Answer:
a) True
Explanation:
There are several possible explanations for this positive charge
* The explanation of the small positive charge in the plant when the bee approaches is like a defense system of the plants,
to prevent the bees from taking the pollen, but the flowers need the bees to transport the pollen for fertilization, so this possibility is not correct
* The air is conductive so the bee indexes a charge in the nearby air, this charge must be negative and this charge induced in the air induces a charge on the flower that must be positive.
When reviewing the different statements we have
a) True, it agrees with the second explanation of the phenomenon
b) False. The earth is a deposit of negative charge
c) false. If this is the case the charge should be negative
d) False. This residual charge from the other bees is quickly neutralized by the charge from the Earth.
Answer:
Explanation:
.
A harmonic oscillator has mass 0.500 kg and an ideal spring with force constant 140 N/m. Find:
a. the period
b. the frequency
c. the angular frequency
Answer:
T = 0.375 s, f = 2.66 Hz and ω = 16.71 rad/s
Explanation:
Given that,
The mass of a harmonic oscillator, m = 0.5 kg
The force constant of the spring, k = 140 N/m
The frequency of a harmonic oscillator is given by :
[tex]f=\dfrac{1}{2\pi }\sqrt{\dfrac{k}{m}}[/tex]
Substitute all the values,
[tex]f=\dfrac{1}{2\pi }\sqrt{\dfrac{140}{0.5}} \\\\f=2.66\ Hz[/tex]
Time period is given by :
[tex]T=\dfrac{1}{f}\\\\T=\dfrac{1}{2.66}\\\\T=0.375\ s[/tex]
The angular frequency is given by :
[tex]\omega=2\pi f\\\\\omega=2\pi \times 2.66\\\\\omega=16.71\ rad/s[/tex]
Hence, this is the required solution.
From rest, we step on the gas of our Ferrari, providing a force F for 4secs, speed up to a final speed v. If the applied force were only 1/2 F, how long would it have to be applied to reach the same final speed?
Answer:
The force must be applied during 8 seconds to reach trhe same final speed.
Explanation:
By Impulse Theorem, a change in the magnitude of linear momentum of a system with constant mass can be done by applying a force during a given time. That is:
[tex]m\cdot (v_{f}-v_{o}) = F \cdot \Delta t[/tex]
Where:
[tex]m[/tex] - Mass, measured in kilograms.
[tex]v_{o}[/tex], [tex]v_{f}[/tex] - Initial and final speed, measured in meters per second.
[tex]F[/tex] - Net external foce, measured in newtons.
[tex]\Delta t[/tex] - Time, measured in seconds.
We can eliminate mass and speeds by constructing the following relationship:
[tex]F_{1}\cdot \Delta t_{1} = F_{2}\cdot \Delta t_{2}[/tex] (2)
If we know that [tex]F_{1} = F[/tex], [tex]\Delta t_{1} = 4\,s[/tex] and [tex]F_{2} = \frac{F}{2}[/tex], then the time is:
[tex]4\cdot F = \frac{F\cdot \Delta t_{2}}{2}[/tex]
[tex]\Delta t_{2} = 8\,s[/tex]
The force must be applied during 8 seconds to reach trhe same final speed.
A ball is thrown straight up into the air. Which of the following best describes the energy present at various stages?
There is more energy at the top of the ball's path than there is at the bottom.
The total amount of energy varies, with more energy at the bottom and less at the top of the path.
At the very top, most of the energy is potential and just before it hits the ground, most of the energy is kinetic.
At the very top, most of the energy is kinetic and just before it hits the ground, most of the energy is potential.
Answer:
Uhh 2 one
Explanation
what is electric potential
Answer:
The electric potential is the amount of work energy needed to move a unit of electric charge from a reference point to the specific point in an electric field with negligible acceleration of the test charge to avoid producing kinetic energy or radiation by test charge.
SI unit: volt
Other units: statvolt
In SI base units: V = kg⋅m^2⋅A^−1⋅s^−3
Dimension: M L^2 T^−3 I^−1
Explanation:
Hope it is helpful....
Explanation:
common symbol - V
SI unit - volt
other unit - statvolt
A boy of mass 60 kg is sledding down a 70 m slope starting from rest. The slope is angled at 15° below the horizontal. After going 20 m along the slope he passes his friend of mass 50 kg, who jumps on the sled. They now move together to the bottom of the slope. The coefficient of kinetic friction between the sled and the snow is 0.12. Ignoring the mass of the sled, find their speed at the bottom.
Albert Bandura emphasized the idea of __________, which is the belief one has in one’s own ability to succeed. A. operant conditioning B. determinism C. self-efficacy D. self-worth
Answer:
C
Explanation:
Albert Bandura emphasized the idea of Self efficacy which is the belief one has in one’s own ability to succeed.
What is Self efficacy?
A person's self-efficacy relates to their confidence in their ability to carry out the behaviors required to achieve particular performance goals (Bandura, 1977, 1986, 1997).
The belief in one's capacity to exercise control over one's own motivation, behavior, and social environment is known as self-efficacy. The goals for which people strive, the amount of effort put out to obtain goals, and the possibility of achieving particular levels of behavioral performance are all influenced by these cognitive self-evaluations.
Self-efficacy beliefs, unlike conventional psychological notions, are anticipated to change according to the operating domain and the environment in which an action occurs.
Therefore, Albert Bandura emphasized the idea of Self efficacy which is the belief one has in one’s own ability to succeed.
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You use a 160 cm plank to lift a large rock. If the rock is 20cm from the fulcrum, what is the plank’s IMA?
Answer:
IMA = 8
Explanation:
Given the following data:
Length of plank = 160cm
Length of resistance = 20cm
To find the ideal mechanical advantage (IMA);
[tex] IMA= \frac {length \; of \; effort}{length \; of \; resistance} [/tex]
[tex] IMA= \frac {160}{20} [/tex]
IMA = 8
Imagine that 10.0 g of liquid helium, initially at 4.20 K, evaporate into an empty balloon that is kept at 1.00 atm pressure. What is the volume of the balloon at (a) 25.0 K and (b) 293 K?
Answer:
(a) The volume of the liquid helium at 25 K is 5.13 L
(b) The volume of the liquid helium at 293 K is 60.14 L.
Explanation:
Given;
mass of the liquid helium, m = 10 g
initial temperature of the liquid helium, T₁ = 4.2 K
pressure of the liquid helium, P = 1.00 atm
Atomic mass of Helium, = 4 g
number of moles of Helium, n = 10 / 4 = 2.5 moles
The initial volume of the liquid helium is calculated as;
[tex]PV_1 = nRT_1\\\\V_1 = \frac{nRT_1}{P} \\\\[/tex]
where;
R is ideal gas constant, = 0.08205 L.atm./mol.K
[tex]V_1 = \frac{2.5 \times 0.08205 \times 4.2}{1 } \\\\V_1 = 0.862 \ L[/tex]
(a) The volume of the liquid helium at 25 K.
Apply Charles law;
[tex]\frac{V_1}{T_1} =\frac{V_2}{T_2} \\\\V_2 = \frac{V_1T_2}{T_1} \\\\V_2 = \frac{0.862 \times 25 }{4.2} \\\\V_2 = 5.13 \ L[/tex]
(b) The volume of the liquid helium at 293 K.
[tex]\frac{V_1}{T_1} =\frac{V_2}{T_2} \\\\V_2 = \frac{V_1T_2}{T_1} \\\\V_2 = \frac{0.862 \times 293 }{4.2} \\\\V_2 = 60.14 \ L[/tex]
Two particles, an electron and a proton, are initially at rest in a uniform electric field of magnitude 570 N/C. If the particles are free to move, what are their speeds (in m/s) after 47.6 ns?
Explanation:
Given that,
Two particles, an electron and a proton, are initially at rest in a uniform electric field of magnitude 570 N/C.
We need to find their speeds after 47.6 ns.
For electron,
The electric force is given by :
[tex]F=qE\\\\F=1.6\times 10^{-19}\times 570\\\\=9.12\times 10^{-17}\ N[/tex]
Let a be the acceleration of the electron. So,
F = ma
m is mass of electron
[tex]a=\dfrac{F}{m}\\\\a=\dfrac{9.12\times 10^{-17}}{9.1\times 10^{-31}}\\\\a=10^{14}\ m/s^2[/tex]
Let v be the final velocity of the electron. So,
v = u +at
u = 0 (at rest)
So,
[tex]v=10^{14}\times 47.6\times 10^{-9}\\\\v=4.76\times 10^6\ m/s[/tex]
For proton,
Acceleration,
[tex]a=\dfrac{9.12\times 10^{-17}}{1.67\times 10^{-27}}\\\\=5.46\times 10^{10}\ m/s^2[/tex]
Now final velocity of the proton is given by :
[tex]v=5.46\times 10^{10}\times 47.6\times 10^{-9}\\\\v=2598.96\ m/s[/tex]
Hence, this is the required solution.
A solar panel is used to collect energy from the sun and change it into other forms of energy. The picture below shows some solar panels on the roof of a building. Which form of energy to collected by the solar panels?
A. Wind
B. sound
C. Magnetic
D. Light
A soccer ball was kicked over the edge of a wall and traveled 35 m horizontally at a speed of 5.6m/s. Calculate the vertical height of the wall.
Answer:
Are you sure it was soccer ball? Or meine hearts
Explanation:
Which statement best compares potential and kinetic energy?
O Objects always have more potentiał energy than kinetic energy.
O Kinetic energy increases and potential energy decreases when the velocity of an object increases
O Only potential energy decreases when an object's height increases.
O Objects always have more kinetic energy than potential energy.
Answer:
Kinetic energy increases and potential energy decrease when velocity of an object increase.
if a current of 5A flows for 2minutes, find the quantity of electricity transfered
Help Please! Been stuck for some time
Answer:
pretty sure A is correct, as they are pushing in opposite directions, hence canceling each other
Explanation:
3. Sodium-24 has a half-life of 15 hours. If a sample of sodium-24 has an
original activity of 800 Bq, what will
its activity be after:
i) 15 hours?
ii) 30 hours?
iii) 45 hours?
iv) 60 hours?
Answer:
See explanation
Explanation:
From the formula;
0.693/t1/2 = 2.303/t log (Ao/A)
t1/2 = half life of Sodium-24
Ao = initial activity of Sodium-24
A= activity of Sodium-24 at time = t
So,
0.693/15 = 2.303/15 log (800/A)
0.0462 = 0.1535 log (800/A)
0.0462/0.1535 = log (800/A)
0.3 = log (800/A)
Antilog(0.3) = (800/A)
1.995 = (800/A)
A = 800/1.995
A = 401 Bq
ii) 0.693/15 = 2.303/30 log (800/A)
0.0462 = 0.0768 log (800/A)
0.0462/0.0768 = log (800/A)
0.6 = log (800/A)
Antilog (0.6) = (800/A)
3.98 = (800/A)
A = 800/3.98
A = 201 Bq
iii)
0.693/15 = 2.303/45 log (800/A)
0.0462 = 0.0512 log (800/A)
0.0462/0.0512 = log (800/A)
0.9 = log (800/A)
Antilog (0.9) = (800/A)
7.94 = (800/A)
A = 800/7.94
A= 100.8 Bq
iv)
0.693/15 = 2.303/60 log (800/A)
0.0462 = 0.038 log (800/A)
0.0462/0.038 = log (800/A)
1.216 = log (800/A)
Antilog(1.216) = (800/A)
16.44 = (800/A)
A = 800/16.44
A = 48.66 Bq
550 nm light passes through a diffraction grating with 3000 lines per centimeter. The screen is 115 cm away from the grating. What is the distance (in cm) between the 2nd order bright fringe and the 2nd dark fringe?
Answer:
69.7 cm
Explanation:
What is the distance (in cm) between the 2nd order bright fringe and the 2nd dark fringe?
For a diffraction grating, dsinθ = mλ where d = grating spacing = 1/(3000 lines per cm) = 1/3000 × 100 m per line = 1/300000 m = 1/3 × 10⁻⁵ m, m = order of fringe and λ = wavelength of light = 550 nm = 550 × 10⁻⁹ m.
Also, tanθ = x/D where x = distance of nth order fringe from central maximum and D = distance of screen from grating = 115 cm = 1.15 m
Now sinθ = d/mλ, Since θ is small, sinθ ≅ tanθ
So, d/mλ = x/D for a second order bright fringe, m = 2.
So, d/2λ = x/D
x = dD/2λ
So, x =
For a dark fringe, we have
d/(m + 1/2)λ = x'/D where x' is the distance of the fringe from the central maximum.
For a second-order dark fringe, m = 2. So,
d/(2 + 1/2)λ = x'/D
d/(5/2)λ = x'/D
2d/5λ = x'/D
x' = 2dD/5λ
So, the distance (in cm) between the 2nd order bright fringe and the 2nd dark fringe is x" = dD/2λ - 2dD/5λ
x" = dD/10λ
Substituting the values of the variables into the equation, we have
x"= 1/3 × 10⁻⁵ m × 1.15 m/(10 × 550 × 10⁻⁹ m)
x" = 1.15/165 × 10² m
x" = 0.00697 × 10² m
x" = 0.697 m
x" = 69.7 cm
The distance (in cm) between the 2nd order bright fringe and the 2nd dark fringe 69.7 cm
What is difraction?Diffraction of light occurs when a light wave passes by a corner or through an opening or slit that is physically the approximate size of, or even smaller than that light's wavelength
For a diffraction grating, dsinθ = mλ
where,
d = grating spacing = 1/(3000 lines per cm) = 1/3000 × 100 m per line = 1/300000 m = 1/3 × 10⁻⁵ m,
m = order of fringe
λ = wavelength of light = 550 nm = 550 × 10⁻⁹ m.
Also, [tex]tan\theta=\dfrac{x}{D}[/tex]t where
x = distance of nth order fringe from central maximum
D = distance of screen from grating = 115 cm = 1.15 m
Now [tex]Sin\theta =\dfrac{d}{m\lambda}[/tex] , Since θ is small, sinθ ≅ tanθ
So, [tex]\dfrac{d}{m\lambda}=\dfrac{x}{D}[/tex]
for a second order bright fringe, m = 2.
So, [tex]\dfrac{d}{2\lambda}=\dfrac{x}{D}[/tex]
[tex]x=\dfrac{dD}{2\lambda}[/tex]
For a dark fringe, we have
[tex]\dfrac{d}{(m+\dfrac{1}{2})\lambda}=\dfrac{X'}{D}[/tex]
where x' is the distance of the fringe from the central maximum.
For a second-order dark fringe, m = 2. So,
[tex]\dfrac{d}{(m+\dfrac{1}{2})\lambda}=\dfrac{X'}{D}[/tex]
[tex]\dfrac{d}{ \dfrac{5}{2}\lambda}=\dfrac{X'}{D}[/tex]
[tex]X'=\dfrac{2dD}{5\lambda}[/tex]
So, the distance (in cm) between the 2nd order bright fringe and the 2nd dark fringe is
[tex]X''=\dfrac{dD}{2\lambda}-\dfrac{2dD}{5\lambda}[/tex]
[tex]X''=\dfrac{dD}{10\lambda}[/tex]
Substituting the values of the variables into the equation, we have
[tex]x''=\dfrac{1}{3\times10^{-5}}\times \dfrac{1.15}{10\times 550\times 10^{-9}}[/tex]x
x" = 0.00697 × 10² m
x" = 0.697 m
x" = 69.7 cm
Hence the distance (in cm) between the 2nd order bright fringe and the 2nd dark fringe 69.7 cm
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1. Suppose the spring in Sample Problem A is replaced with a spring that stretches
36 cm from its equilibrium position.
a. What is the spring constant in this case?
b. Is this spring stiffer or less stiff than the one in Sample Problem A?
Answer:
a. spring constant = 125 N/m
b. This spring is less stiff than the one in Sample Problem A.
Explanation:
P.S - The Sample Problem A is as follows :
Given - Sample Problem A - A load of 45 N attached to a spring that is
hanging vertically stretches the spring 0.14 m. What is the spring
constant?
Suppose the spring in Sample Problem A is replaced with a spring
that stretches 36 cm from its equilibrium position.
To find - a. What is the spring constant in this case?
b. Is this spring stiffer or less stiff than the one in Sample Problem A.
Proof -
As given,
Load = 45 N
Amplitude = 0.14 m
Let the spring constant = k
As we know that,
Load = k (Amplitude)
⇒45 = k(0.14)
⇒k = [tex]\frac{45}{0.14}[/tex] = 321.43
∴ we get
Spring constant in Sample problem A = 321.43
Now,
a.)
Given, Amplitude = 36 cm = 0.36 m
Let the spring constant = k₁
⇒45 = k₁ (0.36)
⇒k₁ = [tex]\frac{45}{0.36}[/tex] = 125 N/m
b.)
AS we can see that k₁ < k
⇒ This spring is less stiff than the one in Sample Problem A.
a. spring constant = 125 N/m
b. This spring is less stiff than the one in Sample Problem A.
what is spring constant?The spring constant generally shows the stiffness of the spring and is the ratio of the force applied to the deflection of the spring.
It is given in the question that:
Sample Problem A - A load of 45 N attached to a spring that is
hanging vertically stretches the spring 0.14 m.
Suppose the spring in Sample Problem A is replaced with a spring
that stretches 36 cm from its equilibrium position.
a. What is the spring constant in this case?
As given,
Load F = 45 N
Amplitude x= 0.14 m
Let the spring constant = k
As we know that spring force will be
[tex]F=k\times x[/tex]
[tex]45=k\times 0.14[/tex]
⇒k = 321.43N/m
∴ we get
Spring constant in Sample problem A is [tex]k=321.43\ \frac{N}{m}[/tex]
Now,
Given, Amplitude = 36 cm = 0.36 m
Let the spring constant = k₁
⇒45 = k₁ (0.36)
⇒k₁= 125 N/m
b.) Is this spring stiffer or less stiff than the one in Sample Problem A.
AS we can see that k₁ < k new spring is less stiff than the one in Sample Problem A.
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This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. Communication satellites are placed in a geosynchronous orbit, i.e., in a circular orbit such that they complete one full revolution about the earth in one sidereal day (23.934 h), and thus appear stationary with respect to the ground. Determine the altitude of these satellites above the surface of the earth in both SI and U.S. customary units. The altitude is SI units in km. The altitude is U.S. customary units in mi.
Answer:
the altitude of these satellites above the surface of the earth;
35790 km ( SI units )
22243.63 miles ( U.S. customary unit )
Explanation:
Given the data in the question;
Time taken by the satellite to complete on revolution is 23.934 hours
= 23.934 × 60 × 60 = 86162.4 seconds
now, let h represent altitude, r represent Orbit radius, v represent Orbit speed.
we know that
v² = GM/r
= gR²/r
= (9.81m/s² × ( 6.37 × 10⁶ m)²) / r
v = 19.95 × 10⁶ / √r ---------- let this be equation
also;
time t = 2πr/v
86162.4 s = 2πr/v -------- let this be equation 2
a) Determine the altitude of these satellites above the surface of the earth in both SI and U.S. customary units
we substitute v in equation into equation 2
so
86162.4 s = 2πr / (19.95 × 10⁶ / √r)
r = 42.16 × 10⁶ m
so, altitude h = r - R
h = 42.16 × 10⁶ m - 6.37 × 10⁶ m
h = 35790000 m
convert to kilometer
h = 35790000 / 1000
h = 35790 km
Convert to miles
h = 35790 / 1.609
h = 22243.63 miles
Therefore, the altitude of these satellites above the surface of the earth;
35790 km ( SI units )
22243.63 miles ( U.S. customary unit )
Define Refraction and give some knowlegde about it
If the diameter of a moose eye is 40 mm, what is the total refractive power of the anterior portion of the eye?
Answer:
-the ratio of the speed of light
in air to the speed of light in the substance.
-speed of light in air 300,000 km/sec, which decreases when it passes through a transparent substance.
-e.g.. speed of light in substance = 200,000 km/sec, R.I. = 300,000/200,000 = 1.5
Explanation:
A dog finds a toy at rest on the floor. The dog pushes the toy horizontally on a frictionless floor with a net force of 2.0 Newtons for 3.0 meters. How much kinetic energy does the toy gain? Round your answer to the nearest whole number.
Answer:
the kinetic energy gained by the toy is 6J.
Explanation:
Given;
net applied to the toy by dog, F = 2 N
distance moved by the toy, d = 3 m
Apply the principle of work-energy theorem to determine the kinetic energy gained by the toy.
ΔK.E = W
= F x d
= 2 x 3
= 6 J
Therefore, the kinetic energy gained by the toy is 6J.
If a car has a momentum of 2.04 x 104 kgm/s and a velocity of 18 m/s, what is its mass?
Answer:
Mass = 1133.33 kg (Approx.)
Explanation:
Given:
Momentum = 2.04 x 10⁴ kg[m/s]
Velocity = 18 m/s
Find:
Mass
Computation:
Mass = Momentum / Velocity
Mass = [2.04 x 10⁴] / 18
Mass = 1133.33 kg (Approx.)
Astronauts need sophisticated spacesuits to protect them from the harsh conditions of space. These spacesuits are very heavy for astronauts to wear on Earth. For an astronaut on the Moon, however, the suit would seem lighter. Which of the following statements explains why the spacesuit would feel lighter on the Moon?
A. There is no gravity on the Moon.
B. Earth’s gravity can’t be felt from the Moon.
C. The force of the Moon’s gravity is less than Earth’s.
D. The force of the Sun’s gravitational attraction varies throughout the Solar System and is stronger closer to Earth.
Answer:
I think it’s c
Explanation:
3% of earth's water is?
Is earth’s freshwater
Only about 3 percent of Earth’s water is fresh water.
Which phrase describes velocity?
u
A. A quantity with direction only
B. A quantity with magnitude only
C. A quantity with no units
D. A quantity with magnitude and direction
SUBMI
Orbital speed of a satellite is dependent of its mass
I'm assuming the question is supposed to be like this:
How does the Orbital velocity of a satellite depends on the mass of the satellite?
Answer
It is independent of mass of satellite.
Static Friction
Now let’s examine the static case. Remain on the “Force graphs” tab at the top of the window. Make sure the box labeled “Ffriction” is checked at the left of the screen, this will allow us to measure to force of friction experienced by an object as it slides down the ramp.
Draw a free body diagram for an object sitting on the incline at rest, assuming the incline is at the maximum angle BEFORE the object starts to move. Be sure to include friction and stipulate whether it is kinetic or static.
What type of bond is CO2?||
Answer:
Lol
Explanation:
Captain Jack Sparrow has been marooned on an island in the Atlantic by his crew, and decides to builda raft to escape. The wind seems quite steady, and first blows him due east for 11km, and then 6km ina direction 6degrees north of east. Confident that he will eventually find himself in safety, he fallsasleep. When he wakes up, he notices the wind is now blowing him gently 11degrees south of east -but after traveling for 21km, he finds himself back on the island.
Variable Name Min Max Step Sample Value
thetab 5 10 1 6
a 10 20 11 1
b 5 15 1 6
c 20 30 1 21
thetac 10 15 11 1
Required:
How far (in km) did the wind blow him while he was sleeping?
Answer:
d₃ = 37,729 km, θ= 5.1º North of West
Explanation:
This is a velocity addition problem, the easiest way to solve it is to decompose the velocities in a Cartesian system, the x-axis coincides with the West-East direction and the y-axis with the South-North direction
* first displacement is
d₁ₓ = 11 km
* second offset is
cos 6 = d₂ₓ / d₂
sin 6 = d_{2y} / d₂
d₂ₓ = d₂ cos 6
d_{2y} = d₂ sin 6
d₂ₓ = 6 cos 6 = 5.967 km
d_{2y} = 6 sin 6 = 0.6272 km
* third displacement is unknown
* fourth and last displacement
cos (-11) = d₄ₓ / d₄
sin (-11) = d_{4y} / d₄
d₄ₓ = d₄ cos (-11)
d_{4y} = d₄ sin (-11)
d₄ₓ = 21 cos (-11) = 20.61 km
d_{4y} = 21 sin (-11) = -4.007 km
They tell us that at the end of the tour you are back on the island, so the displacement must be zero
X axis
x = d₁ₓ + d₂ₓ + d₃ₓ + d₄ₓ
0 = 11 +5.967 + d₃ₓ + 20.61
d₃ₓ = -11 - 5.967 - 20.61
d₃ₓ = -37.577 km
Y axis
y = d_{1y} + d_{2y} + d_{3y} + d_{4y}
0 = 0 + 0.6272 + d_{3y} -4.007
d_{3y} = 4.007 - 0.6272
d_{3y} = 3.3798 km
This distance can be given in the form of module and angle
Let's use the Pythagorean theorem for the module
d₃ = [tex]\sqrt{d_{3x}^2 + d_{3y}^2}[/tex]
d₃ = [tex]\sqrt{37.577^2 + 3.3798^2}[/tex]
d₃ = 37,729 km
Let's use trigonometry for the angle
tan θ = d_{3y} / d₃ₓ
θ = tan⁻¹ [tex]\frac{d_{3y}}{d_{3x}}[/tex]
θ = tan-1 (-3.3798 / 37.577)
θ = 5.1º
Since the y coordinate is positive and the x coordinate is negative, this angle is in the second quadrant, so the direction given in the form of cardinal coordinates is
θ= 5.1º North of West
1. If a radioactive sample has an initial count rate of 600 Bq. What is its count
rate after
i) l half-life?
ii) 2 half-lives
iii) 3 half-lives
iv) 4 half-lives?
Answer:
See explanation
Explanation:
Since the original count rate is 600 Bq,
i) after 1 half life, the count rate decreases to 1/2 of 600 Bq = 300 Bq
ii) after 2 half lives, the count rate decreases to 1/4 of 600 = 150 Bq
iii) after 3 half lives, the count rate decreases to 1/6 of 600 = 100 Bq
iv) after 4 half lives, the count rate decreases to 1/8 of 600 = 75 Bq
