Answer:
6 bits
Explanation:
The quality of digitized signal can be improved by reducing quantizing error. This is done by increasing the number of amplitude levels, thereby minimizing the difference between the levels and hence producing a smoother signal.
Also, Sampling frequently (also known as oversampling) can help in improving signal quality.
To get the number of bits, we use:
2ⁿ = amplitude level
where n is the number of bits.
Given an amplitude level of 64, hence:
2ⁿ = 64
2ⁿ = 2⁶
n = 6 bits
A heavy truck moving with 20 km/hr hits a car at rest. A physics student argued that
the maximum velocity the car suddenly gains is 40 km/hr. Do you agree with it?
Explain with necessary theory
Answer:
Yes
Explanation:
speed of truck = 20 km/h
Initially the car at rest.
maximum velocity of car = 40 km/h
When the truck and the car collide, the momentum of the truck transferred to car.
So, the car can attain the speed of 40 km/h.
A long, straight, vertical wire carries a current upward. Due east of this wire, in what direction does the magnetic field point
The magnetic field of the wire will be directed towards west. Using right thumb rule one can get the direction of field lines.
What is significant about the primary colors of pigments?
They can be mixed together to make almost any other color.
Any two primary colors of pigments combine to make white pigment.
Each primary color of pigment absorbs all other colors.
Any two primary colors of pigments combine to make black pigment.
Answer:
They can be mixed together to make almost any other color.
Explanation:
All the three primary colors can mix to form white color.
Blue and red mix to form a black color.
1.- Que distancia recorrió una carga de 2,5x10-6 coul, generando así un campo eléctrico de 55new/coul.
Answer:
r = 20.22 m
Explanation:
Given that,
Charge,[tex]q=2.5\times 10^{-6}\ C[/tex]
Electric field, [tex]E=55\ N/C[/tex]
We need to find the distance. We know that, the electric field a distance r is as follows :
[tex]E=\dfrac{kq}{r^2}\\\\r=\sqrt{\dfrac{kq}{E}}\\\\r=\sqrt{\dfrac{9\times 10^9\times 2.5\times 10^{-6}}{55}}\\\\r=20.22\ m[/tex]
So, the required distance is 20.22 m.
How many loops are in this circuit?
I see six (6) loops.
I attached a drawing to show where I get six loops from.
Question 8 of 10
What was the name of the book that Ibn al-Haytham wrote?
A. Weather and Air Flow
B. Book of Optics
C. Light and Vision
D. Book of Sound
Answer:b
Explanation:
If 5kg Stone and 1kg stone throw the from the building which will land more fa ster and why?
Answer:
Both stones will land at the same time because both stones will fall with the same acceleration through the same height.
Explanation:
We are given that
Mass of stone ,m1=5 Kg
Mass of stone, m2=1 kg
We have to find which stone more faster will land and why.
[tex]h=u+\frac{1}{2}gt^2[/tex]
Initial velocity of both stones=0
[tex]h=\frac{1}{2}gt^2[/tex]
[tex]t^2=\frac{h}{g}[/tex]
[tex]t=\sqrt{\frac{h}{g}}[/tex]
[tex]t_1=t_2=\sqrt{\frac{h}{g}}[/tex]
Because both stones are thrown from the same height.
Both stones will land at the same time because both stones will fall with the same acceleration through the same height and the acceleration does not depend of its mass.
Find the starting pressure of CCl4 at this temperature that produces a total pressure of 1.1 atm at equilibrium. Express the pressure in atmospheres to three significant figures.
The complete question is as follows: At 700 K, [tex]CCl_{4}[/tex] decomposes to carbon and chlorine. The Kp for the decomposition is 0.76.
Find the starting pressure of [tex]CCl_{4}[/tex] at this temperature that will produce a total pressure of 1.1 atm at equilibrium.
Answer: The starting pressure of [tex]CCl_{4}[/tex] is 0.79 atm.
Explanation:
The equation for decomposition of [tex]CCl_{4}[/tex] is as follows.
[tex]CCl_{4}(g) \rightleftharpoons C(s) + 2Cl_{2}(g)[/tex]
Let us assume that initial concentration of [tex]CCl_{4}[/tex] is 'a'. Hence, the initial and equilibrium concentrations will be as follows.
[tex]CCl_{4}(g) \rightleftharpoons C(s) + 2Cl_{2}(g)[/tex]
Initial: a 0 0
Equilibrium: (a - x) 0 2x
Total pressure = (a - x) + 2x = a + x
As it is given that the total pressure is 1.1 atm.
So, a + x = 1.1
a = 1.1 - x
Now, expression for equilibrium constant for this equation is as follows.
[tex]K_{p} = \frac{P^{2}_{Cl_{2}}}{P_{CCl_{4}}}\\0.76 = \frac{(2x)^{2}}{(a - x)}\\0.76 = \frac{4x^{2}}{1.1 - x - x}\\0.76 = \frac{4x^{2}}{1.1 - 2x}\\x = 0.31 atm[/tex]
Hence, the value of 'a' is calculated as follows.
a + x = 1.1 atm
a = 1.1 atm - x
= 1.1 atm - 0.31 atm
= 0.79 atm
Thus, we can conclude that starting pressure of [tex]CCl_{4}[/tex] is 0.79 atm.
prove mathematically :
1. v = u + at
2. s = ut+1*2 at
Answer:
a.v=u+v/2
a.v=s/t
combining two equation we get,
u+v/2=s/t
(u+v)t/2=s
(u+v)t/2=s
{u+(u+at)}t/2=s
(u+u+at)t/2=s
(2u+at)t/2=s
2ut+at^2/2=s
2ut/2+at^2/2=s
UT +1/2at^2=s
proved
a=v-u/t
at=v-u
u+at=v
The Lamborghini Huracan has an initial acceleration of 0.85g. Its mass, with a driver, is 1510 kg. If an 80 kg passenger rode along, what would the car's acceleration be?
Answer:
7.9 [tex]\frac{m}{s^{2} }[/tex]
Explanation:
Take the fact that mass is inversely proportional to accelertation:
m ∝ a
Therefore m = a, but because we are finding the change in acceleration, we would set our problem up to look more like this:
[tex]\frac{m_{1} }{m_{2} } = \frac{a_{2} }{a_{1} } \\[/tex]
Using algebra, we can rearrange our equation to find the final acceleration, [tex]a_{2}[/tex]:
[tex]a_{2} = \frac{a_{1}*m_{1} }{m_{2} } \\[/tex]
Before plugging everything in, since you are being asked to find acceleration, you will want to convert 0.85g to m/s^2. To do this, multiply by g, which is equal to 9.8 m/s^2:
0.85g * 9.8 [tex]\frac{m }{s^{2} }[/tex] = 8.33 [tex]\frac{m }{s^{2} }[/tex]
Plug everything in:
7.9 [tex]\frac{m }{s^{2} }[/tex] = [tex]\frac{ 8.33\frac{m}{s^{2} }*1510kg }{1590kg}[/tex]
(1590kg the initial weight plus the weight of the added passenger)
Find the volume of cuboid of side 4cm. Convert it in SI form
Answer:
0.000064 cubic meters.
Explanation:
Given the following data;
Length of side = 4 centimeters
Conversion:
100 centimeters = 1 meters
4 cm = 4/100 = 0.04 meters
To find the volume of cuboid;
Mathematically, the volume of a cuboid is given by the formula;
Volume of cuboid = length * width * height
However, when all the sides are equal the formula is;
Volume of cuboid = L³
Volume of cuboid = 0.04³
Volume of cuboid = 0.000064 cubic meters.
I need help with this physics question.
Answer:
5.04 m
Explanation:
You are told that the homeowner wants to increase their fences by 34 percent meaning Original+ 34 percent. If the original is 100 percent, then the new fence size will be 134 % of the original. You are given the original which is 3.76 meters, to find new fence size 1.34 * 3.76m to get 5.0384 meters, rounded to 5.04 m.
Answer:
5.0384m
Explanation:
% increase = 100 x (Final - Initial / | initial | )
( |~~| Bars indicate absolute value since you can't have a negative height)
Action and reaction are equal in magnitude and opposite in direction.Then Why do not balance each other
Answer:
Action and reaction are equal in magnitude and opposite in direction but they do not balance each other because they act on different objects so they don't cancel each other out.
hope this will help you more
A conducting sphere of radius 5.0 cm carries a net charge of 7.5 µC. What is the surface charge density on the sphere?
Answer:
[tex]\sigma=0.014\ C/m^2[/tex]
Explanation:
Given that,
The radius of sphere, r = 5 cm = 0.05 m
Net charge carries, q = 7.5 µC = 7.5 × 10⁻⁶ C
We need to find the surface charge density on the sphere. Net charge per unit area is called the surface charge density. So,
[tex]\sigma=\dfrac{7.5\times 10^{-6}}{\dfrac{4}{3}\pi \times (0.05)^3}\\\\=0.014\ C/m^2[/tex]
So, the surface charge density on the sphere is [tex]0.014\ C/m^2[/tex].
You may have been surprised to learn that Olympic gold medals are not made from solid gold, but instead have a coating of • Saved gold on the outside.
To see a possible reason why, determine the value of the medal the size (not mass) of the Olympic gold medal if it were made of solid gold. Hint: As of mid-2018, the cost of gold is about $40 per gram.
Answer:
A gold medal has the (minimum) dimensions of:
diameter = 60mm
thickness = 3mm
So we will work with those dimensions.
The medal is then a cyinder of diameter
D = 60mm = 6cm
and height:
H = 3mm = 0.3cm
Remember that the volume of a cylinder is:
V = pi*(D/2)^2*H
where pi = 3.14
Then the volume of a medal is:
V = 3.14*(6cm/3)^2*0.3cm = 3.768 cm^3
The density of the gold in g/cm^3 is:
d = 19.3 g/cm^3
And remember that:
density = mass/volume
So, if the volume is 3.768 cm^3
Then the mass will be:
mass = density*volume = 19.3 g/cm^3*3.768 cm^3 = 72.7 g
So, a single gold medal would weight 72.7 grams
And each gram of gold costs $40
Then the total cost of the gold medal would be:
value = $40*72.7 = $2,908
Now, if yo think that in the Olympics there are 35 sports (a lot with a large number of players) and near 50 disciplines, they need a lot of gold medals.
And each gold medal costs $2,908
So the total cost (only for the gold medals, ignoring the others) would be to high.
This is why the gold medals are made mostly of silver.
Characteristics or properties of matter or energy that can be measured
Answer:
Physical properties are properties that can be measured or observed without changing the chemical nature of the substance. Some examples of physical properties are:
color (intensive)
density (intensive)
volume (extensive)
mass (extensive)
boiling point (intensive): the temperature at which a substance boils
melting point (intensive): the temperature at which a substance melts
Explanation:
Describe an imaginary process that satisfies the second law but violates the first law of thermodynamics.
Answer:
Explanation:
First last of thermodynamics, just discusses the changes that a system is undergoing and the processes involved in it. It explains conservation of energy for a system undergoing changes or processes.
Second law of thermodynamics helps in defining the process and also the direction of the processes. It tells about the possibility of a process or the restriction of a process. It states that the entropy of a system always increases.
For this to occur the energy contained by a body has to diminish without converting to work or internal energy. So imagine a machine which works with less than efficiency, this means there are losses but they don’t show up anywhere. But the energy is obtained from a higher energy source to lower.
The easy way to do this is with an imaginary device that extracts zero-point energy to heat a quantity of gas. Energy is being created, so the first law is violated, and the entropy of the system is increasing as the gas heats up.
First law is violated since the energy conversion don't apply but the direction of work is applied so second law is satisfied.
An aircraft has a glide ratio of 12 to 1. (Glide ratio means that the plane drops 1 m in each 12 m it travels horizontally.) A building 45 m high lies directly in the glide path to the runway. If the aircraft dears the building by 12 m, how far from the building does the aircraft touch down on the runway
The aircraft is 12 meters higher than the building so it is at 45 + 12 = 57 meters high.
For every 12 meters it travels it drops 1 m.
Divide the height by 12 to find the distance it travels:
57 / 12 = 4.75
It touches down 4.75 meters from the building.
The building is 684 meters away from the aircraft touching down on the runway.
What are trigonometric functions?A right-angled triangle's side ratios are the easiest way to express a function of an arc or angle, such as the sine, cosine, tangent, cotangent, secant, or cosecant. These functions are known as trigonometric functions.
As given in the problem an aircraft has a glide ratio of 12 to 1. (Glide ratio means that the plane drops 1 m in each 12 m it travels horizontally.) A building 45 m high lies directly in the glide path to the runway. If the aircraft clears the building by 12 m,
the total height of the aircraft when it clears the building = 45 +12
the total height of the aircraft when it clears the building is 57 meters
It is given that the Glide ratio is 12:1,
The distance of the building from touch down on the runway = 12 ×57
The distance of the building from the touch-down on the runway is 684 meters.
Thus, the building is 684 meters away from the aircraft touching down on the runway.
Learn more about the trigonometric functions here,
brainly.com/question/14746686
#SPJ2
You drop two balls of equal diameter from the same height at the same time. Ball 1 is made of metal and has a greater mass than ball 2, which is made of wood. The upward force due to air resistance is the same for both balls. Is the drop time of ball 1 greater than, less than, or equal to the drop time of ball 2? Explain why
Answer:
The drop time ball 1 is less than the drop time of ball 2. A further explanation is provided below.
Explanation:
The net force acting on the ball will be:
⇒ [tex]F_{net}=mg-F_r[/tex]
Here,
F = Force
m = mass
g = acceleration
Now,
According to the Newton's 2nd law of motion, we get
⇒ [tex]F_{net} = ma[/tex]
To find the value of "a", we have to substitute "[tex]F_{net}=ma[/tex]" in the above equation,
⇒ [tex]ma=mg-F_r[/tex]
⇒ [tex]a=g-\frac{F_r}{m}[/tex]
We can see that, the acceleration is greater for the greater mass of less for the lesser mass. Thus the above is the appropriate solution.
Answer:
Both the ball takes equal time to reach to the ground.
Explanation:
Two balls of same diameter
Let the height is h.
Mass of ball 1 is more than the mass of ball 2.
The second equation of motion is
[tex]h = u t +0.5 gt^2[/tex]
Here, the buoyant force due to air is same. So, the time of fall is independent of the mass.
So, both the ball takes equal time to reach to the ground.
Which level of government relies the most on income tax?
OA.
federal
state
OC.
local
Answer:
Its the Federal government
Please show steps as to how to solve this problem
Thank you!
Explanation:
Let x = distance of [tex]F_1[/tex] from the fulcrum and let's assume that the counterclockwise direction is positive. In order to attain equilibrium, the net torque [tex]\tau_{net}[/tex] about the fulcrum is zero:
[tex]\tau_{net} = -F_1x + F_2d_2 = 0[/tex]
[tex] -m_1gx + m_2gd_2 = 0[/tex]
[tex]m_1x = m_2d_2[/tex]
Solving for x,
[tex]x = \dfrac{m_2}{m_1}d_2[/tex]
[tex]\:\:\:\:=\left(\dfrac{105.7\:\text{g}}{65.7\:\text{g}} \right)(13.8\:\text{cm}) = 22.2\:\text{cm}[/tex]
a vechile having a mass of 500kg is moving with a speed of 10m/s.Sand is dropped into it at the rate of 10kg/min.What force is needed to keep the vechile moving with uniform speed
Answer:
1.67 N
Explanation:
Applying,
F = u(dm/dt)+m(du/dt)................ Equation 1
Where F = force, m = mass of the vehicle, u = speed.
Since u is constant,
Therefore, du/dt = 0
F = u(dm/dt)............... Equation 2
From the question,
Given: u = 10 m/s, dm/dt = 10 kg/min = (10/60) kg/s
Substitute these values into equation 2
F = 10(10/60)
F = 100/60
F = 1.67 N
You simultaneously release two balls: one you throw horizontally, and the other you drop straight down. Which one will reach the ground first? Why?
(a) The ball dropped straight down lands first, since it travels a shorter distance.
(b) Neither. Their vertical motion is the same, so they will reach the ground at the same time.
(c) It depends on the mass of the balls—the heavier ball falls faster so lands first
Answer:
Option B.
Explanation:
Remember that we can think on any movement as a sum of a movement in the y-axis, the movement in the x-axis, and the movement in the z-axis. And these are not related, this means that, for example, the movement in x does not affect the movement in y.
So, when we analyze the problem of "how long takes an object to hit the ground"
We do not care for the horizontal motion of the object, we only care for the vertical motion of the object.
So, if an object is dropped, and another has a given initial velocity in the x-axis, in both cases the initial velocity in the y-axis will zero.
And in both cases, the only vertical force acting on the balls will be the gravitational force (so both objects will have the same vertical acceleration and the same vertical initial velocity) with this, we already know that the vertical motion of both objects will be exactly the same.
So, both objects will hit the ground at the same time.
(notice that here we are ignoring things like air resistance and other complex forces)
So here the correct option is b: Neither. Their vertical motion is the same, so they will reach the ground at the same time.
A commuter backs her car out of her garage with an acceleration of . (a) How long does it take her to reach a speed of 2.00 m/s
Question: A commuter backs her car out of her garage with an acceleration of 1.4 m/s² (a) How long does it take her to reach a speed of 2.00 m/s
Answer:
1.43 s
Explanation:
Applying,
a = (v-u)/t........... Equation 1
Where a = acceleration, v = final velocity, u = initial velocity, t = time
make t the subject of the equation
t = (v-u)/a........... Equation 2
From the question,
Given: v = 2 m/s, u = 0m/s (from rest), a = 1.4 m/s²
Substitute into equation 2
t = (2-0)/1.4
t = 1.43 s
Al and Ben are on roller skates and Al rolls into Ben. Al exerts a force of 30 N on Ben when they
collide. Explain what force Ben exerts on AI.
Answer:
Reaction force
Explanation:
Newton´s 3rd law says that every force exerted in nature has an equal and opposite force.
For example here, when Al exerts force on Ben, Ben exerts the same amount of force (30N) on Al.
Al exerts the action force and Ben exerts the reaction force.
As a roller coaster car crosses the top of a 48.01-m-diameter loop-the-loop, its apparent weight is the same as its true weight. What is the car's speed at the top?
Answer:
The speed of the car, v = 21.69 m/s
Explanation:
The diameter is = 48.01 m
Therefore, the radius of the loop R = 24.005 m
Weight at the top is n = mv^2/R - mg
Since the apparent weight is equal to the real weight.
So, mv^2/R - mg = mg
v = √(2Rg)
v = √[2(24.005 m)(9.8 m/s^2)]
The speed of the car, v = 21.69 m/s
Answer:
The speed is 15.34 m/s.
Explanation:
Diameter, d = 48.01 m
Radius, R = 24.005 m
Let the speed is v and the mass is m.
Here, the weight of the car is balanced by the centripetal force.
According to the question
[tex]m g = \frac{mv^2}{R}\\\\v =\sqrt{24.005\times9.8}\\\\v = 15.34 m/s[/tex]
ACCORDING TO NEWTON'S THIRD LAW EVERY ACTION HAS EQUAL AND OPPOSITE REACTION BUT THEN WHY DON'T WE FLY WHEN WE FART??
Answer:
Your fart only has so much force, not nearly enough to launch you into oblivion. Your fart and you still exert a force onto each other, so I guess, hypothetically, you could fly if you really, really try hard enough. Just make sure you don't try too hard and prolapse as a result :)
NEED HELP ASAP. Please show all work.
A point on a rotating wheel (thin hoop) having a constant angular velocity of 200 rev/min, the wheel has a radius of 1.2 m and a mass of 30 kg. ( I = mr2 ).
(a) (5 points) Determine the linear acceleration.
(b) (4 points) At this given angular velocity, what is the rotational kinetic energy?
Answer:
Look at work
Explanation:
a) I am not sure if you want tangential or centripetal but I will give both
Centripetal acceleration = r*α
Since ω is constant, α is 0 so centripetal acceleration is 0m/s^2
Tangential acceleration = ω^2*r
convert 200rev/min into rev/s
200/60= 10/3 rev/s
a= 100/9*1.2= 120/9= 40/3 m/s^2
b) Rotational Kinetic Energy = 1/2Iω^2
I= mr^2
Plug in givens
I= 43.2kgm^2
K= 1/2*43.2*100/9=2160/9=240J
If the source moves, the wavelength of the sound in front of the direction of motion is____than the wavelength behind the direction of motion.
a. the same.
b. smaller than.
c. unrealted to.
d. larger then.
Answer:
B. Smaller than
Explanation:
This question is from the Doppler effect. As the object which is in motion goes off from the other, there's a reduction in the frequency. This is due to the fact that successive soundwave get to be longer. So that the pitch will then be lowered. When the person observing moves towards what is making the sound, each soundwave that follows gets faster than the previous.
High-speed stroboscopic photographs show that the head of a -g golf club is traveling at m/s just before it strikes a -g golf ball at rest on a tee. After the collision, the club head travels (in the same direction) at m/s. Find the speed of the golf ball just after impact.
The question is incomplete. The complete question is :
High-speed stroboscopic photographs show that the head of a 200 g golf club is traveling at 60 m/s just before it strikes a 50 g golf ball at rest on a tee. After the collision, the club head travels (in the same direction) at 40 m/s. Find the speed of the golf ball just after impact.
Solution :
We know that momentum = mass x velocity
The momentum of the golf club before impact = 0.200 x 60
= 12 kg m/s
The momentum of the ball before impact is zero. So the total momentum before he impact is 12 kg m/s. Therefore, due to the conservation of momentum of the two bodies after the impact is 12 kg m/s.
Now the momentum of the club after the impact is = 0.2 x 40
= 8 kg m/s
Therefore the momentum of the ball is = 12 - 8
= 4 kg m/s
We know momentum of the ball, p = mass x velocity
4 = 0.050 x velocity
∴ Velocity = [tex]$\frac{4}{0.050}$[/tex]
= 80 m/s
Hence the speed of the golf ball after the impact is 80 m/s.
