Answer: (d)
Explanation:
Given
15 W is equivalent to 60 W light that is, it save 45 W
So, for 4 hours it is, [tex]4\times 45=180\ W.hr[/tex]
For 30 days, it becomes
[tex]\Rightarrow 180\times 30=5400\ W.hr\\\Rightarrow 5.4\ kWh[/tex]
Thus, [tex]5.4\ kWh[/tex] is saved in 30 days
option (d) is correct.
A gymnast weighs 450 N. She stands on a balance beam of uniform construction which weighs 250 N. The balance beam is 3.0 m long and is supported at each end. If the support force at the right end is four times the force at the left end, how far from the right end is the gymnast
Answer:
x = 9.32 cm
Explanation:
For this exercise we have an applied torque and the bar is in equilibrium, which is why we use the endowment equilibrium equation
Suppose the counterclockwise turn is positive, let's set our reference frame at the left end of the bar
- W l / 2 - W_{child} x + N₂ l = 0
x = [tex]\frac{-W l/2 + n_2 l}{W_{child}}[/tex] 1)
now let's use the expression for translational equilibrium
N₁ - W - W_(child) + N₂ = 0
indicate that N₂ = 4 N₁
we substitute
N₁ - W - W_child + 4 N₁ = 0
5 N₁ -W - W_{child} = 0
N₁ = ( W + W_{child}) / 5
we calculate
N₁ = (450 + 250) / 5
N₁ = 140 N
we calculate with equation 1
x = -250 1.50 + 4 140 3) / 140
x = 9.32 cm
The "Pressure" meter allows you to read the pressure at different depths in the fluid. Place the pressure meter close to the bottom of the pool, and read the pressure. Slowly move the pressure meter toward the surface of the water in the pool and read the pressure at different depths in the pool. What happens to pressure in the fluid as the depth of the fluid decreases?
Answer:
The pressure near the surface of the pool will be less as compared that the bottom of the pool as water has weight. This is in relation to gravity
Explanation:
There is a relationship between volume and pressure. The increase in depth leads to an increase in volume and an increase in the force of gravity near the surface as compared to lifting and rising light pressure as light air rises and heavy air sinks.1 A thing ring has a mass of 6kg and a radius of 20cm. calculate the rotational inertia.
Answer:
2400kgm²
Explanation:
Rotational inertia=mass x radius²
A charge is moving in a magnetic field that points to the
left
What direction can the charge move and experience no
magnetic force? Check all that apply.
O up
O down
Oleft
Oright
O into the screen
O out of the screen
Answer:
Magnetic Forces on Moving Charges. The magnetic force on a free moving charge is perpendicular to both the velocity of the charge and the magnetic field with direction given by the right hand rule.
The direction of the charge where it does not experience magnetic force is left and right. The correct option is C and D.
What is a magnetic field?A magnetic field is a region in space where a magnetic force can be observed. It is created by moving electric charges, such as electrons, and is characterized by the direction and strength of the force it exerts on other magnetic materials or moving charges.
Magnetic field lines are used to visualize the direction and strength of the magnetic field. They represent the path that a small magnetic north pole would follow if placed in the magnetic field. The direction of the magnetic field is given by the direction in which the north pole of a compass needle would point if placed in the field.
Magnetic flux is the measure of the strength of the magnetic field passing through a surface. It is given by the product of the magnetic field strength and the area of the surface, as well as the cosine of the angle between the magnetic field and the surface normal. The unit of magnetic flux is Weber (Wb).
Magnetic flux is important in many applications, such as electric motors and generators, where the interaction between the magnetic field and moving charges produces electrical energy. It is also used in magnetic imaging techniques, such as MRI, to visualize the internal structures of the human body.
Here in the question,
Options A (up), B (down), E (into the screen), and F (out of the screen) are all perpendicular to the direction of the magnetic field and so will experience a magnetic force.
Therefore, options C and D are correct i. e left and right.as they are parallel and anti-parallel to the direction of the magnetic field, respectively.
To learn about Ohm's law click:
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The resistance of the light bulb changed as the voltage (and current) changed. Why does this resistance change occur?
A car can go from 0 to 60 m/s in 5 seconds. What is the acceleration?
A) 50 m/s^2
B) 6 m/s^2
C) 12 m/s^2
D) 300 m/s^2
I think it’s 12 because I did the difference divided by 5 but some places said there was no acceleration
Answer:
12
Explanation:
i agree the answer is 12 because the acceleration is given by the difference in the velocity divided by the time taken
a=v-u/t
60-0/5
=12m/s²
I hope this helps
1. There is a famous intersection in Kuala Lumpur, Malaysia, where thousands of vehicles pass each hour. A 750 kg Tesla Model S traveling south crashes into a 1250 kg Ford F-150 traveling east. What are the initial speeds of each vehicle before collision if they stick together after crashing into each other and move at an angle of 320 and a common velocity of 18 m/s.
Solution :
Let the positive [tex]x-axis[/tex] is along the East and the positive [tex]y[/tex] direction is along the north.
Given :
Mass of the Tesla car, [tex]m_1[/tex] = [tex]750 \ kg[/tex]
Mass of the Ford car, [tex]m_2 = 1250 \ kg[/tex]
Now let the initial velocity of Tesla car in the south direction be = [tex]-v_1j[/tex]
The initial momentum of Tesla car, [tex]p_1 = -750 \ v_1[/tex]
Let the initial velocity of Ford car in the east direction be = [tex]v_2 \ i[/tex]
So the initial momentum of the Ford car is [tex]p_2=1250\ v_2 \ i[/tex]
Therefore, the initial velocity of both the cars is [tex]p_i = p_1+p_2[/tex]
[tex]=1250 \ v_2 \ i - 750\ v_1 \ j[/tex]
Now the final velocity of both the cars is [tex]v = 18 \ m/s[/tex]
So the vector form is :
[tex]v = 18\cos 32\ i-18 \sin 32 \ j[/tex]
[tex]= 15.26 \ i - 9.54 \ j[/tex]
Therefore the momentum after the accident is
[tex]p_f=(m_1+m_2) \times v[/tex]
[tex]=(750+1250) \times (15.26 \ i - 9.54 \ j)[/tex]
[tex]= 30520\ i -19080\ j[/tex]
According to the law of conservation of momentum, we know
[tex]p_i = p_f[/tex]
[tex]1250 \ v_2 \ i - 750\ v_1 \ j[/tex] [tex]= 30520\ i -19080\ j[/tex]
[tex]1250 \ v_2 = 30520[/tex]
[tex]v_2=24.4 \ m/s[/tex]
From, [tex]750\ v_1 = 19080[/tex]
We get, [tex]v_1=25.4 \ m/s[/tex]
Therefore the speed of Tesla car before collision = 25.4 m/s
The speed of ford car before collision = 24.4 m/s
The potential difference between the plates of a capacitor is 234 V. Midway between the plates, a proton and an electron are released. The electron is released from rest. The proton is projected perpendicularly toward the negative plate with an initial speed. The proton strikes the negative plate at the same instant the electron strikes the positive plate. Ignore the attraction between the two particles, and find the initial speed of the proton.
I have tried looking at the cramster.com solution manual and do not like the way it is explained. Simply put, I cannot follow what is going on and I am looking for someone who can explain it in plain man's terms and help me understand and get the correct answer. I am willing to give MAX karma points to anyone who can help me through this. Thank you kindly.
Answer:
The speed of proton is 2.1 x 10^5 m/s .
Explanation:
potential difference, V = 234 V
let the initial speed of the proton is v.
The kinetic energy of proton is
KE = q V
[tex]0.5 mv^2 = e V \\\\0.5\times 1.67\times 10^{-27} v^2 = 1.6\times 10^{-19} \times 234\\\\v=2.1\times 10^5 m/s[/tex]
The resistance of a thermistor over a limited range of temperature is given by the equation R= ( c/T-203 )where c is a constant and T is the absolute temperature. What will be the temperature on the Celsius scale of the thermistor at absolute temperature of T = 300K?
Temperature in kelvin scale=300k
We know
[tex]\boxed{\sf 0°C=273K}[/tex]
[tex] \\ \Large\sf\longmapsto 300K[/tex]
[tex] \\ \Large\sf\longmapsto 300-273[/tex]
[tex] \\ \Large\sf\longmapsto 27°C[/tex]
How can I solve the following?
In (Figure 1), let V = 15.0 V and C1=C2=C3= 24.2 μF.
Part A: How much energy is stored in the capacitor network as shown in (Figure 1)?
Part B: How much energy would be stored in the capacitor network if the capacitors were all in series?
Part C: How much energy would be stored in the capacitor network if the capacitors were all in parallel?
Answer:
Part A - 4.084 mJ
Part B - 0.908 mJ
Part C - 8.168 mJ
Explanation:
Part A: How much energy is stored in the capacitor network as shown in (Figure 1)?
Since capacitors C₂ and C₃ are in series, their equivalent capacitance is C',
1/C' = 1/C₂ + 1/C₃ (Since C₁ = C₂ = C₃ = C)
1/C' = 1/C + 1/C
1/C' = 2/C
C' = C/2
Since C' is in parallel with C₁, the equivalent capacitance for the circuit is C" = C₁ + C' = C + C/2 = 3C/2
C" = 3C/2
The energy stored in the circuit, W = 1/2C"V² where C" = equivalent capacitance = 3C/2 and V = voltage = 15.0 V
W = 1/2C"V²
W = 1/2(3C/2)V²
W = 3CV²/4
since C = 24.2 μF = 24.2 × 10⁻⁶ F
W = 3CV²/4
W = 3 × 24.2 × 10⁻⁶ F (15.0 V)²/4
W = 3 × 24.2 × 10⁻⁶ F × 225 V²/4
W = 16335/4 × 10⁻⁶ FV²
W = 4083.75 × 10⁻⁶ J
W = 4.08375 × 10⁻³ J
W = 4.08375 mJ
W ≅ 4.084 mJ
Part B: How much energy would be stored in the capacitor network if the capacitors were all in series?
If the capacitors are connected in series, their equivalent resistance is C'
and 1/C' = 1/C₁ + 1/C₂ + 1/C₃
Since C₁ = C₂ = C₃ = C
1/C' = 1/C + 1/C + 1/C
1/C' = 3/C
C' = C/3
The energy stored in the circuit, W = 1/2C'V² where C' = equivalent capacitance = C/3 and V = voltage = 15.0 V
W = 1/2C'V²
W = 1/2(C/3)V²
W = CV²/6
since C = 24.2 μF = 24.2 × 10⁻⁶ F
W = CV²/6
W = 24.2 × 10⁻⁶ F (15.0 V)²/6
W = 24.2 × 10⁻⁶ F × 225 V²/6
W = 5445/6 × 10⁻⁶ FV²
W = 907.5 × 10⁻⁶ J
W = 0.9075 × 10⁻³ J
W = 0.9075 mJ
W ≅ 0.908 mJ
Part C: How much energy would be stored in the capacitor network if the capacitors were all in parallel?
If the capacitors are connected in parallel, their equivalent resistance is C'
and C' = C₁ + C₂ + C₃
Since C₁ = C₂ = C₃ = C
C' = C + C + C
C' = 3C
The energy stored in the capacitor network, W = 1/2C'V² where C' = equivalent capacitance = 3C and V = voltage = 15.0 V
W = 1/2C'V²
W = 1/2(3C)V²
W = 3CV²/2
since C = 24.2 μF = 24.2 × 10⁻⁶ F
W = 3CV²/2
W = 3 × 24.2 × 10⁻⁶ F (15.0 V)²/2
W = 3 × 24.2 × 10⁻⁶ F × 225 V²/2
W = 16335/2 × 10⁻⁶ FV²
W = 8167.5 × 10⁻⁶ J
W = 8.1675 × 10⁻³ J
W = 8.1675 mJ
W ≅ 8.168 mJ
Which of the following is not an example of approximate simple harmonic motion
Answer:
where are the options
it's not full question
Xác định ứng lực trong các thanh BC,
CF và FE của hệ giàn và chỉ rõ các thanh
chịu kéo hay nén. Cho P1=P2=600 lb,
P3=800 lb
Answer:
............................................
Explanation:
what is simple machine?
Explanation:
Those tools that helps to make our work easier ,faster and more convenient in our daily life it is called simple Machine.
Hey guys,I hope u r gonna answer this question fast,SI system is extended from of MKS system.Why? I will be waiting for the answer. Good luck thank u
Answer:
Because SI system has fundamental units of MKS System
Answer:
Explanation: the unit of length ,mass , and time are same in both the system , thus, the SI system is the extended from of MKS system.
A car's bumper is designed to withstand a 6.12 km/h (1.7-m/s) collision with an immovable object without damage to the body of the car. The bumper cushions the shock by absorbing the force over a distance. Calculate the magnitude of the average force on a bumper that collapses 0.210 m while bringing a 810 kg car to rest from an initial speed of 1.7 m/s.
Answer:
5572.8 N
Explanation:
Applying,
F = ma.............. Equation 1
Where F = Force, m = mass of the car, a = acceleration.
We can find a by applying,
v² = u²+2as............. Equation 2
Where v = final velocity, u = initial velocity, a = acceleration, = distance.
From the question,
Given: v = 0 m/s (come to rest), u = 1.7 m/s, s = 0.210 m
Substitute these value into equation 2
0² = 1.7²+2×0.21×a
a = -1.7²/(2×0.21)
a = -2.89/0.42
a = -6.88 m/s²
Also given: m = 810 kg
Substitute these value into equation 1
F = 810(-6.88)
F = -5572.8 N
Hence the force on the bumber is 5572.8 N
A proton traveling at 17.6° with respect to the direction of a magnetic field of strength 3.28 mT experiences a magnetic force of 9.14 × 10-17 N. Calculate (a) the proton's speed and (b) its kinetic energy in electron-volts.
Answer:
a) The proton's speed is 5.75x10⁵ m/s.
b) The kinetic energy of the proton is 1723 eV.
Explanation:
a) The proton's speed can be calculated with the Lorentz force equation:
[tex] F = qv \times B = qvBsin(\theta) [/tex] (1)
Where:
F: is the force = 9.14x10⁻¹⁷ N
q: is the charge of the particle (proton) = 1.602x10⁻¹⁹ C
v: is the proton's speed =?
B: is the magnetic field = 3.28 mT
θ: is the angle between the proton's speed and the magnetic field = 17.6°
By solving equation (1) for v we have:
[tex]v = \frac{F}{qBsin(\theta)} = \frac{9.14 \cdot 10^{-17} N}{1.602\cdot 10^{-19} C*3.28 \cdot 10^{-3} T*sin(17.6)} = 5.75 \cdot 10^{5} m/s[/tex]
Hence, the proton's speed is 5.75x10⁵ m/s.
b) Its kinetic energy (K) is given by:
[tex] K = \frac{1}{2}mv^{2} [/tex]
Where:
m: is the mass of the proton = 1.67x10⁻²⁷ kg
[tex] K = \frac{1}{2}mv^{2} = \frac{1}{2}1.67 \cdot 10^{-27} kg*(5.75 \cdot 10^{5} m/s)^{2} = 2.76 \cdot 10^{-16} J*\frac{1 eV}{1.602 \cdot 10^{-19} J} = 1723 eV [/tex]
Therefore, the kinetic energy of the proton is 1723 eV.
I hope it helps you!
Two identical loudspeakers 2.00 m apart are emitting sound waves into a room where the speed of sound is 340 m/s. Abby is standing 5.00 m in front of one of the speakers, perpendicular to the line joining the speakers, and hears a maximum in the intensity of the sound.
Required:
What is the lowest possible frequency of sound for which this is possible?
Answer:
The lowest possible frequency of sound for which this is possible is 1307.69 Hz
Explanation:
From the question, Abby is standing 5.00m in front of one of the speakers, perpendicular to the line joining the speakers.
First, we will determine his distance from the second speaker using the Pythagorean theorem
l₂ = √(2.00²+5.00²)
l₂ = √4+25
l₂ = √29
l₂ = 5.39 m
Hence, the path difference is
ΔL = l₂ - l₁
ΔL = 5.39 m - 5.00 m
ΔL = 0.39 m
From the formula for destructive interference
ΔL = (n+1/2)λ
where n is any integer and λ is the wavelength
n = 1 in this case, the lowest possible frequency corresponds to the largest wavelength, which corresponds to the smallest value of n.
Then,
0.39 = (1+ 1/2)λ
0.39 = (3/2)λ
0.39 = 1.5λ
∴ λ = 0.39/1.5
λ = 0.26 m
From
v = fλ
f = v/λ
f = 340 / 0.26
f = 1307.69 Hz
Hence, the lowest possible frequency of sound for which this is possible is 1307.69 Hz.
You have a 1 W light bulb in your lab. It puts out light of only 1 frequency. The wavelength of this light is 500nm. you set up a detector with a surface area of 1 square centimeter facing the light source at a distance of 100m.
Required:
a. Find the number of photons hitting the detector every second.
b. What is the maximum E field of the E M wave hitting the detector?
c. What is the maximum value of the B field of this E M wave?
d. How far away would you have to place the detector to only receive 1 photon per second from the light bulb?
Answer:
a) # _photon = 2.5 10¹⁸ photons / s, b) E = 10⁻² N / C, c) B = 3 10⁻¹¹ T
d) r= 2 10⁹ m
Explanation:
a) Let's solve this exercise in part, let's start by finding the energy of each photon using the Planck relation
E₀ = h f
c = λ f
E₀ = h c /λ
E₀ = 6.63 10⁻³³⁴ 3 10⁸/500 10⁻⁹
E₀ = 3.978 10⁻⁻¹⁹ J
Let's use a direct ratio rule to find the number of photons
#_foton = E / Eo
#_fototn = 1 / 3.978 10⁻¹⁹
# _photon = 2.5 10¹⁸ photons / s
b) The intensity received by the detector is related to the electric field
I = E²
Let's look for the intensity that the detector receives, suppose that the emission is shapeless throughout the space
I = P / A
P = I A
Let's use index 1 for the point on the bulb and index 2 for the point on the detector.
The area of a sphere is
A = 4π r²
P = I₁ A₁ = I₂ A₂
I₁ r₁² = I₂ r₂²
I₂ = I₁ r₁²/r₂²
I₂ = I₁ 1 / 100²
I₂ = I₁ 10⁻⁴
we must know the intensity at the output of the bulb suppose that I₁ = 1 J
I₂ = 10⁻⁴ J
let's look for the electric field
E =√I
E = √10⁻⁴
E = 10⁻² N / C
c) for the calculation of the magnetic field we use that the field is in phase
E / B = c
B = E / c
B = 10⁻² / 3 10⁸
B = 3 10⁻¹¹ T
d) Let's use a direct proportions rule if we fear 2.5 10¹⁸ photons in an area A = 4π R² where R = 100 m how many photons are there in the area of the detector r = 1 cm, A’= 10⁻⁴ m²
#_photons = 2.5 10¹⁸ A_detector / A_sphere
#_photons = 2.5 1018 10-4 / 4π 10⁴
#_photons = 2 10⁹ photons in the detector area
for the number of photons to decrease to 1, the radius of the sphere must be 2 10⁹ m
Which circuit element is of special importance in AC circuits?
A. Resistor
B. Ammeter
C. Battery
D. Capacitor
Answer:
Explanation:capacitor
Answer:
Ammeter
Explanation:
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In the diagram, disk 1 has a moment of inertia of 3.4 kg · m2 and is rotating in the counterclockwise direction with an angular velocity of 6.1 rad/s about a frictionless rod passing through its center. A second disk rotating clockwise with an angular velocity of 9.3 rad/s falls from above onto disk 1. The two then rotate as one in the clockwise direction with an angular velocity of 1.8 rad/s. Determine the moment of inertia, in kg · m2, of disk 2.
Answer:
I = 3.6 kg•m²
Explanation:
Conservation of angular momentum
Let's assume CW is the positive direction
3.4(-6.1) + I(9.3) = 3.4(1.8) + I(1.8)
I(9.3 - 1.8) = 3.4(1.8 + 6.1)
I(7.5) = 3.4(7.9)
I = 3.4(7.9)/(7.5) = 3.5813333333...
The moment of inertia of the second disk will be [tex]I=3.58\ kg-m^2[/tex]
What is moment of inertia?The moment of inertia is defined as the product of mass of section and the square of the distance between the reference axis and the centroid of the section.
here it is given that
MOI of disk one [tex]I_1=3.4\ kg-m^2[/tex]
Angular velocity [tex]w_1=6.1\ \frac{rad}{s}[/tex]
Angular velocity of disk two [tex]w=1.8\ \frac{rad}{s}[/tex]
MOI of the disk two [tex]I=?[/tex]
The final angular velocity [tex]w_f= 1.8\ \frac{rad}{sec}[/tex]
Now from the conservation of the momentum the angular momentum before collision will be equal to the angular momentum after collision.
[tex]I_1w_1+I_2w_2=(I_1+I_2)w_f[/tex]
Now put the values in the formula
[tex](3.4\times 6.10)+(I_2\times 9.3)=(3.4+I_2)\times 1.8[/tex]
[tex]I_2=3.58\ kg-m^2[/tex]
Thus the moment of inertia of the second disk will be [tex]I=3.58\ kg-m^2[/tex]
To know more about moment of inertia follow
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Tay quay OB quay đều quanh trục cố định đi qua O với vận tốc góc không đổi ω. Con lăn A chuyển động trong rãnh thẳng đứng. Tại vị trí trên hình vẽ thì thanh OB thẳng đứng, OA có phương nằm ngang. Hãy xác định vận tốc góc thanh AB, vận tốc của con lăn A; gia tốc góc của thanh AB, gia tốc của con lăn A. Cho ω = 1,5 rad/s, r = 1 m.
1:
Forces and Motion:Question 2
A car is travelling east, when suddenly a more massive car travelling
north hits it with a greater force. What is likely to happen to the car
that was originally travelling east?
Explanation:
the car will be brought back
An astronaut on the moon drops a rock from rest. The rock falls 0.8m in one second of falling time. If the dropped rock fell for a total of two seconds of time instead of 1 second, then the distance traveled would be:____.
A) The same.
B) Doubled.
C) Tripled.
D) Quadruple.
E) None of the above.
Answer:
D) Quadruple.
Explanation:
We will use the second equation of motion to solve this problem:
[tex]s = v_it + \frac{1}{2}gt^2[/tex]
where,
s = distance travelled by the rock
vi = initial speed of rock = 0 m/s
t = time taken
g = acceleration due to gravity on the surface of the moon
Therefore,
[tex]s = (0\ m/s)t+\frac{1}{2}gt^2\\\\s =\frac{1}{2}gt^2[/tex]----------- equation (1)
Now, we double the time:
[tex]s' = \frac{1}{2}g(2t)^2\\\\s' = 4(\frac{1}{2}gt^2)[/tex]
using equation (1):
s' = 4s
Hence, the correct option is:
D) Quadruple.
Một bơm cánh gạt thủy lực có lưu lượng thực là 20lít/phút, tạo áp suất 230 bar, tốc độ bơm là 1400 vòng/phút. Biết công suất đầu vào là 10kW và hiệu suất cơ là 88%.
a) Tính hiệu suất thể tích của bơm
b) Tính thể tích riêng của bơm (cm/vòng). Câu 3 (2,5đ): Thiết kế hệ thống truyền động khí nén để điều khiển 02 xylanh tác động đơn, sử dụng 02 van đảo chiều 3/2 tác động bằng nút nhấn, 02 van tiết lưu - một chiều. Trình bày nguyên lý làm việc của của hệ thống.
Answer:
English please?
Explanation:
Explain in english?
how many rings does saturn have
Answer:
From far away, Saturn looks like it has seven large rings. Each large ring is named for a letter of the alphabet. The rings were named in the order they were discovered.
why do atom absorb photon since it makes it unstable??
[tex]\textsf{When an electron is hit by a }[/tex] [tex]\textsf{photon of light, it absorbs the quanta}[/tex] [tex]\textsf{of energy the photon was carrying}[/tex] [tex]\textsf{and moves to a higher energya}[/tex] [tex]\textsf{ state. Electrons therefore have to }[/tex] [tex]\textsf{jump around within the atom as }[/tex] [tex]\textsf{they either gain or lose energy. }[/tex]
When an electron is hit by a photon of light, it absorbs the quanta of energy the photon was carrying and moves to a higher energy state. Electrons therefore have to jump around within the atom as they either gain or lose energy.
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An auto mechanic needs to determine the emf and internal resistance of an old battery. He performs two measurements: in the first, he applies a voltmeter to the battery's terminals and reads 11.9 V;11.9 V; in the second, he applies an ammeter to the terminals and reads 16.1 A.16.1 A.
What are the battery's emf E and internal resistance r?
Answer:
Hence the battery's emf E is ε = 11.9 V.
The internal resistance is r = 0.739 ohms.
Explanation:
Now we know that
Voltage V = 11.9 V.
Current I = 16.1 A.
Hence this is an ideal voltmeter there are no current flows when the Voltmeter is applied.
ε = V + I r
∵ I = 0
ε = V
ε = 11.9 V
Then the ammeter is applied.
Let's take ( r ) to be the total resistance which is equal to internal resistance.
V = I r
r = [tex]\frac{V}{I}[/tex]
[tex]= \frac{11.9}{16.1}[/tex]
r = 0.739 ohms
The battery's emf (E) and internal resistance (r) are 11.9 Volts and 0.739 Ampere respectively.
Given the following data:
Voltage = 11.9 Volts.Current = 16.1 Amperes.To determine the battery's emf (E) and internal resistance (r):
How to calculate emf (E).For an ideal voltmeter, there isn't a flow of current and as such the current is equal to 0.
Mathematically, emf (E) is given by this formula:
[tex]E = V + IR[/tex]
Substituting the given parameters into the formula, we have;
[tex]E = 11.9 + 0R\\\\E = 11.9 + 0[/tex]
E = 11.9 Volts.
For the internal resistance (r):
Note: The total resistance is equal to internal resistance.
Applying Ohm's law, we have:
[tex]R = \frac{V}{I} \\\\R = \frac{11.9}{16.1}[/tex]
R = r = 0.739 Ampere.
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I’m a photoelectric effect, which property of the incident light determines how much kinetic energy the ejected electrons have ?
A) brightness
B) frequency
C) size of the beam
D) none of the above
Answer:
b = frequency
When a charged particle moves at an angle of 26.1 with respect to a magnetic field, it experiences a magnetic force of magnitude F. At what angle (less than 90o) with respect to this field will this particle, moving at the same speed?
Answer:
The angle is 153.9 degree.
Explanation:
Let the magnetic field is B and the charge is q. Angle = 26.1 degree
The force is F.
Let the angle is A'.
Now equate the magnetic forces
[tex]q v B sin 26.1 = q v B sin A'\\\\A' = 180 - 26.1 = 153.9[/tex]
You attach a 2.30 kg weight to a horizontal spring that is fixed at one end. You pull the weight until the spring is stretched by 0.500 m and release it from rest. Assume the weight slides on a horizontal surface with negligible friction. The weight reaches a speed of zero again 0.400 s after release (for the first time after release). What is the maximum speed of the weight (in m/s)
Answer: [tex]3.92\ m/s[/tex]
Explanation:
Given
Mass of the attached object is [tex]m=2.3\ kg[/tex]
Spring is stretched by [tex]A=0.5\ m[/tex]
Speed reaches zero after [tex]t=0.4\ s[/tex]
Speed is zero at the extremities of the S.H.M motion that is
[tex]\Rightarrow \dfrac{T}{2}=0.4\\\\\Rightarrow T=0.8\ s[/tex]
Time period of motion is [tex]0.8\ s[/tex] which can also be given by
[tex]\Rightarrow \omega T=2\pi\\\\\Rightarrow \omega=\dfrac{2\pi }{T}\\\\\Rightarrow \omega =\dfrac{2\pi }{0.8}\\\\\Rightarrow \omega=\dfrac{5\pi }{2}[/tex]
Maximum speed for S.H.M. is [tex]v_{max}=A\omega[/tex]
[tex]\Rightarrow v_{max}=0.5\times 2.5\pi\\\Rightarrow v_{max}=3.92\ m/s[/tex]
