Answer:
a. Electric field strength is 1.6*10^6 V/m = 1.6 Megavolt/meter
b. speed of an electron is 5.2*10^14m/s
Explanation:
The electric field strength is given by:
E = ΔV/d
E = electric field strength,
ΔV = potential difference,
d = plate spacin
ΔV = 21×10^3V, d = 1.3×10^-2m
a) E = V / d = (21*10^3)/(1.3×10^-2) = 1.6*10^6 V/m = 1.6 Megavolt/meter
b) Exit energy = V . e
Where 'e' is the charge of electron (1.6 * 10^(-19)
Exit energy = (21*10^3)*(1.6*10^-19) = 3.36*10−15
From the above, speed v = √2*Exit Energy/mass
Where 'm' is the mass of electron (9.1 * 10^(-31)
v = √2*energy/mass
v = √2*(3.36*10^-15)/(9.1*10^-31) = 5.2*10^14m/s
calculate the force needed to push the ball up a 4 m ramp if the work is equal to 16 joules.
how much heat energy is needed to raise the temperature of 2.0 kg of concrete from 10c to 30c
Two motorcycles are traveling due east with different velocities. However, 5.68 seconds later, they have the same velocity. During this 5.68-second interval, motorcycle A has an average acceleration of 3.87 m/s2 due east, while motorcycle B has an average acceleration of 18.2 m/s2 due east. (a) By how much did the speeds differ at the beginning of the 5.68-second interval, and (b) which motorcycle was moving faster?
Answer:
The answer is below
Explanation:
Let a be the initial velocity of motorcycle A and b be the initial velocity of motorcycle B.
After 5.68 seconds, both motorcycle had the same velocity (v), therefore for motorcycle A:
(a - v) / 5.68 = 3.87
a - v = 21.9816
v = a - 21.9816
For motorcycle B:
(b - v) / 5.68 = 18.2
b - v = 103.376
v = b - 103.376
Therefore:
a - 21.9816 = b - 103.376
b - a = -21.9816 + 103.376
b - a = 81.3944
a) The difference between their speeds at the beginning was 81.3944 m/s
b) Since b - a = 81.3944. This means that the initial velocity of motorcycle B is greater than that of motorcycle A by 81.3944 m/s.
Therefore motorcycle B was moving faster
For this assignment, you should mathematically solve and record a video testing your solution for the following prompt: Two rolls of toilet paper, of equal mass and radius, are dropped from different heights so that they hit the ground at the same time. One roll of toilet paper is dropped normally while the other is dropped while a person holds onto a sheet of toilet paper such that the roll unravels as it descends. Determine the ratio of heights h1/h2, where h1 represents the height of the toilet paper dropped normally and h2 represents the height of the toilet paper that unravels, so that both rolls hit the ground at the same time.
Answer:
h1/h2 = [tex]\frac{2R^2}{3R^2 + h^2}[/tex]
Explanation:
Using two rolls of tissue paper : One roll dropped normally while the other drops as some holds onto a sheet of the toilet paper ( I.e. the tissue paper drops rotating about its axis )
Determine the ratio of heights h1/h2
mass of tissues = same
radius of tissues = same
h1 = height of tissue 1
h2 = height of tissue 2
For the first tissue ( Tissue that dropped manually )
potential energy = kinetic energy
mgh = 1/2 mv^2
therefore the final velocity ( v^2 ) = 2gH ----- ( 1 )
second tissue ( Tissue that dropped while rotating )
gh = [tex]\frac{v^2}{u}[/tex] ( 3 + [tex]\frac{u^2}{R^2}[/tex] ) ------ ( 2 )
To determine the ratio of heights we will equate equations 1 and 2
hence :
gh = [tex]\frac{2gH}{u}[/tex] ( 3 + [tex]\frac{u^2}{R^2}[/tex] )
∴ h1/h2 = [tex]\frac{2R^2}{3R^2 + h^2}[/tex]
Two rods, one made of brass and the other made of copper, are joined end to end. The length of the brass section is 0.400 m and the length of the copper section is 0.800 m . Each segment has cross-sectional area 0.00700 m2 . The free end of the brass segment is in boiling water and the free end of the copper segment is in an ice-water mixture, in both cases under normal atmospheric pressure. The sides of the rods are insulated so there is no heat loss to the surroundings.
(a) What is the temperature of the point where the brass and copper segments are joined?
(b) What mass of ice is melted in 5.00 min by the heat conducted by the composite rod?
Answer:
a) 36°
b) 0.109 kg
Explanation:
Heat flows from brass to copper with the brass having its temperature
Length of brass = 0.4
Length of copper = 0.8
Temperature of = 36.15
See attachment for calculation
The temperature at the joint is 36.15°C
The amount of ice melted is 1.086 kg
The rate of transfer of thermal energy,
H = Q/t = KAΔT/L
where, K is the thermal conductivity of the substance, A is cross-sectional area, ΔT is temperature difference at the ends and L is the length
As given in the question,
the length of the brass section [tex]L_{1}[/tex] = 0.4 m
it's thermal conductivity [tex]K_{b}[/tex] = 109 J[tex]s^{-1}m^{-1}K^{-1}[/tex]
the temperature at the brass end [tex]T_{1}[/tex] = 373K
the length of the copper section [tex]L_{2}[/tex] = 0.8 m
it's thermal conductivity [tex]K_{c}[/tex] = 385 J[tex]s^{-1}m^{-1}K^{-1}[/tex]
the temperature at the brass end [tex]T_{1}[/tex] = 273K
cross-sectional area of both the substance is same A = 0.007 [tex]m^{2}[/tex]
Let the temperature at the joint be T
The rate of heat flow must be constant across the whole length of the setup.
Hence at the joint,
[tex]\frac{K_{b}A(T_{1}-T) }{L_{1} } =\frac{K_{c}A(T-T_{2}) }{L_{2} }[/tex]
⇒ [tex]\frac{ 109*A*(373-T)}{0.4} =\frac{385*A(T-273)}{0.8}[/tex] ⇒ T=309.15 K
⇒ T = 36.15°C is the temperature at the joint.
Now we have to calculate the equivalent thermal conductivity K of the setup in order to calculate the amount of heat transfer.
considering equivalent thermal conductivity K throughout the setup we can form the following equation to calculate its value
[tex]\frac{KA(T_{1}-T_{2} ) }{L_{1}+L_{2} } =\frac{K_{c}A(T-T_{2}) }{L_{2} }[/tex]
⇒ [tex]\frac{ K*A*(100)}{1.2} =\frac{385*A(36.15)}{0.8}[/tex]
⇒ K = 208.76 J[tex]s^{-1}m^{-1}K^{-1}[/tex]
the amount of heat transferred at the copper end in ice-water mixture in 5 minutes(300 seconds) :
Q = [tex]\frac{KA(T_{1}-T_{2} ) }{L_{1}+L_{2} }[/tex] × t = [tex]\frac{208.76*0.007*100}{1.2}[/tex] × 300 = 36533 J
latent heat of fusion of ice [tex]L_{f}[/tex] = 33600 J/kg
[tex]Q=mL_{f}[/tex]
[tex]m=\frac{Q}{L_{f} }[/tex]
[tex]m=\frac{36533}{33600}[/tex] ⇒ m = 1.086 kg of ice is melted in 5 minutes
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please help!!!
When a switch is turned from the off to the on position, it is changing the circuit in which of the following ways? O An open circuit is being changed into a closed circuit. A closed circuit is being changed into an open circuit. O A parallel circuit is being changed into a series circuit. A series circuit is being changed into a parallel circuit.
Answer:
i Believe the correct answer is "An open circuit being changed into a closed circuit"
Explanation:
What must the charge (sign and magnitude) of a particle of mass 1.43 g be for it to remain stationary when placed in a downward-directed electric field of magnitude 700 N/C ? Use 9.80 m/s2 for the magnitude of the free-fall acceleration.
Answer:
the sign and magnitude of the charge is - 2 x 10⁻⁵ C.
Explanation:
Given;
mass of the particle, m = 1.43 g = 0.00143 kg
electric field experienced by the particle, E = 700 N/C
The force experienced by the particle is calculated as;
F = mg = EQ
Where;
Q is the magnitude of the charge
[tex]Q = \frac{mg}{E} \\\\Q = \frac{0.00143 \times 9.8}{700} \\\\Q = 2\times 10^{-5} \ C[/tex]
The force must be upward in opposite direction to the electric field. Since the force and the electric field are in opposite direction, the charge must be negative.
Therefore, the sign and magnitude of the charge is - 2 x 10⁻⁵ C.
The particle to remain stationary, when placed in a downward-directed electric the force must be in opposite direction which upward directed.
The charge of the given particle to remain stationary should be [tex]-2\times10^{-5}[/tex] C.
What is electric charge?The electric force experienced by the body when placed it into the electromagnetic field is called electric charge.
Given information-
The mass of the particle is 1.43 g or 0.00143 kg.
The magnitude of the downward-directed electric field is 700 N/C.
The magnitude of the free-fall acceleration is 9.80 meter per second squared.
The electric field is defined as the electric force per unit charge. It can be given as,
[tex]E=\dfrac{F}{q}[/tex]
Rewrite the equation for the charge,
[tex]q=\dfrac{F}{E}[/tex]
Force experienced by the particle is equal to the product of mass and free fall acceleration (gravity). Thus,
[tex]q=\dfrac{0.00143\times9.8}{700}\\q=2\times10^{-5}[/tex]
Thus the magnitude of the charge is [tex]2\times10^{-5}[/tex] C.
The particle to remain stationary, when placed in a downward-directed electric the force must be in opposite direction which upward directed. For the opposite direction the sign of the charge should be negative.
Thus the charge of the given particle to remain stationary should be [tex]-2\times10^{-5}[/tex] C.
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The Earth’s orbit _____.
is an ellipse
goes around the moon
is a circle
causes day and night
A long, straight wire has a uniform constant charge with linear charge density, - 3.60 nC/m. The wire is surrounded by a long nonconducting, thin-walled cylindrical shell that is charged on its outside surface, such that the electric field outside the shell is zero. The shell has a radius of 1.50 cm.
Required:
What uniform area charge density rho is needed on the shell for the electric field to be zero outside the shell?
Answer:
Uniform area charge density rho is needed is 3.82*10^-8 C.m^-2
Explanation:
See the attached files.
To find the rho, I used Gauss law for cylindrical shell which is equation 1 and Gauss law for the rod which is equation 4.
Note that in equation 4, Lamda is the charge per length while L is the length if the rod. Also R is the radius of the shell.
The final answer is 3.82*10^-8 C.m^-2 which is the uniform area charge density rho is needed.
A star's emission line of 400 nm appears shifted to 404 nm in the spectrum. What can you conclude from this shift?
A. The star is approaching you with the speed of 3000 km/s.
B. The star is approaching you with the speed of 30300 km/s.
C. The star is receding from you with the speed of 3000 km/s.
D. The star is receding from you with the speed of 30300 km/s.
Answer:
C. The star is receding from you with the speed of 3000 km/s
Explanation:
To get this answer we use the doppler effect equation . The formula for a receding emissor is given in the attachment.
We solve for V
V = 3x10⁶m/s
V = 3000km/s
We have the wavelength to be shifting towards red. Therefore we conclude that it is receding. We say the star is receding with speed of 3000km/s towards you.
Thank you!
How old do you need to be in order to qualify to be a U.S. Senator
Answer: 30 Years Old
Explanation: The constitution has around three qualifications for service in the U.S. Senate, Your age must be at least 30 years.
I need help will mark brainliest
Answer: ITS 1 TRUST ME MAN BYE K
Explanation: OK BYE TRUST YEAH
Which statement BEST explains why a bouncing basketball will not remain in motion forever?
Group of answer choices
The energy is transferred to sound and heat energy.
The energy is used up and destroyed.
The energy is transferred to light and potential energy.
The energy is transferred to chemical and heat energy.
Answer:
The energy is transferred to chemical and heat energy.
Explanation:
If you define "bouncing" as leaving the ground for any amount of time, the ball stops bouncing when the elastic energy stored in the compression phase of the bounce is not enough to overcome the weight of the ball. This is the proof of the answer i Hope this helps :)
A ball 12 m in 4 seconds and then 2.5 seconds later it rolls 8 m in 2 seconds what is its acceleration
Answer:
If it accelerates at 20 m/s2 for a period of 22 seconds, what is its final velocity? ... How fast is the ball falling after 5 seconds? v = v0 + gt v = 0 + 10(5) v = 50 m/s. 4. ... + ½ 2.5(15)2 x = 281 m. 5. What is the total displacement of the car in question 2? ... 8. A base jumper falls until he reaches a speed of 200 m/s
Explanation:
if the forces on an object are balanced the resultant force is equal to zero true false
Answer:
If the forces are balanced, the resultant force is zero. If the forces on an object are unbalanced, this is what happens: a stationary object starts to move in the direction of the resultant force. a moving object changes speed and/or direction in the direction of the resultant force.
Explanation:
A 28.8 kg child sits on a 6.0 m long teeter-totter at a point 1.5 m from the pivot point (at the center of the teeter-totter). On the other side of the pivot point, an adult pushes straight down on the teeter-totter with a force of 180 N. Determine the direction the teeter-totter will rotate if the adult applies the force at a distance of each of the following from the pivot. (Assume the teeter-totter is horizontal when the adult applies the force and that the child's weight applies a clockwise torque.)
a.
1. 1.0 m
2. counterclockwise
b.
1. 2.0 m
2. clockwise
3. counterclockwise
c.
1. 3.0m
2. clockwise
3. counterclockwise
Answer:
case A) tau_net = -243.36 N m, case B) tau_net = 783.36 N / m, tau_net = -63.36 N m, case C) tau _net = - 963.36 N m,
Explanation:
For this exercise we use Newton's relation for rotation
Σ τ = I α
In this exercise the mass of the child is m = 28.8, assuming x = 1.5 m, the force applied by the man is F = 180N
we will assume that the counterclockwise turns are positive.
case a
tau_net = m g x - F x2
tau_nett = -28.8 9.8 1.5 + 180 1
tau_net = -243.36 N m
in this case the man's force is downward and the system rotates clockwise
case b
2 force clockwise, the direction of
the force is up
tau_nett = -28.8 9.8 1.5 - 180 2
tau_net = 783.36 N / m
in case the force is applied upwards
3) counterclockwise
tau_nett = -28.8 9.8 1.5 + 180 2
tau_net = -63.36 N m
system rotates clockwise
case c
2 schedule
tau_nett = -28.8 9.8 1.5 - 180 3
tau _net = - 963.36 N m
3 counterclockwise
tau_nett = -28.8 9.8 1.5 + 180 3
tau_net = 116.64 Nm
the sitam rotated counterclockwise
77. A drag racing vehicle travels from 0 to 100 mph in 5 seconds north. What is the acceleration?
a).004
s2
b).0056 m/s2 c).0079"
d).01 m/s2
M
m
Answer:
a
Explanation:
i just took the test
A spring has a spring constant of 25 Newtons per meter. The minimum force required to
stretch the spring 0.20 meter from its equilibrium position is approximately
Answer:
6.3N
Explanation:
Guessed it right on castle learning
Answer:
6.3 N
Explanation:
F=kx
F=(25N/m)(0.25m)
6.3 N
A 20 kg box has an initial velocity of 2 m/s starting at the bottom of a 30-degree inclined plane. A person pushes on the box directly up the frictionless inclined plane so that it travels up the inclined plane at a constant velocity of 2 m/s. Calculate the how much is done by the person after 5 seconds have past.
Answer:
Explanation:
The box is moving with constant velocity so acceleration of box is zero . That means net force on the box is zero .
The weight component acting on box parallel to incline plane
= mg sin 30⁰ = 20 x 9.8 x sin 30 = 98 N
This force is acting down the plane , hence to make the net force zero acting on box , force exerted by person will also be 98 N up the incline .
Force exerted by person = 98 N
distance travelled in 5 s
= velocity x time
= 2 x 5 = 10 m
Work done by person
= 98 x 10
= 980 J .
Balance the following equation:
H3B03 →_B203 +_H20
a. 1, 3,2
b. 2,4,6
C. 4, 2, 6
d. 6, 4,2
It's c I think ( 4 , 2 , 6 , )
A river flows at 2m/s the velociy of ferry relative to the shore is 4m/s
Answer:
Explanation:
.....
what is the acceleration of a satellite moving in a circular orbit around the earth of radius 2r
Explanation:
You do the radius times the circumference of the earth
John attaches a ball to a spring. The diagram below shows what happens. Which option shows the direction of the force of the ball on the spring?
Option C shows the direction of the force of the ball on the spring. The direction of the force of the ball on the spring will be downwards.
What is force?Force is defined as the push or pull applied to the body. Sometimes it is used to change the shape, size, and direction of the body.
Force is defined as the product of mass and acceleration. Its unit is Newton.
The spring is extended downward because the weight is always act downwards. The direction of the force of the ball on the spring will be downwards.
Hence, option C shows the direction of the force of the ball on the spring
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What are the two rules that light follows.
ok so i dont know srry5
What is the shortest time that a jet pilot starting from rest can take to reach Mach-3.60 (3.60 times the speed of sound) without graying out? (Use 331 m/s for the speed of sound in cold air.)
Answer:
30.4 s
Explanation:
A pilot , with plane accelerated at 4 g starts greying out . In the problem , the acceleration of jet is 4 g
a = 4 x 9.8 = 39.2 m /s²
initial velocity u = 0
Final velocity = 3.60 times speed of sound
= 3.6 x 331 = 1191.6 m /s
v = u + at
Putting the values
1191.6 = 0 + 39.2 t
t = 30.4 s .
a body accelerates uniformly from rest at 2m/s^2 for 5 seconds. Calculate its averege velocity in this time
HERE IS YOUR ANSWER!
A monk is sitting atop a mountain in complete rest in meditation. What is the kinetic Energy of the monk? (assume mass of 65 kg and the mountain's height was 1000 m)
Answer:
no kinetic energy
hope this helps! :-D
Explanation:
the monk is not moving
A 45945990 prism is immersed in water. A ray of light is incident normally on one of its shorter faces. What is the minimum index of refraction that the prism must have if this ray is to be totally reflected within the glass at the long face of the prism
Answer:
Explanation:
Angle of incidence at the longer face = 45⁰ ( see the figure in the attached file )
For total reflection from longer face
i = C where C is critical angle .
And relation between critical angle and refractive index is as follows .
μ = 1 / sinC
= 1 / sin45
= √2
= 1.414 .
An object with an initial horizontal velocity of 20 ft/s experiences a constant horizontal acceleration due to the action of a resultant force applied for 10 s. The work of the resultant force is 10 Btu. The mass of the object is 55 lb. Determine the constant horizontal acceleration, in ft/s2.
Answer:
a = 7.749 ft/s²
Explanation:
First to all, we need to convert all units, so we can work better in the calculations.
The horizontal acceleration is asked in ft/s² so the units of speed will be the same. The Work is in BTU and we need to convert it in ft.lbf in order to get the acceleration and final speed in ft/s:
W = 10 BTU * 778.15 Lbf.ft / BTU = 7781.5 lbf.ft
Now, to get the acceleration we need to get the final speed of the object first. This can be done, by using the following expression:
W = ΔKe (1)
And Ke = 1/2mV²
So Work would be:
W = 1/2 mV₂² - 1/2mV₁²
W = 1/2m(V₂² - V₁²) (2)
Finally, we need to convert the mass in lbf too, because Work is in lbf, so:
m = 55 lb * 1 lbf.s²/ft / 32.174 lb = 1.7095 lbf.s²/ft
Now, we can calculate the final speed by solving V₂ from (2):
7781.5 = (1/2) * (1.7095) * (V₂² - 20²)
7781.5 = 0.85475 * (V₂² - 441)
7781.5/0.85475 = (V₂² - 400)
9103.83 + 400 = V₂²
V₂ = √9503.83
V₂ = 97.49 ft/s
Now that we have the speed we can calculate the acceleration:
a = V₂ - V₁ / t
Replacing we have:
a = 97.49 - 20 / 10
a = 7.749 ft/s²Hope this helps
A roller coaster moving along its track rolls into a circular loop of radius r. In the loop, it is only affected by its initial velocity, gravity, and the shape of the track. Let v denote the instantaneous speed and a denote the magnitude of the instantaneous acceleration of the roller coaster in the loop. Which of the following is true in the loop?
a. The roller coaster is not in uniform circular motion, but we still have a=v^2/r everywhere on the loop
b. The roller coaster is not in uniform circular motion, but the tangential acceleration is so small that we can approximate a by v^2/r everywhere on the loop
c. The roller coaster is in uniform circular motion
d. The roller coaster is not in uniform circular motion, and a=v^2/r is only applicable at the very top and bottom of the loop where the tangential acceleration vanishes
Answer:
c. The roller coaster is in uniform circular motion
Explanation:
Since the loop is circular with radius r, and its instantaneous speed, v is always constant, and also, its centripetal acceleration, a' = v²/r.
Since the angular speed, ω = v/r does not change, the magnitude of its tangential acceleration is zero although there is a change in its direction because the direction of its initial velocity changes. That is a" = rα and α = Δω/Δt since Δω = 0, α = 0 and a" = r(0) = 0
So, there is no tangential acceleration. Since there is no tangential acceleration, our instantaneous acceleration which is the vector sum of our centripetal acceleration and tangential acceleration is a = √(a'² + a"²) = √(a'² + 0²) = √a'² = a' = v²/r
So, a is always v²/r.
Since the instantaneous acceleration is always (a = v²/r) constant, the motion is uniform. So, it is uniform circular motion.
The roller coaster is not in uniform circular motion, and [tex]a = \frac{v^2}{r}[/tex] is only applicable at the very top and bottom of the loop where the tangential acceleration vanishes.
The given parameters;
radius of the circular path, = r instantaneous speed = v instantaneous acceleration = aThe motion tension on the loop at the lowest point in the circular motion is given as;
[tex]T = mg + \frac{mv^2}{r}[/tex]
The motion tension on the loop at the highest point in the circular motion is given as;
[tex]T = \frac{mv^2}{r} - mg[/tex]
This shows that circular motion is affected by;
acceleration due to gravity, gradius of the circular path, rspeed of the motion, vmass of the object, mThus, we can that the roller coaster is not in uniform circular motion, and [tex]a = \frac{v^2}{r}[/tex] is only applicable at the very top and bottom of the loop where the tangential acceleration vanishes.
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