Que cuerpos celestes observó laika durante su viaje

A rubber bullet of mass m=0.025\,\mathrm{kg}m=0.025kg traveling at velocity v_0 = 50\,\mathrm{m/s}v 0 =50m/s hits an iron bloc

Pavel [41]

Answer: 0.001 m

Explanation:

In order to get the máximum height reached by the iron block, we can use the energy conservation principle, as all the kinetic energy impressed upon the iron blck by the rubber bullet during the collision, becomes gravitational potential energy, as follows:

½ m v2 = m. g. h (1)

We don´t know the value of v, but if we look to the collision, and we asume no external forces act during it, the total momentum must be conserved.

The initial momentum, as the block is at rest, is just the one due to the rubber bullet:

P1 = mb . vb = 0.025 kg. 50 m/s = 1.25 kg. m/s

The final momemtum, is just the sum of the one due to the bullet (after being bounced back) and the one for the iron block:

P2 = mb . vfb + mib . vib= 0.025 Kg. (-35 m/s) + 15 Kg.vib

As we have already said, P1 = P2, so we can write the following equation:

0.025 Kg. (-35 m/s) + 15 Kg. vib = 1.25 Kg. m/s.

Solving for vib, we have:

vib = 0.14 m/s

Now, we can replace this value in the equation (1) above:

½ . 15 Kg. (0.14)2 (m/s)2 = 15 Kg. 9.8 m/s2. H

Solving for H, we have:

H = 0.001 m

3 0

3 years ago

You hold a ruler that has a charge on its tip 6 cm above a small piece of tissue paper to see if it can be picked up. The ruler

scZoUnD [109]

Answer:

q=1.4*10^{-9}C

Explanation:

Given data:

charge on ruler = -14μC

Mass of tissue is 5 g

To Know the minimum charge, equate electrostatic force to weight

we have F = W

so $\frac{KQq}{r^2} =mg$

putting all value in equation,

$=\frac{9*10^9*(14*10^{-6})*q}{0.06^2} = 5* 10^{-3}*9.8$

solving for q

$q =\frac{5* 10^{-3}*9.8 *0.06^2}{9*10^9*(14*10^{-6})}$

or q=1.4*10^{-9}C

5 0

3 years ago

When does the fall equinox occur in the northern hemisphere

aivan3 [116]

Here is your answer

On 23rd September

HOPE IT IS USEFUL

7 0

2 years ago

Light of wavelength 560 nm passes through a slit of width 0. 170 mm. (a) the width of the central maximum on a screen is 8. 00 m

faltersainse [42]

The distance between slit and the screen is 1.214m.

To find the answer, we have to know about the width of the central maximum.

<h3>How to find the distance between slit and the screen?</h3>

It is given that, wavelength 560 nm passes through a slit of width 0. 170 mm, and the width of the central maximum on a screen is 8. 00 mm.