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A ball is dropped from rest from a height h above the ground. another ball is thrown vertically upwards from the ground at the i

Darya [45]

The position of the first ball is

$y_1=h-\dfrac g2t^2$

while the position of the second ball, thrown with initial velocity $v$ , is

$y_2=vt-\dfrac g2t^2$

The time it takes for the first ball to reach the halfway point satisfies

$\dfrac h2=h-\dfrac g2t^2$

$\implies\dfrac h2=\dfrac g2t^2$

$\implies t=\sqrt{\dfrac hg}$

We want the second ball to reach the same height at the same time, so that

$\dfrac h2=v\sqrt{\dfrac hg}-\dfrac g2\left(\sqrt{\dfrac hg}\right)^2$

$\implies h=2v\sqrt{\dfrac hg}-g\left(\dfrac hg\right)$

$\implies h=v\sqrt{\dfrac hg}$

$\implies v=\sqrt{hg}$

8 0

3 years ago

A 2.0 m × 4.0 m flat carpet acquires a uniformly distributed charge of −10 μC after you and your friends walk across it several

mamaluj [8]

Answer:

The charge on the dust particle is $q_d = 6.94 *10^{-13} \ C$

Explanation:

From the question we are told that

The length is $l = 2.0 \ m$

The width is $w = 4.0 \ m$

The charge is $q = -10\mu C= -10*10^{-6} \ C$

The mass suspended in mid-air is $m_a = 5.0 \mu g = 5.0 *10^{-6} \ g = 5.0 *10^{-9} \ kg$

Generally the electric field on the carpet is mathematically represented as

$E = \frac{q}{ 2 * A * \epsilon _o}$

Where $\epsilon _o$ is the permittivity of free space with value $\epsilon_o = 8.85*10^{-12} \ \ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2$

substituting values

$E = \frac{-10*10^{-6}}{ 2 * (2 * 4 ) * 8.85*10^{-12}}$

$E = -70621.5 \ N/C$

Generally the electric force keeping the dust particle on the air equal to the force of gravity acting on the particles

$F__{E}} = F__{G}}$

=> $q_d * E = m * g$

=> $q_d = \frac{m * g}{E}$

=> $q_d = \frac{5.0 *10^{-9} * 9.8}{70621.5}$

=> $q_d = 6.94 *10^{-13} \ C$

4 0

3 years ago

Clara and Eli used materials they could find at home to make a simple seismograph. As Clara pulls the strip of paper through the

tatyana61 [14]

Answer:

The recording that Eli and Clara's Seismograph would show will be very close to that a real Seismograph would show. This is because the mechanism they have constructed is very similar to the real one.

Explanation:

hope this helps

6 0

2 years ago

A small object with momentum 7.0 kg∙m/s approaches head-on a large object at rest. The small object bounces straight back with a

EastWind [94]

Answer:

The magnitude of the large object's momentum change is 3 kilogram-meters per second.

Explanation:

Under the assumption that no external forces are exerted on both the small object and the big object, whose situation is described by the Principle of Momentum Conservation:

$p_{S,1}+p_{B,1} = p_{S,2}+p_{B,2}$ (1)

Where:

$p_{S,1}$ , $p_{S,2}$ - Initial and final momemtums of the small object, measured in kilogram-meters per second.

$p_{B,1}$ , $p_{B,2}$ - Initial and final momentums of the big object, measured in kilogram-meters per second.

If we know that $p_{S,1} = 7\,\frac{kg\cdot m}{s}$ , $p_{B,1} = 0\,\frac{kg\cdot m}{s}$ and $p_{S, 2} = 4\,\frac{kg\cdot m}{s}$ , then the final momentum of the big object is: