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3 years ago

What is the formula of conservation of linear momentum

1 answer:

8 0

Conservation of Linear Momentum. Although the momentum of individual objects may change during a collision, the total momentum of all the objects in an isolated system remains constant. An isolated system is one on which the net force from external sources is zero.

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What state of matter did the earth need to be in order for planetary differentiation to occur?

STatiana [176]

Gas. I know this because first you will need to do the equator so I will say gas

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2 years ago

You are spinning a rock, of mass 0.75 kg, at the end of a string of length 0.86m in a vertical circle in uniform circular motion

Nina [5.8K]

Answer:

v (minimum speed) = 2.90 m/sec.

$\\ \\ maximum speed (v)= 6.57 m/sec.\\$

Maximum value of speed will occur at lowest point of vertical circle.

Explanation:

a) What minimum speed is necessary so that there is no tension in the string at the top of the circle but the rock stays in the same circular path?

Using the force balance expression at the top of the circle,

Gravitational Force + Tension force = Centrifugal force

$m*g + T = m*v^2/R$

Given that : T = 0

R = length of string = 0.86 m

mass of the spinning rock = 0.75 kg

$v = \sqrt{g*R}$

$v = \sqrt{9.81*0.86}$

v (minimum speed) = 2.90 m/sec.

b) what is the maximum speed the rock can have so that the string does not break?

Here the force balance at bottom of circle is represented by the illustration:

$T = m*g + m*v^2/R$

Given that:

maximum tension T = 45 N

maximum speed v = ??

mass m = 0.75 kg

∴

$45 - 0.75*9.81 = 0.75*\frac{v^2}{0.86} \\\\v^2 = 0.86*(45 - 0.75*9.81)/0.75 \\ v = \sqrt{0.86*(45 - 0.75*9.81)/0.75\\ maximum speed (v)= 6.57 m/sec.\\$

At what point in the vertical circle does this maximum value occur?

Maximum value of speed will occur at lowest point of vertical circle.

This is so because at the lowest point; the tension in string will be maximum.

4 0

2 years ago

The speed of light in a vacuum is 3 x 108 miles/hour. true/false?

GarryVolchara [31]

I'm guessing they rounded because it is technically 2.998 X 108 miles/hour

so I would go with true!

8 0

3 years ago

Read 2 more answers

An object is moving with a constant velocity of 278 m/s. How long will it take it to travel 7500 m, using the formula Delta X=Vt

rusak2 [61]

If the velocity is constant then the acceleration of the object is zero.
$a=0 (m/s^2)$
Thus when we apply the equation
$\Delta X=vt+(at^2/2)$
It remains
$\Delta X =vt$
or equivalent
$t=(\Delta X/v) =7500/278 =26.98 (seconds)$

7 0

3 years ago

An electron is accelrated by a unifor electric field (1000v/m) pointing vertically upward. Use energy methods to get the magnitu

ExtremeBDS [4]

Explanation:

In the given situation two forces are working. These are:

1) Electric force (acting in the downward direction) = qE

2) weight (acting in the downward direction) = mg

Therefore, work done by all the forces = change in kinetic energy

Hence, $qE \times S + mg \times S = 0.5 \times mv^{2}$

$1.6 \times 10^{-19} \times 1000 + 9.1 \times 10^{-31} \times 9.8 \times (\frac{0.10}{100}) = 0.5 \times 9.1 \times 10^{-31} \times v^{2}$

It is known that the weight of electron is far less compared to electric force. Therefore, we can neglect the weight and the above equation will be as follows.

$(1.6 \times 10^{-19} \times 1000) \times (\frac{0.10}{100}) = 0.5 \times 9.1 \times 10^{-31} \times v^{2
}$

v = $sqrt{\frac{1.6 \times 10^{-19}}{(0.5 \times 9.1 \times 10^{-31})}$

= 592999 m/s

Since, the electron is travelling downwards it means that it looses the potential energy.

8 0

2 years ago