Answer:
Uncertainty in position (∆x) = 24 × 10⁻¹⁵ mUncertainty in momentum (∆P) = ?Planck's constant (h) = 6.26 × 10⁻³⁴ Js[tex]\longrightarrow \: \: \sf\Delta x .\Delta p = \dfrac{h}{4\pi} [/tex]
[tex]\longrightarrow \: \: \sf24 \times {10}^{ - 15} .\Delta p = \dfrac{6.26 \times {10}^{ - 34}} {4 \times \frac{22}{7} } [/tex]
[tex]\longrightarrow \: \: \sf24 \times {10}^{ - 15} .\Delta p = \dfrac{6.26 \times {10}^{ - 34}} { \frac{88}{7} } [/tex]
[tex]\longrightarrow \: \: \sf24 \times {10}^{ - 15} .\Delta p = \dfrac{6.26 \times {10}^{ - 34} \times 7} { 8 } [/tex]
[tex]\longrightarrow \: \: \sf\Delta p = \dfrac{43.82 \times {10}^{ - 34} } { 8 \times 24 \times {10}^{ - 15} } [/tex]
[tex]\longrightarrow \: \: \sf\Delta p = \dfrac{43.82 \times {10}^{ - 34} } { 192 \times {10}^{ - 15} } [/tex]
[tex]\longrightarrow \: \: \sf\Delta p = \dfrac{43.82 \times {10}^{ - 34} \times {10}^{15} } { 192} [/tex]
[tex]\longrightarrow \: \: \sf\Delta p = \dfrac{43.82 \times {10}^{ -19} } { 192} [/tex]
[tex]\longrightarrow \: \: \sf\Delta p = \dfrac{4382 \times {10}^{ - 2} \times {10}^{ -19} } { 192} [/tex]
[tex]\longrightarrow \: \: \sf\Delta p = \dfrac{4382 \times {10}^{ - 21} } { 192} [/tex]
[tex]\longrightarrow \: \: \sf\Delta p = 22.822\times {10}^{ - 21} [/tex]
[tex]\longrightarrow \: \: \sf\Delta p = 2.2822 \times {10}^{1} \times {10}^{ - 21} [/tex]
[tex]\longrightarrow \: \: \underline{ \boxed{ \red{ \bf\Delta p = 2.2822 \times {10}^{ - 20} \: kg/ms}}}[/tex]
Explain how the gravitational force between the earth and the sun changes as the earth moves from position A to B as shown in the figure. Sun Earth at position B Earth at position A
Answer:
The distance between sun & Earth at position A is less than the earth at position B. The gravitational force of two bodies is inversely proportional to the square of the distance. So At position A gravitational force is more & it decreases as it rotate towards position B.
A driver who does not wear a seat belt continues to move at the initial velocity until she or he hits something solid (e.g the steering wheel) and then comes to rest in a very short distance. Find the net force on a driver without seat belts who comes to rest in 1.1 cm. Fwithout belt
This question is incomplete, the complete question is;
Seatbelts provide two main advantages in a car accident (1) they keep you from being thrown from the car and (2) they reduce the force that acts on your during the collision to survivable levels. This second benefit can be illustrated by comparing the net force encountered by a driver in a head-on collision with and without a seat beat.
1) A driver wearing a seat beat decelerates at roughly the same rate as the car it self. Since many modern cars have a "crumble zone" built into the front of the car, let us assume that the car decelerates of a distance of 1.1 m. What is the net force acting on a 70 kg driver who is driving at 18 m/sec and comes to rest in this distance?
Fwith belt =
2) A driver who does not wear a seat belt continues to move at the initial velocity until she or he hits something solid (e.g the steering wheel) and then comes to rest in a very short distance. Find the net force on a driver without seat belts who comes to rest in 1.1 cm.
Fwithout belt =
Answer:
1) The Net force on the driver with seat belt is 10.3 KN
2) the Net force on the driver without seat belts who comes to rest in 1.1 cm is 1030.9 KN
Explanation:
Given the data in the question;
from the equation of motion, v² = u² + 2as
we solve for a
a = (v² - u²)/2s ----- let this be equation 1
we know that, F = ma ------- let this be equation 2
so from equation 1 and 2
F = m( (v² - u²)/2s )
where m is mass, a is acceleration, u is initial velocity, v is final velocity and s is the displacement.
1)
Wearing sit belt, car decelerates of a distance of 1.1 m. What is the net force acting on a 70 kg driver who is driving at 18 m/sec and comes to rest in this distance.
i.e, m = 70 kg, u = 18 m/s, v = 0 { since it came to rest }, s = 1.1 m
so we substitute the given values into the equation;
F = 70( ((0)² - (18)²) / 2 × 1.1 )
F = 70 × ( -324 / 2.4 )
F = 70 × -147.2727
F = -10309.09 N
F = -10.3 KN
The negative sign indicates that the direction of the force is opposite compared to the direction of the motion.
Fwith belt = 10.3 KN
Therefore, Net force of the driver is 10.3 KN
2)
No sit belt,
m = 70 kg, u = 18 m/s, v = 0 { since it came to rest }, s = 1.1 cm = 1.1 × 10⁻² m
we substitute
F = 70( ((0)² - (18)²) / 2 × 1.1 × 10⁻² )
F = 70 × ( -324 / 0.022 )
F = 70 × -14727.2727
F = -1030909.08 N
F = -1030.9 KN
The negative sign indicates that the direction of the force is opposite compared to the direction of the motion.
Fwithout belt = 1030.9 KN
Therefore, the net force on the driver without seat belts who comes to rest in 1.1 cm is 1030.9 KN
Ayuda Porfavor es URGENTE
Consulta los dibujos adjuntos
Thiết bị trao đổi nhiệt ngược chiều, nhiệt độ chất lỏng nóng vào là 310°C, ra là 115°c. Nhiệt độ chất lỏng lạnh vào là 25°c và ra là 68°c. Hãy tính độ chênh lệch nhiệt độ trung bình ∆t(m)
Answer: 129.5 m
Explanation:
310 + 115 + 25 + 68 = 518
518 / 4 = 129.5 m
i think. Sorry if this is wrong
:)
If you tethered a space station to the earth by a long cable, you could get to space in an elevator that rides up the cable much simpler and cheaper than riding to space on a rocket. There's one big problem, however: There is no way to create a cable that is long enough. The cable would need to reach 36,000 km upward, to the height where a satellite orbits at the same speed as the earth rotates; a cable this long made of ordinary materials couldn't even support its own weight. Consider a steel cable suspended from a point high above the earth. The stress in the cable is highest at the top; it must support the weight of cable below it.
What is the greatest length the cable could have without failing?
Answer:
[tex]l=12916.5m[/tex]
Explanation:
Distance [tex]d=3600km[/tex]
Since
Density of steel [tex]\rho=7900kg/m^3[/tex]
Stress of steel [tex]\mu= 1*10^9[/tex]
Generally the equation for Stress on Cable is mathematically given by
[tex]S=\frac{F}{A}[/tex]
[tex]S=\frac{\rho Alg}{A}[/tex]
Therefore
[tex]l=\frac{s}{\rhog}[/tex]
[tex]l=\frac{ 1*10^9}{7900kg/m^3*9.8}[/tex]
[tex]l=12916.5m[/tex]
Galaxies that are 400 million light years away have a red shift of 0.03 approximately. A radio wave coming from one of these galaxies has an observed wavelength of 125 meters. What is the emitted wavelength in meters and the observed frequency in Megahertz
Answer:
The correct answer is "121.36 meters and 2.40 MHz".
Explanation:
Given:
Red shift,
[tex]\frac{v}{c}=\frac{\lambda - \lambda_0}{\lambda_0} = 0.03[/tex]
Wavelength,
[tex]\lambda = 125 \ meters[/tex]
The observed frequency will be:
⇒ [tex]f = \frac{c}{\lambda}[/tex]
[tex]=\frac{3\times 10^8}{125}[/tex]
[tex]=24\times 10^5[/tex]
[tex]= 2.40\times 10^6[/tex]
[tex]=2.40 \ MHz[/tex]
hence,
The Emitted wavelength will be:
⇒ [tex]\frac{125-\lambda_0}{\lambda_0} =0.03[/tex]
[tex]\frac{125}{\lambda_0}-1 =0.03[/tex]
[tex]125=1.03 \ \lambda_0[/tex]
[tex]\lambda_0=\frac{125}{1.03}[/tex]
[tex]=121.36 \ m[/tex]
Canopus, which is in the constellation of Carina and Argo Navis, is 310 ly away. You plan a sight-seeing vacation for Canopus and book a flight on the fastest spaceship in the universe, which travels at the speed of light. How many meters will you traverse to reach your vacation destination
Answer:
[tex]D'=2.933*10^{18}m[/tex]
Explanation:
From the question we are told that:
Distance [tex]D=310ly[/tex]
Where
[tex]1ly=9.46*10^{15}[/tex]
Generally the equation for meters will you traverse to reach your vacation destination is mathematically given by
[tex]D'=D*1ly[/tex]
[tex]D'=310*9.46*10^{15}[/tex]
[tex]D'=2.933*10^{18}m[/tex]
energy of fossil fuel is also derived from solar energy why
Explanation:
All the energy in oil, gas, and coal originally came from the sun captured through photosynthesis. In the same way that we burn wood to release energy that trees capture from the sun, we burn fossil fuels to release the energy that ancient plants captured from the sun.
Water stands at a depth H in a tank whose side wails are vertical. A hole is made on one of the walls at a depth h below the water surface. Find at what distance from the foot of the wall does the emerging stream of water strike the floor?
Answer:
x = 2h [tex]\sqrt{\frac{H}{h} -1 }[/tex]
Explanation:
This is an exercise in fluid mechanics, let's find the speed of the water in the hole
P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂
where the subscript 1 is for the tank surface and the subscript 2 is for the depth of the hole
the pressure inside and outside the tank is the same
P₁ = P₂
we must measure the distance from the same reference point, let's locate it on the surface of the water
y₁ = 0
y₂ = h
Suppose the gap is small compared to the diameter of the tank
v₁ «v₂
v₂² = 2 g (0-h)
This is the speed of the outlet water in the tank, as the force is horizontal this speed is horizontal.
Let's use the projectile launch ratios
vₓ = [tex]\sqrt{2g |h|}[/tex]
y = y₀ + v₀ t - ½ g t²
the height when reaching the floor is y = 0,
the initial height is measured from the floor therefore y₀ = H-h
0 = (H-h) + 0 - ½ g t²
t = [tex]\sqrt{\frac{2(H-h)}{g} }[/tex]
we look for the distance x traveled
x = vₓ t
x = [tex]\sqrt{2g |h| } \ \sqrt{ 2(H-h)/g}[/tex]
x = [tex]\sqrt{4 h (H-h)}[/tex]
x = 2h [tex]\sqrt{\frac{H}{h} -1 }[/tex]
Put the balloon near (BUT NOT TOUCHING) the wall. Leave about as much space as the width of your pinky finger between the balloon and wall. Does the balloon move, if so which way
Answer:
Move towards the wall.
Explanation:
When the balloon is kept near to the wall not touching the wall, there is a force of electrostatic attraction so that the balloon moves towards the wall and stick to it.
As there is some charge on the balloon and the wall is uncharged so the force is there due to which the balloon moves towards the wall.
What is the difference between muscular endurance and muscular strength? A B or C
Answer:
Muscular endurance is how many times you can move a weight without getting tired.
Muscular strength is the amount force you can put out.
A 2.0-kg block sliding on a rough horizontal surface is attached to one end of a horizontal spring (k = 250 N/m) which has its other end fixed. The block passes through the equilibrium position with a speed of 2.6 m/s and first comes to rest at a displacement of 0.20 m from equilibrium. What is the coefficient of kinetic friction between the block and the horizontal surface?
Suppose the spring begins in a compressed state, so that the block speeds up from rest to 2.6 m/s as it passes through the equilibrium point, and so that when it first comes to a stop, the spring is stretched 0.20 m.
There are two forces performing work on the block: the restoring force of the spring and kinetic friction.
By the work-energy theorem, the total work done on the block between the equilbrium point and the 0.20 m mark is equal to the block's change in kinetic energy:
[tex]W_{\rm total}=\Delta K[/tex]
or
[tex]W_{\rm friction}+W_{\rm spring}=0-K=-K[/tex]
where K is the block's kinetic energy at the equilibrium point,
[tex]K=\dfrac12\left(2.0\,\mathrm{kg}\right)\left(2.6\dfrac{\rm m}{\rm s}\right)^2=6.76\,\mathrm J[/tex]
Both the work done by the spring and by friction are negative because these forces point in the direction opposite the block's displacement. The work done by the spring on the block as it reaches the 0.20 m mark is
[tex]W_{\rm spring}=-\dfrac12\left(250\dfrac{\rm N}{\rm m}\right)(0.20\,\mathrm m)^2=-5.00\,\mathrm J[/tex]
Compute the work performed by friction:
[tex]W_{\rm friction}-5.00\,\mathrm J=-6.76\,\mathrm J \implies W_{\rm friction}=-1.76\,\mathrm J[/tex]
By Newton's second law, the net vertical force on the block is
∑ F = n - mg = 0 ==> n = mg
where n is the magnitude of the normal force from the surface pushing up on the block. Then if f is the magnitude of kinetic friction, we have f = µmg, where µ is the coefficient of kinetic friction.
So we have
[tex]W_{\rm friction}=-f(0.20\,\mathrm m)[/tex]
[tex]\implies -1.76\,\mathrm J=-\mu\left(2.0\,\mathrm{kg}\right)\left(9.8\dfrac{\rm m}{\mathrm s^2}\right)(0.20\,\mathrm m)[/tex]
[tex]\implies \boxed{\mu\approx0.45}[/tex]
The coefficient of kinetic friction between the block and the horizontal surface [tex]\mu =0.45[/tex]
What is coefficient of friction?Coefficient of friction, ratio of the frictional force resisting the motion of two surfaces in contact to the normal force pressing the two surfaces together. It is usually symbolized by the Greek letter mu (μ). Mathematically, μ = F/N, where F is the frictional force and N is the normal force.
Suppose the spring begins in a compressed state, so that the block speeds up from rest to 2.6 m/s as it passes through the equilibrium point, and so that when it first comes to a stop, the spring is stretched 0.20 m.
There are two forces performing work on the block: the restoring force of the spring and kinetic friction.
By the work-energy theorem, the total work done on the block between the equilbrium point and the 0.20 m mark is equal to the block's change in kinetic energy:
[tex]W_{total}=\Delta K[/tex]
or
[tex]W_{friction}+W_{spring}=0-K=-K[/tex]
where K is the block's kinetic energy at the equilibrium point,
[tex]K=\dfrac{1}{2}(2)(2.6)^2=6.76 \ J[/tex]
Both the work done by the spring and by friction are negative because these forces point in the direction opposite the block's displacement. The work done by the spring on the block as it reaches the 0.20 m mark is
[tex]W_{spring}=-\dfrac{1}{2}(250)(0.20)=-5\ J[/tex]
Compute the work performed by friction:
[tex]W_{friction}-5 =-6.76\ J=-1.76\ J[/tex]
By Newton's second law, the net vertical force on the block is
∑ F = n - mg = 0 ==> n = mg
where n is the magnitude of the normal force from the surface pushing up on the block. Then if f is the magnitude of kinetic friction, we have f = µmg, where µ is the coefficient of kinetic friction.
So we have
[tex]W_{friction}=-f(0.20)[/tex]
[tex]-1.76=\mu (2)(9.8)(0.2)[/tex]
[tex]\mu =0.45[/tex]
Thus the coefficient of kinetic friction between the block and the horizontal surface [tex]\mu =0.45[/tex]
To know more about Coefficient of friction follow
https://brainly.com/question/136431
how do I learn French fast for an examination
Explanation:
It isn't possible to learn an entire language fast. For an examination, memorise some of the easy words. Practice mew minutes speaking French and keep speaking it till you know it.
Consider an airplane with a total wing surface of 50 m^2. At a certain speed the difference in air pressure below and above the wings is 4.0 % of atmospheric pressure.
Required:
Find the lift on the airplane.
Answer:
[tex]F=202650N[/tex]
Explanation:
From the question we are told that:
Area [tex]a=50m^2[/tex]
Difference in air Pressure [tex]dP=4.0\% atm=>0.04*101325=>4035Pa[/tex]
Generally the equation for Force is mathematically given by
[tex]F=dP*A[/tex]
[tex]F=4053*50[/tex]
[tex]F=202650N[/tex]
True or false: Increasing the Young’s modulus of a beam in bending will cause it to deflect less.
Answer:
false?
Explanation:
The higher the modulus, the more stress is needed to create the same amount of strain; an idealized rigid body would have an infinite Young's modulus.
Answer:
I think the answer is False.
A pulse traveled the length of a stretched spring the pulse transferred...A)energy only B)mass only C)both energy and mass D) neither energy nor mass
Answer:
A
Explanation:
So a pulse is a part of a mechanical wave, and mechanical waves are energy transfer trough some medium, in this case a stretched spring. So the correct answer is (A) energy only. The pulse cant be transferred into mass.
A possible means for making an airplane invisible to radar is to coat the plane with an antireflective polymer. If radar waves have a wavelength of 3.00 cm and the index of refraction of the polymer is n = 1.50, how thick would you make the coating?
Answer:
[tex]t=0.50cm[/tex]
Explanation:
From the question we are told that:
Wavelength [tex]\lamda=3c[/tex]m
Refraction Index [tex]n=1.50[/tex]
Generally the equation for Destructive interference for Normal incidence is mathematically given by
[tex]2nt=m(\frac{1}{2})\lambda[/tex]
Since Minimum Thickness occurs at
At [tex]m=0[/tex]
Therefore
[tex]t=\frac{\lambda}{2}[/tex]
[tex]t=\frac{3}{4(1.50)}[/tex]
[tex]t=0.50cm[/tex]
What is total resistor formula
Answer:
If you know the current and voltage across the whole circuit, you can find total resistance using Ohm's Law: R = V / I.
Explanation:
Cho điện cực dưới điện cực trên . Hàm thế biến thiên theo qui luật:
Xác định sự phân bố điện tích khối và .
Answer:
Quantum theory gives the concept of
If the average time it takes for the cart from point 1 to point 2 is 0.2 s, calculate the angle θ from the horizontal of the track. Assume the track is frictionless. Hint: use the definitions of acceleration and Newton’s second law.
Answer:
hehe
Explanation:
I dont know because I am a noob ant study
In a robotics circuit, a voltage source of 75V is supplying a current, I to a series circuit of 5
resistances. Resistance, R1 = 5 KΩ and R2 = 10 KΩ. The voltage drops across 3 black boxes of
resistances R3 , R4 and R5 are 15V, 20V and 25V respectively. The current through the black
box of resistance, R5 is measured as 1mA. Calculate the voltage V1 and V2 across the
resistance R1 and R2 using the Voltage Divider Rule.
Answer:
In the given circuit, R
2
,R
6
and R
4
are in series. So,
R
1
′
=7+5+12=24Ω
Now R
1
′
and R
5
′
are in parallel. So,
R
2
′
1
=
8
1
+
24
1
=
24
3+1
=
24
4
=
6
1
R
2
′
=6ohm.
Now R
2
′
,R
1
and R
3
are in series. So,
R=R
2
′
+R
1
+R
3
=6+3+2=11ohm.
We know i=
R+r
E
=
11+1
6
=
12
6
=
2
1
i=0.5amp.
Hi,A body changes its velocity from 60 km/hr to 72 km/hr in 2 sec.Find the acceleration and distance travelled.
Answer:
Initial velocity, u = 60 km/h = 16.7 m/s
Final velocity, v = 72 km/h = 20 m/s
time, t = 2 sec
From first equation of motion:
[tex]{ \bf{v = u + at}}[/tex]
Substitute the variables:
[tex]{ \tt{20 = 16.7 + (a \times 2)}} \\ { \tt{2a = 3.3}} \\ { \tt{acceleration = 1.65 \: {ms}^{ - 2} }}[/tex]
Consult Multiple-Concept Example 11 for background material relating to this problem. A small rubber wheel on the shaft of a bicycle generator presses against the bike tire and turns the coil of the generator at an angular speed that is 40 times as great as the angular speed of the tire itself. Each tire has a radius of 0.330 m. The coil consists of 163 turns, has an area of 3.23 x 10-3 m2, and rotates in a 0.0948-T magnetic field. The bicycle starts from rest and has an acceleration of 0.601 m/s2. What is the peak emf produced by the generator at the end of 6.72 s
Answer:
[tex]Emf=24.4V[/tex]
Explanation:
From the question we are told that:
Angular speed [tex]\omega=40*\omega'[/tex]
Radius [tex]r=0.33m[/tex]
No. Turns [tex]N=163[/tex]
Area [tex]A= 3.23 * 10^{-3} m^2[/tex]
Magnetic field. [tex]B=0.0948T[/tex]
Acceleration [tex]a=0.60m/s^2[/tex]
Time [tex]t=6.72s[/tex]
Generally the equation for momentum is mathematically given by
[tex]\omega'=\omega_o+\frac[a}{r}t[/tex]
[tex]\omega'=0+\frac{0.60}{0.33}*6.72[/tex]
[tex]W=12.22rad/s^2[/tex]
Therefore
[tex]\omega=Aw[/tex]
[tex]\omega=12.22*40[/tex]
[tex]\omega=488.7rads/s[/tex]
Generally the equation for Peak emf is mathematically given by
[tex]Emf=NBA \omega[/tex]
[tex]Emf=163*0.0948* 3.23 * 10^{-3} m^2*488.7rads/s[/tex]
[tex]Emf=24.4V[/tex]
A 0.3-kg object connected to a light spring with a force constant of 19.3 N/m oscillates on a frictionless horizontal surface. Assume the spring is compressed 6 cm and released from rest. (c) Determine the speed of the object as it passes the point 1.9 cm from the equilibrium position
The total work W done by the spring on the object as it pushes the object from 6 cm from equilibrium to 1.9 cm from equilibrium is
W = 1/2 (19.3 N/m) ((0.060 m)² - (0.019 m)²) ≈ 0.031 J
That is,
• the spring would perform 1/2 (19.3 N/m) (0.060 m)² ≈ 0.035 J by pushing the object from the 6 cm position to the equilibrium point
• the spring would perform 1/2 (19.3 N/m) (0.019 m)² ≈ 0.0035 J by pushing the object from the 1.9 cm position to equilbrium
so the work done in pushing the object from the 6 cm position to the 1.9 cm position is the difference between these.
By the work-energy theorem,
W = ∆K = K
where K is the kinetic energy of the object at the 1.9 cm position. Initial kinetic energy is zero because the object starts at rest. So
W = 1/2 mv ²
where m is the mass of the object and v is the speed you want to find. Solving for v, you get
v = √(2W/m) ≈ 0.46 m/s
The motion of a particle is defined by the relation x 5 t 3 2 9t 2 1 24t 2 8, where x and t are expressed in inches and seconds, respectively. Determine (a) when the velocity is zero, (b) the position and the total distance traveled when the acceleration is zero.
Answer:
a) t = 2.0 s, b) x_f = - 24.56 m, Δx = 16.56 m
Explanation:
This is an exercise in kinematics, the relationship of position and time is indicated
x = 5 t³ - 9t² -24 t - 8
a) ask when the velocity is zero
speed is defined by
v = [tex]\frac{dx}{dt}[/tex]
let's perform the derivative
v = 15 t² - 18t - 24
0 = 15 t² - 18t - 24
let's solve the quadratic equation
[tex]t = \frac{18 \pm \sqrt{18^2 + 4 \ 15 \ 24} }{2 \ 15}[/tex]
t = [tex]\frac{18 \pm 42}{30}[/tex]
t1 = -0.8 s
t2 = 2.0 s
the time has to be positive therefore the correct answer is t = 2.0 s
b) the position and distance traveled for a = 0
acceleration is defined by
a = dv / dt
a = 30 t - 18
a = 0
30 t = 18
t = 18/30
t = 0.6 s
we substitute this time in the expression of the position
x = 5 0.6³ - 9 0.6² - 24 0.6 - 8
x = 1.08 - 3.24 - 14.4 - 8
x = -24.56 m
we see that all the movement is in one dimension so the distance traveled is the change in position between t = 0 and t = 0.6 s
the position for t = 0
x₀ = -8 m
the position for t = 0.6 s
x_f = - 24.56 m
the distance
ΔX = x_f - x₀
Δx = | -24.56 -(-8) |
Δx = 16.56 m
An infinite plane lies in the yz-plane and it has a uniform surface charge density.
The electric field at a distance x from the plane
a.) decreases as 1/x^2
b.) increases linearly with x
c.) is undertermined
d.) decreases linearly with x
e.) is constant and does not depend on x
Answer:
So the correct answer is letter e)
Explanation:
The electric field of an infinite yz-plane with a uniform surface charge density (σ) is given by:
[tex]E=\frac{\sigma }{2\epsilon_{0}}[/tex]
Where ε₀ is the electric permitivity.
As we see, this electric field does not depend on distance, so the correct answer is letter e)
I hope it helps you!
A hockey ball is flicked of the ground with initial velocity of 2.0m/s upwards and 10m/s horizontally. Calculate the distance travelled from the point where the ball is flicked and to the point where the ball hits the ground.
Answer:
imma try and fail again and again
Describe how a positively charged object can be used to give another object a negative charge. What is the name of this process
Answer:
induction using ground connection
Explanation:
This process is referred to as induction and it is done using ground process.
The process can be carried out as follows;
-An object that has a positive charge would be brought close to another object that is neutrally charged. then the electrons of the neutral object would be attracted and +ve charges would be repelled. This would cause an alignment at the two opposite corners.
- After this is done, the object would be grounded from the side that has the positive charges. This would bring about electron flowing from the ground to the earth. bringing about a neutralization of the positive charges.
-then the grounding would be taken away resulting in electrons leaving in majority.
-At the end, after the positively charged object has been removed, the charges would reassemble themselves. this would cause the object to be negatively charged.
therefore negative charge can be induced on a different object that is neutral using positive charge, through the induction process and grounding.
The driver provides a constant force on the engine through the foot pedal. Eventually the van stops accelerating and reaches a constant speed.
c Explain why the van reaches a constant speed if the driver provides a constant driving force to the van.
It follows from Newton's second law that there is some counteractive force that cancels out the force exerted by the engine - it's most likely drag due to air resistance in combination with static friction between the tires and the road. The car is moving at constant speed past a certain point, so the net force on the car is
∑ F = (force from engine) - (resistive forces) = 0
A 50.0 kg person is walking horizontally with constant acceleration of 0.25 m/s² inside an elevator. The elevator is also accelerating downward at a rate of 1.0 m/s². Sketch the path of the man as it is observed from someone on the ground. Explain your choice.
Answer:
The acceleration is in 2 D as in between east and south.
Explanation:
mass, m = 50 kg
acceleration, a = 0.25 m/s^2 horizontal
acceleration of elevator, a' = 1 m/s^2 downwards
When a person on the ground the resultant acceleration of the person with respect to the ground is between east and south direction so the path os parabolic in nature. It graph is shown below:
