[tex]\boxed{\sf Q=mc\Delta T}[/tex]
[tex]\\ \sf\longmapsto Q=1000(5000)(3)[/tex]
[tex]\\ \sf\longmapsto Q=15000000J[/tex]
[tex]\\ \sf\longmapsto Q=1.5\times 10^7J[/tex]
Categorize each statement as true or false.
1. Electric field lines radiate away from positive charges and towards negative charges.
2. Electric field is always perpendicular to equipotential lines.
3. The electric field points in the direction of increasing electric potential.
4. The electric field inside a parallel plate capacitor decreases as it approaches the negative plate.
5. The units of electric field are either newtons per coulomb or volts per meter.
Answer:
1. True
2. True
3. False
4. True
5. True
An electron and a proton are initially very far apart (effectively an infinite distance apart). They are then brought together to form a hydrogen atom, in which the electron orbits the proton at an average distance of 5.29x10-11m. What is the change in electric potential energy?
Answer:
[tex]dU=-4.36*10^{-18}J[/tex]
Explanation:
From the question we are told that:
Average distance [tex]d_{avg} =5.29*10^{-11}m[/tex]
Generally the equation for change in electric potential energy is mathematically given by
[tex]dU=u_f-U_1[/tex]
Where
U_1=0 Because of initial lengthy distance apart
And
[tex]U_f=\frac{kq_eq_p}{d}[/tex]
[tex]U_f=\frac{9*10^9*1.6*10^{-19}*-1.6*10^{-19}}{5.29*10^{-11}}[/tex]
[tex]U_f=-4.36*10^{-18}J[/tex]
Therefore
[tex]dU=u_f-U_1[/tex]
[tex]dU=-4.36*10^{-18}J-0[/tex]
[tex]dU=-4.36*10^{-18}J[/tex]
A single-turn circular loop of wire of radius 55 mm lies in a plane perpendicular to a spatially uniform magnetic field. During a 0.10 s time interval, the magnitude of the field increases uniformly from 350 to 450 mT.
Required:
a. Determine the emf induced in the loop (in V). (Enter the magnitude.) V
b. If the magnetic field is directed out of the page, what is the direction of the current induced in the loop?
Answer:
Explanation:
Area of the loop = π x ( 55 x 10⁻³ )²
= 9.5 x 10⁻³ m²
Change in Magnetic flux dφ = 450 x 10⁻³ - 350 x 10⁻³ = 150x 10⁻³ Weber.
time dt =.10 s
emf induced = dφ / dt = 150x 10⁻³ Weber / .10 s
= 1.5 V .
b )
Magnetic field is directed outwards and it is increasing so according to Lenz's law , direction of induced current will be clockwise in the loop.
Answer:
(a) 9.5 mV
(b) clockwise
Explanation:
Radius, r = 55 mm
Time, t = 0.1 s
Change in magnetic field, B = 450 - 350 = 100 mT =0.1 T
(a) induced emf is given by
[tex]e = A \frac{dB}{dt}[/tex]
[tex]e = A \frac{dB}{dt}\\\\e=3.14\times 0.055\times0.055\times \frac{0.1}{0.1}\\\\e= 9.5 \times 10^{-3} V = 9.5 mV[/tex]
(b) According to the Lenz law, the direction of current is clockwise.
Need help! Need help! Need help! Need help! Need help! Need help!
Answer:
i can help you i know this answer
Answer: the side two are 50 then the other two are 140 i thank
Explanation:
Example 2.13 The acceleration a of a particle in a time t is given by the equation a = 2+ 5t^2. Find the instantaneous velocity after 3s. Solution
Answer:
the instantaneous velocity is 51 m/s
Explanation:
Given;
acceleration, a = 2 + 5t²
Acceleration is the change in velocity with time.
[tex]a = \frac{dv}{dt} \\\\a = 2 + 5t^2\\\\The \ acceleration \ (a) \ is \ given \ so \ we \ have \ to \ find \ the \ velocity \ (v)\\\\To \ find \ the \ velocity, \ integrate\ both \ sides \ of \ the \ equation\\\\2 + 5t^2 = \frac{dv}{dt} \\\\\int\limits^3_0 {(2 + 5t^2)} \, dt = dv\\\\v = [2t + \frac{5t^3}{3} ]^3_0\\\\v = 2(3) + \frac{5(3)^3}{3} \\\\v = 6 + 5(3)^2\\\\v = 6 + 45\\\\v = 51 \ m/s[/tex]
Therefore, the instantaneous velocity is 51 m/s
Select the only true statement:
A)A beam in bending experiences tensile stresses on one side and compressive stresses on the other side.
B)A beam in bending experiences tensile stresses along the beam center and compressive stresses along the beam’s edges.
C)A beam in bending experiences only compressive stresses.
D)A beam in bending experiences only tensile stresses.
Answer:
Sorry I dont know this answer sorry
Can an electron be diffracted? Can it exhibit interference?
Answer:
Yeah, it can be diffracted. Though it depends on a diffracting medium.
It must have some magnetic fields .
Forexample; X-ray diffraction where electrons are diffracted to the target filament.
Two beams of coherent light start out at the same point in phase and travel different paths to arrive at point P. If the maximum destructive interference is to occur at point P, what correctly describes the path difference of the two beams
Answer:
Two beams (that we could think as sinusoidal waves) face destructive interference at a point P, if at that point one of the waves is in a peak, and the other wave is in a through, so when we add them, the waves will "cancel" each other (thus, we have destructive interference)
The case of constructive interference happens when at point P we have two peaks or two throughs, so the waves add up.
If both waves started at the same point and with the same phase and both traveled the same distance, they will always have constructive interference. But if the waves traveled different distances, the constructive interference will happen at points where the path difference of the waves is an integer multiple of the wavelength (remember that the distance between a peak and the next one is the wavelength).
If two beams of coherent light start at the same point and are in phase, then we will have the maximum of destructive interference at a point P if the path difference is exactly half of the wavelength (or (n/2) times the wavelength, with n an odd integer), this happens because the distance between a peak and the next thtough is exactly half of the wavelength.
Then we can conclude that the path difference between the two beams is of the form:
(n/2)*λ
where:
n = odd number = (2*k + 1) with k an integer.
λ = wavelength of the beams.
State how you agree or disagree with the following statement. A good circuit cannot have internal resistance.
Answer: I do
Explanation:
Resistance opposes current thereby reducing the amount of current that flows through a circuit. In other words, it leads to a loss of electrical energy.
Ideally speaking, a good circuit should have no internal resistance as this would lead to more energy having to be supplied to overcome that resistance. External resistance however, is not a bad thing. For instance, oxygen being removed from lightbulbs.
once the object is seen clearly (Figure 5).
Cliary muscles
Nea
Obec
image
25 cm
FIGURE 3
dusion : Thus, we observe that the focal length of the eye
ically by the action of cilin
Answer:
Where is the figure ?????
Determine the magnitude as well as direction of the electric field at point A, shown in the above figure. Given the value of k = 8.99 × 1012N/C (figure is in attached file)
Answer:
The magnitude of the electric field is [tex]8.99*10^{12}N/C[/tex] in the r direction.
Explanation:
The equation of the electric field is given by:
[tex]|\vec{E}|=k\frac{q}{r^{2}}[/tex]
Where:
k is the Coulomb constant is [tex]8.99 *10^{9}\: Nm^{2}C^{-2}[/tex]q is the charger is the distance from A to q[tex]|\vec{E}|=8.99*10^{9}\frac{12.5}{0.11^{2}}[/tex]
[tex]\vec{E}=9.29*10^{12} \vac{r} \: N/C[/tex]
Therefore, the magnitude of the electric field is [tex]8.99*10^{12}N/C[/tex] in the r direction.
I hope it helps you!
You throw a Frisbee of mass m and radius r so that it is spinning about a horizontal axis perpendicular to the plane of the Frisbee. Ignoring air resistance, the torque exerted about its center of mass by gravity is: __________
a. 0.
b. mgr
c. 2mgr
d. a function of the angular velocity.
e. small at first, then increasing as the Frisbee loses the torque given it by your hand.
Answer:
the correct answer is a
Explanation:
The torque is
τ = F x r
where the bold letters indicate vectors, in this case the vector of the center of mass is perpendicular to the weight of the body
τ = mg r
in body weight it is applied at the point of the center of mass, therefore as the distance of the force from the axis of rotation (center of amas) is zero, the die is zero
the correct answer is a
You have been called to testify as an expert witness in a trial involving a head-on collision. Car A weighs 690.0 kg and was traveling eastward. Car B weighs 520.0 kg and was traveling westward at 74.0 km/h. The cars locked bumpers and slid eastward with their wheels locked for 6.00 m before stopping. You have measured the coefficient of kinetic friction between the tires and the pavement to be 0.750. 1) How fast (in kilometer per hour) was car A traveling just before the collision
Answer:
The speed of car A before collision is 3.5 km/h.
Explanation:
Mass of car A = 690 kg eastwards
Mass of car B = 520 kg at 74 km/h west wards
Distance, s = 6 m
coefficient of friction = 0.75
Let the speed after collision is v.
Use third equation of motion
[tex]v^2 = u^2 + 2 as \\\\0 =v^2- 2 \times 0.75\times9.8\times 6\\\\v = 9.4 m/s = 33.84 km/h[/tex]
Let the initial speed of car A is v'.
Use conservation of momentum
690 x v' - 520 x 74 = (690 + 520) x 33.8
690 v' + 38480 = 40898
v' = 3.5 km/h
220V a.c. is more dangerous than 220V d.c why?
Answer:
220V a.c is more dangerous than 220V d.c because of the peak voltage of 220V a.c. which is much larger.
the product 17.10 ✕
Explanation:
pls write full question
how does the use of standard units of measurement solve problems in measurement regarding validity and reliabiility? explain it
Answer:
Reliability can be estimated by comparing different versions of the same measurement. Validity is harder to assess, but it can be estimated by comparing the results to other relevant data or theory.
Nhiệt dung riêng của một chất là ?
Answer:
enchantment table language
Explanation:
A series of pulses, each of amplitude 0.1 m, is sent down a string that is attached to a post at one end. The pulses are reflected at the post and travel back along the string without loss of amplitude. What is the net displacement at a point on the string where two pulses are crossing
Answer:
A_resulting = 0.2 m
Explanation:
Let's analyze the impact of the pulse with the pole, this is a fixed obstacle that does not move therefore by the law of action and reluctant, the force that the pole applies on the rope is of equal magnitude to the force of the rope on the pole (pulse), but opposite directional, so the reflected pulse reverses its direction and sense.
With this information we analyze a point on the string where the incident pulse is and each reflected with an amplitude A = 0.1 m, the resulting is
A_res = 2A
A_resultant = 2 .01
A_resulting = 0.2 m
A 0.22LR caliber bullet has a mass of 1.90 g and a muzzle velocity of 500 m/s. The bullet is fired into a door made of a single thickness of pine boards, with a thickness of 0.75 in. The average stopping force exerted by the wood is 960 N. How fast (in m/s) would the bullet be traveling after it penetrated through the door
Answer:
The final speed of the bullet is 480.4 m/s.
Explanation:
mass of bullet, m = 1.9 g
initial speed, u = 500 m/s
thickness, d = 0.75 inch = 0.01905 m
Force, F = 960 N
Let the final speed is v.
According to the work energy theorem,
Work = change in kinetic energy
[tex]W = F d = 0.5 m{\left (v^2 - u^2 \right )}[/tex]
-960 x 0.01905 = 0.5 x 0.0019 x (v^2 - 500 x 500)
-18.288 = 0.00095 (v^2 - 250000)
v = 480.4 m/s
Principal axis is the:________
a. straight line drawn from the center of curvature to the mid point of the mirror.
b. straight line drawn from the center of curvature to a point on the outer edge of the mirror.
c. straight line drawn from the center of curvature to any point of the mirror.
d. straight line joining any two points on the mirror.
e. None of the other answers given is correct.
Answer:
d. straight line joining any two points on the mirror
Explanation:
Principal axis is the straight line that passes through the center of a mirror, which is also perpendicular to the surface of the mirror.
The principal axis connects the principal focus and the center of curvature of the mirror. In other words, it is a straight line or axis on which the center of curvature and principal focus can be found. The principal axis joins these two points; the center of curvature and principal focus.
Therefore, the correct option is "D"
d. straight line joining any two points on the mirror
A marble rolling with speed 20cm/s rolls off the edge of a table that is 80cm high. How far horizontally from the table edge does the marble strike the floor
Answer:
8 cm
Explanation:
From the question given above, the following data were obtained:
Initial velocity (u) = 20 cm/s
Height (h) = 80 cm
Horizontal distance (s) =?
Next, we shall determine the time taken for marble to get to the ground. This can be obtained as follow:
Height (h) = 80 cm
Acceleration due to gravity (g) = 1000 cm/s²
Time (t) =?
t = √(2h/g)
t = √[(2 × 80)/1000]
t = √(160/1000)
T = √0.16
t = 0.4 s
Finally, we shall determine the horizontal distance travelled by the marble. This can be obtained as illustrated below:
Initial velocity (u) = 20 cm/s
Time (t) = 0.4 s
Horizontal distance (s) =?
s = ut
s = 20 × 0.4
s = 8 cm
Thus, the horizontal distance travelled by the marble is 8 cm.
The horizontal distance traveled by the marble is 8 cm.
The given parameters;
speed of the marble, v = 20 cm/sheight of the table, h = 80 cmThe time of motion of the marble is calculated as follows;
[tex]h = ut + \frac{1}{2} gt^2\\\\h = 0 + \frac{1}{2} gt^2\\\\h = \frac{1}{2} gt^2\\\\gt^2 = 2h\\\\t^2 = \frac{2h}{g} \\\\t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2\times 0.8}{9.8} }\\\\t = 0.4 \ s[/tex]
The horizontal distance traveled by the marble is calculated as follows;
[tex]X = v_0_x t\\\\X = (20 \times 0.4)\\\\X = 8 \ cm[/tex]
Thus, the horizontal distance traveled by the marble is 8 cm.
Learn more here:https://brainly.com/question/2411455
Physics question plz help ASAP
We will determine the amount of electric energy stored in a capacitor by discharging it through a light bulb. Light bulbs are rated according to their power output at a given voltage. Considering that power is the rate that energy is converted from one form to another (or, equivalently, work is done) per unit time, the energy stored in an initially-charged capacitor that is hooked up to the light bulb through which the capacitor discharges is approximately
A. the power rating of the light bulb divided by the time that the bulb remains lit.
B. simply the time that the light bulb remains lit.
C. the product of the power rating of the light bulb and the time that it remains lit.
D. the time that the light bulb remains lit divided by the power rating of the bulb.
Answer:
C. the product of the power rating of the light bulb and the time that it remains lit.
Explanation:
The power rating of the light is bulb is defined as the energy supplied to the light bulb divided by the time the bulb is lit up. Therefore,
[tex]P = \frac{E}{t}\\\\E = Pt[/tex]
where,
E = Energy Supplied to the bulb = Energy stored in capacitor = ?
P = Power rating of the bulb
t = time the bulb is lit up
Hence, the correct option is:
C. the product of the power rating of the light bulb and the time that it remains lit.
On the average, in a ferromagnetic domain, permanent atomic magnetic moments are aligned ____ to one another.a. antiparallel b. parallel c. perpendicular d. alternately parallel and antiparallel e. randomly relative
Answer:
b. parallel
Explanation:
Ferromagnetism is a magnetism that is associated with iron and cobalt and nickel. Ferromagnetisms material are magnetics easily and in strong magnetic fields are magnetized by a defined limit called a situation. The force keeps magnetic moments of many atoms parallel to each other.A monochromatic light falls on two narrow slits that are 4.50 um apart. The third destructive fringes which are 35° apart were formed at a screen 2m from the light source. The light source is 0.30 m from the slits. () Calculate Ym. (4 marks) Compute the wavelength of the light. (4 marks)
Answer:
y = 1.19 m and λ = 8.6036 10⁻⁷ m
Explanation:
This is a slit interference problem, the expression for destructive interference is
d sin θ = m λ
indicate that for the angle of θ = 35º it is in the third order m = 3 and the separation of the slits is d = 4.50 10⁻⁶ m
λ = d sin θ / m
let's calculate
λ = 4.50 10⁻⁶ sin 35 /3
λ = 8.6036 10⁻⁷ m
for the separation distance from the central stripe, we use trigonometry
tan θ= y / L
y = L tan θ
the distance L is measured from the slits, it indicates that the light source is at x = 0.30 m from the slits
L = 2 -0.30
L = 1.70 m
let's calculate
y = 1.70 tan 35
y = 1.19 m
What is the electric potential 15 cm above the center of a uniform charge density disk of total charge 10 nC and radius 20 cm?
a) 360 V
b) 450 V
c) 22.5 V
d) 0 V
Answer:
b) 450 V
Explanation:
We are given that
Total charge, q=10nC=[tex]10\times 10^{-9} C[/tex]
[tex]1nC=10^{-9}C[/tex]
Radius, r=20 cm=[tex]\frac{20}{100}=0.2m[/tex]
1 m=100 cm
x=15 cm=0.15 cm
We have to find the electrical potential 15 cm above the center of a uniform charge density disk .
We know that
[tex]\sigma=\frac{q}{A}=\frac{q}{\pi r^2}[/tex]
[tex]\sigma=\frac{10\times10^{-9}}{3.14\times (0.2)^2}[/tex]
Where [tex]\pi=3.14[/tex]
[tex]\sigma=7.96\times 10^{-8}C/m^2[/tex]
Electric potential,[tex]V=\frac{\sigma}{2\epsilon_0}(\sqrt{x^2+r^2}-x)[/tex]
Where [tex]\epsilon_0=8.85\times 10^{-12}[/tex]
Using the formula
[tex]V=\frac{7.96\times 10^{-8}}{2\times 8.85\times 10^{-12}}(\sqrt{(0.15)^2+(0.2)^2}-0.15)[/tex]
[tex]V=449.7 V\approx 450V[/tex]
Hence, option b is correct.
Answer:
The potential is given by 449.7 V.
Explanation:
radius of disc, R = 20 cm = 0.2 m
distance, x = 15 cm = 0.15 m
charge, q = 10 nC
surface charge density
[tex]\sigma = \frac{q}{\pi R^2}\\\\\sigma = \frac{10\times 10^{-9}}{3.14\times 0.2\times 0.2 }\\\\\sigma = 7.96\times 10^{-8} C/m^2[/tex]
The electric potential is given by
[tex]V=\frac{\sigma}{2\varepsilon 0}\left ( \sqrt{R^2 + x^2} - x \right )\\\\V = \frac{7.96\times 10^{-8}}{2\times 8.85\times 10^{-12}}\left ( \sqrt{0.2^2 + 0.15^2} - 0.15 \right )\\\\V = 449.7 V[/tex]
Required information
You are designing a high-speed elevator for a new skyscraper. The elevator will have a mass limit of 2400 kg (including
passengers). For passenger comfort, you choose the maximum ascent speed to be 18.0 m/s, the maximum descent speed
to be 10.0 m/s, and the maximum acceleration magnitude to be 1.80 m/s2. Ignore friction.
What is the minimum upward force that the supporting cables exert on the elevator car?
KN
Answer:
19,224 N
Explanation:
The given parameters are;
The mass limit of the elevator = 2,400 kg
The maximum ascent speed = 18.0 m/s
The maximum descent speed = 10.0 m/s
The maximum acceleration = 1.80 m/s²
Given that the acceleration due to gravity, g ≈ 9.81 m/s²
The minimum upward force that the elevator cable exert on the elevator car, [tex]F_{min}[/tex] , is given in the downward motion as follows;
[tex]F_{min}[/tex] = m·g - m·a
∴ [tex]F_{min}[/tex] = 2,400 kg × 9.81 m/s² - 2,400 kg × 1.80 m/s² = 19,224 N
The minimum upward force that the elevator cable exert on the elevator car, [tex]F_{min}[/tex] = 19,224 N
A mother is pulling a sled at constant velocity by means of a rope at 37°. The tension on the rope is 120 N. Mass of children plus sled is 55 kg. The mother has a mass of 61 kg. Find the static friction acting on the mother.
Answer:
f = 106.3 N
Explanation:
The force applied on the sled must be equal to the static frictional force to move the sled:
Tension Force Horizontal Component = Static Frictional Force
[tex]TCos\theta = \mu W\\TCos\theta = \mu mg[/tex]
where,
T = Tension = 120 N
θ = angle of rope = 37°
μ = coefficient of static friction = ?
m = mass of children plus sled = 55 kg
g = acceleration due to gravity = 9.81 m/s²
Therefore,
[tex](120\ N)Cos\ 37^o = \mu (55\ kg)(9.81\ m/s^2)\\\\\mu = \frac{95.84\ N}{(55\ kg)(9.81\ m/s^2)}\\\\\mu = 0.18[/tex]
Now, the static friction acting on the mother will be:
[tex]f = \mu mg = (0.18)(61\ kg)(9.81\ m/s^2)\\[/tex]
f = 106.3 N
Suppose that 2 J of work is needed to stretch a spring from its natural length of 34 cm to a length of 46 cm. (a) How much work is needed to stretch the spring from 36 cm to 41 cm
Answer:
0.83 J of work
Explanation:
2 J of work is required to stretch a spring from 34cm to 46cm
So that is 12cm stretched with 2 J of work
We can make that 6cm for 1 J of work
So, we need the find the work for stretching 36cm to 41cm
Which is 5cm
So, What is the work required to stretch 5cm?
1 J of work for 6cm
x work for 5cm
So, by proportion method
1 : 6 :: x : 5
6 * x = 1 * 5
6x = 5
x = 5/6
= 0.83
So to stretch 36cm to 41cm we need 0.83 J of work
If you double the current in a long straight wire, the magnetic field at a fixed point will... be cut in half. triple. double. quadruple.
Answer:
the magnetic field must double
Explanation:
For this exercise we use Ampere's law
∫ B . ds = μ₀ I
Where the bold indicate vectors
With this expression we can see that if we double the current, keeping the same trajectory, the magnetic field must double
