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.
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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)
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.
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.
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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]
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
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).
The lamp has a resistance of 10 ohms each resistor has a resistance of 10 ohms what is the total resistance of the circuit
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