Help meh, no links and please give the right answer.

A ball of mass m strikes a wall perpendicularly with a speed v, and it bounces off the wall perpendicularly with a speed v 5 . W

laiz [17]

Answer:

Explanation:

Given

mass of ball is m

initial velocity of ball is $v$

Final velocity of ball is $v_5$

Impulse is the change in momentum of an object when an external force is applied for a given interval of time.

Initial momentum $P_i=mv$

Final momentum $P_f=mv_5$

$Impulse=P_f-P-i=m(v_5-v)$

3 0

3 years ago

A box with a mass of 12.5 kg sits on the floor. how high would you need to lift it for it to have a gpe of 355 j

grandymaker [24]

Gpe=mgh

355=12.5×10×h
h=355/125= 2.84m

3 0

4 years ago

A horizontal beam of electrons initially moving at 4.0×10^7 m/s is deflected vertically by the vertical electric field between o

givi [52]

Answer:

$1.77\times 10^{-7}\ C/m^2$

0.000439077936334 m

Explanation:

q = Charge of electron = $1.6\times 10^{-19}\ C$

E = Electric field = $2\times 10^{4}\ N/C$

$\epsilon_0$ = Permittivity of free space = $8.85\times 10^{-12}\ F/m$

d = Distance between plates = 2 cm (assumed)

m = Mass of electron = $9.11\times 10^{-31}\ kg$

The beam consists of electrons which means it has negative charge this means the upper plates will be positive and the lower plate will be negative.

The direction is upper to lower lower plate.

$\epsilon_0$ = Permittivity of free space = $8.85\times 10^{-12}\ F/m$

Electric flux is given by

$\phi=\epsilon_0E\\\Rightarrow \phi=8.85\times 10^{-12}\times 2\times 10^{4}\\\Rightarrow \phi=1.77\times 10^{-7}\ C/m^2$

The charge per unit area on the plates is $1.77\times 10^{-7}\ C/m^2$

Deflection is given by

$s=\dfrac{1}{2}\dfrac{qE}{m}(\dfrac{d}{v})^2\\\Rightarrow s=\dfrac{1}{2}\dfrac{1.6\times 10^{-19}\times 2\times 10^4}{9.11\times 10^{-31}}(\dfrac{0.02}{4\times 10^7})^2\\\Rightarrow s=0.000439077936334\ m$