v_f [/tex]
[tex] v_f = \frac{7.5 \frac{kg m^2}{s}}{ 0.250 m * 2.00 kg} [/tex]
[tex] v_f = 15 \frac{m}{s} [/tex]
Answer:
15 m/s
Explanation:
L = mvr
Li = (2.00 kg)(0.750 m)(5m/s) = 7.5 kgm^2/s
conservation of angular momentum --> Li=Lf
Lf = 7.5 kgm^2/s
7.5 kgm^2/s = (2.00 kg)(0.250 m)(vf)
vf = 15 m/s
Gas mileage actually varies slightly with the driving speed of a car (as well as with highway vs. city driving). Suppose your car averages 35 miles per gallon on the highway if your average speed is 45 miles per hour, and it averages 19 miles per gallon on the highway if your average speed is 72 miles per hour. Answer parts (a) and (b) below..
(a) What is the driving time for a 2600-mile trip if you drive at an average speed of 45 miles per hour? What is the driving time at 72 miles per hour?
(b) Assume a gasoline price of $3.05 per gallon. What is the gasoline cost for a 2600-mile trip if you drive at an average speed of 45 miles per hour? What is the gasoline cost at 72 miles per hour?
Answers
Answer:
a) Traveling at 45 mph, the driving time is 58 h. Traveling at 72 mph, the driving time will be 36 h.
b) Traveling at 45 mph, the gasoline cost will be $225.7.
Traveling at 72 mpg, the gasoline cost will be $417.9
Explanation:
The average speed can be calculated as the distance traveled over time:
speed = distance / time
Then:
time = distance / speed
a)If you drive at an average speed of 45 mph during a 2600-mile trip, the driving time will be:
time = 2600 mi / 45 mi/h = 58 h
If you drive at 72 mph:
time = 2600 mi / 72 mi/h = 36 h
b) For the 2600-mile trip, you will need ( 2600 mi * (1 gallon/ 35 mi)) 74 gallons if you travel at 45 mph.
If you travel at 72 mph, you will need (2600 mi * (1 gallon /19 mi)) 137 gallons.
Traveling at 45 mph, the gasoline cost will be (74 gallons * ($3,05/gallon)) $225.7
Traveling at 72 mph, the gasoline cost will be (137 gallons * (3.05/gallon)) $417.9
A body-centered cubic lattice has a lattice constant of 4.83 Ă. A plane cutting the lattice has intercepts of 9.66 Å, 19.32 Å, and 14.49 Å along the three cartesian coordi- nates. What are the Miller indices of the plane?
Answers
Along the three Cartesian coordinates. The Miller indices of the plane are [tex](1/2,1/4,1/3)[/tex].
A system known as Miller indices is used to explain how crystal planes and directions inside a crystalline substance are oriented. They serve as a tool to depict the crystal lattice's three-dimensional configuration of atoms or lattice points. In order to describe the surfaces and axes of crystals, Miller indices are frequently used in crystallography.
Miller indices are a crucial tool that crystallographers use to convey crystallographic data and comprehend the geometric arrangement of atoms in crystals. They support the characterization of crystal structures, the prediction of material characteristics, and the comprehension of the microscopic behavior of materials.
Given:
Intercepts = [tex]9.66\ A, 19.32\ A, 14.49\ A[/tex]
Lattice constant, [tex]a = 4.83\ A[/tex]
The reciprocal of intercepts is given as:
[tex]r = (1/9.66),1/19.32,1/14.49)[/tex]
The Miller indices are given as:
[tex]M = r/(1/a)\\M = ((1/9.66),1/19.32,1/14.49))/(1/4.83)\\M = (1/2,1/4,1/3)[/tex]
Hence, along the three Cartesian coordinates. The Miller indices of the plane are [tex](1/2,1/4,1/3)[/tex].
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The Miller indices for a plane intercepting a body-centered cubic lattice with intercepts of 9.66 Å, 19.32 Å, and 14.49 Å are (6,3,4). These are found by taking reciprocals of the intercepts, then multiplying by a common factor to get the smallest set of integers.
Explanation:The student's question is about finding the Miller indices of a plane in a body-centered cubic lattice with given lattice constants and plane intercepts. Miller indices describe the orientation of a plane or set of planes in a crystal lattice.
To find these, we first take the reciprocals of the intercepts, which in this case gives us 1/2, 1/4, and 1/3. We then need to multiply these by a common factor to eliminate any fractions and get the smallest set of integers. Multiplying by 12 gives us the Miller indices of (6,3,4).
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An airplane flies horizontally with a constant speed of 172.0 m/s at an altitude of 1390 m. A package is dropped out of the airplane. Ignore air resistance. The magnitude of the gravitational acceleration is 9.8 m/s2. Choose the RIGHT as positive x-direction. Choose UPWARD as positive y-direction Keep 2 decimal places in all answers
(a) What is the vertical component of the velocity (in m/s) just before the package hits the ground? Pay attention to the direction (the sign).
(b) What is the magnitude of the velocity (in m/s) (including both the horizontal and vertical components) of the package just before it hits the ground?
Answers
Answer:
(a) - 165.032 m/s
(b) 238.37 m/s
Explanation:
initial horizontal velocity, ux = 172 m/s
height, h = 1390 m
g = 9.8 m/s^2
Let it strikes the ground after time t.
Use second equation of motion in vertical direction
[tex]s=ut+\frac{1}{2}at^{2}[/tex]
-1390 = 0 - 0.5 x 9.8 x t^2
t = 16.84 second
(a) Let vy be the vertical component of velocity as it strikes the ground
Use first equation of motion in vertical direction
vy = uy - gt
vy = 0 - 9.8 x 16.84
vy = - 165.032 m/s
Thus, the vertical component of velocity as it strikes the ground is 165.032 m/s downward direction.
(b)
The horizontal component of velocity remains constant throughout the motion.
vx = 172 m/s
vy = - 165.032 m/s
The resultant velocity is v.
[tex]v=\sqrt{172^{2}+165.032^{2}}[/tex]
v = 238.37 m/s
Thus, teh velocity with which it hits the ground is 238.37 m/s.
If the potential due to a point charge is 500 V at a distance of 15.0 m, what are the sign and magnitude of the charge?
Answers
Answer:
[tex]q=+8.34*10^{-7}C}[/tex]
Explanation:
The potential V due to a charge q, at a distance r, is:
[tex]V=k\frac{q}{r}[/tex]
k=8.99×109 N·m^2/C^2 :Coulomb constant
We replace the values in order to find q:
[tex]q=\frac{V*r}{k}=\frac{500*15}{8.99*10^{9}}=8.34*10^{-7}C[/tex]
Answer:
i apolagize im late but yeah bois 700 points
Explanation:
A pendulum on Earth has a period of 1.2 seconds. The same pendulum on Mercury (dark side) has a period of 1.95 seconds. What is the free-fall acceleration of Mercury (in m/s^2)
Answers
Answer:
[tex]g'=3.71\ m/s^2[/tex]
Explanation:
Given that,
Time period of a pendulum on the earth's surface, T₁ = 1.2 s
Time period of the same pendulum on Mercury, T₂ = 1.95 s
The time period of the pendulum is given by :
[tex]T=2\pi \sqrt{\dfrac{l}{g}}[/tex]
On earth :
[tex]T_1=2\pi \sqrt{\dfrac{l}{g}}[/tex]
[tex]1.2=2\pi \sqrt{\dfrac{l}{9.8}}[/tex].............(1)
Let g' is the acceleration due to gravity on Mercury. So,
[tex]1.95=2\pi \sqrt{\dfrac{l}{g'}}[/tex]............(2)
From equation (1) and (2) :
[tex]\dfrac{1.2}{1.95}=\sqrt{\dfrac{g'}{9.8}}[/tex]
[tex]g'=(\dfrac{1.2}{1.95})^2\times 9.8[/tex]
[tex]g'=3.71\ m/s^2[/tex]
So, the acceleration due to gravity on the mercury is [tex]3.71\ m/s^2[/tex]. Hence, this is the required solution.
A 7450 kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.35 m/s2 and feels no appreciable air resistance. When it has reached a height of 520 m , its engines suddenly fail so that the only force acting on it is now gravity.
(a) What is the maximum height this rocket will reach above the launch pad?
(b) How much time after engine failure will elapse before the rocket comes crashing down to the launch pad?
(c) How fast will it be moving just before it crashes?
Answers
Answer:
a) 520m
b) 10.30 s
c) 100,95 m/s
Explanation:
a) According the given information, the rocket suddenly stops when it reach the height of 520m, because the engines fail, and then it begins the free fall.
This means the maximum height this rocket reached before falling was 520 m.
b) As we are dealing with constant acceleration (due gravity) [tex]g=9.8 \frac{m}{s^{2}}[/tex] we can use the following formula:
[tex]y=y_{o}+V_{o} t-\frac{gt^{2}}{2}[/tex] (1)
Where:
[tex]y_{o}=520 m[/tex] is the initial height of the rocket (at the exact moment in which it stops due engines fail)
[tex]y=0[/tex] is the final height of the rocket (when it finally hits the launch pad)
[tex]V_{o}=0[/tex] is the initial velocity of the rocket (at the exact moment in which it stops the velocity is zero and then it begins to fall)
[tex]g=9.8m/s^{2}[/tex] is the acceleration due gravity
[tex]t[/tex] is the time it takes to the rocket to hit the launch pad
Clearing [tex]t[/tex]:
[tex]0=520 m+0-\frac{9.8m/s^{2} t^{2}}{2}[/tex] (2)
[tex]t^{2}=\frac{-520 m}{-4.9 m/s^{2}}[/tex] (3)
[tex]t=\sqrt{106.12 s^{2}[/tex] (4)
[tex]t=10.30 s[/tex] (5) This is the time
c) Now we need to find the final velocity [tex]V_{f}[/tex] for this rocket, and the following equation will be perfect to find it:
[tex]V_{f}=V_{o}-gt[/tex] (6)
[tex]V_{f}=0-(9.8 m/s^{2})(10.30 s)[/tex] (7)
[tex]V_{f}=-100.95 m/s[/tex] (8) This is the final velocity of the rocket. Note the negative sign indicates its direction is downwards (to the launch pad)
Show that a sinusoidal wave propagating to the left along x-axis is a solution of the differential wave equation.
Answers
Answer:
The wave equation is [tex]\frac{d^{2}u }{dt^{2} }[/tex] = [tex]c^{2}[/tex] [tex]\frac{d^{2}u }{dx^{2} }[/tex]
a sinusoidal wave can be u = Acos( ax + bt) + B*sin(ax + bt)
where A, a, B and b are real constants. (here you also can add a phase to the arguments of the sin and cosine)
then [tex]\frac{d^{2}u }{dt^{2} }[/tex] = [tex]b^{2}[/tex]*( -Acos(ax + bt) - B*sin(ax + bt))
and [tex]c^{2}[/tex] [tex]\frac{d^{2}u }{dx^{2} }[/tex]= [tex]ac^{2}[/tex]*( -Acos(ax + bt) - B*sin(ax + bt))
then if a*c = b, this is a solution of the wave equation.
A paratrooper is initially falling downward at a speed of 27.6 m/s before her parachute opens. When it opens, she experiences an upward instantaneous acceleration of 74 m/s^2. (a) If this acceleration remained constant, how much time would be required to reduce the paratrooper's speed to a safe 4.95 m/s? (Actually the acceleration is not constant in this case, but the equations of constant acceleration provide an easy estimate.) (b) How far does the paratrooper fall during this time interval?
Answers
Answer:
a) 0.31 s
b) 19.77 m
Explanation:
We will need the following two formulas:
[tex]V_{f} = V_{0}+at\\\\X=V_{0}t + \frac{at^{2}}{2}[/tex]
We first use the final velocity formula to find the time that it takes to decelerate the paratrooper:
[tex]4.95\frac{m}{s}=27.6\frac{m}{s}-74\frac{m}{s^{2}}t\\\\-22.65\frac{m}{s}=-74\frac{m}{s^{2}}t\\\\t= \frac{22.65\frac{m}{s}}{74\frac{m}{s^{2}}}=0.31s[/tex]
Now that we have the time, we can use the distance formula to calculate the distance travelled by the paratrooper:
[tex]X=27.6\frac{m}{s}*0.31s - \frac{74\frac{m}{s^{2}}*(0.31 s)^{2}}{2}=19.77 m[/tex]
A hollow sphere of inner radius 8.82 cm and outer radius 9.91 cm floats half-submerged in a liquid of density 948.00 kg/m^3. (a) What is the mass of the sphere? (b) Calculate the density of the material of which the sphere is made.
Answers
Answer:
a) 0.568 kg
b) 474 kg/m³
Explanation:
Given:
Inner radius = 8.82 cm = 0.0882 m
Outer radius = 9.91 cm = 0.0991 m
Density of the liquid = 948.00 Kg/m³
a) The volume of the sphere = [tex]\frac{4\pi}{3}\times(0.0991^2-0.0882^2)[/tex]
or
volume of sphere = 0.0012 m³
also, volume of half sphere = [tex]\frac{\textup{Total volume}}{\textup{2}}[/tex]
or
volume of half sphere = [tex]\frac{\textup{0.0012}}{\textup{2}}[/tex]
or
Volume of half sphere =0.0006 m³
Now, from the Archimedes principle
Mass of the sphere = Weight of the volume of object submerged
or
Mass of the sphere = 0.0006× 948.00 = 0.568 kg
b) Now, density = [tex]\frac{\textup{Mass}}{\textup{Volume}}[/tex]
or
Density = [tex]\frac{\textup{0.568}}{\textup{0.0012}}[/tex]
or
Density = 474 kg/m³
A 0.350kg bead slides on a curved fritionless wire,
startingfrom rest at point A. At point B the bead
collideselastically with a 0.530kg ball at rest. find distance call
risesas it moves up the wire. Point A is 2.20 m from ground andfree
fall accel is 9.80 m/s. round answer to 3 significantfigures.
Answers
Answer:
h2 = 0.092m
Explanation:
From a balance of energy from point A to point B, we get speed before the collision:
[tex]m1*g*h-\frac{m1*V_B^2}{2}=0[/tex] Solving for Vb:
[tex]V_B=\sqrt{2gh}=6.56658m/s[/tex]
Since the collision is elastic, we now that velocity of bead 1 after the collision is given by:
[tex]V_{B'}=V_B*\frac{m1-m2}{m1+m2} = \sqrt{2gh}* \frac{m1-m2}{m1+m2}=-1.34316m/s[/tex]
Now, by doing another balance of energy from the instant after the collision, to the point where bead 1 stops, we get the distance it rises:
[tex]m1*g*h2-\frac{m1*V_{B'}^2}{2}=0[/tex] Solving for h2:
h2 = 0.092m
The terminal velocity of a person falling in air depends upon the weight and the area of the person facing the fluid. Find the terminal velocity (in meters per second and kilometers per hour) of an 84.0 kg skydiver falling in a pike (headfirst) position with a surface area of 0.160 m2. (Assume that the density of air is 1.21 kg/m3 and the drag coefficient of a skydiver in a pike position is 0.7.)
Answers
Answer:
110.27 m/s or 396.972 km/h
Explanation:
F = Force
m = Mass = 84 kg
g = Acceleration due to gravity = 9.81 m/s²
C = Drag coefficient = 0.7
ρ = Density of air = 1.21 kg/m³
A = Surface area = 0.16 m²
v = Terminal velocity
F = ma
[tex]F=\frac{1}{2}\rho CAv^2\\\Rightarrow ma=\frac{1}{2}\rho CAv^2\\\Rightarrow v=\sqrt{2\frac{ma}{\rho CA}}\\\Rightarrow v=\sqrt{2\frac{84\times 9.81}{1.21\times 0.7\times 0.16}}\\\Rightarrow v=110.27\ m/s[/tex]
Converting to km/h
[tex]\frac{110.27}{1000}\times 3600=396.972\ km/h[/tex]
Terminal velocity of the skydiver is 110.27 m/s or 396.972 km/h
The sun is 150,000,000 km from earth; its diameter is 1,400,000 km. A student uses a 5.2-cm-diameter lens with f = 10 cm to cast an image of the sun on a piece of paper. Where should the paper be placed relative to the lens to get a sharp image?
Answers
To get a sharp image of the Sun, the paper should be placed at the focal length of the lens, which is 10 centimeters away from the lens. The image is in focus at this point because the light rays from the Sun are effectively parallel when they reach the lens, which then focuses these rays at its focal point.
Explanation:Given the sun is so far away, the light it emits is nearly parallel by the time it reaches Earth. When using a lens to cast an image of the Sun, the point where the image is in focus, that is, the focal point, is also the focal length of the lens.
In this case, the student uses a lens with f = 10 cm, meaning the focal length of the lens is 10 centimeters. To get a sharp image, the paper on which the image is being projected should be placed 10 cm away from the lens, or at the focal length of the lens. This is because the light is in sharp focus at this distance, creating a clear image on the paper.
An important concept here is that the Sun is an astronomical unit away, so the light rays from the Sun are essentially parallel when they reach the lens. The lens then focuses these parallel rays to its focal point, forming a sharp image at a distance equal to its focal length.
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A place-kicker must kick a football from a point 36.0 m (about 40 yards) from the goal. Half the crowd hopes the ball will clear the crossbar, which is 3.05 m high. When kicked, the ball leaves the ground with a speed of 23.6 m/s at an angle of 45.0° to the horizontal. (a) By how much does the ball clear or fall short (vertically) of clearing the crossbar? (Enter a negative answer if it falls short.)
_______m
(b) Does the ball approach the crossbar (and cross above or beneath it) while still rising or while falling?
rising or falling? _________
Answers
Answer:
Part (a) 10.15 m
Part (b) Rising
Explanation:
Given,
Initial speed of the ball = u = 23.6 m/sHeight of the crossbar = h = 3.05 mDistance between the ball and the cross bar = r = 36.0 mAngle of projection = [tex]\theta\ =\ 45.0^o[/tex]Initial velocity of the ball in the horizontal direction = [tex]u_x\ =\ ucos\theta[/tex]Initial velocity of the ball in the vertical direction = [tex]u_y\ =\ usin\theta[/tex]part (a)
Let 't' be the time taken to reach the ball to the cross bar,
In x-direction,
[tex]\therefore r\ =\ u_xt\\\Rightarrow t\ =\ \dfrac{r}{u_x}\ =\ \dfrac{r}{ucos\theta}\\\Rightarrow t\ =\ \dfrac{36.0}{23.6cos45^o}\\\Rightarrow t\ =\ 2.15\ sec[/tex]
Let y be the height attained by the ball at time t = 2.15 sec,
[tex]y\ =\ u_yt\ \ -\ \dfrac{1}{2}gt^2\\\Rightarrow y\ =\ usin\theta t\ -\ \dfrac{1}{2}gt^2\\\Rightarrow y\ =\ 23.6\times sin45^o\times 2.15\ -\ 0.5\times 9.81\ 2.15^2\\\Rightarrow y\ =\ 13.205\ m[/tex]
Now Let H be the height by which the ball is clear the crossbar.
[tex]\therefore H\ =\ y\ -\ h\ =\ 13.205\ -\ 3.05\ =\ 10.15\ m[/tex]
part (b)
At the maximum height the vertical velocity of the ball becomes zero.
i,e, [tex]v_y\ =\ 0[/tex]
Let h be the maximum height attained by the ball.
[tex]\therefore v_y^2\ =\ u_y^2\ -\ 2gh\\\Rightarrow 0\ =\ (usin\theta)^2\ -\ 2gh\\\Rightarrow h\ =\ \dfrac{(usin\theta)^2}{2g}\\\Rightarrow h\ =\ \dfrac{23.6\times sin45.0^o)^2}{2\times 9.81}\\\Rightarrow h\ =\ 14.19\ m[/tex]
Hence at the cross bar the ball attains the height 13.205 m but the maximum height is 14.19 m. Therefore the ball is rising when it reaches at the crossbar.
An object is 30 cm in front of a converging lens with a focal length of 10 cm. Use ray tracing to determine the location of the image. Is the image upright or inverted? Is it real or virtual?
Answers
Answer:
Inverted
Real
Explanation:
u = Object distance = 30 cm
v = Image distance
f = Focal length = 10 cm
Lens Equation
[tex]\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}-\frac{1}{u}=\frac{1}{v}\\\Rightarrow \frac{1}{v}=\frac{1}{10}-\frac{1}{30}\\\Rightarrow \frac{1}{v}=\frac{1}{15}\\\Rightarrow v=15\ cm[/tex]
As, the image distance is positive the image is real and forms on the other side of the lens
[tex]m=-\frac{v}{u}\\\Rightarrow m=-\frac{-15}{30}\\\Rightarrow m=-0.5[/tex]
As, the magnification is negative the image is inverted
By applying the lens equation, we calculate that the image is formed 15 cm behind the lens. This is a real image as it forms on the opposite side of the lens, and in the case of a converging lens, it will be inverted.
Explanation:First, we use the lens equation, which is 1/f = 1/do + 1/di, where f is the focal length, do is the object distance, and di is the image distance. In this case, the object distance 'do' is 30cm and the focal length 'f' is 10cm. Solving for 'di', we find that the image is located 15 cm behind the lens (i.e., on the opposite side from the object).
Since the image forms on the opposite side of the lens from where the object is, this indicates it's a real image. A positive image distance indicates a real image and a negative image distance indicates a virtual image.
For a converging lens, a real image is always inverted, and a virtual image is always upright. Therefore, in this case, the image would be inverted.
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Explain why Earth is not spherical in shape, but bulges at
theEquator?
Answers
Explanation:
Every rotating body experiences centrifugal force. Due to this force the body tends to bulge out around it mid point and gets flattened at the poles. Same is applicable to Earth as well. Since the Earth is rotating at a very high speed, its equator gets bulged out due to centrifugal force. Because of this bulged equator, Earth's pole to pole diameter and equatorial diameter has difference of around 42.76 km. It is flatter on the poles. This also proves that Earth is not a perfect sphere.
Answer and Explanation:
The reason for the not being perfectly spherical ad bulging out at the equator is that The centripetal force acting toward's the earth gravitational center tries to keep the Earth in perfect spherical shape.
Also the angular momentum of the orbiting planet influences the bulge,
The greater angular momentum results in more bulge while the lower value of it results in lesser bulge and more perfect spherical shape.
Also, a greater amount of force directed towards the center and acting on the object at the equator results in the bulges at the equator whereas at poles this force is not required and hence radius is lower in that region.
To provide the pulse of energy needed for an intense bass, some car stereo systems add capacitors. One system uses a 2.4F capacitor charged to 24 V, double the normal 12 V provided by the car's battery. How much energy does the capacitor store at 12 V? How much energy does the capacitor store at 24 V?
Answers
Answer:
Explanation:
Energy stored in a capacitor
= 1/2 CV²
C is capacitance and V is potential of the capacitor .
When capacitor is charged to 24 V ,
E₁ = 1/2 x 2.4 x 24 x24 = 691.2 J
When it is charged to 12 volt
E₂ = 1/2 CV²
.5 X 2.4 X 12 X12
= 172.8 J
If the car’s speed decreases at a constant rate from 71 mi/h to 50 mi/h in 3.0 s, what is the magnitude of its acceleration, assuming that it continues to move in a straight line? What distance does the car travel during the braking period?
Answers
Answer:
The acceleration and the distance are 25200 mi/h² and 0.1008 mi.
Explanation:
Given that,
Initial speed = 71 mi/h
Final speed = 50 mi/h
Time = 3.0 s
(a). We need to calculate the acceleration
Using equation of motion
[tex]v=u+at[/tex]
[tex]a=\dfrac{v-u}{t}[/tex]
Put the value in the equation
[tex]a=\dfrac{(50-71)\times3600}{3}[/tex]
[tex]a=-25200\ mi/h^2[/tex]
Negative sign shows the deceleration.
(b). We need to calculate the distance
Using equation of motion
[tex]v^2=u^2+2as[/tex]
[tex](50)^2=(71)^2+2\times(-25200)\times s[/tex]
[tex]s=\dfrac{(50)^2-(71)^2}{-25200}[/tex]
[tex]s=0.1008\ mi[/tex]
Hence, The acceleration and the distance are 25200 mi/h² and 0.1008 mi.
The distance traveled by the car when the car is constantly deaccelerating at a rate of 25200 miles/h² is 0.0504 miles.
Given to us
Initial Velocity of the car, u = 71 miles/h
Final Velocity of the car, v = 50 miles/h
Time = 3.0 s [tex]=\dfrac{3}{3600}[/tex] hour
What is the acceleration of the car?According to the first equation of motion, acceleration can be written as,
[tex]a=\dfrac{v-u}{t}[/tex]
substituting the values we get,
[tex]a=\dfrac{50-71}{\dfrac{3}{3600}}[/tex]
[tex]a=-25,200\rm\ miles/h^2[/tex]
Thus, the acceleration of the car is -25,200 miles/h².
What distance does the car travel during the braking period?According to the third equation of motion,
[tex]v^2-u^2=2as[/tex]
Substituting the values we get,
[tex](50)^2-(71)^2=2(-25200)s[/tex]
[tex]s = 0.0504 \rm\ miles[/tex]
Thus, the distance car travel during the braking period is 0.0504 miles.
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It’s the 18th century and you are responsible for artillery. Victory hangs in the balance and it all depends on you making a good shot towards the enemy fortress. Thankfully, your physics class gave you all the tools to calculate projectile trajectories. Your cannon launches a cannonball at an initial speed of 100 m/s and you set the angle at 53 degrees from the horizontal. Calculate (a) how far from the fortress should you position your cannon in order to hit it at its foundation?; and (b) how far from the fortress should you position your cannon in order to hit it at its top height (10 m) in order to knock it down? (g = 9.8 m/s^2)
Answers
Answer:
a) You should position the cannon at 981 m from the wall.
b) You could position the cannon either at 975 m or 7.8 m (not recomended).
Explanation:
Please see the attached figure for a graphical description of the problem.
In a parabolic motion, the position of the flying object is given by the vector position:
r =( x0 + v0 t cos α ; y0 + v0 t sin α + 1/2 g t²)
where:
r = position vector
x0 = initial horizontal position
v0 = module of the initial velocity vector
α = angle of lanching
y0 = initial vertical position
t = time
g = gravity acceleration (-9.8 m/s²)
The vector "r" can be expressed as a sum of vectors:
r = rx + ry
where
rx = ( x0 + v0 t cos α ; 0)
ry = (0 ; y0 + v0 t sin α + 1/2 g t²)
rx and ry are the x-component and the y-component of "r" respectively (see figure).
a) We have to find the module of r1 in the figure. Note that the y-component of r1 is null.
r1 = ( x0 + v0 t cos α ; y0 + v0 t sin α + 1/2 g t²)
Knowing the the y-component is 0, we can obtain the time of flight of the cannon ball.
0 = y0 + v0 t sin α + 1/2 g t²
If the origin of the reference system is located where the cannon is, the y0 and x0 = 0.
0 = v0 t sin α + 1/2 g t²
0 = t (v0 sin α + 1/2 g t) (we discard the solution t = 0)
0 = v0 sin α + 1/2 g t
t = -2v0 sin α / g
t = -2 * 100 m/s * sin 53° / (-9.8 m/s²) = 16.3 s
Now, we can obtain the x-component of r1 and its module will be the distance from the wall at which the cannon sould be placed:
x = x0 + v0 t cos α
x = 0 m + 100m/s * 16.3 s * cos 53
x = 981 m
The vector r1 can be written as:
r1 = (981 m ; 0)
The module of r1 will be: [tex]x = \sqrt{(981 m)^{2} + (0 m)^{2}}[/tex]
Then, the cannon should be placed 981 m from the wall.
b) The procedure is the same as in point a) only that now the y-component of the vector r2 ( see figure) is not null:
r2y = (0 ; y0 + v0 t sin α + 1/2 g t² )
The module of this vector is 10 m, then, we can obtain the time and with that time we can calculate at which distance the cannon should be placed as in point a).
module of r2y = 10 m
10 m = v0 t sin α + 1/2 g t²
0 = 1/2 g t² + v0 t sin α - 10 m
Let´s replace with the data:
0 = 1/2 (-9.8 m/s² ) t² + 100 m/s * sin 53 * t - 10 m
0= -4.9 m/s² * t² + 79.9 m/s * t - 10 m
Solving the quadratic equation we obtain two values of "t"
t = 0.13 s and t = 16.2 s
Now, we can calculate the module of the vector r2x at each time:
r2x = ( x0 + v0 t cos α ; 0)
r2x = (0 m + 100m/s * 16.2 s * cos 53 ; 0)
r2x = (975 m; 0)
Module of r2x = 975 m
at t = 0.13 s
r2x = ( 0 m + 100m/s * 0.13 s * cos 53 ; 0)
r2x = (7.8 m ; 0)
module r2x = 7.8 m
You can place the cannon either at 975 m or at 7.8 m (see the red trajectory in the figure) although it could be dangerous to place it too close to the enemy fortress!
A driver has a reaction time of 0.50 s , and the maximum deceleration of her car is 6.0 m/s^2 . She is driving at 20 m/s when suddenly she sees an obstacle in the road 50 m in front of her. What is the distance she passes after noticing the obstacle before fully stopping? Express your answer with the appropriate units.
Answers
Answer:
The car stops after 32.58 m.
Explanation:
t = Time taken for the car to stop
u = Initial velocity = 20 m/s
v = Final velocity = 0
s = Displacement
a = Acceleration = -6 m/s²
Time taken by the car to stop
[tex]v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-20}{-6}\\\Rightarrow t=3.33\ s[/tex]
Total Time taken by the car to stop is 0.5+3.33 = 3.83 s
[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow s=20\times 3.83+\frac{1}{2}\times -6\times 3.83^2\\\Rightarrow s=32.58\ m[/tex]
The car stops after 32.58 m.
Distance between car and obstacle is 50-32.58 = 17.42 m
If an arrow's mass is doubled and the speed is halved, the momentum is changed by a factor of: a) 1 b) 2 c) 0.25 d) 0.5
Answers
Explanation:
The momentum of an object is given by :
[tex]p=m\times v[/tex]............(1)
m is the mas of the object
v is the speed of the object
According to question, arrow's mass is doubled and the speed is halved. So,
m' = 2m
v' = v/2
The new momentum becomes :
[tex]p'=2m\times \dfrac{v}{2}[/tex]
p' = mv
p' = p
So, the momentum remains the same. The momentum is changed by a factor of 1. Hence, this is the required solution.
A proton is released in a uniform electric field, and it experiences an electric force of 2.07 x 10^-14 N toward the south. Part A) What is the magnitude of the electric field? Part B) What is the direction of the electric field? O west O east O south O north
Answers
Answer:
The magnitude of the electric field is 129375 N/C toward south.
Explanation:
Given that,
Electric force [tex]F=2.07\times10^{-14}\ N[/tex]
(A). We need to calculate the magnitude of the electric field
Using formula of electric field
[tex]F = qE[/tex]
[tex]E=\dfrac{F}{q}[/tex]
Where, q = charge of proton
E = electric field
[tex]E=\dfrac{2.07\times10^{-14}}{1.6\times10^{-19}}[/tex]
[tex]E=129375\ N/C[/tex]
(B). The direction of the electric field is toward the direction of the force.
So, The direction of the electric field is toward south
Hence, The magnitude of the electric field is 129375 N/C toward south.
As a science project, you drop a watermelon off the top of the Empire State Building. 320 m above the sidewalk. It so happens that Superman flies by at the instant you release the watermelon. Superman is headed straight down with a constant speed of 30 m/s. A) How much time passes before the watermelon has the same velocity? B) How fast is the watermelon going when it passes Superman?C) How fast is the watermelon traveling when it hits the ground?
Answers
Answer:
3.06 seconds time passes before the watermelon has the same velocity
watermelon going at speed 59.9 m/s
watermelon traveling when it hits the ground at speed is 79.19 m/s
Explanation:
given data
height = 320 m
speed = 30 m/s
to find out
How much time passes before the watermelon has the same velocity and How fast is the watermelon going and How fast is the watermelon traveling
solution
we will use here equation of motion that is
v = u + at ....................1
here v is velocity 30 m/s and u is initial speed i.e zero and a is acceleration i.e 9.8 m/s²
put the value and find time t
30 = 0 + 9.8 (t)
t = 3.06 s
so 3.06 seconds time passes before the watermelon has the same velocity
and
we know superman cover distance is = velocity × time
so distance = 30 × t
and distance formula for watermelon is
distance = ut + 0.5×a×t² .............2
here u is initial speed i.e 0 and a is acceleration i.e 9.8 m/s² and h is 30 × t
30 × t = 0 + 0.5×9.8×t²
t = 6.12 s
so by equation 1
v = u + at
v = 0 + 9.8 ( 6.12)
v = 59.9 m/s
so watermelon going at speed 59.9 m/s
and
watermelon traveling speed formula is by equation of motion
v² - u² = 2as ......................3
here v is speed and u is initial speed i.e 0 and a is acceleration i.e 9.8 m/s² and s is distance i.e 320 m
v² - 0 = 2(9.8) 320
v = 79.19 m/s
so watermelon traveling when it hits the ground at speed is 79.19 m/s
A 50 cm^3 block of iron is removed from an 800 degrees Celsius furnance and immediately dropped into 200 mL of 20 degrees Celsius water. What percentage of the water boils away?
Answers
Answer:
% of water boils away= 12.64 %
Explanation:
given,
volume of block = 50 cm³ removed from temperature of furnace = 800°C
mass of water = 200 mL = 200 g
temperature of water = 20° C
the density of iron = 7.874 g/cm³ ,
so the mass of iron(m₁) = density × volume = 7.874 × 50 g = 393.7 g
the specific heat of iron C₁ = 0.450 J/g⁰C
the specific heat of water Cw= 4.18 J/g⁰C
latent heat of vaporization of water is L_v = 2260 k J/kg = 2260 J/g
loss of heat from iron is equal to the gain of heat for the water
[tex]m_1\times C_1\times \Delta T = M\times C_w\times \Delta T + m_2\times L_v[/tex]
[tex]393.7\times 0.45\times (800-100) = 200\times 4.18\times(100-20) + m_2\times 2260[/tex]
m₂ = 25.28 g
25.28 water will be vaporized
% of water boils away =[tex]\dfrac{25.28}{200}\times 100[/tex]
% of water boils away= 12.64 %
The percentage of the water boils away when the iron block is placed into the water after furnace is 1264%.
What is heat transfer?The heat transfer is the transfer of thermal energy due to the temperature difference.The heat flows from the higher temperature to the lower temperature.
The heat transfer of a closed system is the addition of change in internal energy and the total amount of work done by it.
As the initial volume of the iron block is 50 cm³ and the density of the iron is 7.874 g/cm³. Thus the mass of the iron block is,
[tex]m=50\times7.874\\m=393.7\rm g[/tex]
The temperature of the furnace is 800 degrees Celsius and the specific heat of the iron block is 0.45 J/g-C.
As the boiling point of the water is 100 degree Celsius. Thus the heat loss by the block of iron is,
[tex]Q_L=393.7\times0.45\times(800-100)\\Q_L=124015.5[/tex]
The latent heat of the water is 2260 J/g. Thus the heat gain by vaporized water is,
[tex]Q_v=2260\times m_v\\[/tex]
Now the heat gain by the water is equal to the heat loss by the iron block.
As the specific heat of the water is 4.18 J/g-C and the temperature of the water is 20 degrees and volume of water is 200 ml.
Thus heat gain by water can be given as,
[tex]Q_G=Q_L=200\times4.18(100-20)+2260m_v\\124015.5=200\times4.18(100-20)+2260m_w\\m_v=25.28\rm g[/tex]
Thus the total amount of the water boils away is 25.28 grams.
The percentage of the water boils away is,
[tex]p=\dfrac{25.25}{200}\times100\\p=12.64[/tex]
Thus the percentage of the water boils away when the iron block is placed into the water after furnace is 1264%.
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For an object of mass 1 Kg, estimate the number of protons and neutrons in the material, noting that a proton or neutron weighs 2000 times more than an electron. Assume that the number of neutrons is equal to the number of protons. Also assume that the absolute value of the net charge is less than 1Coulomb. Estimate the number of electrons in this object.
Answers
Answer:
The answer is very close to [tex]N_e=2.989\times10^{26}[/tex], where [tex]N_e[/tex] is the number of electrons.
Explanation:
First we take into account that the block weighs 1Kg, and that the number of protons and electrons is the same. As the electron mass is tiny even compared to that of the proton and neutrons we can neglect it in our considerations.
Let's start by equating the mass of all protons and neutrons to the mass of the of the object:
[tex]N_p m_p+N_n m_n=1[/tex]
Where [tex]N_p[/tex] and [tex]N_n[/tex] is the number of protons and neutrons respectively. [tex]m_p[/tex] and [tex]m_n[/tex] is the mass of an proton and a neutron respectively. Because the number of protons and neutrons is equal we can say the following [tex]N_p=N_n=N[/tex], thus we have:
[tex]N_p m_p+N_n m_n=N(m_e+m_p)=1Kg \implies N=\frac{1}{m_p+m_n} [/tex]
On the other hand we have that the sum of all charges is less than the absolute value of 1C, we can express this by the following:
[tex]-1<N_p\cdot e-N_e\cdot e<1[/tex]
[tex]\implies -1<N\cdot e-N_e\cdot e<1[/tex]
Where [tex]e[/tex] is the proton charge (same as for the electron). We continue with the inequality:
[tex]-N\cdot e-1<N_p\cdot e-N_e\cdot e<-N\cdot e+1[/tex]
[tex]\implies \frac{N\cdot e+1}{e}>N_e>\frac{N\cdot e-1}{e}[/tex]
[tex]\implies \frac{(m_e+m_p)^{-1}\cdot e+1}{e}>N_e>\frac{(m_e+m_p)^{-1}\cdot e-1}{e}[/tex]
We have the estimated number of electrons bound. Because
[tex](m_e+m_p)^{-1}\cdot e>>1[/tex] We can neglect the ones on the rightmost and leftmost parts of the inequality. We then have
[tex]N_e\approx\frac{(m_p+m_n)^{-1}\cdot e}{e}[/tex]
Using the table values of the mass of the proton, mass of the neutron and the electron charge e we get
[tex]N_e\approx\frac{(1.672\times 10^{-27}+1.674\times 10^{-27})^{-1}\cdot e}{e}=(1.672\times 10^{-27}+1.674\times 10^{-27})^{-1}[/tex]
[tex]\, =2.989\times 10^{26}[/tex] electrons
The speed you compute by dividing your car's mileage by the time of travel is the ______ speed of the car, whereas the speed your speedometer reads is the car's ________speed.
Answers
Answer:
average and instant
Explanation:
The average speed is the ratio of the total path traveled and the time it took to travel that path, that is why the first space must be average speed, this because it takes into account the total amount of distance, and the total amount of time.
Instant speed, is the speed an objet (in this case a car) has in a particular moment in time, for this speed it doesn't matter the distance or the time that the car has traveled, it only matters the speed in that moment, that is what the speedometer measures, thus the second blank space must be instant speed.