Does a perfectly elastic colission conserve momentum and impulse

The ink drops have a mass m = 1.00×10^−11 kg each and leave the nozzle and travel horizontally toward the paper at velocity v =

luda_lava [24]

Answer:

9.98 × 10⁻⁹ C

Explanation:

mass, m = 1.00 × 10⁻¹¹ kg

Velocity, v = 23.0 m/s

Length of plates D₀ = 1.80 cm = 0.018 m

Magnitude of electric field, E = 8.20 × 10⁴ N/C

drop is to be deflected a distance d = 0.290 mm = 0.290 × 10⁻³ m

density of the ink drop = 1000 kg/m^3

Now,

Time = $\frac{\textup{Distance}}{\textup{Velocity}}$

or

Time = $\frac{\textup{0.016}}{\textup{23}}$

or

Time = 6.9 × 10⁻⁴ s

Now, force due to the electric field, F = q × E

where, q is the charge

Also, Force = Mass × acceleration

q × E = 1.00 × 10⁻¹¹ × a

or

a = $\frac{q\times8.20\times10^4}{1\times10^{-11}}$

Now from the Newton's equation of motion

$d=ut+\frac{1}{2}at^2$

where,

d is the distance

u is the initial speed

a is the acceleration

t is the time

or

$0.290\times10^{-3}=0\times(6.9\times10^{-4})+\frac{1}{2}\times(\frac{q\times8.20\times10^4}{1\times10^{-11}})\times(6.9\times10^{-4})^2$

or

q = 9.98 × 10⁻⁹ C

4 0

3 years ago

With what speed must you approach a source of sound to observe a 25% change in frequency?

insens350 [35]

Sound source is at rest, you are moving with velocity v, f = frequency, c = speed of sound:

f = f0(1 + v/c)

115 = 100(1 + v/343)
115 = 100 + 100v/343
15 = 100v/343
v = 15*343/100
<span> v = 51,45 m/s </span>

5 0

3 years ago

In Physics, work depends on two factors. What are those two<br> factors?

Aloiza [94]

Answer:

Force and displacement.

Explanation:

Work done is positive when we push table and it move in the direction of applied force.

5 0

3 years ago

15. If an 800.-kg sports car slows to 13.0 m/s to check out an accident scene and the

goldfiish [28.3K]

The final combined velocity after the collision is 20.2 m/s

Explanation:

We can solve this problem by using the law of conservation of momentum: in fact, in absence of external forces, the total momentum of the two car and of the truck must be conserved before and after the collision.

This means that we can write the following equation:

$p_i = p_f\\m_1 u_1 + m_2 u_2 = (m_1+m_2)v$