Answer:
The highest electric field is experienced by a 2 C charge acted on by a 6 N electric force. Its magnitude is 3 N.
Explanation:
The formula for electric field is given as:
E = F/q
where,
E = Electric field
F = Electric Force
q = Charge Experiencing Force
Now, we apply this formula to all the cases given in question.
A) A 2C charge acted on by a 4 N electric force
F = 4 N
q = 2 C
Therefore,
E = 4 N/2 C = 2 N/C
B) A 3 C charge acted on by a 5 N electric force
F = 5 N
q = 3 C
Therefore,
E = 5 N/3 C = 1.67 N/C
C) A 4 C charge acted on by a 6 N electric force
F = 6 N
q = 4 C
Therefore,
E = 6 N/4 C = 1.5 N/C
D) A 2 C charge acted on by a 6 N electric force
F = 6 N
q = 2 C
Therefore,
E = 6 N/2 C = 3 N/C
E) A 3 C charge acted on by a 3 N electric force
F = 3 N
q = 3 C
Therefore,
E = 3 N/3 C = 1 N/C
F) A 4 C charge acted on by a 2 N electric force
F = 2 N
q = 4 C
Therefore,
E = 2 N/4 C = 0.5 N/C
The highest field is 3 N, which is found in part D.
A 2 C charge acted on by a 6 N electric force
A student in the front of a school bus tosses a ball to another student in the back of the bus while the bus is moving forward at constant velocity. The speed of the ball as seen by a stationary observer in the street:_________
a. is less than that observed inside the bus.
b. is the same as that observed inside the bus
c. may be either greater or smaller than that observed inside the bus.
d. may be either greater, smaller, or equal to that observed inside the bus.
e. is greator than that observed inside the bus
Answer:
d
Explanation:
good question. now the bus is moving in constant velocity . a student in front tosses a ball to the student in back. but we dont know the speed at which the student tosses a ball. we have to assume the speed
assume the speed of ball is slightly less than the speed of bus. in this case the stationary observer sees the ball in slower speed than the one inside the bus.
so a is correct
now assume the speed of ball is 1/2 the speed of bus. here stationary observer sees the ball the same speed as the one in bus observe
b is correct
assume the speed of ball is very small than the speed of bus . in this case the stationary observer see in grater speed than the student in bus
e also correct
so correct answer is d. it depends on the speed of ball tossed by the student in front.
A subatomic particle created in an experiment exists in a certain state for a time of before decaying into other particles. Apply both the Heisenberg uncertainty principle and the equivalence of energy and mass to determine the minimum uncertainty involved in measuring the mass of this short-lived particle.
Answer:
Δm Δt> h ’/ 2c²
Explanation:
Heisenberg uncertainty principle, stable uncertainty of energy and time, with the expressions
ΔE Δt> h ’/ 2
h’= h / 2π
to relate this to the masses let's use Einstein's relationship
E = m c²
let's replace
Δ (mc²) Δt> h '/ 2
the speed of light is a constant that we can condense exact, so
Δm Δt> h ’/ 2c²
Gas is contained in a piston-cylinder assembly and undergoes three processes. First, the gas is compressed at a constant pressure of 100 [kPa] from initial volume of 1.0 [m3] to a volume of 0.5 [m3]. Second, the gas pressure is increased by heating at constant volume up to 200 [kPa]. Third, the gas is returned to its initial pressure and volume by a process for which P ∀=constant. All pressures given are absolute. For the gas as a system, is the system best considered open, closed, or isolated? Why?
Complete Question
The complete question is shown on the first uploaded image
Answer:
The correct option is B
Explanation:
The system is best considered a closed system because looking at process we can see that there was no exchange of matter between the system and the surrounding,(as the was no escape of matter from the system to the surrounding )
Secondly we can deduce that there is a variation in the volume. from [tex]1.0 m^3[/tex] to [tex]0.5 m^3[/tex]
A potential difference of 71 mV is developed across the ends of a 12.0-cm-long wire as it moves through a 0.27 T uniform magnetic field at a speed of 6.0 m/s. The magnetic field is perpendicular to the axis of the wire.
Required:
What is the angle between the magnetic field and the wire's velocity?
Answer:
Explanation: please see attached file I attached the answer to your question.
The angle between the magnetic field and the wire's velocity is 33.2 degrees.
Calculation of the angle:Since the potential difference = 71mv = 71 *10 ^-3 V
The length is 12 cm = 0.12m
The magnetic field i.e. B = 0.27T
The speed or v = 4 m/s
here we assume [tex]\theta[/tex] be the angle
So,
e = Bvl sin[tex]\theta[/tex]
So,
[tex]Sin\theta[/tex] = e/bvl
= 71*10^-3 / 0.27 *4*0.12
= 0.5478
= 33.2 degrees
Therefore, the angle should be 33.2 degrees
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Friction is a force that acts in an ___________ direction of movement.
a) similar
b) opposite
c) parallel
d) west
Answer:
the answer is opposite.
plz mark brainliest
Explanation:
The site from which an airplane takes off is the origin. The X axis points east, the y axis points straight up. The position and velocity vectors of the plane at a later time are given by r=(1.21x103i+3.45x104;)m and v= (2 i-3.5j) m/s The magnitude, in meters, of the plane's displacement from the origin is:_________
a. 2.50 x104
b. 1.45 x 104
c. 3.45x104
d. 2.5x103
e. none of the above
Answer:
d = 3.5*10^4 m
Explanation:
In order to calculate the displacement of the airplane you need only the information about the initial position and final position of the airplane. THe initial position is at the origin (0,0,0) and the final position is given by the following vector:
[tex]\vec{r}=(1.21*10^3\hat{i}+3.45*10^4\hat{j})m[/tex]
The displacement of the airplane is obtained by using the general form of the Pythagoras theorem:
[tex]d=\sqrt{(x-x_o)^2+(y-y_o)^2}[/tex] (1)
where x any are the coordinates of the final position of the airplane and xo and yo the coordinates of the initial position. You replace the values of all variables in the equation (1):
[tex]d=\sqrt{(1.12*10^3-0)^2+(3.45*10^4-0)^2}=3.45*10^4m[/tex]
hence, the displacement of the airplane is 3.45*10^4 m
During a baseball game, a baseball is struck at ground level by a batter. The ball leaves the baseball bat with an initial speed v0 = 38 m/s at an angle θ = 35° above horizontal. Let the origin of the Cartesian coordinate system be the ball's position the instant it leaves the bat. Air resistance may be ignored throughout this problem.
Part (a) Express the magnitude of the ball's initial horizontal velocity Or in terms of vo and 20%
Part (b) Express the magnitude of the ball's initial vertical velocity vOy in terms of vo and 0. 20%
Part (c) Find the ball's maximum vertical height Amat in meters above the ground.
Part (d) Create an expression in terms of vo-e, and g for the time-ur İt takes te ball to travel to its maximum vertical height.
Part (e) Calculate the horizontal distance in meters the ball has traveled when it returns to ground level.
Answer:
a) v₀ₓ = v₀ cos θ , b) v_{oy} = v₀ sin θ , c) y = v_{oy}² / 2g, y = 24.25 m
e) R = 138.46 m
Explanation:
This is a projectile launch exercise
a) let's use trigonometry to find the components of the initial velocity
cos θ = v₀ₓ / v₀
v₀ₓ = v₀ cos θ
v₀ₓ = 38 cos 35
v₀ₓ = 31.13 m / s
b) sin θ = [tex]v_{oy}[/tex] / v₀
v_{oy} = v₀ sin θ
v_{oy} = 38 sint 35
v_{oy} = 21 80 m / s
c, d) to find the maximum height, the vertical speed is zero
v_{y}² = v_{oy}² - 2 g y
0 = [tex]v_{oy}[/tex]² - 2 gy
y = v_{oy}² / 2g
let's calculate
y = 21.80 2 / (2 9.8)
y = 24.25 m
e) They ask to find the horizontal distance
for this we can use the expression of reaches
R = v₀² sin 2θ / g
let's calculate
R = 38² sin (2 35) / 9.8
R = 138.46 m
The lowest-pitch tone to resonate in a pipe of length L that is closed at one end and open at the other end is 200 Hz. Which one of the following frequencies will NOT resonate in the same pipe
a. 1800 Hz
b. 1000 Hz
c. 1400 Hz
d. 600 Hz
e. 400 Hz
Answer:
e. 400 Hz
Explanation:
In closed organ pipe, only odd harmonics of fundamental note is possible .
The fundamental frequency is 200 Hz . Then other overtones will be having following frequencies .
200 x 3 , 200 x 5 , 200 x 7 , 200 x 9 etc
600 Hz , 1000 Hz , 1400 Hz , 1800 Hz .
Frequency not possible is 400 Hz .
If you jumped out of a plane, you would begin speeding up as you fall downward. Eventually, due to wind resistance, your velocity would become constant with time. While your velocity is constant, the magnitude of the force of wind resistance is
Answer:
Mg or your weight.
Explanation:
When your velocity is constant, the net force acting on you is 0. That means the upwards force of air resistance must fully balance the downwards force of gravity on you, which is Mg.
A train locomotive is pulling two cars of the same mass behind it. Determine the ratio of the tension in the coupling (think of it as a cord) between the locomotive and the first car (FT1) to that between the first car and the second car (FT2), for any nonzero acceleration of the train
Answer:
The ratio is [tex]\frac{F_{T1}}{F_{T2}} = 2[/tex]
Explanation:
The diagram for this question is shown on the first uploaded image
Here we are assume the acceleration of the train is a
which makes the acceleration of each car a
From the question we are told that
Considering the second car
The force causing it s movement is mathematically represented as
[tex]F_{T2} = ma[/tex]
Considering the first car
The force causing it s movement is mathematically represented as
[tex]F = F_{T1} -F_{T2} = ma[/tex]
=> [tex]F_{T1} -ma = ma[/tex]
=> [tex]F_{T1} = 2 ma[/tex]
=> [tex]\frac{F_{T1}}{ma} = 2[/tex]
=> [tex]\frac{F_{T1}}{F_{T2}} = 2[/tex]
A rod of mass M = 2.95 kg and length L can rotate about a hinge at its left end and is initially at rest. A putty ball of mass m = 45 g, moving with speed v = 4.23 m/s, strikes the rod at angle θ = 46° from the normal at a distance D = 2/3 L, where L = 0.95 m, from the point of rotation and sticks to the rod after the collision.
Required:
What is the angular speed ωf of the system immediately after the collision, in terms of system parameters and I?
Answer:
Explanation:
angular momentum of the putty about the point of rotation
= mvR where m is mass , v is velocity of the putty and R is perpendicular distance between line of velocity and point of rotation .
= .045 x 4.23 x 2/3 x .95 cos46
= .0837 units
moment of inertia of rod = ml² / 3 , m is mass of rod and l is length
= 2.95 x .95² / 3
I₁ = .8874 units
moment of inertia of rod + putty
I₁ + mr²
m is mass of putty and r is distance where it sticks
I₂ = .8874 + .045 x (2 x .95 / 3)²
I₂ = .905
Applying conservation of angular momentum
angular momentum of putty = final angular momentum of rod+ putty
.0837 = .905 ω
ω is final angular velocity of rod + putty
ω = .092 rad /s .
An airplane flies 2500 miles east in 245 seconds what is the velocity of the plane?
Speed = (distance) / (time)
Speed = (
Velocity = speed, and its direction
The velocity of the plane is 10.2 miles per second East.
(about 48 times the speed of sound)
Question 9 of 10
2 Powie
You are riding a bicycle. You apply a forward force of 100 N, and you and the
bicycle have a combined mass of 80 kg. What is the acceleration of the
bicycle?
A. 125 m/s
B. 1.5 m/s2
c. 1.8 m/s?
D. 0.8 m/s
Answer:
1.25 m/s^2
Explanation:
F = m*a ...... force = mass * acceleration
force = 100 N, mass = 80 kg
100 = 80 * a
100/80 = a = 1.25 m/s^2
Answer:
The acceleration is 1.25m/s².
Explanation:
You have to apply Newton's Second Law which is F = m×a where F represents force, m is mass and a is acceleratipn. Then you have to substitute the following values into the formula :
[tex]f = m \times a[/tex]
Let F = 100,
Let m = 80,
[tex]100 = 80 \times a[/tex]
[tex]100 = 80a[/tex]
[tex]a = 100 \div 80[/tex]
[tex]a = 10 \div 8[/tex]
[tex]a = 1.25[/tex]
calculate the volume of marble if its diameter is 10mm
Answer:
The volume of the marble is [tex]523.33\ mm^2[/tex].
Explanation:
Marble is spherical in shape. The diameter of marble is 10 mm. It radius will be 5 mm.
The volume of spherical shaped object is given by :
[tex]V=\dfrac{4}{3}\pi r^3[/tex]
Plugging all the values, we get :
[tex]V=\dfrac{4}{3}\times 3.14\times (5)^3\\\\V=523.33\ mm^2[/tex]
So, the volume of the marble is [tex]523.33\ mm^2[/tex].
Q.1- Find the distance travelled by a particle moving in a straight line with uniform acceleration, in the 10th unit of time.
Answer:
If the acceleration is constant, the movements equations are:
a(t) = A.
for the velocity we can integrate over time:
v(t) = A*t + v0
where v0 is a constant of integration (the initial velocity), for the distance traveled between t = 0 units and t = 10 units, we can solve the integral:
[tex]\int\limits^{10}_0 {A*t + v0} \, dt = ((A/2)10^2 + v0*10) = (A*50 + v0*10)[/tex]
Where to obtain the actual distance you can replace the constant acceleration A and the initial velocity v0.
A 1KW electric heater is switched on for ten minutes
How much heat does it produce?
Explanation:
P=W/T ==> 1000w = Q/600 ==> Q=600000j
If a 1 - kilowatt electric heater is switched on for ten minutes then the heat produced by the electric heater would be 600 - kilo Joules .
What is thermal energy ?It can be defined as the form of the energy in which heat is transferred from one body to another body due to their molecular movements, thermal energy is also known as heat energy .
As given in the problem , we have to find out the heat produced by the 1 - kilo watts electric heater if it is switched on for ten minutes ,
The heat produced by the electric heater = Power × time
= 1000 × 600 Joules
= 600 kilo - Joules
Thus , the heat produced by the electric heater would be 600 - kilo Joules .
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Astrophysicist Neil deGrasse Tyson steps into an elevator on the 29th floor of a skyscraper. For some odd reason, there is a scale on the floor of the elevator. Neil, whose mass is about 115 kg, decides to step on the scale and presses the button for a lower floor. The elevator starts traveling downwards with a constant acceleration of 1.5 m/s2 for 6.0 seconds, and then travels at a constant velocity for 6.0 seconds. Finally, the elevator has an upward acceleration of 1.5 m/s2 for 6.0 seconds as it comes to a stop.
A. If each floor is approximately 4 m tall, which floor does the elevator stop at?
B. If the mass of the elevator is 1,200 kg, what is the maximum tension of the elevator cable?
Answer:
A. Final Floor = 15.5 = 15 (Considering downward portion of elevator)
B. T = 14859.5 N = 14.89 KN
Explanation:
A.
First we calculate distance covered by the elevator during downward motion. Downward motion consists of two parts. First one is uniformly accelerated. For that part we use 2nd equation of motion:
s₁ = Vi t + (0.5)at²
where,
s₁ = distance covered during accelerated downward motion = ?
Vi = initial speed = 0 m/s (since elevator is initially at rest)
t = time taken = 6 s
a = acceleration = 1.5 m/s²
Therefore,
s₁ = (0 m/s)(6 s) + (0.5)(1.5 m/s²)(6 s)²
s₁ = 4.5 m
also we find the final velocity using 1st equation of motion:
Vf = Vi + at
Vf = 0 m/s + (1.5 m/s²)(6 s)
Vf = 9 m/s
Now, the second part of downward motion is with constant velocity. So:
s₂ = vt
where,
s₂ = distance covered during constant speed downward motion = ?
v = Vf = 9 m/s
t = 6 s
Therefore,
s₂ = (9 m/s)(6 s)
s₂ = 54 m
Now for distance covered during upward motion is given by the 2nd equation of motion. Since the values of acceleration and time are same. Therefore, it will be equal in magnitude to s₁:
s₃ = s₁ = 4.5 m
Therefore, the total distance covered by elevator is given by following equation:
s = s₁ + s₂ - s₃ (Downward motion taken positive)
s = 4.5 m + 54 m - 4.5 m
s = 54 m
Therefore, net motion of the elevator was 54 m downwards.
So the final floor will be:
Final Floor = Initial Floor - Distance Covered/Length of a floor
Final Floor = 29 - 54 m/4m
Final Floor = 15.5 = 15 (Considering the downward portion or floor of elevator)
B.
The maximum tension will occur during the upward accelerated motion. It is given by the formula:
T = m(g + a)
where,
T = Max. Tension in Cable = ?
m = total mass of person and elevator = 115 kg + 1200 kg = 1315 kg
g = 9.8 m/s²
a = acceleration = 1.5 m/s²
Therefore,
T = (1315 kg)(9.8 m/s² + 1.5 m/s²)
T = 14859.5 N = 14.89 KN
I really need help with this question someone plz help !
Answer:D
Explanation:
Given
Same force is applied to each ball such that all have different masses
and Force is given by the product of mass and acceleration
[tex]F=m\times a[/tex]
[tex]a=\frac{F}{m}[/tex]
So acceleration of ball A
[tex]a_A=\frac{F}{0.5}=2F[/tex]
acceleration of ball B
[tex]a_B=\frac{F}{0.75}=\frac{4F}{3}=1.33F[/tex]
acceleration of ball C
[tex]a_C=\frac{F}{1}=F[/tex]
acceleration of ball D
[tex]a_D=\frac{F}{7.3}=\frac{F}{7.3}[/tex]
It is clear that acceleration of ball D is least.
Your lab instructor has asked you to measure a spring constant using a dynamic method—letting it oscillate—rather than a static method of stretching it. You and your lab partner suspend the spring from a hook, hang different masses, m, on the lower end, and start them oscillating. One of you uses a meter stick to measure the amplitude, A, and the other uses a stopwatch to time 10 oscillations, t. Your data are as follows:Mass, m(g) Amplitude, A(cm) Time, T(s) 100 6.5 7.8150 5.5 9.8200 6.0 10.9250 3.5 12.4Use the best-fit line of an appropriate graph to determine the spring constant.
Answer:
k = 6,547 N / m
Explanation:
This laboratory experiment is a simple harmonic motion experiment, where the angular velocity of the oscillation is
w = √ (k / m)
angular velocity and rel period are related
w = 2π / T
substitution
T = 2π √(m / K)
in Experimental measurements give us the following data
m (g) A (cm) t (s) T (s)
100 6.5 7.8 0.78
150 5.5 9.8 0.98
200 6.0 10.9 1.09
250 3.5 12.4 1.24
we look for the period that is the time it takes to give a series of oscillations, the results are in the last column
T = t / 10
To find the spring constant we linearize the equation
T² = (4π²/K) m
therefore we see that if we make a graph of T² against the mass, we obtain a line, whose slope is
m ’= 4π² / k
where m’ is the slope
k = 4π² / m'
the equation of the line of the attached graph is
T² = 0.00603 m + 0.0183
therefore the slope
m ’= 0.00603 s²/g
we calculate
k = 4 π² / 0.00603
k = 6547 g / s²
we reduce the mass to the SI system
k = 6547 g / s² (1kg / 1000 g)
k = 6,547 kg / s² =
k = 6,547 N / m
let's reduce the uniqueness
[N / m] = [(kg m / s²) m] = [kg / s²]
The spring-mass system forms a linear graph between the time period and mass. And the value of spring-constant from the given data is 6.46 N/m.
Given data:
Mass suspended by spring is, [tex]m=100 \;\rm g =0.1 \;\rm kg[/tex].
Number of oscillations is, [tex]n =10\;\rm oscillations[/tex].
Time period of oscillation is, [tex]T=7.8 \;\rm s[/tex].
The expression for the angular frequency of spring-mass system is,
[tex]\omega =\drac \sqrt{\dfrac{k}{m} }[/tex] ......................................................(1)
Here, k is the spring constant.
Angular frequency is also expressed as,
[tex]\omega = 2 \pi f[/tex] .........................................................(2)
here, f is the linear frequency of spring-mass system.
And linear frequency is,
[tex]f=\dfrac{n}{T}\\f=\dfrac{10}{7.81}\\f=1.28 \;\rm cycles/sec[/tex]
Then substitute equation (2) in equation (1) as,
[tex]2 \pi f=\drac \sqrt{\dfrac{k}{m} }\\2 \pi \times 1.28=\drac \sqrt{\dfrac{k}{0.1} }\\(2 \pi \times 1.28)^{2}= \dfrac{k}{0.1}\\k = 6.46 \;\rm N/m[/tex]
Thus, the value of spring constant is 6.46 N/m. And the suitable graph for the spring-mass system is given below.
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The inhabitants of a small island export a cloth made from a plant that grows only on their island. A clothier from New York, believing that he can save money by "cutting out the middleman," decides to travel to the island and buy the cloth himself. Ignorant of the local custom where strangers are offered outrageous prices initially, the clothier accepts (much to everyone's surprise) the initial price of 400 tepizes/m^2. The price of this cloth in New York is 120 dollars/yard^2. If the clothing maker bought 500 m^2 of this fabric, how much money did he lose? Use 1tepiz= 0.625dollar and 0.9144m = 1yard.
Answer:
Explanation:
purchase price = 400 tepizes / m²
1 tepiz = .625 dollar
purchase price in terms of dollar = 400 x .625 dollar / m²
= 250 dollar / m²
.9144 m = 1 yard
1 m = 1.0936 yard
1m² = 1.196 yard²
price in terms of dollar / yards²
= 250 / 1.196 dollar / yard²
= 209 dollar / yard²
Price of cloth in New York = 120 dollar / yard²
loss = 209 - 120 = 89 dollar / yard²
500 m² = 500 x 1.196 yard²
= 598 yard²
net loss in purchasing 500 m² cloth
= 598 x 89
= 53222 dollar .
Superman is jogging alongside the railroad tracks on the outskirts of Metropolis at 100 km/h. He overtakes the caboose of a 500-m-long freight train traveling at 50 km/h. At that moment he begins to accelerate at 10 m/s2. How far will the train have traveled before Superman passes the locomotive?
Answer:
d = 41.91 m
Explanation:
In order to calculate the distance traveled by the train while superman passes it, you write the equations of motion for both superman and train:
For train, you have a motion with constant speed. You write the equation of motion of the position of the front of the train:
[tex]x=x_o+v_1t[/tex] (1)
xo: initial position of the front of the train = 500m
v1: speed of the train = 50km/h
For superman, you take into account that the motion is an accelerated motion (you assume superman is at the origin of coordinates):
[tex]x'=v_2t+\frac{1}{2}at^2[/tex] (1)
v2: initial speed of superman = 100km/h
a: acceleration = 10m/s^2
When superman passes the train, both positions x and x' will be equal. Hence, you equal the equations (1) and (2) and you calculate the time t. But before you convert the units of the velocities v1 and v2 to m/s:
[tex]v_1=50\frac{km}{h}*\frac{1000m}{1km}*\frac{1h}{3600s}=13.88\frac{m}{s}\\\\v_2=100\frac{km}{h}=\frac{1000m}{1km}*\frac{1h}{3600s}=27.77\frac{m}{s}[/tex]
Thus, you equal x=x'
[tex]x=x'\\\\x_o+v_1t=v_2t+\frac{1}{2}at^2\\\\500m+(13.88m/s)t=(27.77m/s)t+\frac{1}{2}(10m/s^2)t^2\\\\(50\frac{m}{s^2})t^2+(13.89\frac{m}{s})t-500m=0[/tex]
You solve the last equation for t by using the quadratic formula:
[tex]t_{1,2}=\frac{-13.89\pm \sqrt{(13.89)^2-4(50)(-500)}}{2(50)}\\\\t_{1,2}=\frac{-13.89\pm 316.53}{100}\\\\t_1=3.02s\\\\t_2=-3.30s[/tex]
You only use t1 = 3.02s because negative times do not have physical meaning.
Next, you replace this value of t in the equation (1) to calculate the position of the train (for when superman just passed it):
[tex]x=500m+(13.88m/s)(3.02s)=541.91m[/tex]
x is the position of the front of the train, then, the dstance traveled by the train is:
d = 541.91m - 500m = 41.91 m
Two large insulating parallel plates carry charge of equal magnitude, one positive and the other negative, that is distributed uniformly over their inner surfaces. Rank the points 1 through 5 according to the magnitude of the electric field at the points, least to greatest.
A. 1, 2, 3, 4, 5
B. 2, then 1, 3, and 4 tied, then 5
C. 1, 4, and 5 tie, then 2 and 3 tie
D. 2 and 3 tie, then 1 and 4 tie, then 5
E. 2 and 3 tie, then 1, 4, and 5 tie
Answer:
The correct answer is C 1, 4, and 5 tie, then 2 and 3 tie
Explanation:
Solution
The electric field due to sheets E₁ positive =б/2E₀
E₂ is negative = б/2E₀
Now,
At the point 1, 4, 5 the electric field due to the sheets are in the opposite direction
At the point 1, the net field = -E₁ + E₂ =0
At the point A, the net field = -E₁ - E₂ = 0
Now,
At nay point inside between them, the electric field is seen to be at the same direction.
At the 2, 3 points the field is seen at the right
Thus,
E net = E₁ + E₂
= б/2E₀ + σ/2E₀
=б/E₀
Note: Kindly find an attached copy of the complete question to the solution
The correct answer is option C
The rank of the points according to the magnitude of the electric field is 1, 4, and 5 tie, then 2 and 3 tie
The magnitude of the electric field:
Let sheet 1 has positive surface charge density and sheet 2 has a negative surface charge density
The electric field (without direction) due to sheets will be
E₁ =σ/2E₀
E₂= σ/2E₀
Now,
At the point 1, 4, 5 the electric field due to the sheets is given by:
E = E₁ - E₂
E = σ/2E₀ - σ/2E₀
since the positive charge plate will have electric field lines away from the sheet and the negative charge plate will have electric field lines towards the sheet
E = 0
Now,
At points 2, 3 which are between the plates,
The net electric field is:
E = E₁ + E₂
since the electric field due to both the plates will be from positive to negative ( towards the negatively charged plate)
E = σ/2E₀ + σ/2E₀
E = σ/E₀
Learn more about surface charge density:
https://brainly.com/question/8966223?referrer=searchResults
A uniformly charged ring of radius 10.0 cm has a total charge of 71.0 μC. Find the electric field on the axis of the ring at the following distances from the center of the ring. (Choose the x-axis to point along the axis of the ring.)
(a) 1.00 cm
What is the general expression for the electric field along the axis of a uniformly charged ring? i MN/C
(b) 5.00 cm
i MN/C
(c) 30.0 cm
i MN/C
(d) 100 cm
i MN/C
Answer:
General Expression: E = kql/(l² + r²)^(3/2)
(a) 6.3 MN/C
(b) 22.8 MN/C
(c) 6.1 MN/C
(d) 0.63 MN/C
Explanation:
The general expression for electric field along axis of a uniformly charged ring is:
E = kqL/(L² + r²)^(3/2)
where,
E = Electric Field Strength = ?
k = Coulomb's Constant = 9 x 10⁹ N.m²/C²
q = Total Charge = 71 μC = 71 x 10⁻⁶ C
L = Distance from center on axis
r = radius of ring = 10 cm = 0.1 m
(a)
L = 1 cm = 0.01 m
Therefore,
E = (9 x 10⁹ N.m²/C²)(71 x 10⁻⁶ C)(0.01 m)/[(0.01 m)² + (0.1 m)²]^(3/2)
E = (6390 N.m³/C)/(0.00101 m³)
E = 6.3 x 10⁶ N/C = 6.3 MN/C
(b)
L = 5 cm = 0.05 m
Therefore,
E = (9 x 10⁹ N.m²/C²)(71 x 10⁻⁶ C)(0.05 m)/[(0.05 m)² + (0.1 m)²]^(3/2)
E = (31950 N.m³/C)/(0.00139 m³)
E = 22.8 x 10⁶ N/C = 27.4 MN/C
(c)
L = 30 cm = 0.3 m
Therefore,
E = (9 x 10⁹ N.m²/C²)(71 x 10⁻⁶ C)(0.3 m)/[(0.3 m)² + (0.1 m)²]^(3/2)
E = (191700 N.m³/C)/(0.03162 m³)
E = 6.1 x 10⁶ N/C = 6.1 MN/C
(d)
L = 100 cm = 1 m
Therefore,
E = (9 x 10⁹ N.m²/C²)(71 x 10⁻⁶ C)(1 m)/[(1 m)² + (0.1 m)²]^(3/2)
E = (639000 N.m³/C)/(1.015 m³)
E = 0.63 x 10⁶ N/C = 0.63 MN/C
In each pair, select the body with more internal energy.
Answer:
rt
Explanation:
A certain freely falling object, released from rest, requires 1.85 s to travel the last 26.5 m before it hits the ground. (a) Find the velocity of the object when it is 26.5 m above the ground. (Indicate the direction with the sign of your answer. Let the positive direction be upward.) -2.70 Incorrect: Your answer is incorrect. Your response differs from the correct answer by more than 10%. Double check your calculations. m/s (b) Find the total distance the object travels during the fall.
Answer:
a) -5.26 m/s
b) 27.91 m
Explanation:
a) The acceleration due to gravity makes the velocity increase in magnitude in a linear way. The average velocity over the interval will be equal to the actual velocity halfway through the interval. The velocity at the beginning of the interval will be higher (less negative) by the amount velocity changes in the first half of the interval.
average velocity = (0 -(26.5 m))/(1.85 s) ≈ -14.324 m/s
The change in velocity in the first half of the interval is ...
Δv = (Δt/2)×(-9.8 m/s²) = (1.85 s)(-4.9 m/s²) = -9.065 m/s
So, the initial velocity (at the beginning of the last 1.85 s interval) is ...
v1 = (average velocity) -Δv = (-14.324 m/s) -(-9.065 m/s)
v1 = -5.259 m/s
__
b) The velocity when the object hits the ground is ...
v2 = average velocity +Δv = -14.324 m/s -9.065 m/s = -23.389 m/s
This is related to the distance traveled by ...
v² = 2dg . . . . . where g is the acceleration and d is the distance traveled
d = v²/(2g) = 23.389²/(2·9.8) = 27.911 . . . . meters
The object travels a total distance of about 27.911 meters.
_____
The attached graph shows height vs. time.
A roller coaster car may be approximated by a block of mass m. Thecar, which starts from rest, is released at a height h above the ground and slides along a frictionless track. The car encounters a loop of radius R. Assume that the initial height h is great enough so that the car never losses contact with the track.
Required:
a. Find an expression for the kinetic energy of the car at the top of the loop. Express the kinetic energy in terms of m, g, h, and R.
b. Find the minimum initial height h at which the car can be released that still allows the car to stay in contact with the track at the top of the loop.
Answer:
Explanation:
At height h , potential energy of coaster car having mass m = mgh .
The car will lose potential energy and gain kinetic energy.
height lost by car when it is at the top of loop of radius R
= h - 2R
potential energy lost = mg ( h - 2R )
kinetic energy gained = mg ( h - 2R )
kinetic energy = 0 + mg ( h - 2R )
= mg ( h - 2R )
b )
For the car to remain in contact with the track , if v be the minimum velocity
centripetal force at top = mg
m v² / R = mg
v² = gR
kinetic energy = 1/2 mv²
= 1/2 m x gR
= mgR /
If h be the minimum height that can give this kinetic energy
mg ( h - 2R ) = mgR / 2
h - 2R = R / 2
h = 2.5 R .
In which situation is chemical energy being converted to another form of energy?
Answer:
A burning candle. (chemical energy into energy of heat and light, i.e. thermal and wave)
Explanation:
A tank with a constant volume of 3.72 m3 contains 22.1 moles of a monatomic ideal gas. The gas is initially at a temperature of 300 K. An electric heater is used to transfer 4.5 × 104 J of energy into the gas. It may help you to recall that CV = 12.47 J/K/mole for a monatomic ideal gas, and that the number of gas molecules is equal to Avagadros number (6.022 × 1023) times the number of moles of the gas.
a) What is the temperature of the gas after the energy is added?___K
b) What is the change in pressure of the gas?____Pa
c) How much work was done by the gas during this process?____J
Answer:
a) 463.29 K
b) 8065.65 Pa
c) 0 J
Explanation:
The parameters given are;
Volume of the tank, V = 3.72 m³
Number of moles of gas present in the tank, n = 22.1 moles
Temperature of the gas before heating, T₁ = 300 k
Heat added to the gas, ΔQ = 4.5 × 10⁴ J
Specific heat capacity at constant volume, [tex]c_v[/tex], for monatomic gas = 12.47 J/K/mole
Avogadro's number = 6.022 × 10²³ particles per mole
a) ΔQ = n × [tex]c_v[/tex] × ΔT
Where:
ΔT = T₂ - T₁
T₂ = Final temperature of the gas
Hence, by plugging in the values, we have;
4.5 × 10⁴ = 22.1 × 12.47 × (T₂ - 300)
[tex]T_{2} - 300 = \frac{4.5\times 10^{4}}{22.1\times 12.47}[/tex]
T₂ = 300 + 163.29 = 463.29 K
b) The pressure of the gas is found from the relation;
P×V = n×R×T
[tex]P = \dfrac{n \times R \times T}{V}[/tex]
Where:
P = Pressure of the gas
R = Universal gas constant = 8.3145 J/(mol·K)
T = Temperature of the gas
V = Volume of the gas = 3.72 ³ (constant)
n = Number of moles of gas present = 22.1 moles (constant)
Hence the change in pressure is given by the relation;
[tex]\Delta P = \dfrac{n \times R \times (T_2 - T_1)}{V} = \dfrac{n \times R \times \Delta T}{V}[/tex]
Plugging in the values, we have;
[tex]\Delta P = \dfrac{22.1 \times 8.3145 \times 163.29}{3.72} = 8065.65 \, Pa[/tex]
c) Work done, W, by the gas is given by the area under the pressure to volume graph which gives;
W = f(P) × ΔV
The volume given in the question is constant
∴ ΔV = 0
Hence, W = f(P) × 0 = 0 J
No work done by the gas during the process.
John heats 1 kg of soup from 25 °C to 70 °C for 15 minutes by a heater. How long does the same heater take to heat 1.5 kg of the same kind of soup from 20 °C to 80 °C? The energy output per unit time by the heater is constant.
Answer:
30 minutes
Explanation:
Energy per time is constant, so:
E₁ / t₁ = E₂ / t₂
m₁C₁ΔT₁ / t₁ = m₂C₂ΔT₂ / t₂
(1 kg) C (70°C − 25°C) / 15 min = (1.5 kg) C (80°C − 20°C) / t
(1 kg) (45°C) / 15 min = (1.5 kg) (60°C) / t
3/min = 90 / t
t = 30 min
g: To open a door, you apply a force of 10 N on the door knob, directed normal to the plane of the door. The door knob is 0.9 meters from the hinge axis, and the door swings open with an angular acceleration of 5 radians per second squared. What is the moment of inertia of the door
Answer:
I =1.8 kgm^2
Explanation:
In order to calculate the moment of inertia of the door you use the following formula, which relates the torque applied to the door with its moment of inertia and angular acceleration:
[tex]\tau=I\alpha[/tex] (1)
τ: torque applied to the door
I: moment of inertia of the door
α: angular acceleration = 5 rad/s^2
The torque is also given by τ = Fd, where F is the force applied at a distance of d to the pivot of the door (hinge axis).
F = 10 N
d = 0.9 m
You replace the expression for τ, and solve for I:
[tex]Fd=I\alpha\\\\I=\frac{Fd}{\alpha}\\\\I=\frac{(10N)(0.9m)}{5rad/s^2}=1.8kgm^2[/tex]
The moment of inertia of the door is 1.8 kgm^2
