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3 years ago

15

There is friction between the block and the

1 answer:

Norma-Jean [14]3 years ago

4 0

Answer:

5 no

Explanation:

actually the 4kg lying on table has no influence

it slides towards 4kg weight hung

as it has excess 2kg force

force=miu × m ×g

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In a science fiction novel two enemies, Bonzo and Ender, are fighting in outer spce. From stationary positions, they push agains

Hoochie [10]

Answer:

$\frac{m_B}{m_E}=1.18$

Explanation:

$m_B$ = Mass of Bonzo

$m_E$ = Mass of Ender

$u_B$ = Initial Velocity of Bonzo = 2.2 m/s

$u_E$ = Initial Velocity of Ender = -2.6 m/s

From conservation of linear momentum

$m_Bu_B=m_Eu_E\\\Rightarrow \frac{m_B}{m_E}=\frac{u_E}{u_B}\\\Rightarrow \frac{m_B}{m_E}=\frac{2.6}{2.2}\\\Rightarrow \frac{m_B}{m_E}=1.18$

$\therefore \frac{m_B}{m_E}=1.18$

8 0

3 years ago

Help please help me?

Andrews [41]

Answer:

C. Planet D has the greatest mass and will exert a greater gravitational force

6 0

3 years ago

Listening to the radio, you can hear two stations at once. Describe this wave interaction

Sedaia [141]

Actually says Pantazis, since their frequencies are so wildly different, brain waves don’t interfere with radio waves. Even if that was the case, brain waves are so weak, they are hardly measurable at all. For comparison, says Pantazis, “the magnetic field of the earth is just strong enough to move the needle of a compass. Signals from the brain are a billionth of that strength.”

8 0

3 years ago

Read 2 more answers

A 2-kg ball is thrown at 3 m/s. What is the ball's momentum? *

crimeas [40]

Answer:

Given

mass (m) =2kg

velocity (v) =3m/s

momentum (p) =?

Form

p=mv

2kgx3m/s

p=6kg.m/s

the momentum of ball's =6kg.m/s

8 0

3 years ago

A ball is launched vertically with an initial speed of y˙0= 50 m/s, and its acceleration is governed by y¨=-g-cDy˙2, where the a

stira [4]

Answer:

Explanation:

Given

acceleration is given by

$a=-g-c_Dv^2$

where $\ddot{y}=a$

$\dot{y}=v$

Also acceleration is given by

$a=v\frac{\mathrm{d} v}{\mathrm{d} s}$

$ds=\frac{v}{a}dv$

$\int ds=\int \frac{v}{-g-0.001v^2}dv$

$\Rightarrow Let -g-0.001v^2=t$

$-0.001\times 2vdv=dt$

$vdv=-\frac{dt}{0.002}$

$at\ v_0=50\ m/s,\ t=-g-0.001(50)^2$

$t=-g-2.5$

at $v=0,\ t=-g$

$\int_{0}^{s}ds=\int_{-g}^{-g-2.5}\frac{-dt}{0.002t}$

$\int_{0}^{s}ds=\int^{-g}_{-g-2.5}\frac{dt}{0.002t}$

$s=\frac{1}{0.002}lnt|_{-g}^{-g-2.5}$

$s=\frac{1}{0.002}\ln (\frac{g+2.5}{g})$

$s=113.608\ m$

when air drag is neglected maximum height reached is

$h=\frac{v_0^2}{2g}$

$h=\frac{50^2}{2\times 9.8}$

$h=127.55\ m$

3 0

3 years ago