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elixir [45]
4 years ago
8

What is determined by both mass and velocity

Physics
2 answers:
irga5000 [103]4 years ago
6 0
Power...............................
attashe74 [19]4 years ago
4 0
Kinetic energy is the energy of motion. The mass and velocity of an object determines its kinetic energy
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True or False: Drag and tailwind are examples of a contact force.<br><br> True or False?
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false

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What happens when the temperature of an object decreases?
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Un gallo se pavonea en línea recta en el techo. Su movimiento se muestra en la siguiente gráfica de la posición horizontal, x, c
Svetllana [295]

Answer:

Entonces, velocidad instantánea = | v | = 0,25 m / s

Explanation:

pregunta completa

Por favor, encontrar el archivo adjunto

Solución

La velocidad instantánea está dada por,

v = dx / dt = pendiente de la curva x-t

aquí, del gráfico:

en x = 1 m ---> t = 8 s

en x = 3 m ---> t = 0 s

entonces, dx = 1-3 = -2 m

dt = 8-0 = 8 s

Entonces,

v = -2/8

v = -0,25 m / s

Dado que, pseed es una cantidad escalar.

Entonces, velocidad instantánea = | v | = 0,25 m / s

8 0
3 years ago
Many chemical reactions release energy. Suppose that at the beginning of a reaction, an electron and proton are separated by 0.1
ludmilkaskok [199]

Answer:

1.189eV

Explanation:

The electric potential energy is the potential energy that results from the Coulomb force and is associated with the configuration of two or more charges. For an electron in the presence of an electric field produced by a proton, the electric potential energy is defined as:

U=\frac{kq_{e}q_{p}}{r}

where  q_{e} is the electron charge, q_{p} is the proton charge, r is the separation distance between the charges and k is the coulomb constant.

Knowing this, we can calculate how much electric potential energy was lost:

\Delta U=U_{f}-U{i}\\\Delta U=\frac{kq_{e}q_{p}}{r_{f}}-\frac{kq_{e}q_{p}}{r_{i}}\\\Delta U=kq_{e}q_{p}(\frac{1}{r_{f}}-\frac{1}{r_{i}})\\\Delta U=(8.99*10^9\frac{Nm^2}{C^2})(-1.60*10^{-19}C)(1.60*10^{-19}C)(\frac{1}{0.105*10^{-9}m}-\frac{1}{0.115*10^{-9}m})\\\Delta U=1.90*10^{-19}J*\frac{6.2415*10^{18}eV}{1J}=1.189eV

5 0
4 years ago
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