Answer:
a) x = ⅔ d , b) the charge must be negative, c) Q
Explanation:
a) In this exercise the force is electric between the charges, we are asked that the system of the three charges is in equilibrium, we use Newton's second law. Balance is on the third load that we are placing
∑ F = 0
-F₁₂ + F₂₃ = 0
F₁₂ = F₂₃
let's replace the values
k Q Q / r₁₂² = k Q 4Q / r₂₃²
Q² / r₁₂² = 4 Q² / r₂₃²
suppose charge 3 is placed at point x
r₁₂ = x
r₂₃ = d-x
we substitute
1 / x² = 4 / (d-x) 2
1 / x = 2 / (d-x)
x = 2 (x-d)
x = 2x -2d
3x = 2d
x = ⅔ d
b) The sign of the charge must be negative, to have an attractive charge on the two initial charges
c) Q
Which of the following biotic organisms makes its own energy from inorganic substances?
producers
consumers
decomposers
minerals
Answer:
producers make its own energy frominorganic substances.
A particle of charge = 50 µC moves in a region where the only force on it is an electric force. As the particle moves 25 cm, its kinetic energy increases by 1.5 mJ. Determine the electric potential difference acting on the partice
Answer:
nvbnncbmkghbbbvvvvvvbvbhgggghhhhb
The specific heat of water is about 4200 J/(kg·oC). To heat a cup of water of 0.25 kg from 10 oC to 50 oC, at least how much heat is required?
Answer:
heat required = mass × specific heat × change in temperature
Explanation:
0.25 × 4200 × 40
=42000 J
what tools use cut wood
Answer:
hand saws
power saws
Circular Saw
Explanation:
that is all that i know
5. Sandor fills a bucket with water and whirls it in a vertical circle to demonstrate that the
water will not spill from the bucket at the top of the loop. If the length of the rope from his
hand to the centre of the bucket is 1.24 m, what is the minimum tension in the rope (at the
top of the swing)? How slow can he swing the bucket? Explain your answer.
Given that,
radius = 1.24 m
According to question,
The rope cannot push outwards. It must always have some slight tension or the bucket will fall.
We need to calculate the tension in the rope
At the top the force of gravity is
[tex]F=mg[/tex]
The force needed to move the bucket in a circle is centripetal force.
So, if mg is ever greater than centripetal force then the bucket and the contents will start to fall.
The rope have a tension of less than zero.
We need to calculate the velocity of swing bucket
Using centripetal force
[tex]F=\dfrac{mv^2}{r}[/tex]
[tex]mg=\dfrac{mv^2}{r}[/tex]
[tex]g=\dfrac{v^2}{r}[/tex]
[tex]v^2=gr[/tex]
[tex]v=\sqrt{gr}[/tex]
Put the value into the formula
[tex]v=\sqrt{9.8\times1.24}[/tex]
[tex]v=3.49\ m/s[/tex]
Hence, The minimum tension in the rope is less than zero .
The bucket swings with the velocity of 3.49 m/s.
A wheel starts from rest and rotates with constant angular acceleration to reach an angular speed of 12.9 rad/s in 2.98 s.
(a) Find the magnitude of the angular acceleration of the wheel.
(b) Find the angle in radians through which it rotates in this time interval.
Explanation:
(a) Find the magnitude of the angular acceleration of the wheel.
angular acceleration = angular speed /timeangular acceleration = 12.9/2.98 = 4.329rad/s²(b) Find the angle in radians through which it rotates in this time interval.
angular speed = 2x3.14xf12.9rad = 2 x3.14rad = 6.28/12.9rad = 0.487Now we convert rad to angle
1 rad = 57.296°0.487 = unknown angleunknown angle =57.296 x 0.487 = 27.9°The angle in radians = 27.9°
Approximately what applied force is needed to keep the box moving with a constant velocity that is twice as fast as before? Explain
Complete question:
A force F is applied to the block as shown (check attached image). With an applied force of 1.5 N, the block moves with a constant velocity.
Approximately what applied force is needed to keep the box moving with a constant velocity that is twice as fast as before? Explain
Answer:
The applied force that is needed to keep the box moving with a constant velocity that is twice as fast as before, is 3 N
Force is directly proportional to velocity, to keep the box moving at the double of initial constant velocity, we must also double the value of the initially applied force.
Explanation:
Given;
magnitude of applied force, F = 1.5 N
Apply Newton's second law of motion;
F = ma
[tex]F = m(\frac{v}{t} )\\\\F = \frac{m}{t} v\\\\Let \ \frac{m}{t} \ be \ constant = k\\F = kv\\\\k = \frac{F}{v} \\\\\frac{F_1}{v_1} = \frac{F_2}{v_2}[/tex]
The applied force needed to keep the box moving with a constant velocity that is twice as fast as before;
[tex]\frac{F_1}{v_1} = \frac{F_2}{v_2} \\\\(v_2 = 2v_1, \ and \ F_1 = 1.5N)\\\\\frac{1.5}{v_1} = \frac{F_2}{2v_1} \\\\1.5 = \frac{F_2}{2}\\\\F_2 = 2*1.5\\\\F_2 = 3 N[/tex]
Therefore, the applied force that is needed to keep the box moving with a constant velocity that is twice as fast as before, is 3 N
Force is directly proportional to velocity, to keep the box moving at the double of initial constant velocity, we must also double the value of the applied force.
A stone is thrown towards a wall with an initial velocity of v0=19m/s and an angle = 71 with the horizontal, as illustrated in the figure below. The stone reaches point A at the top of the wall, t=3.5s after being thrown. Determine (a) the height h of the wall, (b) the maximum height H of the path of the stone, (c) the horizontal distance between the launching point and point A and (d) the horizontal reach of the stone if the wall did not exist
Answer:
(a) 2.85 m
(b) 16.5 m
(c) 21.7 m
(d) 22.7 m
Explanation:
Given:
v₀ₓ = 19 cos 71° m/s
v₀ᵧ = 19 sin 71° m/s
aₓ = 0 m/s²
aᵧ = -9.8 m/s²
(a) Find Δy when t = 3.5 s.
Δy = v₀ᵧ t + ½ aᵧ t²
Δy = (19 sin 71° m/s) (3.5 s) + ½ (-9.8 m/s²) (3.5 s)²
Δy = 2.85 m
(b) Find Δy when vᵧ = 0 m/s.
vᵧ² = v₀ᵧ² + 2 aᵧ Δy
(0 m/s)² = (19 sin 71° m/s)² + 2 (-9.8 m/s²) Δy
Δy = 16.5 m
(c) Find Δx when t = 3.5 s.
Δx = v₀ₓ t + ½ aₓ t²
Δx = (19 cos 71° m/s) (3.5 s) + ½ (0 m/s²) (3.5 s)²
Δx = 21.7 m
(d) Find Δx when Δy = 0 m.
First, find t when Δy = 0 m.
Δy = v₀ᵧ t + ½ aᵧ t²
(0 m) = (19 sin 71° m/s) t + ½ (-9.8 m/s²) t²
0 = t (18.0 − 4.9 t)
t = 3.67
Next, find Δx when t = 3.67 s.
Δx = v₀ₓ t + ½ aₓ t²
Δx = (19 cos 71° m/s) (3.67 s) + ½ (0 m/s²) (3.67 s)²
Δx = 22.7 m
In the far future, astronauts travel to the planet Saturn and land on Mimas, one of its 62 moons. Mimas is small compared with the Earth's moon, with mass Mm = 3.75 ✕ 1019 kg and radius Rm = 1.98 ✕ 105 m, giving it a free-fall acceleration of g = 0.0636 m/s2. One astronaut, being a baseball fan and having a strong arm, decides to see how high she can throw a ball in this reduced gravity. She throws the ball straight up from the surface of Mimas at a speed of 43 m/s (about 96 mph, the speed of a good major league fastball)."
Required:
a. Predict the maximum height of the ball assuming g is constant and using energy conservation. Mimas has no atmosphere, so there is no air resistance.
b. Now calculate the maximum height using universal gravitation.
c. How far off is your estimate of part (a)? Express your answer as a percent difference and indicate if the estimate is too high or too low.
Answer:
a)[tex]h_{max}=14536.16 m[/tex]
b)[tex]h = 15687.9 m[/tex]
c)[tex]PD=7.62\%[/tex] The estimate is low.
Explanation:
a) Using the energy conservation we have:
[tex]E_{initial}=E_{final}[/tex]
we have kinetic energy intially and gravitational potential energy at the maximum height.
[tex]\frac{1}{2}mv^{2}=mgh_{max}[/tex]
[tex]h_{max}=\frac{v^{2}}{2g}[/tex]
[tex]h_{max}=\frac{43^{2}}{2*0.0636}[/tex]
[tex]h_{max}=14536.16 m[/tex]
b) We can use the equation of the gravitational force
[tex]F=G\frac{mM}{R^{2}}[/tex] (1)
We have that:
[tex] F = ma [/tex] (2)
at the surface G will be:
[tex]G=\frac{gR^{2}}{M}[/tex]
Now the equation of an object at a distance x from the surface.
is:
[tex]F=\frac{mgR^{2}}{(R+x)^{2}}[/tex]
[tex]m\frac{dv}{dt}=\frac{mgR^{2}}{(R+x)^{2}}[/tex]
Using that dv/dt is vdx/dt and integrating in both sides we have:
[tex]v_{0}=\sqrt{\frac{2gRh}{R+h}}[/tex]
[tex]h=\frac{v_{0}^{2}R}{2gR-v_{0}^{2}}[/tex]
[tex]h=15687.9[/tex]
c) The difference is:
So the percent difference will be:
[tex]PD=|\frac{14536.16-15687.9}{(14536.16+15687.9)/2}*100%[/tex]
[tex]PD=7.62\%[/tex]
The estimate is low.
I hope it helps you!
A 25-coil spring with a spring constant of 350 N/m is cut into five equal springs with five coils each. What is the spring constant of each of the 5-coil springs
1750N/m
Explanation:
According to Hooke's law, the spring constant (k) of an elastic material is the ratio of the force (F) applied to the material to the extension (x) of the material caused by this force. i.e
k = F / x --------------(i)
From the question, the elastic material (spring) of 25 coils has a spring constant of 350N/m. This means that for every 350N force applied to the spring, the spring extends by 1m.
Now if the spring is cut into five equal parts each with five coils, imagine they are attached together such that the same force of 350N is applied to still cause a total extension of 1m. Each spring contributes 1/5 of this extension.
Therefore, the extension caused by each spring is 1/5 of 1m = 0.2m
Since the same force of 350N is applied, substitute F = 350N and x = 0.2m into equation (i) as follows;
k = 350 / 0.2
k = 1750N/m
Therefore, the spring constant of each of the 5-coil spring is 1750N/m
The spring constant of each of the 5-coil springs is 1,750 N/m.
The given parameters;
spring constant, k = 350 N/mnumber of coils, N = 25The total spring constant of the 25 coils is calculated as follows;
[tex]K_t = 25 \times 350 \ N/m\\\\K_t = 8,750 \ N/m[/tex]
The spring constant of each of the given 5 coils is calculated as follows;
[tex]K = \frac{8,750 }{5} \\\\K = 1,750 \ N/m[/tex]
Thus, the spring constant of each of the 5-coil springs is 1,750 N/m.
Learn more here:https://brainly.com/question/22499689
In a ballistic pendulum experiment, a small marble is fired into a cup attached to the end of a pendulum. If the mass of the marble is 0.0215 kg and the mass of the pendulum is 0.250 kg, how high h will the pendulum swing if the marble has an initial speed of 5.15 m/s? Assume that the mass of the pendulum is concentrated at its end so that linear momentum is conserved during this collision.
Answer:
h = 8.48*10^-3m
Explanation:
In order to calculate the height reached by the pendulum with the marble, you first take into account the momentum conservation law, to calculate the speed of both pendulum and marble just after the collision.
The total momentum of the system before the collision is equal to the total momentum after:
[tex]m_1v_1+m_2v_2=(m_1+m_2)v[/tex] (1)
Here you used the fact that the pendulum has its total mass concentrated at the end of the pendulum.
m1: mass of the marble = 0.0215kg
m2: mass of the pendulum concentrated at its end = 0.250kg
v1: horizontal speed of the arble before the collision = 5.15m/s
v2: horizontal speed of the pendulum before the collision = 0m/s
v: horizontal speed of both marble and pendulum after the collision = ?
You solve the equation (1) for v, and replace the values of the other parameters:
[tex]v=\frac{m_1v_1+m_2v_2}{m_1+m_2}\\\\v=\frac{(0.0215kg)(5.15m/s)+(0.250kg)(0m/s)}{0.0215kg+0.250kg}=0.40\frac{m}{s}[/tex]
Next, you use the energy conservation law. In this case the kinetic energy of both marble and pendulum (just after the collision) is equal to the potential energy of the system when both marble and pendulum reache a height h:
[tex]U=K\\\\(m_1+m_2)gh=\frac{1}{2}(m_1+m_2)v^2\\\\h=\frac{v^2}{2g}[/tex]
v = 0.40m/s
g: gravitational acceleration = 9,8m/s^2
[tex]h=\frac{(0.40m/s)^2}{2(9.8m/s^2)}=8.48*10^{-3}m[/tex]
Then, the height reached by marble and pendulum is 8.48*10^-3m
For the last part of the lab, you should have found the mass of the meter stick. So if a mass of 85 g was placed at the 2 cm MARK and the pivot point was moved to the 38.6 cm MARK, what would have been the mass of the meter stick
Answer:
272.89g
Explanation:
Find the diagram to the question in the attachment below;.
Using the principle of moment to solve the question which states that the sum of clockwise moment is equal to the sum of anticlockwise moment.
Moment = Force * Perpendicular distance
Taking the moment of force about the pivot.
Anticlockwise moment:
The 85g mass will move in the anticlockwise
Moment of 85g mass = 85×36.6
= 3111gcm
Clockwise moment.
The mass of the metre stick M situated at the centre (50cm from each end) will move in the clockwise direction towards the pivot.
CW moment = 11.4×M = 11.4M
Equating CW moment to the ACW moment we will have;
11.4M = 3111
M = 3111/11.4
M = 272.89g
The mass of the metre stick is 272.89g
Assume the angular momentum of a diatomic molecule is quantized according to the relation . What are the allowed rotational kinetic energies
Answer:
The answer to this question can be defined as follows:
Explanation:
In the given question, an equation is missing which can be defined as follows:
[tex]I \omega =\sqrt{J(J+1)}h[/tex]
solution:
Angular momentum:
[tex]L=I \omega =\sqrt{J(J+1)}h[/tex]
Convert Angular momentum in terms of kinetic energy:
[tex]K = \frac{L^2}{2I}[/tex]
[tex]= \frac{h^2(J(J+1))}{2I}[/tex]
ou are using a hydrogen discharge tube and high quality red and blue light filters as the light source for a Michelson interferometer. The hydrogen discharge tube provides light of several different wavelengths (colors) in the visible range. The red light in the hydrogen spectrum has a wavelength of 656.3 nm and the blue light has a wavelength of 434.0 nm. When using the discharge tube and the red filter as the light source, you view a bright red spot in the viewing area of the interferometer. You now move the movable mirror away from the beam splitter and observe 146 bright spots. You replace the red filter with the blue filter and observe a bright blue spot in the interferometer. You now move the movable mirror towards the beam splitter and observe 122 bright spots. Determine the final displacement (include sign) of the moveable mirror. (Assume the positive direction is away from the beam splitter.)
Answer:
Explanation:
In interferometer , when the movable mirror is moved away by distance d , there is fringe shift on the screen . If n be number of fringes shifted
2 d = n λ
where λ is wavelength of light
Applying this theory for first case when no of fringes shifted is 146
2 d₁ = 146 x 656.3 nm
d₁ = 47909.9 x 10⁻⁹ m
= .048 x 10⁻³ m
= .048 mm
For second case n = 122
2d₂ = 122 x 434 x 10⁻⁹
d₂ = 26474 x 10⁻⁹ m
= .026 mm
So in the second case , mirror must have been moved towards the beam splitter by .048 - .026 = .022 mm
So movement = - 0 .022 mm ( negative displacement )
The resonance tube used in this experiment produced only one resonance tone. What length of tube would be required to produce a second tone under the same experimental conditions? Explain your answer.
Answer:
the length that would produce a sound tone under the same experimental contditions must be increased by Δl = [tex]\frac{v}{2f}[/tex]
Explanation:
Recall
V = f ×λ
where λ is ⁴/₃l₂ for second resonance
f = [tex]\frac{3v}{4l_{2} }[/tex]
l₂ = [tex]\frac{3v}{4f}[/tex]
where λ is 4l₁ for 1st resonance
f = [tex]\frac{v}{4l_{1} }[/tex]
l₁ = [tex]\frac{v}{4f}[/tex]
∴ Δl = l₂ - l₁ = [tex]\frac{3v}{4f}[/tex] ⁻ [tex]\frac{v}{4f}[/tex]
Δl= [tex]\frac{2v}{4f}[/tex]
Δl = [tex]\frac{v}{2f}[/tex]
Therefore, the length should increase by [tex]\frac{v}{2f}[/tex]
What is the relationship between the magnitudes of the collision forces of two vehicles, if one of them travels at a higher speed?
Explanation:
The collision forces are equal and opposite. Therefore, the magnitudes are equal.
A uniform disk, a uniform hoop, and a uniform solid sphere are released at the same time at the top of an inclined ramp. They all roll without slipping. In what order do they reach the bottom of the ramp
Answer:
sphere, disk, hoop
Explanation:
See attached file
A passenger jet flies from one airport to another 1,233 miles away in 2.4 h. Find its average speed. = ____ m/s
Speed = (distance) / (time)
Speed = (1,233 mile) / (2.4 hour)
Speed = 513.75 mile/hour
Speed = (513.75 mi/hr) x (1609.344 meter/mi) x (1 hr / 3600 sec)
Speed = (513.75 x 1609.344 / 3600) (mile-meter-hour/hour-mile-second)
Speed = 229.7 meter/second
A uniform electric field exists in the region between two oppositely charged plane parallel plates. A proton is released from rest at the surface of the positively charged plate and strikes the surface of the opposite plate, 1.60 cm distant from the first, in a time interval of 3.20×10−6s3.20×10 −6 s. (a) Find the magnitude of the electric field. (b) Find the speed of the proton when it strikes the negatively charged plate.
Answer:
E = 326.17 N/C
Explanation:
(a) In order to calculate the magnitude of the electric field between the parallel plates you first calculate the acceleration of the proton. You use the following formula:
[tex]x=v_ot+\frac{1}{2}at^2[/tex] (1)
vo: initial speed of the proton = 0m/s
t: time that the proton takes to cross the space between the plates = 3.20*10^-6 s
a: acceleration of the proton = ?
x: distance traveled by the proton = 1.60cm = 0.016m
You solve the equation (1) for a, and replace the values of all parameters:
[tex]a=\frac{2x}{t^2}=\frac{2(0.016m)}{(3.20*10^{-6}s)^2}=3.125*10^{10}\frac{m}{s^2}[/tex]
Next, you use the Newton second law for the electric force, to find the magnitude of the electric field:
[tex]F_e=qE=ma[/tex] (2)
q: charge of the proton = 1.6*10^-19C
m: mass of the proton = 1.77*10^-27kg
You solve the equation (2) for E:
[tex]E=\frac{ma}{q}=\frac{(1.67*10^{-27}kg)(3.125*10^{10}m/s^2)}{1.6*10^{-19}C}\\\\E=326.17\frac{N}{C}[/tex]
The magnitude of the electric field in between the parallel plates is 326.17N/C
A stunt man whose mass is 70 kg swings from the end ofa 4.0 m long rope along thearc of a vertical circle. Assuming that he starts from rest whenthe rope is horizontal, find the tensions in the rope that arerequired to make him follow his circular path at each of thefollowing points.
(a) at the beginning of his motion
N
(b) at a height of 1.5 m above the bottom of the circular arc
N
(c) at the bottom of the arc
N
Answer:
a. T= 0
b. T = 1286N
c. T= 2058N
Explanation:
A boat develops a leak and, after its passengers are rescued, eventually sinks to the bottom of a lake. When the boat is at the bottom, what is the force of the lake bottom on the boat?
Answer:
Explanation:
The force of the lake bottom on the boat = force the boat exerts on the bottom
force the boat exerts on the bottom = weight of the boat - buoyant force on the boat
so
The force of the lake bottom on the boat = weight of the boat - buoyant force on the boat
So
The force of the lake bottom on the boat will be less than its weight .
When moving to a new apartment, you rent a truck and create a ramp with a 244 cm long piece of plywood. The top of the moving ramp lies on the edge of the truck bed at a height of 115 cm. You load your textbooks into a wooden box at the bottom of the ramp (the coefficient of kinetic friction between the box and ramp is = 0.2). Then you and a few friends give the box a quick push and it starts to slide up the ramp. A) What angle is made by the ramp and the ground?B) Unfortunately, after letting go, the box only tables 80cm up the ramp before it starts coming back down! What speed was the box initially traveling with just after you stopped pushing it?
Answer:
A) θ = 28.1º , B) v = 2.47 m / s
Explanation:
A) The angle of the ramp can be found using trigonometry
sin θ = y / L
Φ = sin⁻¹ y / L
θ = sin⁻¹ (115/244)
θ = 28.1º
B) For this pate we can use the relationship between work and kinetic energy
W =ΔK
where the work is
W = -fr x
the negative sign is due to the fact that the friction force closes against the movement
Lavariacion of energy cineta is
ΔEm = ½ m v² - mgh
-fr x = ½ m v² - m gh
the friction force has the equation
fr = very N
at the highest part there is no speed and we take the origin from the lowest part of the ramp
To find the friction force we use Newton's second law. Where one axis is parallel to the ramp and the other is perpendicular
Axis y . perpendicular
N- Wy = 0
cos tea = Wy / W
Wy = W cos treaa
N = mg cos tea
we substitute
- (very mg cos tea) x = ½ m v²2 - mgh
v2 = m (gh- very g cos tea x)
let's calculate
v = Ra (9.8 0.80 - 0.2 9.8 0.0 cos 28.1)
v = RA (7.84 -1.729)
v = 2.47 m / s
An ac circuit consist of a pure resistance of 10ohms is connected across an ae supply
230V 50Hz Calculate the:
(i)Current flowing in the circuit.
(ii)Power dissipated
Plz check attachment for answer.
Hope it's helpful
The Huka Falls on the Waikato River is one of New Zealand's most visited natural tourist attractions. On average, the river has a flow rate of about 300,000 L/s. At the gorge, the river narrows to 20-m wide and averages 20-m deep.
(a) What is the average speed of the river in the gorge?
(b) What is the average speed of the water in the river downstream of the falls when it widens to 60 m and its depth increases to an average of 40 m?
Answer:
(a) V = 0.75 m/s
(b) V = 0.125 m/s
Explanation:
The speed of the flow of the river can be given by following formula:
V = Q/A
V = Q/w d
where,
V = Speed of Flow of River
Q = Volume Flow Rate of River
w = width of river
d = depth of river
A = Area of Cross-Section of River = w d
(a)
Here,
Q = (300,000 L/s)(0.001 m³/1 L) = 300 m³/s
w = 20 m
d = 20 m
Therefore,
V = (300 m³/s)/(20 m)(20 m)
V = 0.75 m/s
(b)
Here,
Q = (300,000 L/s)(0.001 m³/1 L) = 300 m³/s
w = 60 m
d = 40 m
Therefore,
V = (300 m³/s)/(60 m)(40 m)
V = 0.125 m/s
The molecules of a gas are in constant random motion. This means that they have energy in what type of energy store?
Answer:
Heat causes the molecules to move faster, (heat energy is converted to kinetic energy ) which means that the volume of a gas increases more than the volume of a solid or liquid.
Explanation:
gg The sound source of a ship’s sonar system operates at a frequency of 22.0 kHzkHz . The speed of sound in water (assumed to be at a uniform 20∘C∘C) is 1482 m/sm/s . What is the difference in frequency between the directly radiated waves and the waves reflected from a whale traveling straight toward the ship at 4.95 m/sm/s ? Assume that the ship is at rest in the water.
Answer:
Δf = 73.72Hz
Explanation:
In order to calculate the difference in frequency between the direct waves and the reflected waves, you first take into account the Doppler's effect for an observer getting closer to the source:
[tex]f'=f(\frac{v+v_o}{v-v_s})[/tex] (1)
You can assume that the reflected waves come from a source "the whale". Then you have:
f': frequency of the reflected waves = ?
f: frequency of the source = 22.0*kHz = 22.0*10^3 Hz
v: speed of the sound in water = 1482m/s
vs: speed of the source = 4.95m/s
vo: speed of the observer = 0m/s
You replace the values of the parameters in the equation (1):
[tex]f'=(22.0*10^3Hz)(\frac{1482m/s}{1482m/s-4.95m/s})=22073.72Hz[/tex]
Then, the difference in frequency is:
[tex]\Delta f = f'-f=22000Hz-22073.72Hz=73.72Hz[/tex]
C.
(11) in parallel
A potentiometer circuit consists of a
battery of e.m.f. 5 V and internal
resistance 1.0 12 connected in series with a
3.0 12 resistor and a potentiometer wire
AB of length 1.0 m and resistance 2.0 12.
Calculate:
(i) The total resistance of the circuit
The current flowing in the circuit
(iii) The lost volt from the internal
resistance of battery across the
battery terminals
(iv) The p.d. across the wire AB
(v) The e.m.f. of a dry cell which can be
balanced across 60 cm of the wire
AB.
Assume the wire has a uniform cross-
sectional area.
Answer:
fggdfddvdghyhhhhggghh
To maximize the percentage of the power from the emf of a battery that is delivered to a device external to the battery, what should the internal resistance of the battery be?
Answer:
the internal resistance of the battery be minimized
Explanation:
Because from
P= I²R+I²r
The amount of emf of battery delivered to the external load depends on internal resistance r such that the smaller the r the higher the emf
The internal resistance of the battery be :
-minimized.
Resistance of BatteryThe internal resistance of the battery be minimized.
Internal resistance refers to the restriction to the stream of current advertised by the cells and batteries themselves coming about within the era of warm.
Internal resistance is measured in Ohms.
The relationship between inside resistance (r) and emf (e) of cell s given by. e = I (r + R).
The sum of emf of battery conveyed to the outside stack depends on inside resistance r such that the littler the r the higher the emf.
Learn more about "resistance":
https://brainly.com/question/20708652?referrer=searchResults
A truck traveling at 60 km / h comes to a complete stop due to a traffic signal. Does this change in velocity violate the law of conservation of energy? Explain
Explanation:
No. Energy is still conserved. The kinetic energy of the truck is converted into heat.
You illuminate a slit with a width of 77.7 μm with a light of wavelength 721 nm and observe the resulting diffraction pattern on a screen that is situated 2.83 m from the slit. What is the width, in centimeters, of the pattern's central maximum
Answer:
The width is [tex]Z = 0.0424 \ m[/tex]
Explanation:
From the question we are told that
The width of the slit is [tex]d = 77.7 \mu m = 77.7 *10^{-6} \ m[/tex]
The wavelength of the light is [tex]\lambda = 721 \ nm[/tex]
The position of the screen is [tex]D = 2.83 \ m[/tex]
Generally angle at which the first minimum of the interference pattern the light occurs is mathematically represented as
[tex]\theta = sin ^{-1}[\frac{m \lambda}{d} ][/tex]
Where m which is the order of the interference is 1
substituting values
[tex]\theta = sin ^{-1}[\frac{1 *721*10^{-9}}{ 77.7*10^{-6}} ][/tex]
[tex]\theta = 0.5317 ^o[/tex]
Now the width of first minimum of the interference pattern is mathematically evaluated as
[tex]Y = D sin \theta[/tex]
substituting values
[tex]Y = 2.283 * sin (0.5317)[/tex]
[tex]Y = 0.02 12 \ m[/tex]
Now the width of the pattern's central maximum is mathematically evaluated as
[tex]Z = 2 * Y[/tex]
substituting values
[tex]Z = 2 * 0.0212[/tex]
[tex]Z = 0.0424 \ m[/tex]