Answer:
Metal or conductors , what they do is that they allow full flow of current that is conduction is due to free electrons only and there is literally no gap between valence and conduction band,so free electrons can easily jump into conduction band from valence band.
And now lets talk about insulators,what they do is that they don’t allow any current to flow i.e they act as strong dielectric,and gap between valence band and conduction band is so big that free electron can never come into conduction band from valence band ever if they try is for eternity…
But semiconductors can act both as an insulator as well as a conductor based on the voltage input. Hence, there is a possibility to control the current flow in semiconductors , so they don’t just relax and let the current pass by, they can control it, and that is why you can design logic circuits with it.
But as the temperature increases free electrons from valence band of insulators can jump to conduction band and can cause a little conductivity, and then the insulator will act as a semi-conductor.
Hi Friends!
please help me with these questions!
SUBJECT: Chemistry, Physics,Biology
Answer:
q.1 : Air near candle gets heated up and after this it rises by convection so the thermometer B will receive more heat than the thermometer A So, according to the given condition thermometer B will show a greater rise in temperature.
q.2 : x is the pure sample of compound . y is the pure sample of element . z is the mixture of different elements
q.3 : the saliva contains an enzyme salivary amylase (ptyalin) which converts starch in roti into maltose, isomaltose and small dextrins called a-dextrin.
An astronaut is traveling in a space vehicle that has a speed of 0.480c relative to Earth. The astronaut measures his pulse rate at 78.5 per minute. Signals generated by the astronaut's pulse are radioed to Earth when the vehicle is moving perpendicularly to a line that connects the vehicle with an Earth observer. (Due to vehicle's path there will be no Doppler shift in the signal.)
(a) What pulse rate does the Earth-based observer measure? beats/min
(b) What would be the pulse rate if the speed of the space vehicle were increased to 0.940c?
beats/min
Explanation:
The heart rate of the astronaut is 78.5 beats per minute, which means that the time between heart beats is 0.0127 min. This will be the time t measured by the moving observer. The time t' measured by the stationary Earth-based observer is given by
[tex]t' = \dfrac{t}{\sqrt{1 - \left(\dfrac{v^2}{c^2}\right)}}[/tex]
a) If the astronaut is moving at 0.480c, the time t' is
[tex]t' = \dfrac{0.0127\:\text{min}}{\sqrt{1 - \left(\dfrac{0.2304c^2}{c^2}\right)}}[/tex]
[tex]\:\:\:\:=0.0145\:\text{min}[/tex]
This means that time between his heart beats as measured by Earth-based observer is 0.0145 min, which is equivalent to 69.1 beats per minute.
b) At v = 0.940c, the time t' is
[tex]t' = \dfrac{0.0127\:\text{min}}{\sqrt{1 - \left(\dfrac{0.8836c^2}{c^2}\right)}}[/tex]
[tex]\:\:\:\:=0.0372\:\text{min}[/tex]
So at this speed, the astronaut's heart rate is 1/(0.0372 min) or 26.9 beats per minute.
Clear umbra can be obtained by .................................
what is the word come to this blank?
we will demonstrate experimentally that light travel in staight line.
OK.
Good luck on your experimental demonstration.
It's a nice exercise.
a. Describe the relationship between the number of batteries and the voltage and explain what you think might be happening
Answer:
Their is a direct relationship between the number of batteries and the increase in power. The voltage is the product of the number of batteries and the voltage which is 9 volts. As the batteries touch ends the voltages of all three combines.
Explanation:
During the data transmission there are chances that the data bits in the frame might get corrupted. This will require the sender to re-transmit the frame and hence it will increase the re-transmission overhead. By considering the scenarios given below, you have to choose whether the packets should be encapsulated in a single frame or multiple frames in order to minimize the re-transmission overhead.
Justify your answer with one valid reason for both the scenarios given below.
Scenario A: Suppose you are using a network which is very prone to errors.
Scenario B: Suppose you are using a network with high reliability and accuracy.
1. Based on Scenario A, multiple frames will minimize re-transmission overhead and should be preferred in the encapsulation of packets.
2. Based on Scenario B, the encapsulation of packets should be in a single frame because of the high level of network reliability and accuracy.
Justification:
There will not be further need to re-transmit the packets in a highly reliable and accurate network environment, unlike in an environment that is very prone to errors. The reliable and accurate network environment makes a single frame economically better.
Encapsulation involves the process of wrapping code and data together within a class so that data is protected and access to code is restricted.
With encapsulation, each layer:
provides a service to the layer above itcommunicates with a corresponding receiving nodeThus, in a reliable and accurate network environment, single frames should be used to enhance transmission and minimize re-transmission overhead. This is unlike in an environment that is very prone to errors, where multiple frames should rather be used to minimize re-transmission overhead.
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the molecule of magnet are independent _____________
Answer:
The first is the electric field, which describes the force acting on a stationary charge and gives the component of the force that is independent of motion. The magnetic field, in contrast, describes the component of the force that is proportional to both the speed and direction of charged particles.
OBJECTI
1. The motion of a liquid inside a U-tube is an
example of what type of motion?
a. Simple Harmonic c. Random
b.Rectilinear
d. Circular
Answer:
option A
Explanation:
simple harmonic motion
Answer:
random motion I think not sure
If a marathon runner runs 9.5 miles in one direction, 8.89 miles in another direction, and 2.333 miles in a third direction, how much distance did the runner run?
We have that the total distance covered by the runner is
[tex]d_t=20.723miles[/tex]
The total distance covered by the runner is a sum of all miles covered by the runner
Therefore
With
[tex]d_t[/tex]=Total distance
[tex]d_t=d_1+d_2+d_3\\\\d_t=9.5+8.89+2.333[/tex]
[tex]d_t=20.723miles[/tex]
in conclusion
The total distance covered by the runner is
[tex]d_t=20.723miles[/tex]
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A block weighting 400kg rests on a horizontal surface and support on top of it another block of weight 100kg placed on the top of it as shown. The block w2 is attached to avertical wall by a string 6m long. If the coefficient of friction between all surface is 0.25 and the system is in equilibrium find the magnitude of the horizontal force applied to the lower block
The horizontal force applied to the block is approximately 1,420.84 N
The known parameters;
The mass of the block, w₁ = 400 kg
The orientation of the surface on which the block rest, w₁ = Horizontal
The mass of the block placed on top of the 400 kg block, w₂ = 100 kg
The length of the string to which the block w₂ is attached, l = 6 m
The coefficient of friction between the surface, μ = 0.25
The state of the system of blocks and applied force = Equilibrium
Strategy;
Calculate the forces acting on the blocks and string
The weight of the block, W₁ = 400 kg × 9.81 m/s² = 3,924 N
The weight of the block, W₂ = 100 kg × 9.81 m/s² = 981 N
Let T represent the tension in the string
The upward force from the string = T × sin(θ)
sin(θ) = √(6² - 5²)/6
Therefore;
The upward force from the string = T×√(6² - 5²)/6
The frictional force = (W₂ - The upward force from the string) × μ
The frictional force, [tex]F_{f2}[/tex] = (981 - T×√(6² - 5²)/6) × 0.25
The tension in the string, T = [tex]F_{f2}[/tex] × cos(θ)
∴ T = (981 - T×√(6² - 5²)/6) × 0.25 × 5/6
Solving, we get;
[tex]T = \dfrac{5886}{\sqrt{6^2 - 5^2} + 28.8} \approx 183.27[/tex]
[tex]Frictional \ force, F_{f2} = \left (981 - \dfrac{5886}{\sqrt{6^2 - 5^2} + 28.8} \times \dfrac{\sqrt{6^2 - 5^2} }{6} \times 0.25 \right) \approx 219.92[/tex]
The frictional force on the block W₂, [tex]F_{f2}[/tex] ≈ 219.92 N
Therefore;
The force acting the block w₁, due to w₂ [tex]F_{w2}[/tex] = 219.92/0.25 ≈ 879.68
The total normal force acting on the ground, N = W₁ + [tex]\mathbf{F_{w2}}[/tex]
The frictional force from the ground, [tex]\mathbf{F_{f1}}[/tex] = N×μ + [tex]\mathbf{F_{f2}}[/tex] = P
Where;
P = The horizontal force applied to the block
P = (W₁ + [tex]\mathbf{F_{w2}}[/tex]) × μ + [tex]\mathbf{F_{f2}}[/tex]
Therefore;
P = (3,924 + 879.68) × 0.25 + 219.92 ≈ 1,420.84
The horizontal force applied to the block, P ≈ 1,420.84 N
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Although human beings have been able to fly hundreds of thousands of miles into outer space, getting inside the earth has proven much more difficult. The deepest mines ever drilled are only about 10 miles deep. To illustrate the difficulties associated with such drilling, consider the following: The density of steel is about 7900 kilograms per cubic meter, and its breaking stress, defined as the maximum stress the material can bear without deteriorating, is about 2.0×1092.0×109 pascals. What is the maximum length of a steel cable that can be lowered into a mine? Assume that the magnitude of the acceleration due to gravity remains constant at 9.8 meters per second per second.
Use two significant figures in your answer, expressed in kilometers.
Answer:
26 km
Explanation:
Let's say our "cable" has a cross section of 1 m²
Then each meter of cable would weight 7900(9.8) = 77420 N
A Pascal is a Newton per square meter
2 x 10⁹ / 77420 = 25840 m or about 26 km or about 16 miles
A rod of length L and electrical resistance R moves through a constant uniform magnetic field ; both the magnetic field and the direction of motion are parallel to the rod. The force that must be applied by a person to keep the rod moving with constant velocity is:
Answer:
don't know what class are you you are using which mobile or laptop
A 1.50 kg book is sliding along a rough horizontal surface. At point A it is moving at 3.21 m/s , and at point B it has slowed to 1.25 m/s .
Part A
How much work was done on the book between A and B ?
Part B
If -0.750J of work is done on the book from B to C , how fast is it moving at point C ?
Part C
How fast would it be moving at C if 0.750J of work were done on it from B to C ?
I assume friction is the only force acting on the book as it slides.
(A) By the work-energy theorem, the total work performed on the book as it slides is equal to the change in its kinetic energy:
W = ∆K
W = 1/2 (1.50 kg) (1.25 m/s)² - 1/2 (1.50 kg) (3.21 m/s)²
W ≈ -6.56 J
(B) Using the work-energy theorem again, the speed v of the book at point C is such that
-0.750 J = 1/2 (1.50 kg) v ² - 1/2 (1.50 kg) (1.25 m/s)²
==> v = 0.750 m/s
(C) Take the left side to be positive, then solve again for v.
0.750 J = 1/2 (1.50 kg) v ² - 1/2 (1.50 kg) (1.25 m/s)²
==> v ≈ 1.60 m/s
Why can't cars be constructed that can magnetically levitate in Earth's magnetic field?
a. The current needed would be excessive.
b. The current within the car must flow in a complete circuit, about half Of which would exert forces against levitation.
c. Magnetic forces are always very weak.
d. Earth's magnetic field is in the wrong direction.
how can you relazie a perfect balck body in pratice
5. A body falls freely from rest. It covers as much distance in the last second of its
motion as covered in the first three seconds. The body has fallen for a time of:
B) 5s
C) 7s
D) 9s
A) 35
Answer:
B 5s
Explanation:
Because of the Displacement in the nth second of the free fall is
Snth=21g(t12−t22)
Given that (tn−tn−1)=1
Displacement in 3 seconds of the free fall
S=21gt2
S=21×10×32
S=45m
Given that: Snth=45
On solving that we get:
t1=5sec
it is easier to drag a stone than to kick it?why.
Answer:
you are going to expend energy to give a lot of velocity (and momentum) to your foot in order to transfer it the stone air drag this time the kicking speed is for superior to walking speed.
Thank You
A car accelerates from 0 m/s to 25 m/s in 5 seconds. What is the average acceleration of the car.
Answer:
5 m/s I hope it will help you
Explanation:
mark me as a brainlist answer
Describe sound and record
Answer:
record is information created, received and maintained as evidence and information by an organization or person.in simpler terms it's a collection of of fields probably of different data types.
sound is however something loud or soft.which can be defined as vibrations that travel through the air or another medium.
I hope this helps
what is the major difference between the natural frequency and the damped frequency of oscillation.
Answer:
This causes the amplitude of the oscillation to decay over time. The damped oscillation frequency does not equal the natural frequency. Damping causes the frequency of the damped oscillation to be slightly less than the natural frequency
local unit of measurement of length : confined to a particular place cise Choose the best answer from the given alterna MKS system stands for ........ i. mass, kilogram and second ii. metre, kilogram and second iii. metre, kilometre and second iv. metre, kilogram and standard
MKS stands for metre , kilogram and second
Which columns are mislabeled?
Answer:
first order date and most recent order date
Explanation:
it was switched. column 5 should be most recent order date because it's 2020 while column 6 should be first order date because it was in 2019
What is the average velocity if the initial velocity of an object is 19 mph and the final velocity of 75 mph ?
Answer:
Hi I hope this is correct!
Explanation:
To find average velocity you can use the formula av = (v1 + v2) / 2
*I converted everything into m/s because that it usually the measurement for velocity*
v1 = initial velocity = 8.49376 m/s , v2 = final velocity = 33.528 m/s
av = 8.49376 + 33.528 / 2
= 21.01088 m/s
*If you were required to leave the final answer in mph here it is
av = 19 + 75 / 2
= 47 mph
Hope this helps! Best of luck <3
Explanation:
hope it helps you
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A 25g rock is rolling at a speed of 5 m/s. What is the kinetic energy of the rock?
Answer:
The answer is 312.5j
Explanation:
The kinetic energy (KE):
KE=1/2*m*v^2
M= mass of the object
v= velocity of the object
We have;
m=25g
v=5m/s
KE=1/2*25g*5^2m/s
KE =312.5j
Using this information...
Determine the velocity of the pebble as it passes over the top of the tree.
[tex]19.2\:\text{m/s}[/tex]
Explanation:
At the top of the tree, the velocity of the pebble is purely horizontal so we can calculate it as
[tex]v_{y} = v_{0y} = v_0\cos 40° = (25\:\text{m/s})(0.766)[/tex]
[tex]\:\:\:\:\:= 19.2\:\text{m/s}[/tex]
The height of the tree is approximately 12.5 meters when velocity of the pebble as it passes over the top of the tree.
Let's calculate the height of the tree step by step:
Given:
Initial velocity (v0) = 25 m/s
Launch angle (θ) = 40° above the horizontal
Time after launch (t) = 2 seconds
Acceleration due to gravity (g) = -9.8 m/s² (negative because it acts downward)
Step 1: Calculate the vertical component of the initial velocity (Vy):
Vy = v0 * sin(θ)
Vy = 25 m/s * sin(40°)
Vy ≈ 25 m/s * 0.6428 ≈ 16.07 m/s (rounded off to two decimal places)
Step 2: Calculate the vertical displacement (change in height) of the pebble after 2 seconds:
d = vot + (1/2)at²
d = (16.07 m/s) * (2 s) + (1/2) * (-9.8 m/s²) * (2 s)²
d ≈ 32.14 m - 19.6 m
d ≈ 12.54 m (rounded off to two decimal places)
Step 3: The height of the tree is equal to the vertical displacement of the pebble:
Height of the tree ≈ 12.54 m ≈ 12.5 m (rounded off to one decimal place)
The height of the tree is approximately 12.5 meters.
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How is centripetal force affected if an object increases its speed?
A. Decreases
B. Increases
C. Cut in half
D. No effect
Answer:
B. Increases.
Explanation:
[tex]{ \bf{F = \frac{m {v}^{2} }{r} }} \\ { \bf{F \: \alpha \: {v}^{2} }}[/tex]
Keeping mass, and radius constant, speed or velocity is directly proportioanal to centripetal force.
,In order to increase the speed of an object on a circular path, YOU have to increase the centripetal force acting on it.
A circular loop of wire 10 cm in radius carries a current of 20 A. The axial magnetic field 15 cm from the center of the loop is approximately:
a. 37 mu-T
b. 13 mu-T
c. 21 mu-T
d. 41 mu-T
e. 18 mu-T]
Answer:
ans is c
Explanation:
chk photo
The axial magnetic field 15 cm from the center of the loop is approximately 21 μT. Hence, option (c) is correct.
What is magnetic field?In the vicinity of a magnet, an electric current, or a shifting electric field, there is a vector field called a magnetic field where magnetic forces can be seen. Electric charges in motion and the intrinsic magnetic moments of elementary particles connected to the fundamental quantum characteristic known as spin create a magnetic field.
Given parameters:
Radius of the circular loop: r = 10 cm = 0.10 m.
Current passing through the loop: I = 20 A.
Axial distance of the point: z = 15 cm = 0.15 m.
Hence, the axial magnetic field at that point:
B = (μ₀/4π) 2πR²I/(z²+R²)^(3/2)
By putting these values, we get B = 21 × 10⁻⁶ T = 21 μT.
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A man is driving a car at speed 25m/s. calculate the distance covered by it in one hour.
Answer:
6.94 km/hr
Explanation:
m/s to km/hr -> Multiply by 18/5
25/(18/5)
=> 25 x 5/18
=> 125/18 km/hr
=> 6.94 km/hr
Answer: 90,000 m = 90 km
Explanation:
Given information
Time = 1 hour
Speed = 25 m/s
Given expression deducted from the given information
Distance = speed × time
Convert units of time
1 hour = 60 minutes
1 minute = 60 seconds
1 hour = 60 × 60 = 3600 seconds
Substitute values into the expression
Distance = 25 × 3600
Simplify by multiplication
Distance = [tex]\boxed{90,000 m=90km}[/tex]
Hope this helps!! :)
Please let me know if you have any questions
A block of mass 2 kg starts from rest at the top of a friction quarter of a circle of radius R. The block then slides over frictionless curved surface in the shape of a eventually comes to rest 8 m from the beginning s a horizontal rough surface where e of the horizontal surface. The coefficient kinetic friction between the rough surface and the block is 0.4 . determine the acceleration of the block over the rough surface length 8m
The acceleration of the block over the rough surface is 1.22625 m/s²
The process through which the acceleration is obtained is presented as follows of approach to
The given parameters are;
Mass of block, m = 2 kg
Nature of the surface of the quarter circle = Frictionless
The length of the horizontal, d = 8 m
The coefficient of friction of the horizontal surface, μ = 0.4
The unknown parameter;
The acceleration of the block over the rough surface
Method;
Find the work done by friction to stop the block and divide the result by the mass of the block
The work done by friction, [tex]W_f[/tex] = (Force of friction) × (Distance the block moves on the rough surface before coming to rest)
[tex]\mathbf{W_f}[/tex] = [tex]\mathbf{F_f}[/tex] × d
[tex]F_f[/tex] = Normal reaction of surface on block, [tex]N_r[/tex] × μ
Normal reaction on block, [tex]\mathbf{N_r}[/tex] = Weight of block
[tex]\mathbf{N_r}[/tex] ≈ 2 kg × 9.81 m/s² = 19.62 N
Therefore;
The work done by friction [tex]\mathbf{W_f}[/tex] = [tex]\mathbf{F_f}[/tex] × d = [tex]\mathbf{N_r}[/tex] × μ × d
[tex]\mathbf{W_f}[/tex] = 19.62 N × 0.4 × 8 m = 62.784 J
The work done by the block, W = Force, F × d
Force, F = m × a
Where;
a = The acceleration of the block
According to the principle of conservation of energy, we have;
[tex]\mathbf{W_f}[/tex] = W
∴ 19.62 J = 2 kg × a × 8 m
a = 19.62/(2 kg × 8 m) = 1.22625 m/s²
The acceleration of the block over the rough surface, a = 1.22625 m/s²
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what is the area of velocity time graph
