Answer:
a) Average speed is 24.04 m/s
b) the rms speed is 25.84 m/s
c) the most probable speed is 16 m/s
Explanation:
Given the data in the question;
a) Determine the average speed.
To determine the average speed, we simply divide total some of speed by number of particles;
Average speed = [(2×11 m/s)+(7×16 m/s)+(4×19 m/s)+(3×26 m/s)+(6×31 m/s)+(1×37 m/s)+(2×45 m/s)] / 25
= 601 / 25
= 24.04 m/s
Therefore, Average speed is 24.04 m/s
b) Determine the rms speed
we know that (rms speed)² = sum of square speed / total number of particles
so
(rms speed)² = [(2×11²)+(7×16²)+(4×19²)+(3×26²)+(6×31²)+(1×37²)+(2×45²)] / 25
(rms speed)² = 16691 / 25
(rms speed)² = 667.64
(rms speed) = √ 667.64
(rms speed) = 25.84 m/s
Therefore, the rms speed is 25.84 m/s
c) Determine the most probable speed.
Most particles (7) have velocity 16 m/s
i.e 7 is the maximum number of particle for a particular speed ,
Therefore, the most probable speed is 16 m/s
Like charges will exert a force of
a. positive
b. negative
c. attraction
d. repulsion
e. neutral
Answer:
D- Repulsion
Explanation:
A positively charged object will exert a repulsive force upon a second positively charged object.
Consider a swimmer that swims a complete round-trip lap of a 50 m long pool in 100 seconds. The swimmer's... average speed is 0 m/s and average velocity is 0 m/s. average speed is 0.5 m/s and average velocity is 0.5 m/s. average speed is 1 m/s and average velocity is 0 m/s. average speed is 0 m/s and average velocity is 1 m/s.What is the swimmers average speed and average velocity?
Answer:
average speed is 1 m/s and average velocity is 0 m/s.
Explanation:
Given that :
Length of round trip = 50 m
Time taken = 100 seconds
The average speed :
Total distance / total time taken
Length of complete round trip :
(50 + 50) m, total. Distance = 100 m
100 / 100 = 1m/s
The average velocity :
Total Displacement / total time taken
Total Displacement of round trip = end point - start point = 0
0 / 100 = 0
Average speed is 1 m/s and average velocity is 0 m/s.
The average speed is defined as the ratio of distance to time. Speed is a scalar quantity hence it does not take direction into account while velocity is a vector quantity hence it takes direction into account.
The speed is obtained from;
Speed = Distance/time = 2(50 m)/100 s = 1 m/s.
The velocity is 0 m/s since it is complete round-trip lap.
Learn more about speed: https://brainly.com/question/7359669
A ratio is another name for a decimal true or false
An RC car is carrying a tiny slingshot with a spring constant of 85 N/m at 0.2 m off the ground at 5.6 m/s. The sling shot is pulled back 3.5 cm from a relaxed state and shoots a 25 g steel pellet in the same direction the car is moving. What is the velocity of the steel pellet relative to the ground as it leaves the sling shot
Answer:
The velocity of the steel pellet relative to the ground when it leaves the sling shot is approximately 5.960 meters per second.
Explanation:
Let suppose that RC car-slingshot-steelpellet is a conservative system, that is, that non-conservative forces (i.e. friction, air viscosity) can be neglected. The velocity of the steel pellet can be found by means of the Principle of Energy Conservation and under the consideration that change in gravitational potential energy is negligible and that the RC car travels at constant velocity:
[tex]\frac{1}{2}\cdot (m_{C}+m_{P})\cdot v_{o}^{2} + \frac{1}{2}\cdot k\cdot x^{2} = \frac{1}{2}\cdot m_{C}\cdot v_{o}^{2} + \frac{1}{2}\cdot m_{P}\cdot v^{2}[/tex]
[tex]\frac{1}{2}\cdot m_{P}\cdot v_{o}^{2} + \frac{1}{2}\cdot k\cdot x^{2} = \frac{1}{2}\cdot m_{P}\cdot v^{2}[/tex]
[tex]m_{P}\cdot v_{o}^{2} + k\cdot x^{2} = m_{P}\cdot v^{2}[/tex]
[tex]v^{2} = v_{o}^{2} + \frac{k}{m_{P}}\cdot x^{2}[/tex]
[tex]v = \sqrt{v_{o}^{2}+\frac{k}{m_{P}}\cdot x^{2} }[/tex] (1)
Where:
[tex]v_{o}[/tex] - Initial velocity of the steel pellet, measured in meters per second.
[tex]v[/tex] - Final velocity of the steel pellet, measured in meters per second.
[tex]k[/tex] - Spring constant, measured in newtons per meter.
[tex]m_{P}[/tex] - Mass of the steel pellet, measured in kilograms.
[tex]m_{C}[/tex] - Mass of the RC car, measured in kilograms.
[tex]x[/tex] - Initial deformation of the spring, measured in meters.
If we know that [tex]v_{o} = 5.6\,\frac{m}{s}[/tex], [tex]k = 85\,\frac{N}{m}[/tex], [tex]m_{P} = 0.025\,kg[/tex] and [tex]x = 0.035\,m[/tex], then the velocity of the steel pellet relative to the ground when it leaves the sling shot is:
[tex]v = \sqrt{\left(5.6\,\frac{m}{s} \right)^{2}+\frac{\left(85\,\frac{N}{m} \right)\cdot (0.035\,m)^{2}}{0.025\,kg} }[/tex]
[tex]v \approx 5.960\,\frac{m}{s}[/tex]
The velocity of the steel pellet relative to the ground when it leaves the sling shot is approximately 5.960 meters per second.
A small sphere of reference-grade iron with a specific heat of 447 J/kg K and a mass of 0.515 kg is suddenly immersed in a water-ice mixture. Fine thermocouple wires suspend the sphere, and the temperature is observed to change from 15 to 14C in 6.35 s. The experiment is repeated with a metallic sphere of the same diameter, but of unknown composition with a mass of 1.263 kg. If the same observed temperature change occurs in 4.59 s, what is the specific heat of the unknown material
Answer:
The specific heat of the unknown material is 131.750 joules per kilogram-degree Celsius.
Explanation:
Let suppose that sphere is cooled down at steady state, then we can estimate the rate of heat transfer ([tex]\dot Q[/tex]), measured in watts, that is, joules per second, by the following formula:
[tex]\dot Q = m\cdot c\cdot \frac{T_{f}-T_{o}}{\Delta t}[/tex] (1)
Where:
[tex]m[/tex] - Mass of the sphere, measured in kilograms.
[tex]c[/tex] - Specific heat of the material, measured in joules per kilogram-degree Celsius.
[tex]T_{o}[/tex], [tex]T_{f}[/tex] - Initial and final temperatures of the sphere, measured in degrees Celsius.
[tex]\Delta t[/tex] - Time, measured in seconds.
In addition, we assume that both spheres experiment the same heat transfer rate, then we have the following identity:
[tex]\frac{m_{I}\cdot c_{I}}{\Delta t_{I}} = \frac{m_{X}\cdot c_{X}}{\Delta t_{X}}[/tex] (2)
Where:
[tex]m_{I}[/tex], [tex]m_{X}[/tex] - Masses of the iron and unknown spheres, measured in kilograms.
[tex]\Delta t_{I}[/tex], [tex]\Delta t_{X}[/tex] - Times of the iron and unknown spheres, measured in seconds.
[tex]c_{I}[/tex], [tex]c_{X}[/tex] - Specific heats of the iron and unknown materials, measured in joules per kilogram-degree Celsius.
[tex]c_{X} = \left(\frac{\Delta t_{X}}{\Delta t_{I}}\right)\cdot \left(\frac{m_{I}}{m_{X}} \right) \cdot c_{I}[/tex]
If we know that [tex]\Delta t_{I} = 6.35\,s[/tex], [tex]\Delta t_{X} = 4.59\,s[/tex], [tex]m_{I} = 0.515\,kg[/tex], [tex]m_{X} = 1.263\,kg[/tex] and [tex]c_{I} = 447\,\frac{J}{kg\cdot ^{\circ}C}[/tex], then the specific heat of the unknown material is:
[tex]c_{X} = \left(\frac{4.59\,s}{6.35\,s} \right)\cdot \left(\frac{0.515\,kg}{1.263\,kg} \right)\cdot \left(447\,\frac{J}{kg\cdot ^{\circ}C} \right)[/tex]
[tex]c_{X} = 131.750\,\frac{J}{kg\cdot ^{\circ}C}[/tex]
Then, the specific heat of the unknown material is 131.750 joules per kilogram-degree Celsius.
If a cyclist travels 30 km in 2 h, What is her average speed?
Answer:
15km/h
Explanation:
→ Speed = Distance ÷ Time
30 ÷ 2 = 15km/h
Two children, each with a mass of 25.4 kg, are at fixed locations on a merry-go-round (a disk that spins about an axis perpendicular to the disk and through its center). One child is 0.78 m from the center of the merry-go-round, and the other is near the outer edge, 3.14 m from the center. With the merry-go-round rotating at a constant angular speed, the child near the edge is moving with translational speed of 11.5 m/s.
a. What is the angular speed of each child?
b. Through what angular distance does each child move in 5.0 s?
c. Through what distance in meters does each child move in 5.0 s?
d. What is the centripetal force experienced by each child as he or she holds on?
e. Which child has a more difficult time holding on?
Answer:
a) ω₁ = ω₂ = 3.7 rad/sec
b) Δθ₁ = Δθ₂ = 18.5 rad
c) d₁ = 14.5 m d₂ = 57.5 m
d) Fc1 = 273.9 N Fc2 = 1069.8 N
e) The boy near the outer edge.
Explanation:
a)
Since the merry-go-round is a rigid body, any point on it rotates at the same angular speed.However, linear speeds of points at different distances from the center, are different.Applying the definition of angular velocity, and the definition of angle, we can write the following relationship between the angular and linear speeds:[tex]v = \omega*r (1)[/tex]
Since we know the value of v for the child near the outer edge, and the value of r for this point, we can find the value of the angular speed, as follows:[tex]\omega = \frac{v_{out} }{r_{out} } = \frac{11.5m/s}{3.14m} = 3.7 rad/sec (2)[/tex]
As we have already said, ωout = ωin = 3.7 rad/secb)
Since the angular speed is the same for both childs, the angle rotated in the same time, will be the same for both also.Applying the definition of angular speed, as the rate of change of the angle rotated with respect to time, we can find the angle rotated (in radians) as follows:[tex]\Delta \theta = \omega * t = 3.7 rad/sec* 5.0 sec = 18.5 rad (3)[/tex]⇒ Δθ₁ = Δθ₂ = 18.5 rad.
c)
The linear distance traveled by each child, will be related with the linear speed of them.Knowing the value of the angular speed, and the distance from each boy to the center, we can apply (1) in order to get the linear speeds, as follows:[tex]v_{inn} = \omega * r_{inn} = 3.7 rad/sec * 0.78 m = 2.9 m/s (4)[/tex]
vout is a given of the problem ⇒ vout = 11. 5 m/s
Applying the definition of linear velocity, we can find the distance traveled by each child, as follows:[tex]d_{inn} = v_{inn} * t = 2.9m/s* 5.0 s = 14.5 m (5)[/tex]
[tex]d_{out} = v_{out} * t = 11.5 m/s* 5.0 s = 57.5 m (6)[/tex]
d)
The centripetal force experienced by each child is the force that keeps them on a circular movement, and can be written as follows:[tex]F_{c} = m*\frac{v^{2}}{r} (7)[/tex]
Replacing by the values of vin and rin, since m is a given, we can find Fcin (the force on the boy closer to the center) as follows:[tex]F_{cin} = m*\frac{v_{in}^{2}}{r_{in}} = 25.4 kg* \frac{(2.9m/s)^{2} }{0.78m} = 273.9 N (8)[/tex]
In the same way, we get Fcout (the force on the boy near the outer edge):[tex]F_{cout} = m*\frac{v_{out}^{2}}{r_{out}} = 25.4 kg* \frac{(11.5m/s)^{2} }{3.14m} = 1069.8 N (9)[/tex]
e)
The centripetal force that keeps the boys in a circular movement, is not a different type of force, and in this case, is given by the static friction force.The maximum friction force is given by the product of the coefficient of static friction times the normal force.Since the boys are not accelerated in the vertical direction, the normal force is equal and opposite to the force due to gravity, which is the weight.As both boys have the same mass, the normal force is also equal.This means that for both childs, the maximum possible static friction force, is the same, and given by the following expression:[tex]F_{frs} = \mu_{s} * m* g (10)[/tex]If this force is greater than the centripetal force, the boy will be able to hold on.So, as the centripetal force is greater for the boy close to the outer edge, he will have a more difficult time holding on.A truck travelling down the street suddenly brakes, applying a 14 N force over 3.5 seconds. What was the impulse over the given time.
Answer:
49 Ns
Explanation:
Given data
Force= 14N
time = 3.5seconds
Applying the expression for impulse
P= Ft
substitute
P=14*3.5
P=49 Ns
Hence the impulse is 49 Ns
Car À moves at a speed of 8m/s for 43 seconds. Car B moves at a speed of 7 m/s for 50 seconds. Which car traveled a longer distance
Please show working
Distance = (speed) x (time)
Car A: Distance = (8 m/s) x (43 s) = 344 meters
Car B: Distance = (7 m/s) x (50 s) = 350 meters
350 meters is a longer distance than 344 meters.
Car-B traveled a longer distance than Car-A did.
Answer:
[tex]\boxed {\boxed {\sf Car \ B : 350 \ meters }}[/tex]
Explanation:
Distance is equal to the product of speed and time.
[tex]d=s*t[/tex]
1. Car A
Car A has a speed of 8 meters per second and travels for 43 seconds.
[tex]s= 8 \ m/s \\t= 43 \ s[/tex]
Substitute the values into the formula.
[tex]d= 8 \ m/s *43 \ s[/tex]
Multiply and note that the seconds will cancel out.
[tex]d= 8 \ m*43= 344 \ m[/tex]
2. Car B
Car B has a speed of 7 meters per second and travels for 50 seconds.
[tex]s= 7 \ m/s \\t= 50 \ s[/tex]
Substitute the values in and multiply.
[tex]d= 7 \ m/s * 50 \ s[/tex]
[tex]d= 7 \ m * 50 = 350 \ m[/tex]
350 meters is a longer distance than 344 meters, so Car B traveled the longer distance.
An 8.00 kg mass moving east at 15.4 m/s on a frictionless horizontal surface collides with a 10.0 kg object that is initially at rest. After the collision, the 8.00 kg object moves south at 3.90 m/s. (a) What is the velocity of the 10.0 kg object after the collision
Answer:
9.3m/s
Explanation:
Based on the law of conservation of momentum
Sum of momentum before collision = sum of momentum after collision
m1u1 +m2u2 = m1v1+m2v2
m1 = 8kg
u1 = 15.4m/s
m2 = 10kg
u2 = 0m/s(at rest)
v1 = 3.9m/s
Required
v2.
Substitute
8(15.4)+10(0) = 8(3.9)+10v2
123.2=31.2+10v2
123.2-31.2 = 10v2
92 = 10v2
v2 = 92/10
v2 = 9.2m/s
Hence the velocity of the 10.0 kg object after the collision is 9.2m/s
How much force is needed to accelerate a 65 kg rider AND her 215 kg motor scooter 8 m/s?? (treat
the masses as like terms)
Answer:
Force = 2240 Newton.
Explanation:
Given the following data;
Mass A= 65kg
Mass B = 215kg
Acceleration = 8m/s²
To find the force;
Force is given by the multiplication of mass and acceleration.
Mathematically, Force is;
[tex] F = ma[/tex]
Where;
F represents force.
m represents the mass of an object.
a represents acceleration.
First of all, we would have to find the total mass.
Total mass = Mass A + Mass B
Total mass = 65 + 215
Total mass = 280kg
Substituting into the equation, we have
[tex] Force = 280 * 8 [/tex]
Force = 2240 Newton.
1. (6x + 8)(5x - 8)
a. 30x2 + 49x + 20
2. (5x + 6(5x - 5)
b. 24x3 + 8x2 + 6x + 4
3. (6x + 3)(6x - 4)
c. 25x2 + 5x - 30
4. (6x + 5)(5x + 5)
d. 30x2 - 8x - 64
e. 36x2 - 6x - 1
5. (4x + 2) (6x2 - x + 2)
Answer:
form 1 question??????????
One of the disadvantages of experimental research is that __________.
A.
it isn’t easily replicated
B.
it doesn’t often reflect reality
C.
the results aren’t generalizable
D.
conditions are not controllable
Please select the best answer from the choices provided
Answer:
B
Explanation:
Two objects travel the same distance. The one that is moving faster will:
Take more time to go the distance
Take less time to go the same distance
Take the same time as the slower object
None of the above
Answer: take less time to go the same distance
Explanation:
Because if it is going faster let’s say mph 60 mph is 60 miles per hour if you are going 40 miles per hour it will take you longer to get to your destination.
a wooden block is cut into two pieces, one with three times the mass of the other. a depression is made in both faces of the cut so that a fire cracker can be placed in it and the block is reassembled. the reassembled block is set on rough surface and the fuse is lit. when the fire cracker explodes, the two blocks separate. what is the ratio of distances traveled by blocks?
Answer:
1/9
Explanation:
Let A denote the bigger piece and let B denote the smaller piece.
We are told that one with three times the mass of the other.
Therefore, we have;
M_a = 3M_b
Firecracker is placed in the block and it explodes and thus, momentum is conserved.
Thus;
V_ai = V_bi = 0
Where V_ai is initial velocity of piece A and V_bi is initial velocity of piece B.
Since initial momentum equals final momentum, we have;
P_i = P_f
Thus;
0 = (M_a × V_af) + (M_b × V_bf)
Since M_a = 3M_b, we have;
(3M_b × V_af) + (M_b × Vbf) = 0
Making V_af the subject, we have;
V_af = -⅓V_bf
The kinetic energy gained by each block during the explosion will later be lost due to the negative work done by friction. Thus;
W_f = -½M_b•(v_bf)²
Now, let's express the work is in terms of the force and the distance.
Thus;
W_f = F_f × Δx × cos 180°
Frictional force is also expressed as μmg
Thus;
W_f = -μM_b × g × Δx
Earlier, we saw that;
W_f = -½M_b•(v_bf)²
Thus;
-½M_b•(v_bf)²= -μM_b × g × Δx
Δx = (v_bf)²/2μg
Let the distance travelled by block A be Δx_a and that travelled by B be Δx_b
Thus;
Δx_a/Δx_b = ((v_ba)²/2μg)/((v_bf)²/2μg)
Δx_a/Δx_b = ((v_af)²/((v_bf)²)
Δx_a/Δx_b = (-⅓V_bf)²/(V_bf)²
Δx_a/Δx_b = 1/9
3. What is the SI unit of force? What is this unit equivalent to in terms of fundamental units?
4. Why is force a vector quantity?
Answer:
force = mass * acceleration
therefore the SI unit is kg*m/s2 or newton's
it's a vector quantity because it has both direction(acceleration) and size (mass)
A violin has a string of length
0.320 m, and transmits waves at
622 m/s. At what frequency does
it oscillate?
Answer:
1.9kHz
Explanation:
Given data
wavelength [tex]\lambda= 0.32m[/tex]
velocity [tex]v= 622 m/s[/tex]
We know that
[tex]v= f* \lambda\\\\f= v/ \lambda[/tex]
substitute
[tex]f= 622/ 0.32\\\\f= 1943.75\\\\f= 1.9kHz[/tex]
Hence the frequency is 1.9kHz
Answer:
971.2
Explanation:
It was right on acellus :)
"45 meters north" is an example of
Answer:
Displacement
Explanation:
The quantity 45m north is a typical example of displacement.
Displacement is the distance traveled by a body in a specific direction. Displacement is a vector quantity with both magnitude and direction.
When we are specifying the displacement of a body, the direction must be indicated accurately. Therefore, the quantity given is displacementAs a person pushes a box across a floor, the energy from the person's moving arm is transferred to the box, and the box and the floor becomes warm. During the process, what happens to energy
Answer:
isnt heat transfer
Explanation:
sorry if im wrong
A mass m is gently placed on the end of a freely hanging spring. The mass then falls 33 cm before it stops and begins to rise. What is the frequency of the oscillation
Answer:
Explanation:
The mass falls by .33 m before it begins to rise . At that point loss of potential energy is equal to gain of elastic energy .
1/2 k x² = mgx
.5 x k x .33² = m x 9.8 x .33
k / m = 59.4
frequency of oscillation = [tex]\frac{1}{2\pi} \times\sqrt{\frac{k}{m} }[/tex]
= [tex]\frac{1}{2\pi} \times\sqrt{59.4}[/tex]
= 1.22 per second .
25 points!
A 6 kg object accelerates from 5 m•s to 25 m•s in 30 seconds. What was the net force acting on the
object? Give your answer in Newtons to one significant figure and without a unit.
(Show Work)
Answer:
6N
Explanation:
Given parameters:
Mass of object = 6kg
Initial velocity = 5m/s
Final velocity = 25m/s
Time = 30s
Unknown:
Net force acting on the object = ?
Solution:
From Newton's second law of motion:
Force = mass x acceleration
Acceleration is the rate of change of velocity with time
Acceleration = [tex]\frac{Final velocity - Initial velocity }{time}[/tex]
Force = mass x [tex]\frac{Final velocity - Initial velocity }{time}[/tex]
So;
Force = 6 x [tex]\frac{25 - 5}{30}[/tex] = 6N
It has been argued that power plants should make use of off-peak hours (such as late at night) to generate mechanical energy and store it until it is needed during peak load times, such as the middle of the day. One suggestion has been to store the energy in large flywheels spinning on nearly frictionless ball-bearings. Consider a flywheel made of iron, with a density of 7800 kg/m^3 , in the shape of a uniform disk with a thickness of 11.3 cm.
Required:
a. What would the diameter of such a disk need to be if it is to store an amount of kinetic energy of 14.1 MJ when spinning at an angular velocity of 93.0 rpm about an axis perpendicular to the disk at its center?
b. What would be the centripetal acceleration of a point on its rim when spinning at this rate?
Answer:
Explanation:
kinetic energy = 14.1 MJ = 14.1 x 10⁶ J
Let radius of flywheel be r .
volume of flywheel = π r² x t where t is thickness
= 3.14 x r² x .113 m³
= .04 r² m³
mass = volume x density
= .04 r² x 7800 = 312.73 r²kg
moment of inertia I = 1 / 2 mass x radius²
= .5 x 312.73 r² x r²
= 156.37 r⁴ kg m²
angular velocity ω = 2π x 93/60
= 9.734 rad /s
kinetic energy = 1/2 Iω² where ω is angular velocity
= .5 x 156.37 r⁴ x 9.734²
= 7408.08 r⁴
Given
7408.08 r⁴ = 14.1 x 10⁶
r⁴ = .19 x 10⁴
r = .66 x 10
= 6.60 m .
Diameter = 13.2 m
b )
centripetal acceleration of a point on its rim = ω² r
= 9.734² x 6.6
= 625.35 m /s²
HELP PLEASE!!!
Running at 3.0 m/s, Burce, the 50.0 kg quarterback, collides with Max, the 100.0 kg tackle, who is traveling at 6.0 m/s in the other direction. Upon collision, Max continues to travel forward at 2.0 m/s.If the collision between the players lasted for 0.04 s, determine the impact force on either during the collision
Answer:
10kN
Explanation:
Given data
m1= 50kg
u1= 3m/s
m2= 100kg
u2= 6m/s
v1= 2m/s
time= 0.04s
let us find the final velocity of Bruce v1
from the conservation of linear momentum
m1u1+m2u2=m1v1+m2v2
substitute
50*3+100*6= 50*v1+100*2
150+600=50v1+200
750-200=50v1
550= 50v1
divide both sides by 50
v1= 550/50
v1=11 m/s
From
F= mΔv/t
for Bruce
F=50*(11-3)/0.04
F=50*8/0.04
F=400/0.04
F=10000
F=10kN
for Max
F=100*(6-2)/0.04
F=100*4/0.04
F=400/0.04
F=10000
F=10kN
Two spherical objects are separated by a distance that is 1.08 x 10-3 m. The objects are initially electrically neutral and are very small compared to the distance between them. Each object acquires the same negative charge due to the addition of electrons. As a result, each object experiences an electrostatic force that has a magnitude of 3.89 x 10-21 N. How many electrons did it take to produce the charge on one of the objects
Answer:
the charge on the object is 71.043×10^-20 C and the number of electron is 4.44
Explanation:
from coulumbs law, The force that is acting over both charge can be computed as
F=( kq1q2)/r^2..............eqn(1)
Where
F=electrostatic force= 3.89 x 10-21 N
k= column constant= 9 x 10^9 Nm^2/C^2
q1 and q2= magnitude of the charges
r= distance between the charges= 1.08 x 10-3 m.
Since both charges are experiencing the same force, eqn(1) can be written as
F=( kq^2)/r^2.
We can make q subject of the formula
q= √(Fr^2)/k
= √[(3.89 x 10^-21× (1.08 x 10^-3)^2]/8.99 x 10^9
q= 71.043×10^-20 C
Hence, the charge is 71.043×10^-20 C
From quantization law, the number of electron can be computed as
N=q/e
Where
N= number of electron
q= charges
=1.6×10^-19C
N=71.043×10^-20/1.6×10^-19
=4.44
Hence, the charge on the object is 71.043×10^-20 C and the number of electron is 4.44
What kind of scattering (Rayleigh, Mie, or non-selective) would you expect to be most important when radiation of the specified wavelength encounters the following natural or anthropogenic particles?
Slides 16-31, Lecture 2 ought to help - slides 19, 24, and 31 are key.
Wavelength O2 molecules Smoke particles Cloud droplets Rain droplets
(size 10^-10 m) (size 0.3 (μm) (20 μm) (size 3 mm)
550 nm
11 μm
1600 nm
1 cm
Solution :
1. Rayleigh scattering takes place when the particle size is smaller than the wavelength (λ).
2. Mie scattering takes place when particle size is nearly equal to the wavelength (λ).
3. Non-selective scatter takes place when particle size in greater than the wavelength (λ).
We have the sizes of different particles :
[tex]$O_2 \rightarrow 10^{10} \ m $[/tex]
Smoke particles [tex]$\rightarrow 3 \times 10^{-7} \ m$[/tex]
Cloud droplets [tex]$\rightarrow 2 \times 10^{-5} \ m$[/tex]
Rain droplets [tex]$\rightarrow 3 \times 10^{-3} \ m$[/tex]
Wavelength [tex]$ O_2 $[/tex] Smoke particles Cloud droplets Rain droplets
[tex]$10^{-10} \ m$[/tex] [tex]$ 3 \times 10^{-7} \ m$[/tex] [tex]$ 2 \times 10^{-5} \ m$[/tex] [tex]$ 3 \times 10^{-3} \ m$[/tex]
[tex]$5500 \times 10^{-4} \ m$[/tex] Rayleigh Non-selective Non-selective Non-selective
[tex]$11 \times 10^{-6} \ m $[/tex] Rayleigh Rayleigh Non-selective Non-selective
[tex]$1600 \times 10^{-10} \ m $[/tex] Rayleigh Non-selective Non-selective Non-selective
[tex]$10^{-2} \ m $[/tex] Rayleigh Rayleigh Rayleigh Mie
A sinusoidal wave is traveling on a string with speed 19.3 cm/s. The displacement of the particles of the string at x = 6.0 cm is found to vary with time according to the equation y = (2.6 cm) sin[1.8 - (5.8 s-1)t]. The linear density of the string is 5.0 g/cm. What are (a) the frequency and (b) the wavelength of the wave? If the wave equation is of the form
Answer:
Explanation:
equation of wave is given by the following equation
y = (2.6 cm) sin[1.8 - (5.8 s-1)t].
Comparing it with standard form of wave
y = A sin ( ωt - kx )
we get
ω = 5.8
2πn = 5.8
n = .92 per second
kx = 1.8
k x 6 = 1.8
k = 0.3
[tex]\frac{2\pi}{\lambda}[/tex] = 0.3
λ = 20.9 cm
Which landform is produced at location E where the Mississippi River enters the Gulf of
Mexico?
a delta a drumlin an out wash an escarpment
Answer:
a delta
Explanation:
The landform produced at the location E where the Mississippi River enters the Gulf of Mexico is a delta.
A delta is a depositional landform where a smaller body of water enters into a larger one.
The Gulf of Mexico contains a larger body of water and as the Mississippi river enters into it, it splits up into many distributaries.
So, this feature is a delta.
How could a change in straight line motion due to unbalanced forces be predicted from an understanding of inertia?
Answer:
If the force goes in the direction of movement, the speed must increase and if the net force goes in the opposite direction, the speed must decrease.
Explanation:
The principle of inertia or Newton's first law states that every body remains static or with constant velocity if there is no net force acting on it.
Based on this principle, if we have a net force, the velocity of the body changes by having an unbalanced force acting.
If the force goes in the direction of movement, the speed must increase and if the net force goes in the opposite direction, the speed must decrease.
Which of the following is a mixture?
a air
biron
Chydrogen
d nickel
Answer:
it will option option A hope it helps
A student is driving through a mountainous region where the road is at some times flat, at some times inclined upward, and at some time inclined downward. The student maintains a speed of 20 m/s on the roadway, but is required to make an emergency stop on the three sepearte occasions. On levels roadway, it takes 25 m to stop. On a downward-sloping roadway, it takes 40 m to stop. On an upward-sloping roadway, it takes 18 m to stop. Explain why the stopping distances are different. (Focus answer using work and energy, other concepts may be used as well but be sure work and energy are included.)
Answer:
Explanation:
It is frictional force of the ground that helps in bringing the vehicle to stop . In the process of stopping , negative work is done on the car by friction force to overcome its kinetic energy .
At levelled road , for stoppage
Kinetic energy of vehicle = Work done by frictional force . = friction force x displacement .
At upward slopping road , gravitational force acting downward also helps the vehicle to stop do friction has to do less work .
At upward inclined road , for stoppage
Kinetic energy of vehicle = Work done by frictional force + work done by gravitational force = (friction force + gravitational force ) x displacement .
Hence displacement is less .
At downward slopping road , friction has to do more work because friction has to do work against gravitational force acting downwards wards and kinetic energy of the vehicle also .
At downward inclined road , for stoppage
Kinetic energy of vehicle + work done by gravitational force = Work done by frictional force = friction force x displacement .
Hence displacement is more .
Hence displacement is more in the downward slopping.
What is Displacement?Displacement is defined as the change in position of an object. It is a vector quantity and has a direction and magnitude.
It is frictional force of the ground that helps in bringing the vehicle to stop . In the process of stopping , negative work is done on the car by friction force to overcome its kinetic energy .
At levelled road , for stoppage
Kinetic energy of vehicle = Work done by frictional force . = friction force x displacement .
At upward slopping road , gravitational force acting downward also helps the vehicle to stop do friction has to do less work .
At upward inclined road , for stoppage
Kinetic energy of vehicle = Work done by frictional force + work done by gravitational force = (friction force + gravitational force ) x displacement .
Hence displacement is less .
At downward slopping road , friction has to do more work because friction has to do work against gravitational force acting downwards wards and kinetic energy of the vehicle also .
At downward inclined road , for stoppage
Kinetic energy of vehicle + work done by gravitational force = Work done by frictional force = friction force x displacement .
Hence displacement is more in the downward slopping.
To learn more about displacement refer to the link:
brainly.com/question/11934397
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