Suppose you are asked to compute the tangent of 5.00 meters. Is this possible? Why or why not?
Answer:
Generally it is only possible to compute the tangent of angle in their various unit but because 5 meter is not an angle then it is impossible to compute the tangent of 5 meters.
Explanation:
An athlete completes a round of circular tract with diameter 150 m in 5 min. What is the average speed V of the athlete in m/s? Answer with two decimal places
Answer:
Explanation:
From the question we are told that
The diameter is [tex]d = 150 \ m[/tex]
The time taken is [tex]t = 5 \ min = 300\ s[/tex]
Generally the circumference is mathematically represented as
[tex]C = d * \pi[/tex]
[tex]C = 150 * 3.142[/tex]
[tex]C = 471.3 \ m[/tex]
The average speed is mathematically represented as
[tex]v = \frac{C}{ t}[/tex]
[tex]v = \frac{471.3}{300}[/tex]
[tex]v = 1.571 \ m/s[/tex]
A catapult launches a large stone from ground level at a speed of 45.2 m/s at an angle of 57.0° with the horizontal. The stone returns to ground level shortly thereafter.
(a) How long is it in the air
s
(b) What maximum height does the stone reach? (Neglect air friction.)
Im
Answer:
a
[tex]T =7.7 \ s [/tex]
b
[tex] s = 73.3 \ m [/tex]
Explanation:
From the question we are told that
The speed is [tex]u = 45.2 \ m/s[/tex]
The angle is [tex]\theta = 57 .0^o[/tex]
Generally the velocity of the stone in the y-axis is mathematically evaluated as
[tex]v_y = v * sin(\theta )[/tex]
=> [tex]v_y = 45.2 * sin(57 )[/tex]
=> [tex]v_y = 37.9 \ m/s [/tex]
Now the time taken for the first flight is mathematically represented as
[tex]t = \frac{v_y}{g}[/tex]
=> [tex]t = \frac{37.9}{9.8}[/tex]
=> [tex]t = 3.87 \ s [/tex]
The time taken for the total flight is mathematically represented as
[tex]T = 2 * t[/tex]
[tex]T = 2 * 3.87[/tex]
[tex]T =7.7 \ s [/tex]
From the equation of kinematics we have that
[tex]v^2 = v_y^2 + 2gs[/tex]
At maximum heigth v = 0
[tex] = 37.9 ^2 + 2 *( -9.8) * s[/tex] The negative sign is because it is moving against gravity to attain the maximum height
So
[tex] s = 73.3 \ m [/tex]
The time a projectile is in the air and the maximum height reached is required.
The projectile is in the air for 7.73 s.
The maximum height reached by the projectile is 73.24 m.
u = Initial velocity = 45.2 m/s
[tex]\theta[/tex] = Angle of inclination = [tex]57^{\circ}[/tex]
g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]
Time of flight is given by
[tex]t=\dfrac{2u\sin\theta}{g}\\\Rightarrow t=\dfrac{2\times 45.2\sin 57^{\circ}}{9.81}\\\Rightarrow t=7.73\ \text{s}[/tex]
Maximum height is given by
[tex]h=\dfrac{u^2\sin^2\theta}{2g}\\\Rightarrow h=\dfrac{45.2^2\sin^257^{\circ}}{2\times 9.81}\\\Rightarrow h=73.24\ \text{m}[/tex]
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What does the law of universal gravitation describe? The motion of the planets around the sun The observation that matter cannot be created or destroyed The observation that all objects that have mass attract each other The amount of force acting on an object being equal to its mass times its acceleration
yes its The observation that all objects that have mass attract each other
Which of these statements partially defines law?
Answer: I dont see the option choices
Explanation:
[tex]11 \times 4[/tex]
Answer:
44 is right answer
Explanation:
Step 1 :
11+11+11+11+11
step 2:
=22+22
step 3:
=44
hope it helped you:)
Thanks!
A communications satellite orbiting the earth has solar panels that completely absorb all sunlight incident upon them. The total area A of the panels is 10m2.1) The intensity of the sun's radiation incident upon the earth is about I=1.4kW/m2. Suppose this is the value for the intensity of sunlight incident upon the satellite's solar panels. What is the total solar power P absorbed by the panels?Express your answer numerically in kilowatts to two significant figures.2) What is the total force F on the panels exerted by radiation pressure from the sunlight?Express the total force numerically, to two significant figures, in units of newtons.
Answer:
0.000047N
Explanation:
We know that
intensity (I) = P/ A
Where
P= power
A= Area
So lets say that the power absorbed
Will be = Intensity x Area
Which Is = 1.4 x 10^3 x(10)
So
14000 Watt = 14 kWatt
However we know that radiation pressure is equal to
time-averaged intensity all over the speed of light in free space
So
P = (1.4 x 1000)/c
But
F= P x A
So
((1.4 x 1000)/(3 x1 0^8)) x 10
Which is
=0.000046666N
To two SIG figures we have
=0.000047 N
(a) The total solar power P absorbed by the panels is 14 kW.
(b) The total force F on the panels exerted by radiation pressure from the sunlight is [tex]4.7 \times 10^{-5} \ N[/tex].
The given parameters;
area of the panel, A = 10 m²intensity of the sun radiation, I = 1.4 kW/m^2The total solar power P absorbed by the panels is calculated as follows;
[tex]P = I A\\\\P = (1.4 \ kW /m^2 ) \times 10\ m^2\\\\P = 14 \ kW[/tex]
The total force F on the panels exerted by radiation pressure from the sunlight is calculated as follows;
[tex]Pressure, P = \frac{I}{c} \\\\Force, F = PA\\\\F = (\frac{I}{c} )A\\\\F = \frac{IA}{c} \\\\F = \frac{(1.4 \times 10^3) \times (10)}{3\times 10^8} \\\\F = 4.7 \times 10^{-5} \ N[/tex]
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The work done when a force moves a body through a distance of 15m is 1800j. What is the value of the force applied
Answer:
120
Work :
W = Fd (work = force x distance)
Force :
F = W/d
Distance :
d = W/F
A 7.52nC charge is located 1.65m from a 4.10nC point charge.a) Find the magnitude of the force that the -2.5 nC charge exerts on the -8.0 nC charge.b) Is the force attractive or repulsive?
Answer:
a. [tex]1.019 \times 10^{-7} N[/tex]
b. Repulsive force
Explanation:
The computation is shown below:
a. The magnitude of the force is
Given that
Charge = 7.52nc
Location = 1.65m
Point charge = 4.10nC
Based on the above information,
As we know that
[tex]F= \frac{kqQ}{r^2} \\\\= \frac{9\times10^{9} \times 7.52 \times 10^{-9} \times 4.1 \times 10^{-9}}{1.652} \\\\= 1.019 \times 10^{-7} N[/tex]
b. Now as it can be seen that both contains the positive charges so this represents the repulsive force and the same is to be considered
A physics student runs along a line tangent to the edge of a motionless merry-go-round and jumps on at the very outside. The merry-go-round has the shape of a uniform disk.
A) Make an angular momentum chart to help you keep track of what is and is not changing. Consider the two objects, the student and merry-go-round, with the total angular momentum of these two objects conserved. Must the student have angular momentum before jumping on the merry-go-round?
B) Is linear momentum conserved in this interaction if you just consider the merry-go-round and the student?
Answer:
We know that angular momentum is mvr
So that of the students initially will be
A1= mvr
and the final of the student and the merrygo round will be
A2= (1/2Mr²+mr)omega
So
Angular velocity (omega)= mvr/(1/2Mr²+mr)
This is the final angular velocity of the system so yea the student must have angular momentum before jumping on the merry-go-round
B. Because of directional changes at every edge of the Merry go round that result in changes in velocity, this results in changes in momentum too thus linear momentum is not conserved
A mass is tied to a spring and begins vibrating periodically the distance between its highest and lowest position is 48cm what is the amplitude of the vibrating?
Answer 24cm
Explanation:
The amplitude of vibration of the spring is "24 cm"
The periodic vibrating body's motion follows a sinusoidal path. This sinusoidal path is illustrated in the attached picture.
From the picture, it can be clearly seen that the amplitude of the periodic vibration motion is the distance from its mean position to the highest point.
Since the distance of both the highest and the lowest points from the mean position is the same. Therefore, the distance between the lowest and the highest point must be equal to two times the amplitude of the wave.
[tex]Distance\ between\ highest\ and\ lowest\ point = 2(Amplitude)\\\\Amplitude = \frac{Distance\ between\ highest\ and\ lowest\ point}{2}\\\\Amplitude = \frac{48\ cm}{2}\\\\[/tex]
Amplitude = 24 cm
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The current through a 0.2-H inductor is i(t) = 10te–5t A. What is the energy stored in the inductor?
Answer:
E = 10t^2e^-10t Joules
Explanation:
Given that the current through a 0.2-H inductor is i(t) = 10te–5t A.
The energy E stored in the inductor can be expressed as
E = 1/2Ll^2
Substitutes the inductor L and the current I into the formula
E = 1/2 × 0.2 × ( 10te^-5t )^2
E = 0.1 × 100t^2e^-10t
E = 10t^2e^-10t Joules
Therefore, the energy stored in the inductor is 10t^2e^-10t Joules
A metre rule balances when the 50 cm mark is directly above a pivot. State where in the rule its centre of mass is located?
Answer:
The ruler's center of mass is located on the vertical line defined by the 50 cm mark.
Explanation:
Recall that in order to have equilibrium as that described by this example, the net Torque in the ruler should be zero. which means in our case that the weight due to the right half of the ruler, should equal in absolute value of the weight of the left side of the ruler, thus defining where the ruler's center of mass is.
Since this happens at the mark 50 cm, then the center of mass of the ruler is located somewhere inside the ruler somewhere above where the pivot is located (50 cm mark)
differences between Constant velocity and constant acceleration
Answer:
Traveling with a constant velocity means you're going at the same speed in the same direction continuously. If you have a constant velocity, this means you have zero acceleration. ... If you travel with a constant acceleration, your velocity is always changing, but it's changing by a consistent amount each second.
the pressure of a gas is 100.0kpa and its volume is 500.0ml if the volume increase to 1000.0ml what is the new pressure of the gas
50Kpa
Explanation:
P1V1 = P2V2
Where;
P1= 100. P2= ?
V1 = 500. V2 = 1000
100 × 500 = P2 × 1000
50000 = 1000P2
50000/1000 = P2
50 = P2
P2 = 50Kpa
change-22^0f to kelvin?
Answer:
The formula is kelvin = (tempature in farenheit-32) ÷ 1.8 +273.15
So it will be (-22-32) ÷ 1.8 +273.15= 243.15 kelvin
Answer:
T(K) = (T(°F) + 459.67)× 5/9
answer is 243.15
The position-time equation for a certain train is
2.9m + (8.8m/s)t + (2.4m/s2)+2
What is it’s acceleration?
Answer:
[tex]a=4.8m/s^2[/tex]
Explanation:
Hello,
In this case, since the acceleration in terms of position is defined as its second derivative:
[tex]a=\frac{d^2x(t)}{dt^2}=\frac{d^2}{dt^2}(2.9+8.8t+2.4t^2)[/tex]
The purpose here is derive x(t) twice as follows:
[tex]a=\frac{d^2x(t)}{dt^2}=\frac{d}{dt}(8.8+2*2.4*t)\\ \\a=4.8m/s^2[/tex]
Thus, the acceleration turns out 4.8 meters per squared seconds.
Best regards.
A clock mechanism at rest has a period of exactly 1 second. When this clock is observed while moving with a speed of 0.80 c, what is its period?
Answer:
1.67 s
Explanation:
Since the clock is moving at relativistic speed, the period dilates. So, we use the equation
t = t₀/√[1 - (v/c)²].
Given that t = the period of the clock = 1 s and v = speed of the clock = 0.80c, substituting these values into t, we have
t = t₀/√[1 - (v/c)²]
t = 1/√[1 - (0.8c/c)²]
t = 1/√[1 - (0.8)²]
t = 1/√[1 - 0.64]
t = 1/√0.36
t = 1/0.6
t = 1.67 s
So the period of the clock when it is observed while moving with a speed of 0.8c is 1.67 s
A sinusoidal electromagnetic wave from a radio station passes perpendicularly through an open window that has area of 0.700 m2 . At the window, the electric field of the wave has an rms value 0.0400 V/m .
How much energy does this wave carry through the window during a 30.0-s commercial?
Express your answer with the appropriate units.
Answer:
The value is [tex]E = 8.9 *10^{-5} \ J[/tex]
Explanation:
From the question we are told that
The area is [tex]A = 0.700 \ m^2[/tex]
The root mean square value is [tex]E_{rms} = 0.0400 \ V/m[/tex]
The time taken is [tex]t = 30.0 \ s[/tex]
Generally the energy is mathematically represented as
[tex]E = c * \sepsilon_o * A * t * E_{rms}^2[/tex]
=> [tex]E = 3.0*10^{8} * 8.85*10^{-12} * 0.700 * 30 * (0.04)^2[/tex]
=> [tex]E = 8.9 *10^{-5} \ J[/tex]
What mirror diameter gives 0.1 arc second resolution for infrared radiation of wavelength 2 micrometers?
Answer:
The diameter of mirror [tex] 5\times10^{-6}\ m[/tex]
Explanation:
Given that,
Wavelength = 2 μm
Resolution = 0.1
We need to calculate the diameter of mirror
Using formula of resolution
[tex]resolution=0.25\times\dfrac{wavelength}{diameter}[/tex]
[tex]diameter =0.25\times\dfrac{wavelength}{resolution}[/tex]
Put the value into the formula
[tex]diameter=0.25\times\dfrac{2\times10^{-6}}{0.1}[/tex]
[tex]diameter = 0.000005\ m[/tex]
[tex]diameter = 5\times10^{-6}\ m[/tex]
Hence, The diameter of mirror [tex] 5\times10^{-6}\ m[/tex]
How much ice at a temperature of -22.3 ∘C must be dropped into the water so that the final temperature of the system will be 30.0 ∘C ?How much ice at a temperature of -22.3 ∘C must be dropped into the water so that the final temperature of the system will be 30.0 ∘C ?
Answer:
111.6 g or 0.112 kg
Explanation:
specific heat of liquid water = 4190 J/kg⋅K
specific heat of ice = 2100 J/kg⋅K
heat of fusion for water = 3.34*10^5 J/kg
You didn't state the mass of the beaker, so, I will be assuming that the mass is negligible.
Assuming that the mass of ice required is m kg
Then the heat gained by the ice to attain zero degree will be
= m * 22.3 * 2100
= 46830m J
The heat gained by the ice to melt
= m * 3.34*10⁵ J
= 334000m J
The heat gained by water at zero degree to warm up to 30° =
m * 4190 * 30 = 125700m J
Total heat gained = 506530m J
Note: You didn't state the mass of the water and it's temperature, so I will be assuming that the mass of water is 0.3 kg, and it's temperature was 75° C
The heat lost by hot water to cool up to 30°
= .3 * 4190 * (75 - 30)
= 1257 * 45
= 56565 J
Using the relation, Heat lost = heat gained
506530m = 56565
m = 56565 / 506530 kg
m = 0.112 kg or 111.6 g
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A typical atom has a diameter of about 1.32 x 10-10 m.
A) What is this in inches?
B) How many atoms are there along a 21.0cm line?
Answer:
A) The value of the diameter in inches is 5.19685 × 10⁻⁹ inches
B) There are 1,590,909,091 atoms on a 21.0cm line
Explanation:
A) To determine what its diameter is in inches
We will convert 1.32 × 10⁻¹⁰ m to inches
First convert this to centimeters (cm)
1.32 × 10⁻¹⁰m = 1.32 × 10⁻¹⁰ × 10² cm
= 1.32 × 10⁻⁸cm
Now, since 1 inch = 2.54 cm
∴ 1 cm = [tex]\frac{1}{2.54}[/tex] inches
If 1 cm = [tex]\frac{1}{2.54}[/tex] inches
Then 1.32 × 10⁻⁸ cm will be
1.32 × 10⁻⁸ × [tex]\frac{1}{2.54}[/tex] inches
= 5.19685 × 10⁻⁹ inches
This is the value of the diameter in inches
B) For the number of atoms on a 21.0cm line
The diameter of 1 atom is 1.32 × 10⁻¹⁰m
1.32 × 10⁻¹⁰m = 1.32 × 10⁻⁸cm
Hence, on a 21.0 cm line, we will have
(21.0 / 1.32 × 10⁻⁸) atoms
= 1590909091 atoms
Hence, there are 1,590,909,091 atoms on a 21.0cm line
Suppose the U.S. national debt is about $15 trillion. If payments were made at the rate of $1,500 per second, how many years would it take to pay off the debt, assuming no interest were charged? Note: Before doing these calculations, try to guess at the answers. You may be very surprised. yr (b) A dollar bill is about 15.5 cm long. How many dollar bills attached end to end would it take to reach the Moon? The Earth-Moon distance is 3.84 108 m. dollar bills
Answer:
This question has already been answered.
Explanation:
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The sun appears to move through the background stars. This apparent motion would not exist if:_________.
a) the moon did not orbit the earth
b) the earth did not orbit the sun
c) the earth did not turn on its axis once a day
Answer:
b) the earth did not orbit the sun
Explanation:
The sun appears to move through the background stars due to parallax . Parallax is a phenomenon when the near object appears to move faster than the distant object . If we travel in a train , the near object like electric poles on rail track or trees and fields nearby appear to move faster against the distant background .
Hence when the earth moves around the sun , the sun appears to move against the background the stars which are far away . Had earth remained stationary at a place on its orbit around the sun , the sun would have appeared stationary against the background the stars .
1. A signal source has an open-circuit voltage of 1V, and a short-circuit current of10mA. What is the source resistance
Answer:
100 Ω
Explanation:
Given that
Open circuit voltage, V = 1 V
Short circuit current, I = 10 mA
Source resistance R, = ?
This is rather a straight forward question. Remember Ohms Law? Current being directly proportional to the voltage and inversely proportional to the resistance?
Yeah, that's the formula we'd be using.
Ohms Law states that V = IR, and thus, if we make R subject of the formula, we have
R = V / I, on substituting the values, we have
R = 1 / 10*10^-3
R = 1 / 0.01
R = 100 Ω
(4) A stainless steel storage tank contains 5 kg of carbon dioxide gas and 7 kg of argon gas. How many kmoles are in the tank?
Answer:
0.2886 Kmoles
Explanation:
Number of moles is calculated by dividing mass of element/molecular mass of element
Mass of Carbon Di oxide = 5000 g or 5 kg
Molecular mass of CO2 = 12 + 16×2 = 44 g
No of moles = 5000/44 = 113.6 moles = 0.1136 Kmoles
Mass of argon = 7Kg or 7000 kg
Molecular mass of Argon(Ar) = 40g
No of moles = 7000/40 = 175 moles = 0.175 Kmoles
Total Kmoles = 0.175+0.1136=0.2886 Kmoles
Manganese-52 has a half-life of 6 days. How many days would a scientist have to wait for the radioactivity to be 12.5% the starting amount?
Answer:
Let N = N0 where N0 is the number of atoms originally present.
In 6 days N = N0 / 2
In 12 days N = N0 / 4
In 18 days N = N0 / 8 = .125 N0
So it would take 18 days.
Manganese-52 has a half-life of 6 days. a scientist has to wait 18 days for the radioactivity to be 12.5% of the starting amount
What is radioactivity?The ability of some unstable atoms to emit nuclear radiation spontaneously, typically in the form of alpha or beta particles frequently accompanied by gamma rays, is known as radioactivity. This radiation is released when a nucleus undergoes radioactive decay and transforms into a different isotope that may be radioactive (unstable) or non-radioactive depending on the number of neutrons and protons in the nucleus (stable).
As given in the problem statement Manganese-52 has a half-life of 6 days. this means after 6 days Manganese-52 will reduce to its half value
similarly, after 12 days it further reduces to 50% of the previous value which would be 25 %
Similarly, after 18 days, it will reduce 50% from the previous value which comes out to be 12.5 %
Thus, The half-life of manganese-52 is 6 days. It takes 18 days for the radioactivity to drop to 12.5% of the initial level.
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Which law of motion is this an example of?
Newton's 1st Law of Motion
Newton's 2nd Law of Motion
Newton's 3rd Law of Motion
An airplane flies east with a velocity of 150 km/h. It encounters a tail wind of 30 km/h. How fast is the airplane traveling? (convert answer in m/s)
Answer:
50 m/s
Explanation:
Hi.
just for little clarification, I'd explain what a tail wind is, if you don't mind.
A tail wind is a wind that moves in the direction of the aircraft. If an aircraft is moving in a direction northwards, then a tail wind is also a wind moving northwards.
Now to the question, the airplane is flying at, 150 km/h with a tail wind of 30 km/h.
The airplane's speed is then,
150 + 30 = 180 km/h
And to convert to m/s, we have
180 km/h = 180 * 1000/3600 m/s
180 km/h = 50 m/s
If a space probe moves 20 m in 4 s, how fast is it moving?
Answer:
Speed =distance/time.
=20/4
=5m/s.
Formula: s = d/t
s = speed
d = distance
t = time
Solve using the values we are given.
s = 20/4
s = 5m/s
best of Luck!