Ask question

Ask question

All categories

3 years ago

10

PLEASE HELP !!

1 answer:

Paraphin [41]3 years ago

7 0

Distance from the reference point

You might be interested in

Is an electron an antiparticle, boson, lepton, or hadron?

Masteriza [31]

Answer:

An electron is a lepton

4 0

3 years ago

According to thomas gage's spies, in which town had the minutemen been collecting a large amount of arms?

Citrus2011 [14]

The correct answer for the given question above would be option B. CONCORD. According to Thomas Gage's spies, the town in which the minutemen had been collecting a large amount of arms is in Concord. The minutemen are civilian colonists who established an independent organization forming a <span> well-</span><span>prepared militia. Hope this answers your question.</span>

7 0

3 years ago

Read 2 more answers

how much current is in a circuit that includes a 9-volt battery and a bulb with a resistance of 3 ohms?

ExtremeBDS [4]

3 Amper 9 divided by 3

8 0

3 years ago

The variable ______________ describes how quickly something moves.

IrinaK [193]

it's up in Gogle trust me

8 0

3 years ago

A small block of mass 20.0 grams is moving to the right on a horizontal frictionless surface with a speed of 0.68 m/s. The block

Usimov [2.4K]

Answer:

a) $v'=-0.227\ m.s^{-1}$

b) $v=1.36\ m.s^{-1}$

Explanation:

Given:

mass of the lighter block, $m'=0.02\kg$

velocity of the lighter block, $u'=0.68\ m.s^{-1}$

mass of the heavier block, $m=0.04\ kg$

velocity of the heavier block, $u=0\ m.s^{-1}$

a)

Using conservation of linear momentum:

$m'.u'+m.u=m'.v'+m.v$

where:

$v'=$ final velocity of the lighter block

$v=$ final velocity of the heavier block

$m'.u'=m'.v'+m.v$

$m'(u'-v')=m.v$ ........................(1)

Since kinetic energy is conserved in elastic collision:

$\frac{1}{2}m'.u'^2=\frac{1}{2}m'.v'^2+\frac{1}{2}m.v^2$

$m'(u'^2-v'^2)=m.v^2$

$m'(u'-v')(u'+v')= m.v^2$

divide the above equation by eq. (1)

$v=u'+v'$ .............................(2)

now we substitute the value of v from eq. (2) in eq. (1)

$m'(u'-v')=m(u'+v')$

$\frac{m'+m}{m'-m} =\frac{u'}{v'}$

$\frac{0.02+0.04}{0.02-0.04} =\frac{0.68}{v'}$

$v'=-0.227\ m.s^{-1}$ (negative sign denotes that the direction is towards left)

b)

now we substitute the value of v' from eq. (2) in eq. (1)

$m'(u'-v+u')=m.v$

$2m'.u'=(m-m')v$

$2\times 0.02\times 0.68=(0.04-0.02)\times v$

$v=1.36\ m.s^{-1}$

6 0

3 years ago