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3 years ago

Given that the collision is elastic and glider 2 is initially at rest (v2,i =0), please use below Eqs. to explain why

Physics

1 answer:

Morgarella [4.7K]3 years ago

5 0

Answer:

Explanation:

1 )

Put v2,i =0, in second equation

v2,f= (m2-m1)v2,i + 2m1v1,i/m1+m2

v2,f = 0 + 2m1v1,i/m1+m2

v2,f = 2m1v1,i/m1+m2

In this equation coefficient of v1,i is positive so v2,f and v1,i have the same sign.

2 )

Put m1 < m2 and v2,i =0 in first equation

v1,f= (m1-m2)v1,i + 2m2v2,1/m1+m2

v1,f= (m1-m2)v1,i

As m1-m2 is negative , v1f and v1i will have opposite sign.

3 )

Put m1 > m2 and v2,i =0 in first equation

v1,f= (m1-m2)v1,i + 2m2v2,1/m1+m2

v1,f= (m1-m2)v1,i

m1 - m2 is positive so v1f and v1i will have same sign.

4 )

Put m1 = m2 and v2,i =0 in first equation

v1,f= (m1-m2)v1,i

= 0 because m1 = m2

So glider 1 will stop because v1,f = 0 .

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Rufina [12.5K]

Answer:

24.79 m/s2

Explanation:

Let us assume that there is an object with a mass of 'm' on the

Jupiter. Jupiter will attract this object:

$mg_j=G\frac{mM_j}{r_j^{2} }$

$g_j=G\frac{M_j}{r_j^{2} }$

$g_j=6.67*10^{-11} \frac{1.9*10^{27} }{71492000^{2} }$

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4 years ago

Suppose astronomers built a 20-meter telescope. How much greater would its light-collecting area be than that of the 10-meter Ke

nirvana33 [79]

Answer:

4 times greater

Explanation:

Step 1: Calculate light-collecting area of a 20-meter telescope (A₁) by using area of a circle.

Area of circle = π*r² = $\frac{\pi d^{2}}{4}$

Where d is the diameter of the circle = 20-m

$A_{1} = \frac{\pi d^{2}}{4}$

$A_{1} = \frac{\pi (20^{2})}{4}$

A₁ = 314.2 m²

Step 2: Calculate light-collecting area of a 10-meter Keck telescope (A₂)

$A_{2} = \frac{\pi d^{2}}{4}$

Where d is the diameter of the circle = 10-m

$A_{2} = \frac{\pi (10^{2})}{4}$

A₂ = 78.55 m²

Step 3: divide A₁ by A₂

$= \frac{314.2 m^2}{78.55 m^2}$

= 4

Therefor, the 20-meter telescope light-collecting area would be 4 times greater than that of the 10-meter Keck telescope.

5 0

4 years ago

A 15.0-μF capacitor is charged by a 130.0-V power supply, then disconnected from the power and connected in series with a 0.280-

SVETLANKA909090 [29]

The resonant frequency of a circuit is the frequency $\omega_0$ at which the equivalent impedance of a circuit is purely real (the imaginary part is null).

Mathematically this frequency is described as

$f = \frac{1}{2\pi}(\sqrt{\frac{1}{LC}})$

Where

L = Inductance

C = Capacitance

Our values are given as

$C = 15*10^{-6}\mu F$

$L = 0.280*10^{-3}mH$

Replacing we have,

$f = \frac{1}{2\pi}(\sqrt{\frac{1}{LC}})$

$f = \frac{1}{2\pi}(\sqrt{\frac{1}{(15*10^{-6})(0.280*10^{-3})}})$

$f= 2455.81Hz$

From this relationship we can also appreciate that the resonance frequency infers the maximum related transfer in the system and that therefore given an input a maximum output is obtained.

For this particular case, the smaller the capacitance and inductance values, the higher the frequency obtained is likely to be.

7 0

4 years ago

A dart is thrown horizontally at a target's center that is 5.00 m, away. The dart hits the target 0.150 m below the targets cent

boyakko [2]

Answer:

v₀ₓ = 28.6 m / s

Explanation:

This is a missile throwing exercise

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