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Ilia_Sergeevich [38]
3 years ago
7

One particle has mass m and a second particle has mass 2m. The second particle is moving with speed v and the first with speed 2

v. How do their kinetic energies compare?
Physics
1 answer:
choli [55]3 years ago
3 0

Answer:

Explanation:

The formula for kinetic energy to be used here is 1/2mv².

If the first particle is "particle a" and the second particle is "particle n"; there kinetic energies (K.E) will be

K.Eₐ = 1/2.m2v² = mv²

K.Eₙ = 1/2.2mv² = mv²

From the above, <u>it can be said that there kinetic energies are the same</u>.

NOTE that the m and v used in the question means mass and velocity respectively.

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A capacitor is formed from two concentric spherical conducting shells separated by vacuum. The inner sphere has radius 12.0 cm ,
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The radius of curvature of both sides of a converging lens is 18 cm. One side of the lens is coated withsilver so that the inner
Dahasolnce [82]

Answer:

n = 1.4

Explanation:

Given,

R1 = 18 cm, R2 = -18 cm

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Now, this lens is in with conjunction with a concave mirror which then can be thought of as to be in conjunction with another thin lens

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Net power of the combination = 2P + P' = 2(n-1)/0.09 + 1/0.09 = 1/0.05

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7 0
3 years ago
A particle executes simple harmonic motion with an amplitude of 1.69 cm. At what positive displacement from the midpoint of its
m_a_m_a [10]

Answer: 0.0146m

Explanation: The formula that defines the velocity of a simple harmonic motion is given as

v = ω√A² - x²

Where v = linear velocity, A = amplitude = 1.69cm = 0.0169m, x = displacement.

The maximum speed of a simple harmonic motion is derived when x = A, hence v = ωA

One half of maximum speed = speed of motion

3ωA/2 = ω√A² - x²

ω cancels out on both sides of the equation, hence we have that

A/2 = √A² - x²

(0.0169)/2 = √(0.0169² - x²)

0.00845 = √(0.0169² - x²)

By squaring both sides, we have that

0.00845² = 0.0169² - x²

x² = 0.0169² - 0.00845²

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5 0
4 years ago
A lawn roller in the form of a thin-walled, hollow cylinder with mass M is pulled horizontally with a constant horizontal force
olga nikolaevna [1]

Answer:

a_{cm}=\frac{F}{2M}\\\\F_{fr}=\frac{F}{2}

Explanation:

Given the mass as M, the rotational inertia of the mower is;

I_{cm}=MR^2

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v_{cm}=wR, a_{cm}=\alpha R

\sum F_x=Ma_x, -F+F_{fr}=-Ma_{cm}\\\\a_{cm}=\frac{F-Fr}{M}    \ \ \ \ \ \ \ \ eqtn1\\\\\sum \tau_{cm}=I_{cm}\alpha, F_{fr}(R)}=(MR^2)(\frac{a_{cm}}{R}), ->F_{fr}=Ma_c_m\ \ \  \ \ \ \ eqtn2\\\\a_{cm}=\frac{F-Ma_{cm}}{M}, ->F=2Ma_{cm}\\\\a_{cm}=\frac{F}{2M}\\\\\\\therefore F_{fr}=M(\frac{F}{2M})\\\\F_{fr}=\frac{F}{2}

6 0
3 years ago
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