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.).

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

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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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]

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).

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)

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