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Luda [366]
3 years ago
15

Assume that the particle has initial speed viviv_i. Find its final kinetic energy KfKfK_f in terms of viviv_i, MMM, FFF, and DDD

.
Physics
1 answer:
NeX [460]3 years ago
3 0

Answer:

KE= 1/2mv²

Explanation:

The kinetic energy of a body is the energy possessed by virtue of the body in motion

Given the parameters

m which is the mass of the body

v which is the velocity of the body too

K.E = kinetic energy

The expression for the kinetic energy of a body is given as

KE= 1/2mv²

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7200 grams is what equivalen to 7.s kilograms


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A 30 g particle is undergoing simple harmonic motion with an amplitude of 2.0 ✕ 10-3 m and a maximum acceleration of magnitude 8
KiRa [710]

Answer:

Explanation:

By using the Newton second law and the position equation for a simple harmonic motion we have

F=ma\\a_{max}=\omega^{2}A\\x=Acos(\omega t+ \phi)\\

where a is the acceleration, w is the angular frecuency and \phi is the phase constant. We can calculate w from the equation for the maximum acceleration

\omega=\sqrt{\frac{a_{max}}{A}}=\sqrt{\frac{8*10^{3}m/s^{2}}{2*10^{-3}}}=2000rad/s

(a).

F=ma=m\omega^{2}Acos(\omega t + \phi)\\F=(30*10^{-3}kg)(2*10^{-3}m)(2000\frac{rad}{s})^{2}cos(2000\frac{rad}{s} t - \frac{\pi}{2})=240N*cos(2000\frac{rad}{s} t - \frac{\pi}{2})

(b). T=\frac{2\pi}{\omega}=\frac{2\pi}{2000}=3.14*10^{-13} s

(c). v_{max}=A\omega=(2*10^{-3}m)(2000\frac{rad}{s})=4\frac{m}{s}

(d). The mecanical energy is the kinetic energy when the velocity is a maximum

E_{m}=E_{k}(v_{max})=\frac{mv_{max}^{2}}{2}=\frac{30*10^{-3}kg(4\frac{m}{s})^{2}}{2}=0.024J

3 0
3 years ago
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Give an example of a mixture that contains more than one state of matter
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GEL IT IS COMBINATION OF SOLID AND LIQUID
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3 years ago
Given a temperature of 300 Kelvin, what is the approximate temperature in degrees Celsius?
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4 years ago
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You are trying to determine the specific gravity of a solid object that floats in water. If m is the mass of your object, mS is
Alisiya [41]

Answer:

Specific Gravity = m/[m(s)-m(os)]

Explanation:

Specific gravity, also called relative density, is the ratio of the density of a substance to the density of a reference substance. By this definition we need to find out the ratio of density of the object of mass m to the density of the surrounding liquid.

m = mass of the object

<u>Weight in air</u>

W (air) = mg, where g is the gravitational acceleration

<u>Weight with submerged with only one mass</u>

m(s)g + Fb = mg + m(b)g, <em>consider this to be equation 1</em>

where Fb is the buoyancy force

Weight with submerged with both masses

m(os)g + Fb’ = mg + m(b)g, <em>consider this to be equation 2</em>

<u>equation 1 – equation 2 would give us</u>

m(s)g – m(os)g = Fb’ – Fb

where Fb = D x V x g, where D is the density of the liquid the object is submerged in, g is the force of gravity and V is the submerged volume of the object

m(s)g – m(os)g = D(l) x V x g

m(s) – m(os) = D(l) x V

we know that Mass = Density x V, which in our case would be, D(b) x V, which also means

V = Mass/D(b), where D(b) is the density of the mass

<u>Substituting V into the above equation we get</u>

m(s) – m(os) = [D(l) x m)/ D(b)]

Rearranging to get the ratio of density of object to the density of liquid

D(b)/D(l) = m/[m(s)-m(os)], where D(b)/D(l) denotes the specific gravity

8 0
3 years ago
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