The electric potential V(z) on the z-axis is : V = [tex](\frac{Q}{a^2} ) [ (a^2 + z^2)^{\frac{1}{2} } -z[/tex]
The magnitude of the electric field on the z axis is : E = kб 2[tex]\pi[/tex]( 1 - [z / √(z² + a² ) ] )
Given data :
V(z) =2kQ / a²(v(a² + z²) ) -z
Determine the electric potential V(z) on the z axis and magnitude of the electric fieldConsidering a disk with radius R
Charge = dq
Also the distance from the edge to the point on the z-axis = √ [R² + z²].
The surface charge density of the disk ( б ) = dq / dA
Small element charge dq = б( 2πR ) dr
dV [tex]\frac{k.dq}{\sqrt{R^2+z^2} } \\\\= \frac{k(\alpha (2\pi R)dR}{\sqrt{R^2+z^2} }[/tex] ----- ( 1 )
Integrating equation ( 1 ) over for full radius of a
∫dv = [tex]\int\limits^a_o {\frac{k(\alpha (2\pi R)dR)}{\sqrt{R^2+z^2} } } \,[/tex]
V = [tex]\pi k\alpha [ (a^2+z^2)^\frac{1}{2} -z ][/tex]
= [tex]\pi k (\frac{Q}{\pi \alpha ^2})[(a^2 +z^2)^{\frac{1}{2} } -z ][/tex]
Therefore the electric potential V(z) = [tex](\frac{Q}{a^2} ) [ (a^2 + z^2)^{\frac{1}{2} } -z[/tex]
Also
The magnitude of the electric field on the z axis is : E = kб 2[tex]\pi[/tex]( 1 - [z / √(z² + a² ) ] )
Hence we can conclude that the answers to your question are as listed above.
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Two particles are 15 meters apart.
Particle A has a charge of 6.0*10^-4 C
Particle B has a charge of 5.0*10^-4 C.
The resulting Coulomb force is 12 N. At the same distance, what combination of charges would yield the same Coulomb force?
A. Particle A: 8.0*10^-4 C, and Particle B: 3.0*10^-4 C
B. Particle A: 12.0*10^-4 C, and Particle B: 2.5*10^-4 C
C. Particle A: 9.0*10^-4 C, and Particle B: 8.0*10^-4 C
D. Particle A: 9.0*10^-4 C, and Particle B: 3.0*10^-4 C
Considering the Coulomb's Law, the combination of particle A: 12.0×10⁻⁴ C and particle B: 2.5×10⁻⁴ C would yield the same Coulomb force of 12 N.
Coulomb's LawCharged bodies experience a force of attraction or repulsion on approach.
From Coulomb's Law it is possible to predict what the electrostatic force of attraction or repulsion between two particles will be according to their electric charge and the distance between them.
From Coulomb's Law, the electric force with which two point charges at rest attract or repel each other is directly proportional to the product of the magnitude of both charges and inversely proportional to the square of the distance that separates them:
[tex]F=k\frac{Qq}{d^{2} }[/tex]
where:
F is the electrical force of attraction or repulsion. It is measured in Newtons (N).Q and q are the values of the two point charges. They are measured in Coulombs (C).d is the value of the distance that separates them. It is measured in meters (m).K is a constant of proportionality called the Coulomb's law constant. It depends on the medium in which the charges are located. Specifically for vacuum k is approximately 9×10⁹ [tex]\frac{Nm^{2} }{C^{2} }[/tex].The force is attractive if the charges are of opposite sign and repulsive if they are of the same sign.
This caseIn this case, you know that two particles are 15 meters apart and the resulting Coulomb force is 12 N.
On the other side, you know:
Particle A: 8.0×10⁻⁴ C, and Particle B: 3.0×10⁻⁴ CParticle A: 12.0×10⁻⁴ C, and Particle B: 2.5×10⁻⁴ CParticle A: 9.0×10⁻⁴ C, and Particle B: 8.0×10⁻⁴ CParticle A: 9.0×10⁻⁴ C, and Particle B: 3.0×10⁻⁴ CReplacing in the Coulomb's Law, you get:
[tex]F=9x10^{9} \frac{Nm^{2} }{C^{2} }\frac{(8x10^{-4}C) (3x10^{-4} C )}{(15 m)^{2} }[/tex] → F= 9.6 N[tex]F=9x10^{9} \frac{Nm^{2} }{C^{2} }\frac{(12x10^{-4}C) (2.5x10^{-4} C )}{(15 m)^{2} }[/tex] → F= 12 N[tex]F=9x10^{9} \frac{Nm^{2} }{C^{2} }\frac{(9x10^{-4}C) (8x10^{-4} C )}{(15 m)^{2} }[/tex] → F= 28.8 N[tex]F=9x10^{9} \frac{Nm^{2} }{C^{2} }\frac{(9x10^{-4}C) (3x10^{-4} C )}{(15 m)^{2} }[/tex] → F=10.8 NFinally, the combination of particle A: 12.0×10⁻⁴ C and particle B: 2.5×10⁻⁴ C would yield the same Coulomb force of 12 N.
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A crate of oranges is shoved across a grocery store floor with a force of 100 N for a distance of 1 meter. The crate travels an additional 1 meter after the shove and stops just short of the display. You can conclude: the magnitude of work done by frictional forces on the crate is 100 J the magnitude of work done by the applied force on the crate is 100 J the magnitude of the force of friction on the crate is 50 N all of the above conclusions are correct none of the above conclusions are correct
The statements that are true are;
the magnitude of work done by frictional forces on the crate is 100 Jthe magnitude of work done by the applied force on the crate is 100 JWhat is work done?Work is said to be done when te force applied travels a distance in the direction of the force. Now we can see that when the force is applied by shoving the crate, work is done, an additional work is done by friction to bring the crate to a stop.
Hence, the following are true;
the magnitude of work done by frictional forces on the crate is 100 Jthe magnitude of work done by the applied force on the crate is 100 JLearn more about work: https://brainly.com/question/18094932?
A source of light emits photons with a wavelength of 8.1 x 10-8 meters. What is the frequency of this light
Answer:
Explanation:
Speed of light v = 3 x 10⁸ m/s
wavelength λ = 8.1 x 10⁻⁸ m
frquency f = v/λ = 3.7 x 10¹⁵ Hz
If a source of light emits photons with a wavelength of 8.1 x 10⁻⁸ meters, then the frequency of the light would be 3.7 × 10¹⁵ Hz, as the wavelength and the frequency of the photon are inversely proportional to each other.
What is Wavelength?It can be understood in terms of the distance between any two similar successive points across any wave for example wavelength can be calculated by measuring the distance between any two successive crests.
C = λν
As given in the problem if a source of light emits photons with a wavelength of 8.1 x 10⁻⁸ meters, then we have to find out the frequency of the light,
The frequency of the light = 3 × 10⁸ / 8.1 x 10⁻⁸
=3.7 × 10¹⁵ Hz
Thus, the frequency of the light would be 3.7 × 10¹⁵ Hz
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cornvet 500000grams in short form of using suitable prefix.
Answer:
0.5 mega grams
Explanation:
PLEASE HELP!!!!!!! MT TIME FOR MY TEST IS ALMOST OVER!!!
An ideal spring, with a pointer attached to its end, hangs next to a scale. With a 100-N weight attached, the pointer indicates "40" on the scale as shown. Using a 200-N weight instead results in "60" on the scale. Using an unknown weight X instead results in "30" on the scale. The weight of X is:
Answer:
50 N
Explanation:
Let the natural length of the spring = L
so
100 = k(40 - L) (1)
200 = k(60 - L) (2)
(2)/(1): 2 = (60 - L)/(40 - L)
60 - L = 2(40 - L)
60 - L = 80 - 2L
2L - L = 80 - 60
L = 20
Sub it into (1):
100 = k(40 - 20) = 20k
k = 100/20 = 5 N/in
Now
X = k(30 - L) = 5(30 - 20) = 50 N
The two graphs shown represent the motion of two blocks with different masses, m1 and m2. The blocks are oscillating on identical springs. Which of the following statements correctly describes the relationship between m1 and m2 and provides evidence from the graphs?
Answer:
M1 would seem to be slower because of a larger mass
x1 = A1 sin ω1 t1 describes the displacement
ω1 / ω2 = ((k1 / k2) / (m1 / m2))^1/2 = (m2 / m1)^1/2 since k's are equal
ω1 / ω2 = 1/2 from graph (frequency of 2 is greater)
(m1 / m2)^1/2 = ω2 / ω1 from above
m1 / m2 = 2^2 = 4 so m1 would have 4 times the mass of m2
M1 would seem to be slower because of a larger mass
x1 = A1 sin ω1 t1
ω1 / ω2 = ((k1 / k2) / (m1 / m2))^1/2 = (m2 / m1)^1/2 since k's are equal
ω1 / ω2 = 1/2 from graph (frequency of 2 is greater)
(m1 / m2)^1/2 = ω2 / ω1 from above
m1 / m2 = 2^2 = 4 so m1 would have 4 times the mass of m2.
What is the graph represents?The two graphs shown represent the motion of two blocks with different masses, m1 and m2. The blocks are oscillating on identical springs. For the system consisting of the two blocks, the change in the kinetic energy of the system is equal to work done by gravity on the system. For the system consisting of the two blocks, the pulley and the Earth, the change in the total mechanical energy of the system is zero.
The two graphs shown represent the motion of two blocks with different masses, m1 and m2. The blocks are oscillating on identical springs. For the system consisting of the two blocks, the change in the kinetic energy of the system is equal to work done by gravity on the system.
Therefore, M1 would seem to be slower because of a larger mass
x1 = A1 sin ω1 t1
ω1 / ω2 = ((k1 / k2) / (m1 / m2))^1/2 = (m2 / m1)^1/2 since k's are equal
ω1 / ω2 = 1/2 from graph (frequency of 2 is greater)
(m1 / m2)^1/2 = ω2 / ω1 from above
m1 / m2 = 2^2 = 4 so m1 would have 4 times the mass of m2.
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A proton's speed as it passes point A is 4.00×104 m/s . It follows the trajectory shown in the figure. What is the proton's speed at point B?
We know that:
[tex]0.5m(vB^{2} -vA^{2} )=q*[VB-VA]\\== > vB^{2} =vA^{2} +(\frac{2q}{m})*[VB-VA]\\ =(4.30*10^{4} )^{2} +(\frac{2*1.6*10^{-19} }{2*1.67*10^{-27} }) [-10-30]\\[/tex]
[tex]=18.49*10^{8} -76.64*10^{8}[/tex]
[tex]=-58.15*10^{8} \\== > vB=-7.62*10^{4} m/s[/tex]
The proton's speed at point B is -7.62 x ×10⁴ m/s.
What is proton?The proton is a subatomic particle lies inside the nucleus of an atom of element. The number of proton gives an idea of the atomic number of the element.
The proton's speed as it passes point A is 4.00×10⁴ m/s . It follows the trajectory shown.
1/2m (vb² - va²) = q (Vb - Va)
vb² = va² +2q/m (Vb - Va)
where v is the velocity and V is the potential at point A and B, q ia the charge on proton and m is the mass of proton.
Substitute the given values, we get
vb² = (4.00×10⁴ ) + (2x1.6 x 10⁻¹⁹/1.67 x 10⁻²⁷) (-10 - 30)
vb = -7.62 x ×10⁴ m/s
Thus, proton's speed at point B is -7.62 x ×10⁴ m/s.
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Which of the following would NOT cause acceleration? A. An object falling towards the earth. B. The wind catching the sail of a sailboat. C. A car traveling at a constant velocity. D. A pitcher throwing a baseball.
A student uses a motion detector to study the kinematics of a block suspended on a vertical spring that obeys Hooke’s law. The student pulls the block a distance y from equilibrium, releases it from rest, and records the speed v of the block as it passes the equilibrium position. The student repeats this process several times for different values of y. Which variables should be plotted on the horizontal and vertical axes to yield a linear graph?
When the speed of the block is plotted on the vertical axis and displacement of the block plotted on the horizontal axis, a linear graph will be obtained showing the inverse relationship between the speed of the block and the displacement of the block.
Hooke's lawHooke's law states that force or load applied to elastic material is directly proportional to the extension produced in the material. This law can be written as;
F = kx
where;
F is the applied forcek is spring constantx is extension of the materialVelocity of the blockThe speed of the block increases with decreasing displacement of the bock. This is because the kinetic energy of the block is maximum at zero displacement and minimum at maximum displacement of the block.
Thus, when the speed of the block is plotted on the vertical axis and displacement of the block plotted on the horizontal axis, a linear graph will be obtained showing the inverse relationship between the speed of the block and the displacement of the block.
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. Radiation travels at the speed of light
T or F?
Answer:
electromagnetic radiation moves at the speed of light
CAN SOMEONE PLS HELP ME!!!
Answer:
c
Explanation:
the question answered itself.
At an altitude of 1.3x10^7 m above the surface of the earth an incoming meteor mass of 1x10^6 kg has a speed of 6.5x10^3 m/s. What would be the speed just before impact with the surface of earth?Ignore air resistance.
Show all steps.
Answer:
Approximately [tex]1.1 \times 10^{4}\; {\rm m\cdot s^{-1}}[/tex] if air friction is negligible.
Explanation:
Let [tex]G[/tex] denote the gravitational cosntant. Let [tex]M[/tex] denote the mass of the earth. Lookup the value of both values: [tex]G \approx 6.67 \times 10^{-11}\; {\rm N\cdot m^{2}\cdot kg^{-2}}[/tex] while [tex]M \approx 5.697 \times 10^{24}\; {\rm kg}[/tex].
Let [tex]m[/tex] denote the mass of the meteor.
Let [tex]v_{0}[/tex] denote the initial velocity of the meteor. Let [tex]r_{0}[/tex] denote the initial distance between the meteor and the center of the earth.
Let [tex]r_{1}[/tex] denote the distance between the meteor and the center of the earth just before the meteor lands.
Let [tex]v_{1}[/tex] denote the velocity of the meteor just before landing.
The radius of planet earth is approximately [tex]6.371 \times 10^{6}\; {\rm m}[/tex]. Therefore:
At an altitude of [tex]1.3 \times 10^{7}\; {\rm m}[/tex] about the surface of the earth, the meteor would be approximately [tex]r_{0} \approx 6.371 \times 10^{6}\; {\rm m} + 1.3 \times 10^{7}\; {\rm m} \approx 1.9 \times 10^{7}\; {\rm m}[/tex] away from the surface of planet earth. The meteor would be only [tex]r_{1} \approx 6.371 \times 10^{6}\; {\rm m}[/tex] away from the center of planet earth just before landing.Note the significant difference between the two distances. Thus, the gravitational field strength (and hence acceleration of the meteor) would likely have changed significant during the descent. Thus, SUVAT equations would not be appropriate.
During the descent, gravitational potential energy ([tex]\text{GPE}[/tex]) of the meteor was turned into the kinetic energy ([tex]\text{KE}[/tex]) of the meteor. Make use of conservation of energy to find the velocity of the meteor just before landing.
Initial [tex]\text{KE}[/tex] of the meteor:
[tex]\displaystyle (\text{Initial KE}) = \frac{1}{2}\, m\, {v_{0}}^{2}[/tex].
Initial [tex]\text{GPE}[/tex] of the meteor:
[tex]\displaystyle (\text{Initial GPE}) &= -\frac{G\, M\, m}{r_{0}}[/tex].
(Note the negative sign in front of the fraction.)
Just before landing, the [tex]\text{KE}[/tex] and the [tex]\text{GPE}[/tex] of this meteor would be:
[tex]\displaystyle (\text{Final KE}) = \frac{1}{2}\, m\, {v_{1}}^{2}[/tex].
[tex]\displaystyle (\text{Final GPE}) &= -\frac{G\, M\, m}{r_{1}}[/tex].
If the air friction on this meteor is negligible, then by the conservation of mechanical energy:
[tex]\begin{aligned}& (\text{Initial KE}) + (\text{Initial GPE}) \\ =\; & (\text{Final KE}) + (\text{Final GPE})\end{aligned}[/tex].
[tex]\begin{aligned}& \frac{1}{2}\, m\, {v_{0}}^{2} - \frac{G\, M\, m}{r_{0}} \\ =\; & \frac{1}{2}\, m\, {v_{1}}^{2} - \frac{G\, M\, m}{r_{1}}\end{aligned}[/tex].
Rearrange and solve for [tex]v_{1}[/tex], the velocity of the meteor just before landing:
[tex]\begin{aligned}{v_{1}} &= \sqrt{\frac{\displaystyle \frac{1}{2}\, m\, {v_{0}}^{2} - \frac{G\, M\, m}{r_{0}} + \frac{G\, M\, m}{r_{1}}}{(1/2)\, m}} \\ &= \sqrt{{v_{0}}^{2} - \frac{G\, M}{r_{0}} + \frac{G\, M}{r_{1}}} \\ &= \sqrt{{v_{0}}^{2} - G\, M\, \left(\frac{1}{r_{1}} - \frac{1}{r_{0}}\right)}\end{aligned}[/tex].
Substitute in the values and evaluate:
[tex]\begin{aligned}v_{1} &= \sqrt{{v_{0}}^{2} - G\, M\, \left(\frac{1}{r_{1}} - \frac{1}{r_{0}}\right)} \\ &\approx \sqrt{\begin{aligned}(& 6.5 \times 10^{3}\; {\rm m \cdot s^{-1}}) \\ & - [6.67 \times 10^{-11}\; {\rm N \cdot {m}^{2}\cdot {kg}^{2} \times 5.697\; {\rm kg}}\\ &\quad\quad \times (1 / (6.371 \times 10^{6}\; {\rm m}) - 1 / (1.9371 \times 10^{7}\; {\rm m}))]\end{aligned}} \\ &\approx 1.1 \times 10^{4}\; {\rm m\cdot {s}^{-1}}\end{aligned}[/tex].
(Note that assuming a constant acceleration of [tex]g = 9.81\; {\rm m\cdot s^{-2}}[/tex] would give [tex]v_{1} \approx 1.7\times 10^{4}\; {\rm m\cdot s^{-1}}[/tex], an inaccurate approximation.
A 4500 kg Aston Martin traveling at 102 m/s has to stop short because some ducklings
hazard onto the road. The Aston Martin was able to stop in 1.77 seconds. How much
force was placed on the car?
Answer:
-259322.03N
Explanation:
[tex]F=m*(\frac{v}{t})\\ F=4500kg*(\frac{0-102m/s}{1.77s} )\\F=-259322.033898\\\\[/tex]
match each term to its definition.
the vocabulary
Volt
electric potential energy
electric potential difference
electric potential
the definitions
the SI unit of electric potential difference
difference in electric potential between two positions
potential energy an electric charge has due
to its location in a field
electric potential energy of a charged particle divided by its charge
here are the answers
Electric potential difference is the difference in electric potential between two positions and it is measured in volts.
What is electric potential energy?This is the potential energy of an electric charge has due to its location in a field.
What is electric potential?The electric potential of a charged particle is the electric potential energy of the charged particle divided by its charge magnitude.
What is electric potential difference?Electric potential difference is the difference in electric potential between two positions.
What is volt?Volt is the SI unit of electric potential difference.
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Pls helppppp my last question!
Answer:
load
a generator, a light bulb (load) and a closed switch
Explanation:
as explained in the other question, the fan is using generated electric energy to create mechanical movement. as such it is a load on the grid or circuit or net.
and electric power can only flow, if there is a closed (uninterupted) circuit from the power source to a load and back.
any open switch is an interruption of the circuit.
a buzzer is a kind of switch. it closes the circuit (and puts a load on) only when somebody presses it.
by the way, a closed circuit without a load will "destroy" (short circuit) the power source or at least the wires (burn through).
state the precautions that is taken when charging a metal objectexplain why a rubber balloon rubbed will often stick to the wall when it has been rubbed
The balloon will attach to the wall because the balloon's negative charges will drive the electrons in the wall to shift to the other side of their atoms, leaving the wall's surface positively charged.
The block falls for a time t0, but the string does not completely unwind. What is the change in angular momentum of the pulley-block system from the instant that the block is released from rest until time t0?
The change in angular momentum of the pulley-block system is given as [tex]\omega = Rm_ogt_o[/tex]
Data;
time = t0mass = MoTorqueThis is the force that makes object rotate about an axis. The formula is given as the product between the radius about the axis and the force acting upon it.
The torque about point o is given as
[tex]\tau = R * m_og[/tex]
The change in angular momentum about point 'o' is the product between the torque and time.
[tex]\omega = \tau * t_o\\\omega = Rm_ogt_o[/tex]
The change in angular momentum of the pulley-block system is given as
[tex]\omega = Rm_ogt_o[/tex]
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If the range of a projectile's trajectory is six times larger than the height of the trajectory, then what was the angle of launch with respect to the horizontal? (Assume a flat and horizontal landscape.)
Answer:
H = 1/2 g t^2 where t is time to fall a height H
H = 1/8 g T^2 where T is total time in air (2 t = T)
R = V T cos θ horizontal range
3/4 g T^2 = V T cos θ 6 H = R given in problem
cos θ = 3 g T / (4 V) (I)
Now t = V sin θ / g time for projectile to fall from max height
T = 2 V sin θ / g
T / V = 2 sin θ / g
cos θ = 3 g / 4 (T / V) from (I)
cos θ = 3 g / 4 * 2 sin V / g = 6 / 4 sin θ
tan θ = 2/3
θ = 33.7 deg
As a check- let V = 100 m/s
Vx = 100 cos 33.7 = 83,2
Vy = 100 sin 33,7 = 55.5
T = 2 * 55.5 / 9.8 = 11.3 sec
H = 1/2 * 9.8 * (11.3 / 2)^2 = 156
R = 83.2 * 11.3 = 932
R / H = 932 / 156 = 5.97 6 within rounding
Adrian wants to take a personality test to help her decide which college major to choose. How can the test help her
Answer:
It could point her in a direction that best suits her
Explanation:
Hope that helps!
Calculate the average velocity of the cart for each fan speed. Round your answers to the nearest tenth. The cart with Low fan speed has an average velocity of cm/s. The cart with Medium fan speed has an average velocity of cm/s. The cart with High fan speed has an average velocity of cm/s.
A 4-column table with 3 rows. Column 1 has entries Elapsed time to finish line(s); Total distance (c m); Average velocity (c m/s). Column 2 is labeled Cart Speed (Low Fan Speed) (c m/s) with entries 7.4; 500; unknown. Column 3 is labeled Cart Speed (medium Fan Speed) (c m/s) with entries 6.4; 500; unknown. Column 4 is labeled Cart Speed (High Fan Speed) (c m/s) with entries 5.6; 500; unknown.
This is defined as the rate of change of position of an object with respect to time. The unit is metres per second(m/s).
Velocity = distance/ time
The velocity for column 1 = 500m/7.4s = 67.57m/s.
The velocity for column 2 = 500m/6.4s = 78.13m/s.
The velocity for column 3 = 500m/5.6s = 89.29m/s.
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Answer:
67.6
78.1
89.3
Explanation:
A women runs a kilometer using a of 210N and a pwower output of 500W how long does it take this woman to complete 1 kilometer
Answer:
7 minutes
Explanation:
Explanation:
Power = work / time
Power = force × distance / time
500 W = 210 N × 1000 m / t
t = 420 s
t = 7 min
What potential difference is required across an 32 -Ω resistor to cause 33.72 A to flow through it?
[tex]\text{Voltage,}~ V = IR =33.72\times 32 = 1079.04~ \text{volt}[/tex]
A 330-ohm resistor is connected to a 5-volt battery. The current through the resistor is
Question :-
A 330 Ohm Resistor is connected to a 5 Volt Battery . What is the Current through the Resistor ?Answer :-
Current of the Battery is 66 Ampere.Explanation :-
As per the provided information in the given question, The Resistance is given as 330 Ohm . The Voltage is given as 5 Volt . And, we have been asked to calculate the Current .
For calculating the Current , we will use the Formula :-
[tex] \bigstar \: \: \: \boxed{ \sf{ \: Current \: = \: \dfrac{Voltage}{Resistance} \: }} [/tex]
Therefore , by Substituting the given values in the above Formula :-
[tex] \dag \: \: \: \sf {Current \: = \: \dfrac {Voltage}{Resistance} } [/tex]
[tex] \longmapsto \: \: \: \sf {Current \: = \: \dfrac {5}{330} } [/tex]
[tex] \longmapsto \: \: \: \sf {Current \: = \: \dfrac {1}{66} } [/tex]
[tex] \longmapsto \: \: \: \textbf {\textsf {Current \: = \: 66 }} [/tex]
Hence :-
Current = 66 Ampere .[tex] \underline {\rule {180pt} {4pt}} [/tex]
Additional Information :-
[tex] \Longrightarrow \: \: \: \sf {Voltage \: = \: Current \: \times \: Resistance} [/tex]
[tex] \Longrightarrow \: \: \: \sf {Current \: = \: \dfrac {Voltage}{Resistance} } [/tex]
[tex] \Longrightarrow \: \: \: \sf {Resistance \: = \: \dfrac {Voltage}{Current} } [/tex]
The frictional force between two surfaces at rest depends on which of the following?
Answer:
the two types of materials interacting (described by the coefficient μk) and how hard these two surfaces are pushed together (the normal force)
hope this helps and is it needs more let me know
acar start from rest and go with velocity of 4m/s for 4second what is acceleration of car
Answer:
1 m/s^2
Explanation:
The formula for accleration is a=Δv/Δt
where, Δv = final velocity - initial velocity = 4 - 0 = 4
initial velocity = 0 since the car starts form rest and final velovity is 4 as the car goes from rest till 4 m/s
Δt = 4 since the car takes 4 seconds to reach a velocity of 4m/s
Hence, a = 4 m/s / 4 s = 1 m/s^2
Answer:
1m/s
Explanation:
Given
initial velocity(u)=0
final velocity(v)=4m/s
timetaken (t)=4
Aceleration(a)=v-u÷t
now
=4-0÷4
=1m/s
An electron with an initial speed of 700,000 m/s is brought to rest by an electric field. What was the potential difference that stopped the electron? What was the initial kinetic energy of the electron, in electron volts?
Answer:
See below.
Explanation:
According to the question, we know that,
work done is given by, [tex]W=qV[/tex]
and change in kinetic energy is, Δ [tex]KE=W=1=1/2[mv^{2} ][/tex]
therefore equating both the equations we get,
[tex]qV=1/2[mv^{2} ][/tex] ⇒ [tex]V=\frac{mv^{2} }{2q}[/tex]
m= mass of electron = [tex]9.1*10^{-31} kg[/tex]
q= charge on an electron = [tex]1.6*10^{-19} C[/tex]
v= speed of electron= 700000m/s
substituting the values in the above equation, we get
[tex]V=\frac{9.1*10^{-31} *(700000)^{2} }{2*1.6*10^{-19} } =1.39V[/tex]
(1). the potential difference that stopped the electron is 1.39 volts.
now the kinetic energy equation is : 2 ways[tex]KE=1/2[mv^{2} ]=\frac{9.1*10^{-31} *700000^{2} }{2} =2.22*10^{-19} J\\[/tex]
or [tex]KE=\frac{2.22*10^{-19} }{1.6*10^{-19} } =1.39eV[/tex]
(2). the initial kinetic energy of the electron is 1.39eV.
10 N
30 N
4 kg
what’s the net force & acceleration
Answer:
The question is not complete. Since force is a vector quantity we need to consider the direction of the two forces.
hence if they are acting opposite on the body . then the net force will be 30-10=20N20N
Acceleration=Force/mass
=20/4
=5m/s^2
Explanation:
A stage has a radius R = 2.00 m and a mass M = 100 kg. The stage, which has a circular disk
shape, rotates in a horizontal plane, without any friction, about a vertical axle. Sandra has a
mass of m = 60.0 kg and she walks from the rim of the stage toward the center. Calculate the
angular speed when she reaches a point r = 0.500 m from the center, if the angular speed of the
system is 2.0 rad/s when Sandra is at the rim
i want the answer of this exercise
A 15kg penguin slides on its belly down an icy, frictionless glacier. If the penguin started from rest and reaches a speed of 11m/S when at the bottom, how high is the glacier? (Metres)
Answer:
Mechanical Energy (initial) = Mechanical energy (final)
Ep(initial) + Ek(initial) = Ep(final) + Ek(final)
mgh(initial) + 1/2mv²(initial) = mgh(final) + 1/2mv²(final)
m being the mass of the penguin (kg)
g being gravitational acceleration (9.80 m/s²)
v(initial) being zero
h(final) being zero,i.e. final height is zero
v(final) being final velocity
solve for h(initial) being initial height at the top of the glacier
answer will be in metres
NB: don't forget to square the velocity
Explanation:
Since they mentioned frictionless, we know that this is a closed system therefore we are free to use this equation of conservation of mechanical energy
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A tire on a car is rolling smoothly. Its center of mass is moving at 18 m/s. How fast is the top of the tire moving in m/s
For a tire on a car is rolling smoothly and It's center of mass is moving at 18 m/s, the speed of the top of the tire is mathematically given as
Vtop=36m/s
What is the speed of the top of the tire?Generally, the equation for the relationship between the top and the core is mathematically given as
Vtop=2*(vcore)
Threrefore
Vtop=2*18
Vtop=36m/s
In conclusion, the speed of the top
Vtop=36m/s
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