The amount of load lifted by the pulley when the efficiency is 60% and effort is 50 N is determined as 30 N.
What is the amount of load lifted?The amount of load lifted by the pulley is calculated by applying the formula for efficiency of a machine as follows;
E = L / E x 100%
where;
L is the load overcome or output workE is the effort applied or the input workThe amount of load lifted by the pulley is calculated as;
60 = L / 50 x 100%
60 = 100L / 50
100L = 50 x 60
100 L = 3000
L = 3000 / 100
L = 30 N
Thus, the amount of load lifted by the pulley is 30 N.
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Brainlist!! Help!! Atom A consists of 10 protons, 12 neutrons, and 10 electrons.
Atom B consists of 10 protons, 10 neutrons, and 12 electrons.
The atoms are isotopes of each other.
The atoms are not isotopes of each other.
Atom A has 10 protons, 12 neutrons, and 10 electrons, while Atom B has 10 protons, 10 neutrons, and 12 electrons.
Atom A and Atom B are not isotopes of each other. Isotopes are atoms of the same element that differ in the number of neutrons but have the same number of protons. In this case, Atom A and Atom B have different numbers of protons, which means they are not isotopes of each other.
The number of protons determines the element, and since Atom A and Atom B have different numbers of protons, they belong to different elements.
Isotopes, on the other hand, have the same number of protons but differ in the number of neutrons.
This variation in the number of neutrons gives isotopes different atomic masses while retaining the same chemical properties.
However, Atom A and Atom B do not fulfill this criterion, so they cannot be considered isotopes of each other.
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PLS HELP!! QUICK!
Given below are two images formed by a lens and a mirror, one of which is real and the other virtual. Identify the real and virtual images and use the diagram to describe atleast 3 ways by which you can identify and describe a real and virtual image.
The image formed by A is real and the image formed by B is virtual.
The 3 ways are broken lines, inverted lines and upright arrow.
What is real and virtual images?A real image is formed when light rays converge at a specific point after passing through an optical system. It is formed by the actual intersection of light rays.
A virtual image is formed when light rays appear to intersect or diverge but do not actually converge at a specific point. It is an apparent image that cannot be projected onto a screen.
From the ray diagram we can observer the following;
A. the image formed by the convex lens is real
B. The image formed by the concave lens is virtual.
The three ways in which we can identify the real and virtual images;
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pls help!!
An object is placed 3.0 cm away from a convex lens of focal length 2.0 cm as shown in fig.
1. Use the lens equation to calculate the image distance.
2. Is the image real or virtual? how do you know ?
(1) The distance of the image formed by the lens is determined as 6 cm.
(2) The image formed is real.
What is the image distance?The distance of the image formed by the lens is calculated by applying the following formula as follows;
1/f = 1/v + 1/u
where;
v is the image distanceu is the object distancef is the focal length of the lensThe distance of the image formed by the lens is calculated as;
1/v = 1/f - 1/u
1/v = 1/2 - 1/3
1/v = 1/6
v = 6 cm
Since the sign of the image of the image is positive, the image formed is real.
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I need help with (c) please
Answer:
3*10^8 ÷ 475*10^-9
= 6.3 x 10^14
Brainlist!! Help!! Atom A consists of 10 protons, 12 neutrons, and 10 electrons.
Atom B consists of 10 protons, 10 neutrons, and 12 electrons.
The atoms are isotopes of each other.
The atoms are not isotopes of each other.
Answer:
They are isotopes of each other
Explanation:
Isotopes are atoms of the same element that have the same number of protons but different numbers of neutrons.
For atom A and atom B, their number of protons is the same (10)
But their number of neutrons is different (12) and (10)
Give examples of stochastic and non-stochastic effects of radiation and explain why this information is essential in our field of study
Stochastic impacts of radiation allude to those that happen arbitrarily and are not reliant upon the portion got. These impacts are related to the likelihood of events and incorporate disease and hereditary changes. Non-stochastic impacts, then again, have a limit, and their seriousness increments with expanding portions.
Models incorporate radiation consumption and intense radiation conditions. Understanding the qualification among stochastic and non-stochastic impacts of radiation is significant in fields like radiation security, atomic medication, and radiobiology.
It assists in setting radiation with dosing limits, creating well-being rules, and carrying out suitable radiation safeguarding measures. By separating these impacts, experts can evaluate and deal with the dangers related to openness to ionizing radiation all the more successfully.
This information guides choices in regard to radiation wellbeing conventions, word-related openness limits, and the improvement of radiation therapy systems in medication.
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please ans. Ill mark brainliest.
a. The value of t is 5 seconds.
b. The maximum velocity of the particle is 20 m/s.
c. The distance traveled with uniform velocity is 50 meters.
How to calculate the valueFrom the graph, we can see that the velocity (V) decreases linearly with time (t) until it reaches zero at t = 10 seconds. After that, the velocity increases linearly until t = 20 seconds.
a) We can use the equation of motion:
V = U + at,
0 = 20 + (-4)t,
-4t = -20,
t = 5 seconds.
Therefore, the value of t is 5 seconds.
b) From the graph, we can see that the maximum velocity occurs at t = 20 seconds, where the velocity is 20 m/s. This is the maximum velocity during the journey. Therefore, the maximum velocity of the particle is 20 m/s.
c) The distance traveled with uniform velocity: During the acceleration phase (0 seconds to 5 seconds):
We can calculate the distance using the equation:
d = (U + V) * t / 2,
d = (0 + 20) * 5 / 2,
d = 50 meters.
Therefore, the distance traveled with uniform velocity is 50 meters.
Overall, the particle travels a total distance of 50 meters with uniform velocity during the motion.
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3. A car with a mass of 1600 kg has a kinetic energy of 125 000 J. How fast is it moving?
The car is moving at approximately 12.5 meters per second.
The kinetic energy (KE) of an object can be calculated using the formula:
KE = 1/2 * m * [tex]v^2[/tex]
where
KE = kinetic energy,
m =Mass of the object, and
v = velocity.
In this case, we are given the mass (m) of the car as 1600 kg and the kinetic energy (KE) as 125,000 J. To find the velocity .
Substituting the values , we have:
125,000 J = 1/2 * 1600 kg *[tex]v^2[/tex]
Now, we can solve for v by rearranging the equation:
[tex]v^2[/tex] = (2 * 125,000 J) / 1600 kg
[tex]v^2[/tex] = 156.25 [tex]m^2/s^2[/tex]
Taking the square root, we find:
v = √156.25[tex]m^2/s^2[/tex]
v ≈ 12.5 m/s
Therefore, the car is moving at approximately 12.5 meters per second.
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An airbys A380 airliner lands at 30 m/s. Partially loaded, its mass is 480000 kg. The engines apply reverse thrust for 12s to slow the plane to 25 m/s.How much thrust did the engines apply?
To determine the thrust applied by the engines, we can use Newton's second law of motion, which states that force (thrust) is equal to mass times acceleration. In this case, we need to calculate the force required to decelerate the plane from 30 m/s to 25 m/s in 12 seconds.
First, we calculate the change in velocity (∆v):
[tex]\displaystyle\sf \Delta v=25\,m/s-30\,m/s=-5\,m/s[/tex]
Next, we calculate the acceleration (∆a) using the formula:
[tex]\displaystyle\sf \Delta a=\frac{\Delta v}{\Delta t}[/tex]
where ∆t is the change in time, which is 12 seconds in this case.
[tex]\displaystyle\sf \Delta a=\frac{-5\,m/s}{12\,s}[/tex]
Now, we can determine the force (thrust) applied by the engines using Newton's second law:
[tex]\displaystyle\sf F=m\cdot a[/tex]
where m is the mass of the airplane, which is 480000 kg.
[tex]\displaystyle\sf F=480000\,kg\cdot \left(\frac{-5\,m/s}{12\,s}\right)[/tex]
Calculating the result:
[tex]\displaystyle\sf F=-200000\,N[/tex]
Therefore, the engines applied a thrust of -200000 Newtons (N) to decelerate the plane. The negative sign indicates that the thrust is in the opposite direction of the motion.
At certain distances from the Sun, asteroids in the main asteroid belt are in an orbital resonance with Jupiter, creating gaps in the belt called Kirkwood gaps. Jupiter orbits the Sun with a period of 11.9 years.
Which of the following is not a period of an asteroid that is in an orbital resonance with Jupiter?
Answer:
The answer is 3.40
Explanation:
To determine which of the given periods is not in an orbital resonance with Jupiter, we need to calculate the resonant periods using the formula:
Resonant Period = (jupiter_period * asteroid_period) / |2 * jupiter_period - asteroid_period|
Let's calculate the resonant periods for the given options:
a) Resonant Period = (11.9 * 4.38) / |2 * 11.9 - 4.38| ≈ 47.082 / 19.44 ≈ 2.423
b) Resonant Period = (11.9 * 5.95) / |2 * 11.9 - 5.95| ≈ 70.805 / 5.95 ≈ 11.905
c) Resonant Period = (11.9 * 3.97) / |2 * 11.9 - 3.97| ≈ 47.083 / 19.86 ≈ 2.371
d) Resonant Period = (11.9 * 3.40) / |2 * 11.9 - 3.40| ≈ 40.46 / 20.8 ≈ 1.947
Comparing these results to the given periods:
a) 4.38 - In an orbital resonance with Jupiter
b) 5.95 - In an orbital resonance with Jupiter
c) 3.97 - In an orbital resonance with Jupiter
d) 3.40 - Not in an orbital resonance with Jupiter
Therefore, the answer is 3.40
Answer:
3.97
Explanation: i got it right on the hw
I need help with (c) please
The frequency of light with a wavelength of 655 nm is 4.58 x 1014 Hz, the frequency of light with a wavelength of 515 nm is 5.83 x 1014 Hz, and the frequency of light with a wavelength of 475 nm is 6.32 x 1014 Hz. The frequency of light can be calculated using the equation c = λν, where c is the speed of light, λ is the wavelength, and ν is the frequency.
The visible spectrum of electromagnetic radiation has a range of wavelengths that varies from the longest (red) to the shortest (violet) that our eyes can detect. The frequencies of light with wavelengths of 655 nm, 515 nm, and 475 nm can be calculated using the equation c = λν, where c is the speed of light, λ is the wavelength, and ν is the frequency.(a) For light with a wavelength of 655 nm, the frequency can be calculated as: c = λν ==> ν = c/λ where c = 3.00 x 108 m/s (speed of light in a vacuum)λ = 655 nm = 6.55 x 10-7 m. Therefore, ν = (3.00 x 108 m/s) / (6.55 x 10-7 m) = 4.58 x 1014 Hz(b) For light with a wavelength of 515 nm, the frequency can be calculated as: c = λν ==> ν = c/λ where c = 3.00 x 108 m/s (speed of light in a vacuum)λ = 515 nm = 5.15 x 10-7 m. Therefore, ν = (3.00 x 108 m/s) / (5.15 x 10-7 m) = 5.83 x 1014 Hz(c) For light with a wavelength of 475 nm, the frequency can be calculated as: c = λν ==> ν = c/λ where c = 3.00 x 108 m/s (speed of light in a vacuum)λ = 475 nm = 4.75 x 10-7 m. Therefore, ν = (3.00 x 108 m/s) / (4.75 x 10-7 m) = 6.32 x 1014 Hz.For more questions on frequency
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I'm not sure about how can solve this problem. Please help me!!
The magnitude of the power dissipated in resistor R4 is approximately 10,028 watts.
How to find magnitude?To find the power dissipated in resistor R4, use the formula:
P = I² × R
where P = power, I = current flowing through the resistor, and R = resistance of the resistor.
The total resistance, Rt, can be calculated using the formula:
1/Rt = 1/R1 + 1/R2 + 1/R3
Substituting the given values:
1/Rt = 1/3 + 1/0.8 + 1/2
Simplifying the equation:
1/Rt ≈ 1.6667
Rt ≈ 0.6 Ω
Next, calculate the total voltage, Vt, by summing the individual voltage sources:
Vt = ε1 + ε2 + ε3
Substituting the given values:
Vt = 9 + 6 + 4
Vt = 19 V
Now calculate the current flowing through resistor R4 using Ohm's Law:
I = Vt / Rt
Substituting the calculated values:
I = 19 / 0.6
I ≈ 31.6667 A
Finally, calculate the power dissipated in resistor R4:
P = I² × R4
Substituting the calculated values:
P = (31.6667)² × 10
P ≈ 10,028 W
Therefore, the magnitude of the power dissipated in resistor R4 is approximately 10,028 watts.
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