Difference between scissors and nut cracker
Force, work,power and energy
Answer:
pull or push is called force. those activities in which a force is applied along distance is called work.the rate of doing work and some conversion of energy by a machine or person is called power.the capacity or ability to do work is called energy.this is the answer if u are asking for definition.the rate of doing work is called power...
the capacity of doing work is called energy...
work is said to be done when the body moves in the direction of force applied....
Explanation:
i don't know what you want of those topics.i was confused and I write definition of them. I hope this will help you
1 of 3 : please help got an extra day for a test and i don’t get this (must show work) points and brainliest!
Answer:
y = 1/2at^2
we could also write it as-
y = (at^2)/2
2y = at^2
2y/a = t^2
√2y/a = t
hope it helps
Read the following sentence from the article. Life can evolve into complex and specialized forms that exploit every possible niche in their surroundings. What is the definition of "exploit" as it is used in this sentence? A to turn to advantage B to invent a new application C to draw on an earlier experience D to narrow down options
Answer:
C. To draw on an earlier experience.
Three children are riding on the edge of a merry-go-round that is a solid disk with a mass of 102 kg and a radius of 1.53 m. The merry-go-round is initially spinning at 9.71 revolutions/minute. The children have masses of 31.7 kg, 29.0 kg and 31.9 kg. If the child who has a mass of 29.0 kg moves to the center of the merry-go-round, what is the new angular velocity in revolutions/minute
Three children of masses and their position on the merry go round
M1 = 22kg
M2 = 28kg
M3 = 33kg
They are all initially riding at the edge of the merry go round
Then, R1 = R2 = R3 = R = 1.7m
Mass of Merry go round is
M =105kg
Radius of Merry go round.
R = 1.7m
Angular velocity of Merry go round
ωi = 22 rpm
If M2 = 28 is moves to center of the merry go round then R2 = 0, what is the new angular velocity ωf
Using conservation of angular momentum
Initial angular momentum when all the children are at the edge of the merry go round is equal to the final angular momentum when the second child moves to the center of the merry go round Then,
L(initial) = L(final)
Ii•ωi = If•ωf
So we need to find the initial and final moment of inertia
NOTE: merry go round is treated as a solid disk then I= ½MR²
I(initial)=½MR²+M1•R²+M2•R²+M3•R²
I(initial) = ½MR² + R²(M1 + M2 + M3)
I(initial) = ½ × 105 × 1.7² + 1.7²(22 + 28 + 33)
I(initial) = 151.725 + 1.7²(83)
I(initial) = 391.595 kgm²
Final moment of inertial when R2 =0
I(final)=½MR²+M1•R²+M2•R2²+M3•R²
Since R2 = 0
I(final) = ½MR²+ M1•R² + M3•R²
I(final) = ½MR² + (M1 + M3)• R²
I(final)=½ × 105 × 1.7² + ( 22 +33)•1.7²
I(final) = 151.725 + 158.95
I(final) = 310.675 kgm²
Now, applying the conservation of angular momentum
L(initial) = L(final)
Ii•ωi = If•ωf
391.595 × 22 = 310.675 × ωf
Then,
ωf = 391.595 × 22 / 310.675
ωf = 27.73 rpm
Answer: So, the final angular momentum is 27.73 revolution per minute
how much is need to lift a load of 100n placed at a distance of 29 cm from fulcrum if effort is applied at 60cm from the fulcrum on opposite side of the load? calculate mechanical advantage and velocity ratio of the lever
Answer:
206.8965517 n
Explanation:
First, we need to see that 60:29 is 2.078965517:1. Then we need to multiply the energy put 29 cm from the fulcrum by 2.078965517, giving us the end result of our answer.
What's the minimum Out PUT WORK
required to raise 14,0m3 of water 26.0m?
Answer:
3.57 MJ
Explanation:
ASSUMING it's fresh water with density of 1000 kg/m³
W = ΔPE = mgΔh = 14.0(1000)(9.81)(26.0) = 3,570,840 J
Salt water would require more.
A single-slit diffraction pattern is formed on a distant screen. Assuming the angles involved are small, by what factor will the width of the central bright spot on the screen change if the slit width is doubled?
a. It will become four times as large.
b. It will double.
c. It will be cut in half.
d. It will become eight times as large.
e. It will be cut to one-quarter its original size.
Answer:
c : it wil be cut in half.
The pattern is formed on a distant screen so we can use the Fraunhofer difracction for a single slit. The formula of the width of the central bright spot is given by Δx = (2λz)/a, where λ is the wavelength and a is the width of the slit. So if the inicial width (a_1) is doubled (a_2= 2 x a_1),the width of the central spot will be cut in half Δx = (2λz)/a_2 = (2λz)/2xa_1 .
A mysterious constant force of 10 N acts horizontally on everything. The direction of the force is found to be always pointed toward a wall in a big hall. Find the potential energy of a particle due to this force when it is at a distance x from the wall, assuming the potential energy at the wall to be zero.
Answer:
it will be 10x
Explanation:
workdone(potential energy before it hits the wall)= horizontal force × distance
=10× x = 10x joules
A mysterious constant force of 10 N acts horizontally on everything. The direction of the force is found to be always pointed toward a wall in a big hall.The potential energy of a particle due to this force is 10x.
What is force?A force is an effect that can alter an object's motion according to physics. An object with mass can change its velocity, or accelerate, as a result of a force. An obvious way to describe force is as a push or a pull. A force is a vector quantity since it has both magnitude and direction.
Given in the question a mysterious constant force of 10 N acts horizontally on everything. The direction of the force is found to be always pointed toward a wall in a big hall the potential energy,
Work done (potential energy before it hits the wall)
= horizontal force × distance
=10× x = 10x joules
The potential energy of a particle due to this force is 10x.
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if C is the vector sum of A and B C=A+B what must be true about directions and magnitude of A and B if C=A+B? what must be true about the directions and magnitude of A and B if C=0
The vector sum is the algebraic sum if the two vectors have the same direction.
The sum vector is zero if the two vectors have the same magnitude and opposite direction
Vector addition is a process that can be performed graphically using the parallelogram method, see attached, where the second vector is placed at the tip of the first and the vector sum goes from the origin of the first vector to the tip of the second.
There are two special cases where the vector sum can be reduced to the algebraic sum if the vectors are parallel
case 1. if the two vectors are parallel, the sum vector has the magnitude of the sum of the magnitudes of each vector
case 2. If the two vectors are antiparallel and the magnitude of the two vectors is the same, the sum gives zero.
In summary in the sum of vectors If the vectors are parallel it is reduced to the algebraic sum, also in the case of equal magnitude and opposite direction the sum is the null vector
a) Magnitudes: [tex]\| \vec A\| \ge 0[/tex], [tex]\|\vec B\| \ge 0[/tex], [tex]\|\vec C\| \ge 0[/tex]; Directions: [tex]\theta_{A} \in (-\infty, +\infty)[/tex] for [tex]\|\vec A\|\ne 0[/tex]. Undefined for [tex]\|\vec A\| = 0[/tex], [tex]\theta_{B} \in (-\infty, +\infty)[/tex] for [tex]\|\vec B\|\ne 0[/tex]. Undefined for [tex]\|\vec B\| = 0[/tex], [tex]\theta_{C} \in (-\infty, +\infty)[/tex] for [tex]\|\vec C\|\ne 0[/tex]. Undefined for [tex]\|\vec C\| = 0[/tex].
b) Magnitudes: [tex]\|\vec A\| \ge 0[/tex], [tex]\|\vec B\| \ge 0[/tex], [tex]\|\vec C\| = 0[/tex]; Directions: [tex]|\theta_{A}-\theta_{B}| = 180^{\circ}[/tex], [tex]\theta_{C}[/tex] is undefined.
a) Let suppose that [tex]\vec A \ne \vec O[/tex], [tex]\vec B \ne \vec O[/tex] and [tex]\vec C \ne \vec O[/tex], where [tex]\vec O[/tex] is known as Vector Zero. By definitions of Dot Product and Inverse Trigonometric Functions we derive expression for the magnitude and directions of [tex]\vec A[/tex], [tex]\vec B[/tex] and [tex]\vec C[/tex]:
Magnitude ([tex]\vec A[/tex])
[tex]\|\vec A\| = \sqrt{\vec A\,\bullet\,\vec A}[/tex]
[tex]\| \vec A\| \ge 0[/tex]
Magnitude ([tex]\vec B[/tex])
[tex]\|\vec B\| = \sqrt{\vec B\,\bullet\,\vec B}[/tex]
[tex]\|\vec B\| \ge 0[/tex]
Magnitude ([tex]\vec C[/tex])
[tex]\|\vec C\| = \sqrt{\vec C\,\bullet \,\vec C}[/tex]
[tex]\|\vec C\| \ge 0[/tex]
Direction ([tex]\vec A[/tex])
[tex]\vec A \,\bullet \,\vec u = \|\vec A\|\cdot \|u\|\cdot \cos \theta_{A}[/tex]
[tex]\theta_{A} = \cos^{-1} \frac{\vec A\,\bullet\,\vec u}{\|\vec A\|\cdot \|u\|}[/tex]
[tex]\theta_{A} = \cos^{-1} \frac{\vec A\,\bullet\,\vec u}{\|\vec A\|}[/tex]
[tex]\theta_{A} \in (-\infty, +\infty)[/tex] for [tex]\|\vec A\|\ne 0[/tex]. Undefined for [tex]\|\vec A\| = 0[/tex].
Direction ([tex]\vec B[/tex])
[tex]\vec B\,\bullet \, \vec u = \|\vec B\|\cdot \|\vec u\| \cdot \cos \theta_{B}[/tex]
[tex]\theta_{B} = \cos^{-1} \frac{\vec B\,\bullet\,\vec u}{\|\vec B\|\cdot \|\vec u\|}[/tex]
[tex]\theta_{B} = \cos^{-1} \frac{\vec B\,\bullet\,\vec u}{\|\vec B\|}[/tex]
[tex]\theta_{B} \in (-\infty, +\infty)[/tex] for [tex]\|\vec B\|\ne 0[/tex]. Undefined for [tex]\|\vec B\| = 0[/tex].
Direction ([tex]\vec C[/tex])
[tex]\vec C \,\bullet\,\vec u = \|\vec C\|\cdot\|\vec u\|\cdot \cos \theta_{C}[/tex]
[tex]\theta_{C} = \cos^{-1}\frac{\vec C\,\bullet\,\vec u}{\|\vec C\|\cdot\|\vec u\|}[/tex]
[tex]\theta_{C} = \cos^{-1} \frac{\vec C\,\bullet\,\vec u}{\|\vec C\|}[/tex]
[tex]\theta_{C} \in (-\infty, +\infty)[/tex] for [tex]\|\vec C\|\ne 0[/tex]. Undefined for [tex]\|\vec C\| = 0[/tex].
Please notice that [tex]\vec u[/tex] is the Vector Unit.
b) Let suppose that [tex]\vec A \ne \vec O[/tex] and [tex]\vec B \ne \vec O[/tex] and [tex]\vec C = \vec O[/tex]. Hence, [tex]\vec A = -\vec B[/tex]. In other words, we find that both vectors are antiparallel to each other, that is, that angle between [tex]\vec A[/tex] and [tex]\vec B[/tex] is 180°. From a) we understand that [tex]\|\vec A\| \ge 0[/tex], [tex]\|\vec B\| \ge 0[/tex], but [tex]\|\vec C\| = 0[/tex].
Then, we have the following conclusions:
Magnitude ([tex]\vec A[/tex])
[tex]\|\vec A\| \ge 0[/tex]
Magnitude ([tex]\vec B[/tex])
[tex]\|\vec B\| \ge 0[/tex]
Magnitude ([tex]\vec C[/tex])
[tex]\|\vec C\| = 0[/tex]
Directions ([tex]\vec A[/tex], [tex]\vec B[/tex]):
[tex]|\theta_{A}-\theta_{B}| = 180^{\circ}[/tex]
Direction ([tex]\vec C[/tex]):
Undefined
A sinewave has a period (duration of one cycle) of 645 μs (microseconds). What is the corresponding frequency of this sinewave, in kHz
The corresponding frequency of this sinewave, in kHz, expressed to 3 significant figures is: 155 kHz.
Given the following data:
Period = 645 μsNote: μs represents microseconds.
Conversion:
1 μs = [tex]1[/tex] × [tex]10^-6[/tex] seconds
645 μs = [tex]645[/tex] × [tex]10^-6[/tex] seconds
To find corresponding frequency of this sinewave, in kHz;
Mathematically, the frequency of a waveform is calculated by using the formula;
[tex]Frequency = \frac{1}{Period}[/tex]
Substituting the value into the formula, we have;
[tex]Frequency = \frac{1}{645 * 10^-6}[/tex]
Frequency = 1550.39 Hz
Next, we would convert the value of frequency in hertz (Hz) to Kilohertz (kHz);
Conversion:
1 hertz = 0.001 kilohertz
1550.39 hertz = X kilohertz
Cross-multiplying, we have;
X = [tex]0.001[/tex] × [tex]1550.39[/tex]
X = 155039 kHz
To 3 significant figures;
Frequency = 155 kHz
Therefore, the corresponding frequency of this sinewave, in kHz is 155.
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DESCRIBE THE FORMATION OF THE SOLAR SYSTEM ACCORDING TO THE NEBULAR THEORY
When a cloud of gas and dust in space was disturbed, maybe by the explosion of a nearby star.This explosion made waves in space which squeezed the cloud of gas & dust.
Two buses 'A' and 'B' are moving in the same direction with the velocities 30m/s and 40m/s respectively. Find the relative velocity of the bus 'A' with respect to the bus 'B'. Calculate the relative displacement between them after 4 minutes.:)
Hope who are reading this may have a good life;)
Lots of love from Nepal <3
Relative Velocity, aVb:
[tex]{ \bf{ _{A} V_{B} =V_{A} - V_{B} }} \\ _{A} V_{B} = 30 - 40 \\ _{A} V_{B} = - 10 \: {ms}^{ - 1} \\ |_{A} V_{B}| = 10 \: {ms}^{ - 1} [/tex]
Relative displacement:
[tex] = \frac{ |_{A} V_{B}| }{time} \\ \\ = \frac{10}{4} \\ = 2.5 \: metres[/tex]
a person lifts 60kg on the surface of the earth, how much mass can he lift on the surface of the moon if he applies same magnitude of force
Explanation:
Hey there!
According to the question;
A person can lift mass of 60 kg on earth.
mass(m1) = 60kg
acceleration due to gravity on earth (a) = 9.8m/s²
Now;
force (f) = m.a
= 60*9.8
= 588 N
Since, there is application of same magnitude of force on moon,
mass(m) =?
acceleration due to gravity on moon (a) = 1.67m/s²
Now;
force (f) = m.a
588 = m*1.67
m = 352.09 kg
Therefore, the person who can lift the mass of 60 kg on earth can lift mass of 352 kg on moon.
Hope it helps!
the plane of a 5.0 cm by 8.0 cm rectangular loop wire is parallel to a 0.19 t magnetic field. if the loop carries a current of 6.2 amps, what is the magnitude of the torque on the loop
What happens to the temperature of matter as distance is increased? Explain why.
Answer:
The tempature of the matter increases.
Explanation:
The matter is heated by friction, weather that is through the air or if its touching the ground like a car.
please mark me as brainliest of this is right, thanks
Why is Newton's law of gravitation called universal law?
Answer:
Newtons law of gravitation is called the universal law of gravitation because it is applicable for all masses at all distances, independent of the medium.
Explanation:
[tex]\sf{}[/tex]
For two masses m1 and m2 placed at a distance d, the gravitational force between them will always be given by: Fg=Gd2m1m2. That's why it is a "universal" law.
(b) State the unit of volume in S.I. system and define it. 4.
Answer:
the cubic meter (m3)
Explanation:
Volume is the measure of the 3-dimensional space occupied by matter, or enclosed by a surface, measured in cubic units. The SI unit of volume is the cubic meter (m3),
Answer:
please answer all component's
Explanation:
please answer all component's
Tectonic processes are fueled by heat from __________, while atmospheric convection is fueled by heat from ___________.
A. the core, the sun
B. asteroid impacts, radiation
C. the sun, the core
D. radiation, asteroid impacts
Tectonic processes are fueled by heat from the core, while atmospheric convection is fueled by heat from the sun. That is option A.
What are tectonic processes?The tectonic processes is defined as the process catalysed by heat from the earth core that results to the structuring and properties of the earth crust. This results to the following:
convergent, where plates move into one another;divergent, where plates move apart; and transform, where plates move sideways in relation to each other.Learn more about tectonic processes here:
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1. Why do only some people get addicted to
drugs?
Answer:
When drugs are taken in are body are brain release dopamine: which make us feel so pleasure and good, and for this some people are addicted to drugs which makes them feel good. on other hand damaging their health.
What is the solution to this equation?
X- 14 = -9
O A. x = 5
O B. x= -23
O c. x = 23
O D. x= -5
Answer:
X-14=9
x-14+14=9+14
x=23
Explanation:
Your answer is 23.
1.2miles=__________km
Answer:
1.931 kilometres is the answer of 1.2 miles
Answer and Explanation:
1 mile = 1.609 km
Set up a fraction to cancel the miles to get the kilometers.
[tex]\frac{1.2mi}{?km} *\frac{1.609}{1mi} = 1.9308km[/tex] <- This is the answer.
#teamtrees #PAW (Plant And Water)
Two infinitely long parallel wires carry current in opposite directions. Wire 1 has current 15.0 A and wire 2 has current 19.9 A. If they are separated by a distance 4.3 m, at what location between the wires is the net magnetic field be twice the strength of the magnetic field from wire 1
Answer:
x= 2*I1*d/(I1+I2) meter
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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A horizontal board of negligible thickness and area 1.0 m2 hangs from a spring scale that reads 50 N when a 4.0 m/s wind moves above the board. The air below the board is stationary. What does the scale read when the wind stops? The density of air is 1.25 kg/m3 .
the 2kg block slids down a firctionless curved ramp starting from rest at heiht of 3m what is the speed of the block at the bottemvof the ramp
A
Explanation:
1qdeeeeeeeeeeehhhhhhhhhwilffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffff.
What did Edwin Hubble discover by observing absorption spectrum
blue light
red light
grey matter
ultra-violet light
Answer:
According to different sources Edwin Hubble observed red light of galaxies directly proportional to the distance of the galaxy from earth.
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30 POINTS HELP PLEASE
A galvanometer consists of a loop of wire in a magnetic field.
Question 19 options:
True
False
Answer:
A galvanometer consists of a loop of wire in a magnetic field.
TRUE
A force of 16.88 N is applied tangentially to a wheel of radius 0.340 m and gives rise to an angular acceleration of 1.20rad / (s ^ 2) . Calculate the rotational inertia of the wheel. A. 2.77 kg - m ^ 2 B. 0.73 kg - m ^ 2 C. 4.41 kg - m ^ 2 O. 4.78 kg - m ^ 2
Given.
force = 16.88 N is
radius = 0.340m
an angular acceleration = 1.20rad/s^2
the formula for torque is
F*r = I*a
where I is moment of inertia
16.88*.34 = I*1.2
I = 4.78Kg-m^2
so rotational inertia I = 4.78Kg-m^2
A diode has a power rating of 5W. If the diode voltage is 3.6V and the diode current is 1.75A, what is the power dissipation?
Answer:
Explanation:
Power = current * voltage
Power = 3.6 * 1.75
Power = 6.3 watts.
I hope this diode does not have to do this for very long. The power dissipation is above the diode's rated power -- a condition that won't last forever.
The power dissipation is 6.3 watts.
