Answer: C. The case on the inclined surface had the least decrease intotal mechanical energy.
Explanation:
First and foremost, it should be noted that the mechanical energy is the addition of the potential and the kinetic energy.
From the information given, it should be known that when the block is projected with the same speed v up an incline where is slides to a stop due to friction, the box will lose its kinetic energy but there'll be na increase in the potential energy as a result of the veritcal height. This then brings about an increase in the mechanical energy.
Therefore, the total mechanical energy of the block will decrease the least when the case on the inclined surface had the least decrease intotal mechanical energy.
If a force of 50 N stretches a spring 0.40 m, what is the spring constant?
Answer:
The spring constant = 125 N/m
Explanation:
Given that :
Force = 50 N
distance (x) = 0.40 m
Recall that, From Hooke's law
Force = kx
where;
k = spring constant.
∴
50 N = k × 0.40 m
k = 50 N/0.40m
k = 125 N/m
What is the main form of energy present in the initial state? (Initial state is when the rubber band is stretched , just before it is released)
O gravitational potential
O kinetic
O thermal
O elastic potential
HELPPP PLSS!!!!!!
What is the chemical formula for iodine trichloride?
A. 12C|
B. ICI3
C. 3ICI
D. |1C13
Answer:
B
Explanation:
A 25.0 kg child slides down a long slide in a playground. She starts from rest at a height h1 of 21.00 m. When she is partway down the slide, at a height h2 of 8.00 m, she is moving at a speed of 7.80 m/s. Calculate the mechanical energy lost due to friction (as heat, etc.).
Answer:
The mechanical energy lost due to friction is 2,424.5 J
Explanation:
Given;
mass of the child, m = 25 kg
intial velocity of the child, u = 0
final velocity of the child, v = 7.8 m/s
initial position of the child, h₁ = 21 m
final position of the child, h₂ = 8 m
Let the energy lost due to heat = ΔE
ΔE + ΔK.E + ΔP.E = 0
ΔE + ¹/₂m(v² - u²) + mg(h₂ - h₁) = 0
ΔE + ¹/₂ x 25(7.8² - 0) + 25 x 9.8(8 - 21) = 0
ΔE + 760.5 J - 3185 J =
ΔE - 2,424.5 J = 0
ΔE = 2,424.5 J
Therefore, the mechanical energy lost due to friction is 2,424.5 J
2. A Plate 0.02 mm distance from a fixed Plate moves at a velocity Of 0.6mls and requires a force of 1.962 N Per unit area to maitain this Speed. Determine the viscosity of the fluid between the plates?
Answer:
6.54 × 10⁻⁵ Pa-s
Explanation:
Since the shear force, F = μAu/y where μ = viscosity of fluid between plates, A = area of plates, u = velocity of fluid = 0.6 m/s and y = separation of plates = 0.02 mm = 2 × 10⁻⁵ m
Since F = μAu/y
F/A = μu/y where F/A = force per unit area
Since we are given force per unit area, F/A = 1.962 N per unit area = 1.962 N/m²
So, μ = F/A ÷ u/y
substituting the values of the variables into the equation, we have
μ = F/A ÷ u/y
μ = 1.962 N/m² ÷ 0.6 m/s/2 × 10⁻⁵ m
μ = 1.962 N/m² ÷ 0.3 × 10⁵ /s
μ = 6.54 × 10⁻⁵ Ns/m²
μ = 6.54 × 10⁻⁵ Pa-s
Calculate the RMS voltage of the following waveforms with 10 V peak-to-peak:
a. Sine wave;
b. Square wave,
c. Triangle wave.
Calculate the period of a waveform with the frequency of:
a. 100 Hz,
b. 1 kHz,
c. 100 kHz.
Answer:
a) [tex]T=0.01s[/tex]
b) [tex]T=0.001s[/tex]
c) [tex]T=0.00001s[/tex]
Explanation:
From the question we are told that:
Given Frequencies
a. 100 Hz,
b. 1 kHz,
c. 100 kHz.
Generally the equation for Waveform Period is mathematically given by
[tex]T=\frac{1}{f}[/tex]
Therefore
a)
For
[tex]T=100 Hz[/tex]
[tex]T=\frac{1}{100}[/tex]
[tex]T=0.01s[/tex]
b)
For
[tex]F=1kHz[/tex]
[tex]T=\frac{1}{1000}[/tex]
[tex]T=0.001s[/tex]
c)
For
[tex]F=100kHz[/tex]
[tex]T=\frac{1}{100*100}[/tex]
[tex]T=0.00001s[/tex]
Which electromagnetic waves have the greatest frequencies?
Answer:
Gamma rays
Explanation:
Gamma rays have the highest frequency in the electro magnetic spectrum
calculate the electrical potential at a point P a distance of 1 m from either two to charge of +10 micro coulomb and -5 micro coulomb which are 10 cm apart calculate also the potential energy of a +2 micro coulomb charge placed at a point p
Answer:
a) V = 45 10³ V, b) U = 4.59 J
Explanation:
a) The electric potential for a series of point charges is
V = k ∑ [tex]\frac{q_i}{r_i}[/tex]
in this case point P is at a distance of 1 m from each charge, so the point is located perpendicular to the charges at its midpoint
V = k ( [tex]\frac{q_1}{r} + \frac{q_2}{r}[/tex])
V = 9 10⁹ (10 - 5/ 1) 10⁻⁶
V = 45 10³ V
b) the potential energy is
U = k ( [tex]\frac{q_1q}{r} + \frac{q_2q}{r} + \frac{q_1q_2}{r_2}[/tex] )
where r = 1m and r₂ is the distance between the two charges r₂ = 0.10 m
U = 9 10⁹ (10 2 / 1 - 5 2/1 - 10 5 /0.10) 10⁻¹²
U = 9 10⁻³ 510
U = 4.59 J
You are at a furniture store and notice that a Grandfather clock has its time regulated by a physical pendulum that consists of a rod with a movable weight on it. When the weight is moved downward, the pendulum slows down; when it is moved upward, the pendulum swings faster. If the rod has a mass of 1.23 kg and a length of 1.25 m and the weight has a mass of [10] kg, where should the mass be placed to give the pendulum a period of 2.00 seconds
Answer:
The distance is 1.026 m.
Explanation:
mass of rod, M = 1.23 kg
Length, L = 1.25 m
mass, m = 10 kg
Time period, T = 2 s
Let the distance is d.
The formula of the time period is given by
[tex]T = 2\pi\sqrt\frac{\frac{1}{3}ML^2+md^2}{(M +m)g}\\\\2\times 2 = 4\pi^2\times \frac{\frac{1}{3}\times1.23\times1.25\times 1.25+ 10d^2}{(1.23 + 10)\times9.8}\\\\11.16 = 0.64 + 10d^2\\\\d= 1.026 m[/tex]
The sound from a trumpet radiates uniformly in all directions in 20C air. At a distance of 5.00 m from the trumpet the sound intensity level is 52.0 dB. The frequency is 587 Hz. (a) What is the pressure amplitude at this distance
Answer:
The answer is below
Explanation:
The intensity level (B) of a sound wave is given by:
B = 10log(I/I₀);
where I₀ is the threshold intensity = 1 * 10⁻¹² W/m², I is the intensity at distance 5 m, B is the intensity level = 52 dB
Substituting gives:
[tex]52=10log(\frac{I}{10^{-12}} )\\\\log(\frac{I}{10^{-12}} )=5.2\\\\I=1.58*10^{-7}\ W/m^2[/tex]
The pressure is given by:
[tex]I=\frac{p_{max}^2}{2\rho v} \\\\\rho=air\ density=1.2\ kg/m^3,v=speed\ of\ sound\ in\ air=344\ m/s,p_{max}=pressure:\\\\p_{max}=\sqrt{2\rho vI}=\sqrt{2*1.58*10^{-7}*1.2*344} =1.14*10^{-2}Pa[/tex]
Two identical ambulances with loud sirens are driving directly towards you at a speed of 40 mph. One ambulance is 2 blocks away and the other is 10 blocks away. Which of the following is true? [Note that pitch = frequency.]a) The siren from the closer ambulance sounds higher pitched to you.b) The siren from the farther ambulance sounds higher pitched to you.c) The pitch of the two sirens sounds the same to you.d) The siren from the farther ambulance sounds higher pitched, until the closer ambulance passes you.
Answer:
c) The pitch of the two sirens sounds the same to you
Explanation:
The pitch does not depend on the distance of the object from the observer.
As per the given data
pitch = frequency
Frequency = [tex]f_{0}[/tex] [tex]\frac{V +- V_{0}}{V +- V_{s}}[/tex]
[tex]f^{'}[/tex] = [tex]f_{0}[/tex] [tex]\frac{V }{V - V_{s}}[/tex]
Hence, the pitch of the two sirens remains the same for the observer.
Answer:
c) The pitch of the two sirens sounds the same to you
Explanation:
Mark pushes his broken car 140 m down the block to his friend's house. He has to exert a 110 N horizontal force to push the car at a constant speed. How much thermal energy is created in the tires and road during this short trip?
Answer:
Explanation:
This is a work problem...energy is created and used in the form of work.
W = FΔx where W is work, F is the force needed to move the object Δx in meters.
W = 110(140) ∴
W = 15000 J
The moon Phobos orbits Mars
(mass = 6.42 x 1023 kg) at a distance
of 9.38 x 106 m. What is its period of
orbit?
Answer:
The moon Phobos orbits Mars (m = 6.42 x 1023 kg) at a distance of 9.38 x 106 m.
You have a simple pendulum that oscillates with a period of 2 s as you stand on the surface of Earth. Your friend, an astronaut standing on the surface of the Moon, has a pendulum of the same length. What would be the period of oscillation of your friend’s pendulum?
a. Less than 2 s
b. The answer depends on whether the amplitudes are the same
c. More than 2 s
d. Exactly 2 s
Answer:
c. More than 2 s
Explanation:
First, we will find the length of the pendulum:
[tex]T = 2\pi \sqrt{\frac{l}{g}}\\\\2\ s = 2\pi \sqrt{\frac{l}{9.81\ m/s^2}}\\\\4\ s^2 = 4\pi^2 (\frac{l}{9.81\ m/s^2})\\\\l = \frac{(4\ s^2)(9.81\ m/s^2)}{4\pi^2} \\\\l = 0.99\ m[/tex]
Now, the value of g becomes 1.625 m/s² on the surface of the moon. So the time period will be:
[tex]T = 2\pi \sqrt{\frac{l}{g}}\\\\T = 2\pi \sqrt{\frac{0.99\ m}{1.625\ m/s^2}}\\\\[/tex]
T = 4.9 s
Therefore, the correct option is:
c. More than 2 s
What is the Y-component of a vector A, which is of magnitude
16-12 and at a 45° angle to the horizontal?
Explanation:
the answer is in the image above
The Y-component of a vector A, which is of magnitude 16√2 and at a 45° angle to the horizontal would be 16
What is a vector quantity?The quantities that contain the magnitude of the quantities along with the direction are known as the vector quantities.
Examples of vector quantities are displacement, velocity acceleration, force, etc.
As given in the problem we have to find out the Y-component of a vector A, which is of magnitude 16√12 and at a 45° angle to the horizontal,
Y component of the vector A = 16√2 sin45°
=16√2 ×1/√2
=16
Thus, the Y component of vector A would be 16.
To learn more about the vector quantity here, refer to the link;
https://brainly.com/question/15516363
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A ball with a mass of 275 g is dropped from rest, hits the floor, and rebounds upward. If the ball hits the floor with a speed of 2.60 m/s, rebounds with a speed of 1.84 m/s, and is in contact with the floor for 1.40 ms,determine the following.
(a) magnitude of the change in the ball's momentum (Let up be in the positive j hat direction.)
_______________kg
Answer:
-0.209 kg.m/s
Explanation:
The mass of the ball, m = 275g or 0.275 kg
Speed or velocity, v = 2.60 m/s
Momentum, P = mv
Momentum when velocity is 2.60 = 0.275 x 2.60 = 0.715 kg.m/s
Speed or velocity, v = 1.84 m/s
Momentum, P = mv
Momentum when velocity is 1.84= 0.275 x 1.84 = 0.506 kg.m/s
Change in magnitude = 0.506 - 0.715 = -0.209 kg.m/s
Describes the relationship between the free energy change, the reaction quotient, and the equilibrium constant.
Explanation:
Reaction quotient is defined as the ratio of the concentration of the products and reactants of a reaction at any point of time with respect to some unit. It is represented by the symbol Q.
The ratio of the concentration of products and reactants of a reaction in equilibrium with respect to some unit is said to be equilibrium constant expression. It is represented by the symbol K.
The relationship between Gibbs free energy change and reaction quotient of the reaction is:
[tex]\Delta G=\Delta G^o+RT ln Q[/tex] ......(1)
where,
[tex]\Delta G[/tex] = Gibbs free energy change
[tex]\Delta G^o[/tex] = Standard Gibbs free energy change
R = Gas constant
T = Temperature
At equilibrium, the free energy change of the reaction becomes 0 and standard Gibbs free energy change can be related to the equilibrium constant by the equation:
[tex]\Delta G^o=-RT ln Q[/tex] ...(2)
Which of the following changes would double the force between two charged particles?
A. Doubling the amount of charge on each particle
B. Increasing the distance between the particles by a factor of 2
C. Decreasing the distance between the particles by a factor of 2
D. Doubling the amount of charge on one of the particles
Answer:
Doubling the amount of charge on one of the particles.
Explanation:
The force between two charges is given by :
[tex]F=\dfrac{kq_1q_2}{r^2}[/tex]
Where
r is the distance between charges
or
[tex]F\propto \dfrac{1}{r^2}[/tex]
On doubling the charge on one of the particle,
F' = 2F
So, the force gets doubled. Hence, the correct option is (d).
A toy car rolls down a slope. If it takes 5.94 s to accelerate from 3.22 m/s to 12.4 m/s, what is the value of the acceleration?
2.01 m/s2
1.35 m/s2
1.55 m/s2
0.219 m/s2
Answer:
a = 1.55m/s²
Explanation:
a = (v_f-v_0)/t
a = (12.4m/s-3.22m/s)/5.94s
a = 1.55m/s²
Charlotte throws a paper airplane into the air, and it lands on the ground. Which best explains why this is an example of projectile motion? The paper airplane’s motion is due to horizontal inertia and the vertical pull of gravity. A force other than gravity is acting on the paper airplane. The paper airplane’s motion can be described using only one dimension. A push and a pull are the primary forces acting on the paper airplane.
highschool physics, not college physics
Answer:
Answer:
A). The paper airplane’s motion is due to horizontal inertia and the vertical pull of gravity.
Explanation:
Edge.
Answer:
The motion of the paper airplane is best explained by horizontal inertia and vertical pull of gravity.
Explanation:
What is horizontal inertia and vertical pull of gravity?Inertia is the property by which the body wants to remain in its position unless any external for is applied. Here horizontal inertia is inertia of motion which is acting horizontally .
While vertical pull is due to the earth .
In a paper airplane , four forces act .these forces provide it flight.These forces are horizontal inertia , vertical pull downwards , lift by air and drag.Hence horizontal inertia and vertical pull best explain the projectile motion of paper airplane.
Also read it;
https://brainly.com/question/11049671
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An empty parallel plate capacitor is connected between the terminals of a 8.85-V battery and charges up. The capacitor is then disconnected from the battery, and the spacing between the capacitor plates is doubled. As a result of this change, what is the new voltage between the plates of the capacitor
Answer:
The new voltage is 17.7 V.
Explanation:
Voltage, V = 8.85 V
The spacing is doubled.
When it is disconnected, the charge remains same,
q = C V ..... (1)
where, C is the capacitance, V is the voltage.
The capacitance is inversely proportional to the distance between the two plates.
So, when the spacing is doubled, the capacitance is halved.
Let the new voltage is V'.
C V = C' V'
C x 8.85 = C/2 x V'
V' = 17.7 V
The lamp has a resistance of 10 ohms each resistor has a resistance of 10 ohms what is the total resistance of the circuit
PLEASE HELP!! Newton's second law of motion states that force equals mass times acceleration.
How would this law apply to a runner in a track and field event?
The runner needs to take a few steps to slow down after crossing the finish line.
To win the race, the runner must slow down on the turns and speed up on the straight sections.
The shoes of the runner help increase his mass so he can run faster.
Newton's second law of motion does not apply to track and field.
Define reversible change
Answer:
Reversible changes are changes that can be undone or reversed. Melting, freezing, boiling, evaporating, condensing, dissolving and also, changing the shape of a substance are examples of reversible changes. If playback doesn't begin shortly, try restarting your device.
Explanation:
Answer:
A reversible change is a change that can be undone or reversed.
Explanation:
Diwn unscramble the word
Answer:
WIND Is what you're looking for
Explanation:
The word is WIND
In an experiment, a shearwater (a seabird) was taken from its nest, flown a distance 5120 km away, and released. It found its way back to its nest 12.5 days after release. If we place the origin at the nest and extend the + x-axis to the release point, what was the bird’s average velocity in m/s:
a. for the return flight
b. for the whole episode, from leaving the nest to returning
Answer:
a) v = -4.74 m / s, b) v = 0 m / s
Explanation:
Average speed is
v = Δx /Δt
a) the variation of the distance on the return trip is
Δx = -5120 km = -5120 10³ km
the negative sign is because the bird is going back
Δt = 12.5 days (24 h / 1 day) (3600 s / 1 h) = 1.08 10⁶ s
v = -5120 10³ / 1.08 10⁶
v = -4.74 m / s
b) the displacement for the round trip is zero, therefore the average velocity is
v = 0 m / s