Answer:
hi your pinterest I'd or Twitter I'd pls.
How to calculate voltage U1 ?
Please help!
Answer:
he is a baby art and design
A cosmic ray proton moving toward Earth at 5.00 x 107 m/s experiences a magnetic force of 1.7 x 10-16 N. What is the strength of the magnetic field if there is a 45o angle between it and the proton's velocity
Answer:
the strength of the magnetic field is 3 x 10⁻⁵ T
Explanation:
Given;
velocity of the cosmic ray, v = 5 x 10⁷ m/s
force experienced by the ray, f = 1.7 x 10⁻¹⁶ N
angle between the ray's velocity and the magnetic field, θ = 45⁰
The strength of the magnetic field is calculated as;
[tex]F = qvB \ sin(\theta)\\\\B = \frac{F}{qv\times sin(\theta)} \\\\where;\\\\B \ is \ the \ strength \ of \ the \ magnetic \ field\\\\q \ is \ the \ charge \ of \ the \ cosmic \ ray \ proton = 1.602 \times 10^{-19} \ C\\\\B = \frac{1.7\times 10^{-16}}{(1.602 \times 10^{-19})\times (5\times 10^7) \times sin \ (45)} \\\\B = 3 \times 10^{-5} \ T[/tex]
Therefore, the strength of the magnetic field is 3 x 10⁻⁵ T
With the frequency set at the mid-point of the slider and the amplitude set at the mid-point of the slider, approximately how many grid marks is the wavelength of the wave (use the pause button and step button as you need to in order to get a good measure, and round to the nearest whole grid mark)?
Answer:
The wavelength stays the same.
Explanation:
When the amplitude is increased, the wavelength stays the same.
Here the wavelength doesn't depend upon the amplitude.
Describe the change in motion and kinetic energy of the particles as thermal energy is added to a liquid. Which change of state might happen?
please ill put brainliest!!!
Answer:
If a liquid is heated the particles are given more energy and move faster and faster expanding the liquid. The most energetic particles at the surface escape from the surface of the liquid as a vapour as it gets warmer. Liquids evaporate faster as they heat up and more particles have enough energy to break away
What is the feature known as the "Great Dark Spot" of Neptune? It is an apparently permanent feature about five times the size of Earth, similar to the Great Red Spot of Jupiter, near Neptune's south pole. It was a dark hole in the upper atmosphere left by the collision of the comet Shoemaker-Levy 9. It was an apparently temporary feature about the size of Earth, similar to the Great Red Spot of Jupiter, but disappeared within a few years. It is a dark surface feature on the surface snow layers caused by radiation discoloration of the older layers. It is a permanent discoloration of the north polar region of Neptune caused by locally prevailing lower surface temperatures there.
Answer:
It was an apparently temporary feature about the size of Earth, similar to the Great Red Spot of Jupiter, but disappeared within a few years.
Explanation:
The Great Dark Spot of Neptune was an immense spinning storm in the southern atmosphere of Neptune. The size of the entire Earth, it had the strongest winds ever recorded on any planet in the solar system. It was discovered by the Voyager 2 spacecraft in 1989, but by 1994 the Hubble Space Telescope saw it was gone.
The Great Red Spot is a storm found in Jupiter's southern hemisphere, with similar characteristics to the Great Dark Spot.
The paper dielectric in a paper-and-foil capacitor is 0.0785 mm thick. Its dielectric constant is 2.35, and its dielectric strength is 49.5 MV/m. Assume that the geometry is that of a parallel-plate capacitor, with the metal foil serving as the plates.
Required:
a. What area of each plate is required for for a 0.300 uF capacitor?
b. If the electric field in the paper is not to exceed one-half the dielectric strength, what is the maximum potential difference that can be applied across the compactor?
Answer:
a) required area is 1.1318 m²
b) the maximum potential difference that can be applied across the compactor is 1931.1 V
Explanation:
Given the data in the question;
dielectric constant εr = 2.35
distance between plates ( thickness ) d = 0.0785 mm = 7.85 × 10⁻⁵ m
dielectric strength = 49.5 MV/m
a)
given that capacity capacitor C = 0.3 uF = 0.3 × 10⁻⁶ F
To find the Area, we use the following the expression.
C = ε₀εrA / d
we know that The permittivity of free space, ε₀ = 8.854 x 10⁻¹² (F/m)
we substitute
0.3 × 10⁻⁶ = [ (8.854 x 10⁻¹²) × 2.35 × A ] / 7.85 × 10⁻⁵
A = [ (0.3 × 10⁻⁶) × (7.85 × 10⁻⁵) ] / [ 2.35 × (8.854 x 10⁻¹²) ]
A = 2.355 × 10⁻¹¹ / 2.08069 × 10⁻¹¹
A = 1.1318 m²
Therefore, required area is 1.1318 m²
b)
the maximum potential difference that can be applied across the compactor.
We use the following expression;
⇒ 1/2 × dielectric strength × thickness d
we substitute
⇒ 1/2 × ( 49.5 × 10⁶ V/m ) × ( 7.85 × 10⁻⁵ m )
⇒ 1931.1 V
Therefore, the maximum potential difference that can be applied across the compactor is 1931.1 V
Which nucleus completes the following equation?
39 17 CI-> 0 -1 e+?
Answer:
[tex]_{18}^{39} } Ar[/tex]
Explanation:
The given equation shows the disintegration of an unstable isotope of chlorine to beta particle and Argon nucleus. The nucleus undergoes the emission of a beta particle to form a more stable nucleus of Argon.
[tex]_{17} ^{39} Cl[/tex] ⇒ [tex]_{-1}^{0} e[/tex] + [tex]_{18}^{39} } Ar[/tex]
Argon is a stable gas and is found in the group 8 on the periodic table of elements.
Answer:
Answer is below
Explanation:
39 18 Ar
I REALLY NEED HELP WITH PHYSICS ASAP!!!
Vf^2 = v0^2 + 2a (xf - x0)
Solve for a
Answer:
a. solve for a
[tex]vf ^{2} = vo ^{2} + 2a(xf - xo) \\ 2a(xf - xo) = vf^{2} - vo ^{2} \\ a = \frac{vf^{2} - vo^{2} }{2(xf - xo)} \\ a = \frac{vf ^{2} - vo ^{2} }{2xf - 2xo} [/tex]
I hope I helped you ^_^
The exponent of the exponential function contains RC for the given circuit, which is called the time constant. Use the units of R and C to find units of RC. Write ohms in terms of volts and amps and write farads in terms of volts and coulombs. Simplify until you get something simple. Show your work below.
Answer:
The unit of the time constant RC is the second
Explanation:
The unit of resistance, R is the Ohm, Ω and resistance, R = V/I where V = voltage and I = current. The unit of voltage is the volt, V while the unit of current is the ampere. A.
Since R = V/I
Unit of R = unit of V/unit of I
Unit of R = V/A
Ω = V/A
Also, The unit of capacitance, C is the Farad, F and capacitance, F = Q/V where Q = charge and V = voltage. the unit of charge is the coulomb, C while the unit of voltage is the volt, V
Since C = Q/V
Unit of C = unit of Q/unit of V
Unit of C = C/V
F = C/V
Now the time constant equals RC.
So, the unit of the time constant = unit of R × unit of C = Ω × F = V/A × C/V = C/A
Also. we know that the 1 Ampere = 1 Coulomb per second
1 A = 1 C/s
So, substituting 1 A in the denominator, we have
unit of RC = C/A = C ÷ C/s = s
So, the unit of RC = s = second
So, the unit of the time constant RC is the second
Select the correct answer.
What is abstraction?
OA. the concept that software architecture can be separated into modules and that each module can be examined independently
OB. the process of containing information within a module, preventing any crossover or access to Irrelevant information
OC. the process of splitting a program both horizontally and vertically
OD. the process of cutting down irrelevant information so only the information that is useful for a particular purpose remains
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Answer:
OD. The process of cutting down irrelevant information so only the information that is useful for particular purpose remains
Abstraction is the process of cutting down irrelevant information so only the information that is useful for a particular purpose remains.
What is abstraction?Abstraction is the practice of removing anything from a set of core features by eliminating or deleting attributes.
One of the three core ideas of object-oriented programming is abstraction order to decrease complexity and maximize efficiency, a programmer uses abstraction to conceal all but the important facts about an object.
Abstraction is the process of cutting down irrelevant information so only the information that is useful for a particular purpose remains.
Hence option D is correct.
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write a use of magnetic force and frictional force each
The time delay between transmission and the arrival of the reflected wave of a signal using ultrasound traveling through a piece of fat tissue was 0.13 ms. At what depth did this reflection occur? (The average propagation speed for sound in body tissue is 1540 m/s)
Answer:
10.01 cm
Explanation:
Given that,
The time delay between transmission and the arrival of the reflected wave of a signal using ultrasound traveling through a piece of fat tissue was 0.13 ms.
The average propagation speed for sound in body tissue is 1540 m/s.
We need to find the depth when the reflection occur. We know that, the distance is double when transmitting and arriving. So,
[tex]v=\dfrac{2d}{t}\\\\d=\dfrac{vt}{2}\\\\d=\dfrac{1540\times 0.13\times 10^{-3}}{2}\\\\d= $$0.1001\ m[/tex]
or
d = 10.01 cm
So, the reflection will occur at 10.01 cm.
Để sử dụng nguồn điện xoay chiều 220V/50Hz thắp sáng bóng đèn 12V/3W, ta chọn điện trở giảm áp có giá trị:
Explanation:
Hi Linda,
How's it going?
Sorry I haven't been in touch for such a long time but I've had exams so I've been studying every free minute. Anyway, I'd love to hear all your news and I'm hoping we can get together soon to catch up. We just moved to a bigger flat so maybe you can come and visit one weekend?
How's the new job?
Looking forward to hearing from you!
Helga
two bodies A and B with some asses 20 kg and 30 kg respectively above the ground which have greater potential
Answer:
B has greater potential
Explanation:
We know;
Potential Energy (PE) = mgh
where, m=mass of body
g=acceleration due to gravity
h=height of body
From the formula,
PE is directly proportional to the mass of the body
so the body with greater mass has greater potential.
Calculate area moment of inertia for a circular cross-section with 3 mm diameter:
Answer:
circles
A=7.07multiple 10-6 m2
I hope you understand and help
The area of the circular cross-section will be 7068 × [tex]10^{-6}m^{2}[/tex].
What is the area?The measurement that represents the size of a region on a plane or curved surface is called an area.
What is cross-section?A cross-section would be the non-empty point where a solid body intersects a plane in three dimensions or its equivalent in higher dimensions.
Given data:
Diameter = 3 × [tex]10^{-3} m[/tex].
It is known that. Diameter = 2 radius.
The area can be calculated by using the formula:
A = 1/4 [tex]\pi[/tex][tex]d^{2}[/tex] = 1/4 (3.14) [tex](3 * 10^{-3})^{2}[/tex]= 7068 × [tex]10^{-6}m^{2}[/tex].
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You are a detective investigating why someone was hit on the head by a falling flowerpot. One piece of evidence is a smartphone video taken in a 4th-floor apartment, which happened to capture the flowerpot as it fell past a window. In a span of 8 frames (captured at 30 frames per second), the flowerpot falls 0.84 of the height of the window. You visit the apartment and measure the window to be 1.27 m tall.
Required:
Assume the flowerpot was dropped from rest. How high above the window was the flowerpot when it was dropped?
Answer:
0.37 m
Explanation:
Given :
Window height, [tex]h_1[/tex] = 1.27 m
The flowerpot falls 0.84 m off the window height, i.e.
[tex]h_2[/tex] = (1.27 x 0.84 ) m in a time span of [tex]$t=\frac{8}{30}$[/tex] seconds.
Assuming that the speed of the pot just above the window is v then,
[tex]h_2=ut+\frac{1}{2}gt^2[/tex]
[tex]$(1.27 \times 0.84) = v \times \left( \frac{8}{30} \right) + \frac{1}{2} \times 9.81 \times \left( \frac{8}{30} \right)^2$[/tex]
[tex]$v=\left(\frac{30}{8}\right) \left[ (1.27 \times 0.84) - \left( \frac{1}{2} \times 9.81 \times \left( \frac{8}{30 \right)^2 \right) \right]}$[/tex]
[tex]$v= 2.69$[/tex] m/s
Initially the pot was dropped from rest. So, u = 0.
If it has fallen from a height of h above the window then,
[tex]$h = \frac{v^2}{2g}$[/tex]
[tex]$h = \frac{(2.69)^2}{2 \times 9.81}$[/tex]
h = 0.37 m
a model car moves round a circular path of radius 0.3m at 2 revolutions per secs what is its angular speed, the period of the car and the speed of the car
Answer:
a) T = 0.5 s
b) v = 1.2π m/s ≈ 3.77 m/s
Explanation:
It makes two revolutions in one second so makes one revolution in ½ second
circumference of the circle is
C = 2πr = 0.6π m
which it traverses in one time period
0.6π m / 0.5 s = 1.2π m/s
To solve this, we must be knowing each and every concept related to speed and its calculations. Therefore, the angular speed of a model car moves round a circular path of radius 0.3m at 2 revolutions per secs is 3.77 m/.
What is speed?Speed may be defined as the distance traveled by an item in the amount of time it requires to travel that distance. In other words, it measures how rapidly an item travels but does not provide direction.
Speed may be calculated in Science. The speed equation is a scientific formula that is used to calculate various types of speed.
Mathematically, the formula for speed can be given as
speed= distance/time
Values that are given
Time period= 0.5 s
Circumference = 2πr = 0.6π m
substituting all the given values in the above equation, we get
speed =0.6π m / 0.5 s
On calculations, we get
= 1.2π m/s
=3.77 m/s
Therefore, the angular speed of a model car moves round a circular path of radius 0.3m at 2 revolutions per secs is 3.77 m/.
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A seesaw has an irregularly distributed mass of 30 kg, a length of 3.0 m, and a fulcrum beneath its midpoint. It is balanced when a 60-kg person sits on one end and a 78-kg person sits on the other end.
Required:
Find a displacement of the center of mass of the system relatively to the seesaw's midpoint.
Answer:
x = 0.9 m
Explanation:
For this exercise we must use the rotational equilibrium relation, we will assume that the counterclockwise rotations are positive
∑ τ = 0
60 1.5 - 78 1.5 + 30 x = 0
where x is measured from the left side of the fulcrum
90 - 117 + 30 x = 0
x = 27/30
x = 0.9 m
In summary the center of mass is on the side of the lightest weight x = 0.9 m
The speed of a car decreases uniformly as it passes a curve point where normal component of acceleration is 4 ft/sec2. If the car total acceleration of 5ft/sec2 is the same as it passes a hump, the tangential component of acceleration is _______________ ft/sec2.
Answer:
45
Explanation:
ft/sec2
which option is correct n why?
6. The projectile motion is a good example of
A. one dimensional motion.
B. two dimensional motion.
C. three dimensional motion.
D. four dimensional motion.
2. two dimensional motion
Because it has just 2 dimensions x and y
write state of matter with 5 example of each
There are broadly 3 states of matter (there are 5, but they don't teach 2 of them at school).
1. Solid
Examples: Iron, wood, steel, ice, paper
2. Liquid
Examples: Water, mercury, milk, soup, juice
3. Gas
Examples: Oxygen, Chlorine, Carbon dioxide, Sulphur dioxide, Nitrogen
If a full wave rectifier circuit is operating from 50 Hz mains, the fundamental frequency in the ripple will be
Hz 50
Hz 70.7
Hz 100
Hz 25
Answer:
100Hz
Explanation:
In a full wave rectifier, the fundamental frequency of the ripple is twice that of input frequency. Given the input frequency of 50 Hz, the fundamental frequency will be 2 × 50 = 100Hz
Answer:
HZ 100 is the right answer hope you like it
A ball is thrown upward from the edge of a cliff with an initial velocity of 6 m/s. (a) How fast is it moving 0.5 s later? In what direction? (b) How fast is it moving 2 s later? In what direction?
Answer:
Explanation:
Kinematic equation
v = u + at
If UP is assumed to be the positive direction and we let gravity be 10 m/s² which will be in the downward direction so will be negative.
a) v = 6 + (-10)(0.5) = 1 m/s the result is positive, so upward
b) v = 6 + (-10)(2) = -14 m/s the result is negative, so downward
A mass of 240 grams oscillates on a horizontal frictionless surface at a frequency of 2.5 Hz and with amplitude of 4.5 cm.
a. What is the effective spring constant for this motion?
b. How much energy is involved in this motion?
Answer:
(a) The spring constant is 59.23 N/m
(b) The total energy involved in the motion is 0.06 J
Explanation:
Given;
mass, m = 240 g = 0.24 kg
frequency, f = 2.5 Hz
amplitude of the oscillation, A = 4.5 cm = 0.045 m
The angular speed is calculated as;
ω = 2πf
ω = 2 x π x 2.5
ω = 15.71 rad/s
(a) The spring constant is calculated as;
[tex]\omega = \sqrt{\frac{k}{m} } \\\\\omega ^2 = \frac{k}{m} \\\\k = m\omega ^2\\\\where;\\\\k \ is \ the \ spring \ constant\\\\k = (0.24) \times (15.71)^2\\\\k = 59.23 \ N/m[/tex]
(b) The total energy involved in the motion;
E = ¹/₂kA²
E = (0.5) x (59.23) x (0.045)²
E = 0.06 J
The double bond between two oxygen atoms (a molecule of oxygen air) has
two characteristics. What are they?
A. Four valence electrons are shared.
B. A metallic bond is formed.
C. Valence electrons are shared between oxygen atoms.
D. An ionic bond is formed.
Answer:
valance electrons are shared between oxygen atoms.. making them have eight in the outer most shells.
I hope this helps
Infrared and ultraviolet waves have different frequencies.
Both types of wave can have harmful effects on human beings.
Describe the harmful effects of infrared and ultraviolet waves, relating them to the frequencies of the waves.
Answer:
For infrared and ultraviolet waves have different frequencies. Both types of wave can have harmful effects on human beings. Describe the harmful effects of infrared and ultraviolet waves, relating them to the frequencies of the waves. Medical studies indicate that prolonged IR exposure can lead to lens, cornea and retina damage, including cataracts, corneal ulcers and retinal burns, respectively. To help protect against long-term IR exposure, workers can wear products with IR filters or reflective coatings.When you look at the EM spectrum, UV waves are quite a bit smaller in wavelength than infrared, and x-rays/gamma rays are even smaller. Therefore, UV waves are probably causing more harm than infrared waves, and x-rays/gamma rays are probably doing even more damage.
Thank You
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Infrared and ultraviolet waves have different frequencies. Infrared waves have lower frequencies and longer wavelengths, while ultraviolet waves have higher frequencies and shorter wavelengths.
Harmful effects of Infrared waves:
Infrared waves have lower frequencies and are often associated with heat radiation. Prolonged exposure to intense infrared radiation can lead to thermal burns and damage to the skin and eyes. Infrared radiation can also cause dehydration and overheating of the body, especially in hot environments. While infrared radiation is not as harmful as ultraviolet radiation, excessive exposure can still lead to health issues.
Harmful effects of Ultraviolet waves:
Ultraviolet waves have higher frequencies and shorter wavelengths, making them more energetic than infrared waves. UV radiation from the sun is a well-known harmful agent. Short-term exposure to intense UV radiation can cause sunburn, skin redness, and eye irritation. Long-term exposure to UV radiation can lead to more serious health problems such as skin aging, cataracts, and an increased risk of skin cancer. UV radiation can also damage DNA in skin cells, leading to mutations and potential carcinogenesis.
It is essential to protect ourselves from both infrared and ultraviolet waves to prevent harmful effects. Using sunscreen and wearing protective clothing can help shield the skin from UV radiation. Limiting exposure to intense sources of infrared radiation, such as hot objects or infrared heaters, can help reduce the risk of thermal burns and overheating. Understanding the differences in the frequencies of these waves allows us to implement appropriate safety measures and protect our health.
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The force of gravity is an inverse square law. This means that, if you double the distance between two large masses, the gravitational force between them Group of answer choices weakens by a factor of 4. strengthens by a factor of 4. weakens by a factor of 2. also doubles. is unaffected.
Answer:
the force decreases by a factor of 4
Explanation:
The expression for the law of universal gravitation is
F = [tex]G \frac{m_1m_2}{r^2}[/tex]
let's call the force Fo for the distance r
F₀ = [tex]G \frac{m_1m_2}{r^2}[/tex]
They indicate that the distance doubles
r ’= 2 r
we substitute
F = [tex]G \frac{m_1m_2}{(r')^2}[/tex]
F = [tex]G \frac{m_1m_2}{r^2} \ \frac{1}{4}[/tex]
F = ¼ F₀
consequently the correct answer is that the force decreases by a factor of 4
If the distance between two large masses are doubled, the gravitational force between them weakens by a factor of 4.
Let the initial force be F
Let the initial distance apart be r
Thus, we can obtain the final force as follow:
Initial force (F₁) = F
Initial distance apart (r₁) = r
Final distance apart (r₂) = 2r
Final force (F₂) =?F = GM₁M₂ / r²
Fr² = GM₁M₂ (constant)
Thus,
F₁r₁² = F₂r₂²
Fr² = F₂(2r)²
Fr² = F₂4r²
Divide both side by 4r²
F₂ = Fr² /4r²
F₂ = F / 4From the illustration above, we can see that when the distance (r) is doubled, the force (F) is decreased (i.e weakens) by a factor of 4
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Tres personas, A, B y C jalan una caja con ayuda de cuerdas cuya masa es despreciable. Si la persona A aplica −3 en dirección horizontal y la persona B aplica a su vez 5 en dirección horizontal, ¿Cuál es el valor de la fuerza que debe ejercer la persona C, para que la caja esté en equilibrio físico?
Answer:
Un objeto se encuentra en equilibrio físico si la fuerza neta que se le aplica es igual a 0.
En este caso solo se aplican fuerzas en el eje horizontal, por lo que las podremos sumar directamente.
La persona A aplica una fuerza:
Fa = -3N
La persona B aplica una fuerza:
Fb = 5N
La persona C aplica una fuerza Fc, la cual aún no conocemos.
Pero sabemos que la caja está en equilibrio físico, por lo que:
Fa + Fb + Fc = 0N
reemplazando los valores que conocemos, obtenemos:
-3N + 5N + Fc = 0N
Ahora podemos resolver esto para Fc, la fuerza que aplica la persona C.
Fc = 0N + 3N - 5N
Fc = -2N
Podemos concluir que la persona C aplica una fuerza horizontal de -2N
A brass road is 2cm long at instance to what is the lense for a temperature rise of 100k, If the expansivity of brass is 18x10^-6/k^-1
The length of the brass at a temperature rise of 100 K is 2.0036 m
From the question given above, the following data were obtained:
Original length (L₁) = 2 m
Temperature rise (ΔT) = 100 K
Coefficient of linear expansion (α) = 18×10¯⁶ K¯¹
Final length (L₂) =?The final length of the brass can be obtained as follow:
α = L₂ – L₁ / L₁ΔT
18×10¯⁶ = L₂ – 2 / (2 × 100)
18×10¯⁶ = L₂ – 2 / 200
Cross multiply
L₂ – 2 = 18×10¯⁶ × 200
L₂ – 2 = 0.0036
Collect like terms
L₂ = 0.0036 + 2
L₂ = 2.0036 m
Thus, the length of the brass at a temperature rise of 100 K is 2.0036 m
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Outline five ways of varying the force on a current-carrying conductor in a magnetic field. (7 marks)
