Answer:
a. Moon around Earth.
Explanation:
Charon orbit takes around 6.4 earth days to complete its orbit. Charon does not rises or sets, it hovers over same spot around the Pluto. The same side of Charon faces the Pluto, this is called Tidal Locking.
The moon orbit takes around 27 days to complete its orbit. The moon has different sides that are faced with sun which creates light or dark face of moon on the earth. Moon has 384,400 km distance from the earth.
The object that should exhibit the longest orbital period is option a. Moon around Earth.
What is Charon's orbit?Charon's orbit takes around 6.4 earth days to finish its orbit. Charon does not rise or sets, it hovers over similar spot around Pluto. The same side of Charon faces Pluto, this we called Tidal Locking. Here the moon orbit should take approx 27 days to finish its orbit. The moon has various sides that are faced with the sun which developed the light or dark face of the moon on the earth. Also, Moon has 384,400 km distance from the earth.
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A 5.0-µC point charge is placed at the 0.00 cm mark of a meter stick and a -4.0-µC point charge is placed at the 50 cm mark. At what point on a line joining the two charges is the electric field due to these charges equal to zero?
Answer:
Electric field is zero at point 4.73 m
Explanation:
Given:
Charge place = 50 cm = 0.50 m
change q1 = 5 µC
change q2 = 4 µC
Computation:
electric field zero calculated by:
[tex]E1 =k\frac{q1}{r^2} \\\\E2 =k\frac{q2}{R^2} \\\\[/tex]
Where electric field is zero,
First distance = x
Second distance = (x-0.50)
So,
E1 = E2
[tex]k\frac{q1}{r^2}=k\frac{q2}{R^2} \\\\[/tex]
[tex]\frac{5}{x^2}=\frac{4}{(x-50)^2} \\\\[/tex]
x = 0.263 or x = 4.73
So,
Electric field is zero at point 4.73 m
Some radar systems detect the size and shape of objects such as aircraft and geological terrain. Approximately what is the smallest observable detail (in m) utilizing 495 MHz radar?
Answer:
0.61 m
Explanation:
The smallest observable length by the radar must be at least equal to or greater than the wavelength of the radar.
using the relationship
c = fλ
where
c is the speed of light in vacuum = 3 x 10^8 m/s
f is the frequency of the wave = 495 MHz = 4.95 x 10^8 Hz
λ is the wavelength = ?
λ = c/f = (3 x 10^8)/(4.95 x 10^8) = 0.61 m
answer to your question is 0.6m
A fisherman in a stream 39 cm deep looks downward into the water and sees a rock on the stream bed. How deep does the stream appear to the fisherman
Answer:
30cm
Explanation:
assume that the eyes are substantially above the water so that sin(theta) is approximately theta.
( small angle approximation).
The point at which a ray leaving the fish hits the surface of the water is x to the side of the centreline and the depth of the water is d
x/d = sin( angle of incidence)
if the apparent depth of the water is h then
x/h = sin( angle of refraction)
and applying snells law
1 sin ( theta air) = 1.33 sin( theta water)
1 * x/h = 1.33 * x/d
d/h = 1.33
or h/d = 1/1.33
h/39 = 1.33
h = 39 /1.33 so that is the apparent depth of the stream assuming:-
1. Your eyes are almost directly overhead
and
2. your eyes are a significant distance above the surface of the water.
x/d = 1.33 x/h
h/d =39/1.3
= 30cm
A 3200-lb car is moving at 64 ft/s down a 30-degree grade when it runs out of fuel. Find its velocity after that if friction exerts a resistive force with magnitude proportional to the square of the speed with k
Answer:
The velocity is 40 ft/sec.
Explanation:
Given that,
Force = 3200 lb
Angle = 30°
Speed = 64 ft/s
The resistive force with magnitude proportional to the square of the speed,
[tex]F_{r}=kv^2[/tex]
Where, k = 1 lb s²/ft²
We need to calculate the velocity
Using balance equation
[tex]F\sin\theta-F_{r}=m\dfrac{d^2v}{dt^2}[/tex]
Put the value into the formula
[tex]3200\sin 30-kv^2=m\dfrac{d^2v}{dt^2}[/tex]
Put the value of k
[tex]3200\times\dfrac{1}{2}-v^2=m\dfrac{d^2v}{dt^2}[/tex]
[tex]1600-v^2=m\dfrac{d^2v}{dt^2}[/tex]
At terminal velocity [tex]\dfrac{d^2v}{dt^2}=0[/tex]
So, [tex]1600-v^2=0[/tex]
[tex]v=\sqrt{1600}[/tex]
[tex]v=40\ ft/sec[/tex]
Hence, The velocity is 40 ft/sec.
Do an Internet search to determine what minerals are extracted from the ground in order to manufacture the following products:
a. Stainless steel utensils
b. Cat litter
c. Tums brand antacid tablets
d. Lithium batteries
e. Aluminum beverage cans
Answer:
Raw materials are most times gotten from the earth through various forms of extraction procedures.
A) Stainless steel utensils is made up of mainly Iron and other elements such as chromium , carbon etc.
B) Cat litter comprises of ceramic products which is made up of clay.
C) Tums brand antacid tablets comprises of calcium carbonate, magnesium hydroxide, aluminum hydroxide and sodium bicarbonate which could be extracted from the earth.
D)Lithium batteries are made up of elements in the earth such as lithium and carbon.
E)Aluminum beverage cans are made up of aluminum extracted from the ground.
A nearsighted person has a far point that is 4.2 m from his eyes. What focal length lenses in diopters he must use in his contacts to allow him to focus on distant objects?
Answer:
-0.24diopters
Explanation:
The lens is intended that makes an object at infinity appear to be 4.2 m away, so do=infinity, dI = - 4.2m (minus sign because image is on same side of lens as object)
So 1/do +1/di = 1/f
1/infinity + 1/-4.2 = 1/f
1/f = 1/-4.2 = -0.24diopters
A spring is hung from the ceiling. When a block is attached to its end, it stretches 2.5 cm before reaching its new equilibrium length. The block is then pulled down slightly and released. What is the frequency of oscillation?
Answer:
0.99Hz
Explanation:
Using F= -mx ( spring force)
At equilibrium the gravitational force will be balanced by the spring force so mg= kx
K= mg/ 0.25 N/m
But
Frequency f= 1/2pi √g/0.25
Frequency is 0.99Hz
The block is pulled down slightly and released so, Frequency of oscillation is 3.15 Hz
Frequency of oscillation based problem:What information do we have?
Length starched = 2.5 cm
F = Kx
We know that
F = mg
So,
mg = Kx
K/m = g/x
[tex]f=\frac{1}{2\pi}\sqrt{\frac{g}{x} }\\f=\frac{1}{2\pi}\sqrt{\frac{9.8}{0.025} }[/tex]
Frequency of oscillation = 3.15 Hz
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A foot is 12 inches and a mile is 5280I ft exactly. A centimeter is exactly 0.01m or mm. Sammy is 5 feet and 5.3tall. what is Sammy's height in inches?
Answer:
65.3 Inches tall
Explanation:
If Sammy is 5 feet and 5.3 inches tall, we simply need to convert the feet to inches, and sum the remaining inches from his height to determine his overall height in inches.
So, 5 feet = (12 inches/1foot) * (5 feet) = 60 inches
And 60 inches + 5.3 inches = 65.3 inches.
Hence, Sammy is 65.3 inches tall.
Cheers.
What is the answer?
Answer: i think it is d. none of them.
Explanation: The speed of light in a vacuum is 186,282 miles per second and so when you look and the answer choices and the question it doesnt make any since.
A third wire of the same material has the same length and twice the diameter as the first. How far will it be stretched by the same force
Complete question is;
A force stretches a wire by 0.60 mm. A second wire of the same material has the same cross section and twice the length.
a) How far will it be stretched by the same force?
b) A third wire of the same material has the same length and twice the diameter as the first. How far will it be stretched by the same force?
Answer:
0.15 mm
Explanation:
According to Hooke's Law,
E = Stress(σ)/Strain(ε)
Where E is youngs modulus
Formula for stress is;
Stress(σ) = Force(F)/Area(A)
Formula for strain is;
Strain(ε) = Change in length/original length = (Lf - Li)/Li
We are also told that a second wire of the same material has the same cross section and twice the length.
Thus;
Rearranging Hooke's Law to get the constants on one side, we have;
F/(AE) = ε
Thus from the conditions given;
ε1 = 0.6/Li
ε2 = (Change in length)/(2*Li)
And ε1 = ε2
Thus;
0.6/Li = Change in length/(2*Li)
Li will cancel out and we now have;
Change in length = 2 × 0.6 = 1.2 mm
Finally, we are told A third wire of the same material has the same length and twice the diameter as the first.
Area of a circle;A1 = πd²/4
Now, we are told d is doubled.
Thus, new area of the new circle is;
A2 = π(2d)²/4 = πd²
Rearranging Hooke's Law,we have;
F/A = εE
Since F and E are now constants, we have;
F/E = constant = Aε
Thus;
A1(ε1) = A2(ε2)
A1 = πd²/4
e1 = 0.60/Li
A2 = πd²
e2 = Change in length/Li
Thus;
((πd²/4) × 0.6)/Li = (πd² × Change in length)/ Li
Rearranging, Li and πd² will cancel out to give;
0.6/4 = Change in length
Change in length = 0.15 mm
Unpolarized light is passed through three successive Polaroid filters, each with its transmission axis at 45.0° to the preceding filter. What percentage of light gets through?
Answer:
The percentage is [tex]k = 12.5 \%[/tex]
Explanation:
From the question we are told that
The axis is is at [tex]\theta = 45 ^o[/tex]
Generally the of intensity light emerging from the first polarizer is mathematically represented as
[tex]I_{1} = \frac{I_o}{ 2}[/tex]
Where [tex]I_o[/tex] is the intensity of unpolarized light
Now the light emerging from the second polarizer is mathematically represented as
[tex]I_2 = I_ 1 * cos ^2(\theta )[/tex]
[tex]I_2 = \frac{I_o}{2} * cos ^2(45 )[/tex]
[tex]I_2 = \frac{I_o}{2} * \frac{1}{2} = \frac{I_o}{4}[/tex]
Now the light emerging from the third polarizer is mathematically represented as
[tex]I_3 = I_ 2 * cos ^2(\theta )[/tex]
[tex]I_3 = \frac{I_o}{4} * cos ^2(45 )[/tex]
[tex]I_3 = \frac{I_o}{8}[/tex]
Now the percentage of the intensity of light that emerged with respect to the intensity of the unpolarized light is
[tex]k = \frac{\frac{I_o}{8} }{I_o } * 100[/tex]
[tex]k = 12.5 \%[/tex]
The percentage of light that gets through the three successive Polaroid filters is; 12.5%
We are given;
Angle of transmission axis; θ = 45°
Formula for intensity of light from first polarizer is;
I₁ = ¹/₂I₀
Formula for intensity of light from second polarizer is;
I₂ = I₁cos²θ
Formula for intensity of light from third polarizer is;
I₃ = I₂cos²(90 - θ)
Combining the 3 equations;
Put ¹/₂I₀ for I₁ in second formula to get;
I₂ = ¹/₂I₀cos²θ
Put ¹/₂I₀cos²θ for I₂ in third formula to get;
I₃ = ¹/₂I₀cos²θ*cos²(90 - θ)
Plugging in 45° for θ gives;
I₃ = ¹/₂I₀cos²45*cos²(90 - 45)
⇒ I₃ = ¹/₂I₀cos²45*cos²45
⇒ I₃ = ¹/₂I₀cos⁴45
Now, cos 45 in surd form is 1/√2. Thus;
I₃ = ¹/₂I₀(1/√2)⁴
I₃ = ¹/₂I₀(¹/₄)
I₃ = ¹/₈I₀
I₃/I₀ = ¹/₈
I₃/I₀ = 0.125
In percentage form, we have;
I₃/I₀ = 12.5%
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15. Food chain always start with
a. Photosynthesis
Decay
b. Respiration
d. N2 Fixation
C.Photosynthesis
A diffraction grating with 161 slits per centimeter is used to measure the wavelengths emitted by hydrogen gas. At what angles in the first-order spectrum would you expect to find the two violet lines of wavelength 434 nm and of wavelength 410 nm
Answer:
[tex]\theta_1 = 0.400^o[/tex]
[tex]\theta_2 =0.378^o[/tex]
Explanation:
From the question we are told that
The number of slits per cm is k = [tex]161\ slits\ per\ cm = 161 \ slits\ per\ 0.01 m[/tex]
The order of the maxima is n = 1
The wavelength are [tex]\lambda_1 = 434 nm = 434 *10^{-9} \ m \ \ \ , \lambda_2 = 410nm = 410 *10^{-9} \ m[/tex]
The spacing between the slit is mathematically represented as
[tex]d = \frac{ 0.01}{k}[/tex]
=> [tex]d = \frac{ 0.01}{161}[/tex]
=> [tex]d = 6.211 *10^{-5} \ m[/tex]
Generally the condition for constructive interference is
[tex]n\lambda = d \ sin \theta[/tex]
At [tex]\lambda_1[/tex]
[tex]\theta _1 = sin^{-1} [ \frac{1 * 434 *10^{-9}}{6.211 *10^{-5}} ][/tex]
[tex]\theta_1 = 0.400^o[/tex]
At [tex]\lambda_2[/tex]
[tex]\theta _2 = sin^{-1} [ \frac{1 * 410 *10^{-9}}{6.211 *10^{-5}} ][/tex]
[tex]\theta_2 =0.378^o[/tex]
It's nighttime, and you've dropped your goggles into a 3.2-m-deep swimming pool. If you hold a laser pointer 1.2 m above the edge of the pool, you can illuminate the goggles if the laser beam enters the water 2.0 m from the edge.
How far are the goggles from the edge of the pool?
Answer:
Explanation:
Laser angle with water surface is given by: Tan α = 1/2.0= 0.5/
α = 26.56°
Laser angle with Normal = 90 - 26.56 = 63.44 °
Assuming a red laser, refractive index in water is 1.331.
Angle of refraction in water is given by:
Ref Ind = Sin i / Sin r
1.331 = Sin 63.44 / Sin r
Sin r = 0.8945 / 1.331 = 0.6721
Angle r = 42.22°
For the path in water:
Tan 42.22 = x / 3.2
x = 2.9m where x is the lateral displacement of the laser ince it hits the water
So the goggles are 2.0 + 2.9 = 4.9 m from edge of pool
A bullet is fired from a rifle pointed 45 degrees above horizontal. The bullet leaves the muzzle traveling 1400 m/s. How many seconds does it take the bullet to reach the high point of its trajectory?
The bullet's vertical velocity at time [tex]t[/tex] is
[tex]v=1400\dfrac{\rm m}{\rm s}-gt[/tex]
where [tex]g=9.80\frac{\rm m}{\mathrm s^2}[/tex] is the acceleration due to gravity.
At its highest point, the bullet's vertical velocity is 0, which happens
[tex]0=1400\dfrac{\rm m}{\rm s}-gt\implies t=\dfrac{1400\frac{\rm m}{\rm s}}g\approx\boxed{142.857\,\mathrm s}[/tex]
(or about 140 s, if you're keeping track of significant figures) after being fired.
A cylinder rotating about its axis with a constant angular acceleration of 1.6 rad/s2 starts from rest at t = 0. At the instant when it has turned through 0.40 radian, what is the magnitude of the total linear acceleration of a point on the rim (radius = 13 cm)?
a. 0.31 m/s^2
b. 0.27 m/s^2
c. 0.35 m/s^2
d. 0.39 m/s^2
e. 0.45 m/s^2
Answer:
The magnitude of the total linear acceleration is 0.27 m/s²
b. 0.27 m/s²
Explanation:
The total linear acceleration is the vector sum of the tangential acceleration and radial acceleration.
The radial acceleration is given by;
[tex]a_t = ar[/tex]
where;
a is the angular acceleration and
r is the radius of the circular path
[tex]a_t = ar\\\\a_t = 1.6 *0.13\\\\a_t = 0.208 \ m/s^2[/tex]
Determine time of the rotation;
[tex]\theta = \frac{1}{2} at^2\\\\0.4 = \frac{1}{2} (1.6)t^2\\\\t^2 = 0.5\\\\t = \sqrt{0.5} \\\\t = 0.707 \ s\\\\[/tex]
Determine angular velocity
ω = at
ω = 1.6 x 0.707
ω = 1.131 rad/s
Now, determine the radial acceleration
[tex]a_r = \omega ^2r\\\\a_r = 1.131^2 (0.13)\\\\a_r = 0.166 \ m/s^2[/tex]
The magnitude of total linear acceleration is given by;
[tex]a = \sqrt{a_t^2 + a_r^2} \\\\a = \sqrt{0.208^2 + 0.166^2} \\\\a = 0.266 \ m/s^2\\\\a = 0.27 \ m/s^2[/tex]
Therefore, the magnitude of the total linear acceleration is 0.27 m/s²
b. 0.27 m/s²
A quantity of an ideal gas is compressed to half its initial volume. The process may be adiabatic, isothermal or occurring at constant pressure. Rank those three processes in order of the work required of an external agent, least to greatest. A : adiabatic, isothermal, constant pressure B : constant pressure, isothermal, adiabatic C : adiabatic, constant pressure, isothermal D : isothermal, adiabatic, constant pressure E : constant pressure, adiabatic, isothermal
Answer:
B. constant pressure, isothermal, adiabatic
Explanation:
A quantity of an ideal gas is compressed to half its initial volume.
The process may be adiabatic, isothermal or occurring at constant pressure.
Adiabatic-constant heat
Constant pressure or isobaric
Isothermal or constant temperature
An external agent is a system that does work on a system or a machine.
This external agent applies force , or changes the state of the body it is acting on.
In order of the work required of an external agent, least to greatest
The following processes will be arranged.
constant pressure, isothermal, adiabatic
Three capacitors C1 = 10.7 µF, C2 = 23.0 µF, and C3 = 29.3 µF are connected in series. To avoid breakdown of the capacitors, the maximum potential difference to which any of them can be individually charged is 125 V. Determine the maximum energy stored in the series combination.
Answer:
E = 1336.71875 J
Explanation:
We are given;. Capacitance of Capacitor 1; C1 = 10.7 µF
Capacitor 2; C2 = 23.0 µF
Capacitor 3; C3 = 29.3 µF
Supply voltage;V = 125 V
Formula for capacitance in series is;
Capacitors in series circuit: C(eq) = 1/C(1) +1/C(2) +1/C(3) .......
Thus, equivalent capacitance is;
C(eq) = (1/10.7) + (1/23) + (1/29.3) = 0.1711 µF = 0.1711 × 10^(6) F
Now, the formula for maximum energy stored is;
E = ½ × C(eq) × V²
E = ½ × 0.1711 × 10^(-6) × 125²
E = 1336.71875 J
The elastic portion of the downward-sloping straight-line demand curve lies:_______
a. at the intersection with the supply curve.
b. anywhere to the right of the current market price.
c. above the point of unit elasticity.
d. below the point where total revenue is maximized.
Answer:
c. above the point of unit elasticity.
Explanation:
The elastic portion of the downward-sloping straight-line demand curve lies above the point of unit elasticity. Supply and demand are fundamental concept in economics. The demand curve shows how much of a good people will want at a different prices. The demands curves illustrates the intuition why people purchase a good for a lower price. For the demand curve, the price is always shown on the vertical axis and the demand curve is shown on the horizontal axis. Thus , the quantity demanded increases as the price gets lower. However, the price elasticity of the demand curve varies along the demand curve. This is because there is a key distinction between the gradient and the elasticity. The gradient which is the slope of the line is always the same in the demand curve but elasticity of the demand changes in the percentage of the quantity demand. Therefore, elasticity will vary along the downward-sloping straight - line demand curve. So, in a downward-sloping straight-line demand curve, the elastic portion is usually above the point of unit elasticity
Can a person break a wall with a punch?(I mean technically,maybe with inhuman speed)
Answer:
yes, shaolin monks can do. you can find tutorials on Youtb
A father and his son want to play on a seesaw. Where on the seesaw should each of them sit to balance the torque?
Answer:
A The father should sit closer to the pivot.
C The longer wrench makes the job easier because less force is needed when there is more distance from the pivot.
A As far from the head of the hammer as possible because this will maximize torque.
D at the opposite side of the seesaw towards the middle
:) gl
Explanation:
If a father and his son want to play on a seesaw then to balance the torque of the seesaw the father should sit near the pivot as he had more weight as compared to his son, while the son should sit a little farther from the pivot point as compared to his father.
What is the mechanical advantage?
Mechanical advantage is defined as a measure of the ratio of output force to input force in a system, It is used to analyze the forces in simple machines like levers and pulleys.
Mechanical advantage = output force(load) /input force (effort)
As given in the problem statement If a father and his son wish to play on a seesaw,
The father should sit close to the pivot because he weighs more than his son, and the son should sit a little farther away from the pivot point than his father. This will help balance the torque of the seesaw.
Thus, the father should sit near the pivot on the one side and the son should sit a little farther from the pivot of a seesaw on the other side.
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A balloon contains 1.21 x 105 L of ideal gas at 265K. The gas is then cooled to 201 K. What is the volume (L) assuming no gas enters or exits the balloon
Answer:
The new volume will be 0.918 x 10^5 L
Explanation:
initial volume [tex]V_{1}[/tex] = 1.21 x 10^5 L
Initial temperature [tex]T_{1}[/tex] = 265 K
Final volume [tex]V_{2}[/tex] = ?
Final temperature [tex]T_{2}[/tex] = 201 K
Th gas is an ideal gas.
For ideal gases, the equation [tex]V_{1}[/tex]/[tex]T_{1}[/tex] = [tex]V_{2}[/tex]/[tex]T_{2}[/tex] = constant
substituting value, we have
(1.21 x 10^5)/265 = [tex]V_{2}[/tex]/201
[tex]V_{2}[/tex] = 24321000/265 = 91777.4 L
= 0.918 x 10^5 L
Several books are placed on a table. These books have a combined weight of 25 N and cover an area of 0.05 m2. How much pressure do the books exert on the table? The pressure the books apply to the table top is __ Pa.
Answer:
500 PascalsExplanation:
[tex]Force = 25N\\Area = 0.05m^2\\\\Pressure = \frac{Force}{Area}\\ \\Pressure = \frac{25}{0.05}\\\\ Pressure = 500 Pascals[/tex]
A motorist traveling with a constant speed of 15 m/s (about 34 mi/h) passes a school-crossing corner, where the speed limit is 0 m/s (about 22 mi/h). Just as the motorist passes, a police officer on a motorcycle stopped at the comer starts off in pursuit with constant acceleration of 3.0 m/S2. (a) How much time elapses before the officer catches up with the motorist? (b) What is the officer's speed at that point? (c) What is the total distance each vehicle has traveled at that point? Please help me
Answer:
(a) 10 s
(b) 30 m/s
(c) 150 m
Explanation:
The motorist's position at time t is:
x = 15t
The officer's position at time t is:
x = ½ (3) t² = 1.5 t²
(a) When they have the same position, the time is:
15t = 1.5 t²
t = 0 or 10 s
(b) The officer's speed is:
v = 3t
v = 30 m/s
(c) The position is:
x = 15t = 150 m
A satellite of mass m circles a planet of mass M and radius R in an orbit at a height 2R above the surface of the planet. What minimum energy is required to change the orbit to one for which the height of the satellite is 3R above the surface of the planet
Answer:
ΔE = GMm/24R
Explanation:
centripetal acceleration a = V^2 / R = 2T/mr
T= kinetic energy
m= mass of satellite, r= radius of earth
= gravitational acceleration = GM / r^2
Now, solving for the kinetic energy:
T = GMm / 2r = -1/2 U,
where U is the potential energy
So the total energy is:
E = T+U = -GMm / 2r
Now we want to find the energy difference as r goes from one orbital radius to another:
ΔE = GMm/2 (1/R_1 - 1/R_2)
So in this case, R_1 is 3R (planet's radius + orbital altitude) and R_2 is 4R
ΔE = GMm/2R (1/3 - 1/4)
ΔE = GMm/24R
I MIND TRICK PLZ HELP LOL
Troy and Abed are running in a race. Troy finishes the race in 12 minutes. Abed finishes the race in 7 minutes and 30 seconds. If Troy is running at an average speed of 3 miles per hour and speed varies inversely with time, what is Abed’s average speed for the race?
Answer:
Explanation:
Let the race be of a fixed distance x
[tex]Average Speed = \frac{Total Distance}{Total Time}[/tex]
Troy's Average speed = 3 miles/hr = x / 0.2 hr
x = 0.6 miles
Abed's Average speed = 0.6 / 0.125 = 4.8 miles/hr
Lamar has been running sprints to prepare for his next football game.He has found that he can maintain his maximum speed for 45 yards.He’s thinking of running in a 5km race in a few months,but doesn’t know if he can maintain his maximum speed for the entire 5 km.Can you help him determine how far he can?
Answer:
Kindly check explanation
Explanation:
Length of race = 5km
Maximum speed = 45 yards
Converting from yards to kilometer :
1km = 1093.613 yards
x = 45 yards
(1093.613 * x) = 45
x = 45 / 1093.613
x = 0.0411480 km
Where x = maximum length for which he can maintain his maximum speed expressed in kilometers.
Therefore, with the available information, it can be concluded that Lamar cannot maintain his maximum speed for the entire 5km race and will only be able maintain his maximum speed for 0.0411 kilometers.
Lamar cannot maintain his maximum speed for the entire 5km race and will only be able maintain his maximum speed for 0.0411 kilometers.
The calculation is as follows;
Length of race = 5km
Maximum speed = 45 yards
Converting from yards to kilometer :
1km = 1093.613 yards
x = 45 yards
[tex](1093.613 \times x) = 45[/tex]
[tex]x = 45 \div 1093.613[/tex]
x = 0.0411480 km
here x represent maximum length for which he can maintain his maximum speed expressed in kilometers.
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A city of punjab has 15 percemt chance of wet weather on any given day. What is probability that it will take a week for it three wet weather on 3 sepaprate days? Also find it standard deviation
Answer:
The probability that it will take a week for it three wet weather on 3 separate days is 0.06166 and its standard deviation is 0.9447
Explanation:
We are given that A city of Punjab has 15 percent chance of wet weather on any given day.
So, Probability of wet weather = 0.15
Probability of not being a wet weather = 1-0.15 =0.85
We are supposed to find probability that it will take a week for it three wet weather on 3 separate days
Total number of days in a week = 7
We will use binomial over here
n = 7
p =probability of failure = 0.15
q = probability of success=0.85
r=3
Formula :[tex]P(r=3)=^nC_r p^r q ^{n-r}[/tex]
[tex]P(r=3)=^{7}C_{3} (0.15)^3 (0.85)^{7-3}\\P(r=3)=\frac{7!}{3!(7-3)!} (0.15)^3 (0.85)^{7-3}\\P(r=3)=0.06166[/tex]
Standard deviation =[tex]\sqrt{n \times p \times q}[/tex]
Standard deviation =[tex]\sqrt{7 \times 0.15 \times 0.85}[/tex]
Standard deviation =0.9447
Hence The probability that it will take a week for it three wet weather on 3 separate days is 0.06166 and its standard deviation is 0.9447
As you finish listening to your favorite compact disc (CD), the CD in the player slows down to a stop. Assume that the CD spins down with a constant angular acceleration. If the CD rotates clockwise (let's take clockwise rotation as positive) at 500 rpm (revolutions per minute) while the last song is playing, and then spins down to zero angular speed in 2.60 s with constant angular acceleration, the angular acceleration of the CD, as it spins to a stop at -20.1 rad/s 2. How many revolutions does the CD make as it spins to a stop?
Answer:
10.8rev
Explanation:
Using
Wf²-wf = 2 alpha x theta
0²- 56.36x56.36/ 2(-20.13) x theta
Theta = 68.09 rad
But 68.09/2π
>= 10.8 revolutions
Explanation:
The sentence, "The popcorn kernels popped twice as fast as the last batch," is a(n) _____. experiment hypothesis observation control
The correct answer is C. Observation
Explanation:
An observation is a statement a describes a phenomenon, which is the result of measuring the phenomenon or using the senses to collect information about it. Additionally, observations are part of the Scientific method because through observations it is possible to understand phenomena.
The sentence presented is an observation because this statement is the result of the researcher observing or measuring how fast kernels pops, which means the statement derives from studying a phenomenon. Also, this cannot be classified as a hypothesis because a hypothesis is a probable explanation, and it cannot be classified as an experiment because the experiment is the general method to prove or disprove a hypothesis.
