A particle, mass 0.25 kg is at a position (-7i + 7j + 5k) m, has a velocity (6i - j + 4k) m/s, and is subject to a force (-5i + 0j - k) N. What is the magnitude of the torque on the particle about the origin?
Answer:
47.94Nm
Explanation:The torque (τ) on a particle subject to a force (represented as force vector F) at a position (represented as position vector r) about the origin is given by the cross product of the position vector r for the point of application of a force and the force F. i.e
τ = r x F
Given:
r = (-7i + 7j + 5k) m
F = (-5i + 0j - k) N
| i j k |
r x F = | -7 7 5 |
| -5 0 -1 |
r x F = i(-7 - 0) - j(7+25) + k(0+35)
r x F = i(-7) - j(32) + k(35)
r x F = -7i - 32j + 35k
Therefore the torque τ = -7i - 32j + 35k
The magnitude of the torque is therefore;
|τ| = [tex]\sqrt{(-7)^2 + (-32)^2 + (35)^2}[/tex]
|τ| = [tex]\sqrt{49 + 1024 + 1225}[/tex]
|τ| = [tex]\sqrt{2298}[/tex]
|τ| = 47.94Nm
The magnitude of the torque on the particle about the origin is 47.94Nm
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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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 ^_^
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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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
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 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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Two point charges, Q1 and Q2, are separated by a distance R. If the magnitudes of both charges are doubled and their separation is halved, what happens to the electrical force that each charge exerts on the other one
Answer:
F' = 16 F
Hence, the electric force between charges becomes sixteen times its initial value.
Explanation:
The electric force between the two charges is given by the Colomb's Law:
[tex]F = \frac{KQ_1Q_2}{R^2}[/tex] ------------------- eq(1)
where
F = electric force
K = Colomb's Constant
Q₁ = magnitude of the first charge
Q₂ = magnitude of the second charge
R = Distance between charges
Now the magnitudes of the charges are doubled and the distance between them is halved. Therefore:
[tex]F' = \frac{K(2Q_!)(2Q_2)}{(\frac{R}{2})^2}\\\\F' = 16 \frac{KQ_1Q_2}{R^2}[/tex]
using equation (1):
F' = 16 F
Hence, the electric force between charges becomes sixteen times of its initial value.
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
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
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
Twin skaters approach each other with identical speeds. Then, the skaters lock hands and spin. Calculate their final angular velocity, given each had an initial speed of 2.50 m/s relative to the ice. Each has a mass of 70.0 kg, and each has a center of mass located 0.800 m from their locked hands. You may approximate
Answer:
[tex]\omega=3.135rad/s[/tex]
Explanation:
From the question we are told that:
initial Speed [tex]V_1=2.50[/tex]
Mass [tex]m=70.0kg[/tex]
Center of mass [tex]d=0.0.800m\[/tex]
Generally the equation for angular velocity is is mathematically given by
[tex]\omega=\frac{v}{r}\\\\\omega=\frac{2.50}{0.0800}[/tex]
[tex]\omega=3.135rad/s[/tex]
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
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
write a use of magnetic force and frictional force each
Integrated Concepts Fusion probability is greatly enhanced when appropriate nuclei are brought close together, but mutual Coulomb repulsion must be overcome. This can be done using the kinetic energy of high-temperature gas ions or by accelerating the nuclei toward one another.
(a) Calculate the potential energy of two singly charged nuclei separated by 1.00×10–12 m by finding the voltage of one at that distance and multiplying by the charge of the other.
(b) At what temperature will atoms of a gas have an average kinetic energy equal to this needed electrical potential energy?
Answer:
(a) 2.3 x 10^-16 J
(b) 1.1 x 10^7 K
Explanation:
charge, q = 1.6 x 10^-19 C
distance, r = 10^-12 m
(a) Let the potential energy is U.
[tex]U = \frac{k q q}{r^2}\\\\U = \frac{9\times 10^{9}\times 1.6\times 1.6\times 10^{-38}}{10^{-12}}\\\\U = 2.3\times 10^{-16} J[/tex]
(b) Let the temperature is T.
[tex]U = K = \frac{3}{2} kT\\\\2.3 \times 10^{-16} = 1.5\times 1.38\times 10^{-23} T\\\\T = 1.1\times 10^7 K[/tex]
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
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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PLEASE HELP ME WITH THIS PHYSICS QUESTION PLSSS!!!
Vf^2 = v0^2 + 2a (xf -x0)
Solve for v0
b. solve for Vo
[tex] vf ^{2} = vo^{2} + 2a(xf - xo) \\ vf ^{2} = vo ^{2} + 2axf - 2axo\\ vo ^{2} = vf ^{2} - 2axf + 2axo \\ vo = \sqrt{vf ^{2} - 2axf + 2axo } \\ vo = - \sqrt{vf^{2} - 2axf + 2axo } [/tex]
I hope I helped you ^_^
Air contained in a rigid, insulated tank fitted with a paddle wheel, initially at 300 K, 2 bar, and a volume of 2 m3, is stirred until its temperature is 500 K. Assuming the ideal gas model, for the air, and ignoring kinetic and potential energy, determine
Answer:
The final pressure in bar will be "[tex]\frac{10}{3} \ Bar[/tex]".
Explanation:
As we know,
PV = nRT
[tex]\frac{P_1}{T_1} =\frac{P_2}{T_2} =CONST[/tex]
then,
⇒ [tex]\frac{2 \ bar}{300 \ K} = \frac{P_2}{500 \ K}[/tex]
⇒ [tex]P_2=(\frac{2}{300}\times 500 )Bar[/tex]
[tex]=\frac{10}{3} \ Bar[/tex]
Thus the above is the correct answer.
What would be the consequences if the animated structures suddenly become non-polar
Answer:
the lipid bilayer would not be able to hold its shape in water and the cell membrane would disassemble. Only a lipid monolayer would be possible. The fatty acid "tails' would fall off The phospholipids would flip upside down.
Explanation:
hope this helps
Outline five ways of varying the force on a current-carrying conductor in a magnetic field. (7 marks)
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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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.
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.
If "FRICTION" means breaking up of relationships, then how can you reduce friction among friends?
A. Respect the opinion of your friends
B. Learn how to be understanding
C. Avoid bullying your friends
D. Force your friends to do things they don't like
Answer:
A
Explanation:
respect the opinion of your friends
which one of the following is a product of an acid base reaction? A. Base B. Acid C. Salt D. Fire
Answer:
salt
Explanation:
salt is a component for many acid base reactions
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
What is a measure between the difference in start and end positions?
Answer:
Displacement
General Formulas and Concepts:
Kinematics
Displacement vs Total DistanceExplanation:
Displacement is the difference between the start position and end position.
Total Distance is the entire distance traveled between the start and end position.
Topic: AP Physics 1 Algebra-Based
Unit: Kinematics
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.
A bullet 2cm log is fired at 420m/s and passes straight a 10cm thick board exiting at 280m/s
a) what is the average acceleration of the bullet through the board?
b)what is the total time the bullet is in contact with the board?
c)what minimum thickness could the board have if it was supposed to bring the bullet to a stop?
Answer:
Explanation:
(a)Solving for the acceleration of the bullet
acceleration = (vf^2 – vi^2) / 2d
acceleration = ((280 m/s)^2 – (420 m/s)^2) / (2 * 0.12 m)
acceleration = (78400 - 176400) / 0.24 m
acceleration = -98000 / 0.24
acceleration = -408333 m/s^2
(a)Solving for contact time with board
t^2 = 2d/a
t^2 = 2 * 0.12 m / 408333 m/s^2
t^2 = 0.24 m / 408333 m/s^2
t^2 = 5.8775558 x 10^-7
t = 0.0007666 s or 767 microseconds
