Answer: Biosensors are modern electronic devices which are made up of biological recognition system and a transducer, for processing of signals and to quantify a particular analyte.
Explanation:
In modern medicine, a Biosensors are used to replace some clinical laboratory investigations of biological fluids with the advantages of being easy to use, rapid and robust as well as offering multianalyte testing capability. They are classified based on their biological recognition elements which includes:
--> enzymatic biosensors
--> immuno biosensors
--> DNA and whole cell biosensors.
They can also be classified according to their signal transduction methods. These includes:
--> electrochemical Biosensors,
--> optical biosensors,
--> thermal and mass-based biosensors;
Some of the HEALTH APPLICATIONS of a modern biosensor includes:
--> Monitoring of glucose levels in diabetic patients which makes use of enzyme based sensors to monitor concentration of glucose in patients.
--> Detection of general metabolic status of bacteria, fungi, yeast, animal or plant cells.
--> Monitoring of cholesterol levels to prevent cardiovascular diseases. Cardiac troponin and. C - Reactive Proteins are biomarkers that are easily detected using biosensors.
--> They are used for Cancer clinical testings
The total yearly world consumption of energy is approximately 4.0 × 1020 J. How much mass would have to be completely converted into energy to provide this amount of energy?
Answer:
m = 4.4 × 10³ kg
Explanation:
Given that:
The total yearly energy is 4.0 × 10²⁰ J
The amount of mass that provides this energy can be determined by using the formula:
E = mc²
where;
c = speed of light in free space = (3 × 10⁸)
4.0 × 10²⁰ = m × (3 × 10⁸)²
[tex]m = \dfrac{4.0 \times 10^{20} }{(3\times 10^8)^2}[/tex]
m = 4.4 × 10³ kg
g A frequently quoted rule of thumb in aircraft design is that wings should produce about 1000 N of lift per square meter of wing. (The fact that a wing has a top and bottom surface does not double its area.) a)At takeoff the aircraft travels at 61.1 m/s, so that the air speed relative to the bottom of the wing is 61.1 m/s. Given the sea level density of air to be 1.29 kg/m3, how fast (in m/s) must it move over the upper surface to create the ideal lift
Answer:
The value is [tex]u = 72.69 \ m/s[/tex]
Explanation:
From the question we are told that
The amount of force a square meter of an aircraft wing should produce is [tex]F = 1000 \ N[/tex]
The air speed relative to the bottom of the wing is [tex]v = 61.1 \ m/s[/tex]
The air level density of air is [tex]\rho_s = 1.29\ kg/m^3[/tex]
Gnerally this force per square meter of an aircraft wing is mathematically represented as
[tex]F = \frac{1}{2} * \rho_s * A * [ u^2 - v^2 ][/tex]
Here u is the speed air need to go over the top surface to create the ideal lift
A is the area of a square meter i.e [tex]A = 1 \ m^2[/tex]
So
[tex]1000 = \frac{1}{2} * 1.29 * 1 * [ u^2 - 61.1 ^2 ][/tex]
=> [tex]u = 72.69 \ m/s[/tex]
The arrow strikes a deer in the woods with the speed of 55 m/sec at an angle of 315 degrees. Calculate the Horizontal and vertical components of the arrow’s velocity.
Answer:
100 m
Explanation:
The mass of Earth is 5.972 x 1024 kg and its orbital radius is an average of 1.496 x 1011 m. Calculate its linear momentum, given the period of one rotation is 3.15 x 107 s
Answer: its linear momentum is 1.78 × 10²⁹ kg.m/s
Explanation:
Given that;
mass of Earth m = 5.972 x 10²⁴ kg
radius r = 1.496 x 10¹¹ m
period t = 3.15 x 10⁷ s
now we know that Earth rotates in a circular path so the distance travelled per rotation is;
d = 2πr we substitute
d = 2π × 1.496 x 10¹¹ m
= 9.4 × 10¹¹ m
Now formula for speed v is;
v = d/t
we substitute
v = 9.4 × 10¹¹ m / 3.15 x 10⁷ s
v = 2.98 × 10⁴ m/s
now we determine the linear momentum p
linear momentum p = mv
we substitute
p = (5.972 x 10²⁴ kg) × (2.98 × 10⁴ m/s)
p = 1.78 × 10²⁹ kg.m/s
Therefore its linear momentum is 1.78 × 10²⁹ kg.m/s
The linear momentum of earth is 17.8 * 10²⁸ kgm/s
The orbital radius (r) of earth is 1.496 x 10¹¹ m. Hence the distance covered by one rotation is:
Distance = 2πr = 2π(1.496 x 10¹¹) m
The period of one rotation is 3.15 x 10⁷ s.
The velocity of earth (v) = distance/time = 2π(1.496 x 10¹¹) m/ 3.15 x 10⁷ s = 298840 m/s
Linear momentum = mass * velocity = 5.972 x 10²⁴ kg * 298840 m/s = 17.8 * 10²⁸ kgm/s
Therefore the linear momentum of earth is 17.8 * 10²⁸ kgm/s
Find out more at: https://brainly.com/question/12194595
A red laser with a wavelength of 670 nm and a blue laser with a wavelength of 450 nm emit laser beams with the same light power. How do their rates of photon emission compare
E=hf C=wavelength*F
E=hC/wavelength
E=(6.626*10^-34)*(3.00*10^8)/670*10^-9
E=(6.626*10^-34)*(3.00*10^8)/450*10^-9
A 3 Kg exercise ball is held 2m above the ground. What is the gravitational potential energy?
Answer:
58.8
Explanation:
we should apply formula
m*g*h
3*9.81*2
A 5-kg projectile is fired over level ground with a velocity of 200 m/s at an angle of 25
above the horizontal. Just before it hits the ground its speed is 150 m/s. Over the entire trip the
change in the thermal energy of the projectile and air is:
Answer: 43.8 kJ
Explanation:
Given;
mass of the object, m = 5kg
initial velocity of the projectile, v₁ = 200 m/s
final velocity of the projectile, v₂ = 150 m/s
To determine the the change in the thermal energy of the projectile and air, we consider change in potential and kinetic enrgy of the projectile. Since the projectile was fired over level ground, change in potential energy is zero.
Then, change in thermal energy of the projectile, KE = Δ¹/₂mv²
KE = Δ¹/₂mv² = ¹/₂m(v₁²-v₂²)
KE = ¹/₂ × 5(200²-150²) = 2.5(17500) = 43750 J = 43.8 kJ
Therefore, the change in thermal energy of the projectile is 43.8 kJ
If you place 48 cm object 15 m in front of a convex mirror of focal length 5 m, what is the magnification of the image?
Answer:
m = 0.25
Explanation:
Given that,
Object distance, u = -15cm
Height of the object, h = 48
Focal length, f = cm
We need to find the magnification of the image.
Let v is the image distance. Using mirror's equation.
[tex]\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}\\\\\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}\\\\=\dfrac{1}{5}-\dfrac{1}{(-15)}\\\\=3.75\ cm[/tex]
Magnification,
[tex]m=\dfrac{-v}{u}\\\\=\dfrac{-3.75}{-15}\\\\=0.25[/tex]
Hence, the magnification of the image is 0.25.
Which of the following types of energy are present at some point in the energy transfer process in a nuclear power plan? Select all that apply.
Heat energy
Geothermal energy
Tidal energy
Hydroelectric energy
Nuclear energy
Electrical energy
Solar Energy
Solar Energy is the answer to the question tell me if i`m wrong
Which statement describes one way that nuclear fission differs from nuclear
fusion?
A. Nuclear fission occurs naturally only in the Sun and othling stars.
B. Nuclear fission occurs only at very high temperatures.
C. Nuclear fission produces too little energy for practical
applications.
O D. Nuclear fission is used to power submarines and aircraft carriers.
Answer:
D. Nuclear fission is used to power submarines and aircraft carriers.
Explanation:
A and B describe nuclear fusion. C doesn't apply to nuclear fusion or nuclear fission.
Find the quantity of heat needed
to melt 100g of ice at -10 °C
into water at 10 °C. (39900 J)
(Note: Specific heat of ice is
2100 Jkg 'K', specific heat
of water is 4200 Jkg K',
Latent heat of fusion of ice is
336000 Jkg ').
Answer:
Approximately [tex]3.99\times 10^{4}\; \rm J[/tex] (assuming that the melting point of ice is [tex]0\; \rm ^\circ C[/tex].)
Explanation:
Convert the unit of mass to kilograms, so as to match the unit of the specific heat capacity of ice and of water.
[tex]\begin{aligned}m&= 100\; \rm g \times \frac{1\; \rm kg}{1000\; \rm g} \\ &= 0.100\; \rm kg\end{aligned}[/tex]
The energy required comes in three parts:
Energy required to raise the temperature of that [tex]0.100\; \rm kg[/tex] of ice from [tex](-10\; \rm ^\circ C)[/tex] to [tex]0\; \rm ^\circ C[/tex] (the melting point of ice.)Energy required to turn [tex]0.100\; \rm kg[/tex] of ice into water while temperature stayed constant.Energy required to raise the temperature of that newly-formed [tex]0.100\; \rm kg[/tex] of water from [tex]0\; \rm ^\circ C[/tex] to [tex]10\;\ rm ^\circ C[/tex].The following equation gives the amount of energy [tex]Q[/tex] required to raise the temperature of a sample of mass [tex]m[/tex] and specific heat capacity [tex]c[/tex] by [tex]\Delta T[/tex]:
[tex]Q = c \cdot m \cdot \Delta T[/tex],
where
[tex]c[/tex] is the specific heat capacity of the material,[tex]m[/tex] is the mass of the sample, and[tex]\Delta T[/tex] is the change in the temperature of this sample.For the first part of energy input, [tex]c(\text{ice}) = 2100\; \rm J \cdot kg \cdot K^{-1}[/tex] whereas [tex]m = 0.100\; \rm kg[/tex]. Calculate the change in the temperature:
[tex]\begin{aligned}\Delta T &= T(\text{final}) - T(\text{initial}) \\ &= (0\; \rm ^\circ C) - (-10\; \rm ^\circ C) \\ &= 10\; \rm K\end{aligned}[/tex].
Calculate the energy required to achieve that temperature change:
[tex]\begin{aligned}Q_1 &= c(\text{ice}) \cdot m(\text{ice}) \cdot \Delta T\\ &= 2100\; \rm J \cdot kg \cdot K^{-1} \\ &\quad\quad \times 0.100\; \rm kg \times 10\; \rm K\\ &= 2.10\times 10^{3}\; \rm J\end{aligned}[/tex].
Similarly, for the third part of energy input, [tex]c(\text{water}) = 4200\; \rm J \cdot kg \cdot K^{-1}[/tex] whereas [tex]m = 0.100\; \rm kg[/tex]. Calculate the change in the temperature:
[tex]\begin{aligned}\Delta T &= T(\text{final}) - T(\text{initial}) \\ &= (10\; \rm ^\circ C) - (0\; \rm ^\circ C) \\ &= 10\; \rm K\end{aligned}[/tex].
Calculate the energy required to achieve that temperature change:
[tex]\begin{aligned}Q_3&= c(\text{water}) \cdot m(\text{water}) \cdot \Delta T\\ &= 4200\; \rm J \cdot kg \cdot K^{-1} \\ &\quad\quad \times 0.100\; \rm kg \times 10\; \rm K\\ &= 4.20\times 10^{3}\; \rm J\end{aligned}[/tex].
The second part of energy input requires a different equation. The energy [tex]Q[/tex] required to melt a sample of mass [tex]m[/tex] and latent heat of fusion [tex]L_\text{f}[/tex] is:
[tex]Q = m \cdot L_\text{f}[/tex].
Apply this equation to find the size of the second part of energy input:
[tex]\begin{aligned}Q_2&= m \cdot L_\text{f}\\&= 0.100\; \rm kg \times 3.36\times 10^{5}\; \rm J\cdot kg^{-1} \\ &= 3.36\times 10^{4}\; \rm J\end{aligned}[/tex].
Find the sum of these three parts of energy:
[tex]\begin{aligned}Q &= Q_1 + Q_2 + Q_3 = 3.99\times 10^{4}\; \rm J\end{aligned}[/tex].
a car accelerates from rest at 2 m/s. what is the speed after 8 sec?
Answer:
16m/s
Explanation:
[tex]v_{f}=v_{i}+at[/tex]
[tex]v_{f}=0+2\cdot8[/tex]
[tex]v_{f}=16\ \frac{m}{s}[/tex]
Therefore, the speed after 8 seconds is 16m/s
9. Who was the FIRST person to propose that the continents might fit together
Answer:
Alfred Wegener
Explanation:
A 0.0600-kilogram ball traveling at 60.0 meters per second hits a concrete wall. What speed must a 0.0100-kilogram bullet have in order to hit the wall with the same magnitude of momentum as the ball?
Answer:
the speed that have 0.0100 kilogram bullet with the similar momentum magnitude is 360 m/s
Explanation:
The computation of the speed that have 0.0100 kilogram bullet with the similar momentum magnitude is as follows:
The ball momentum is
[tex]\Delta Q = mv\\\\\Delta Q = 6 \times 1^-2 \times 60\\\\\Delta Q = 3.6 kg \times m/s[/tex]
As there is a similar momentum
So,
[tex]\Delta Q = MV\\\\3.6 = 10^-2V\\\\V = 360 m/s[/tex]
Hence, the speed that have 0.0100 kilogram bullet with the similar momentum magnitude is 360 m/s
Can someone please help?
Answer:
Gold = 19.3g/cm³
Explanation:
Given the following data;
Mass = 38.6g
Volume = 2cm³
To find the density;
Density can be defined as mass all over the volume of an object.
Simply stated, density is mass per unit volume of an object.
Mathematically, density is given by the equation;
[tex]Density = \frac{mass}{volume}[/tex]
Substituting into the equation, we have;
[tex]Density = \frac{38.6}{2}[/tex]
Density = 19.3g/cm³
Therefore, from the table the material is Gold.
Gold is a chemical element and it is the element 79 on the periodic table. Thus, it has an atomic number of 79. The chemical symbol for Gold is Au, it is chemically classified as a transition metal and as a solid at room temperature.
Generally, the chemical element Gold is known to have the following physical properties;
I. Malleable.
II. Ductile.
III. A good conductor of electricity and heat.
Also, it is a non-toxic chemical element with a beautiful lustrous sheen.
Why do we consider eyes, ears, and fingers as physical sensor
Answer:
because
Explanation:
because we need eye to see if you don't have eyes how you regnise some one
if you have ears how can you know who's voice is who's
if don't have fingers how can you know what your are holding this is why they are called physical sensors
Answer:
because they stimulate our brain and the sensor of human body detect stimuli.
Explanation: