During an eclipse the sun the earth and the moon act as a light source,an obstacle or screen=

Physics

1 answer:

Zigmanuir [339]3 years ago

6 0

Answer:

Un eclipse solar se produce cuando la luna se interpone en el camino de la luz del sol y proyecta su sombra en la Tierra. Eso significa que durante el día, la luna se mueve por delante del sol y se pone oscuro. ... Este eclipse total se produce aproximadamente cada año y medio en algún lugar de la Tierra.

Explanation:

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Skater 1 has a mass of 45 kg and is at rest. Skater 2 has a mass of 50 kg and is moving slowly at a constant velocity of 3.2 m/s

algol13

Answer:

None

Explanation:

Force, F is given by ma where m is the mass of an object and a is acceleration

Acceleration is the rate of change in velocity per unit time. Since skaters with mass of 75 kg and 50 kg are moving at a constant speed, there is no acceleration hence F=50*0=0 and F=75*0=0

For skater of 45 kg, he is at rest to mean the initial and final velocitu of the skater is zero hence no acceleration, the force will be 45*0=0

Therefore, none of the skaters will experience a greater net force.

3 0

3 years ago

A 75-g bullet is fired from a rifle having a barrel 0.540 m long. choose the origin to be at the location where the bullet begin

lyudmila [28]

Part a) The work done by the gas on the bullet is the integral of the force in dx, where x is the distance covered by the bullet inside the barrel with respect to the origin:
$W= \int\limits^{0.540m}_{0} {F} \, dx = \int\limits^{0.540m}_{0} {(16000+10000x-26000x^2)} \, dx =$
$=16000x+10000 \frac{x^2}{2} - 26000 \frac{x^3}{3}$
By substituting the length of the barrel, L=0.540 m, we find the total work done by the gas on the bullet:
$W=16000(0.540m)+10000 \frac{(0.540m)^2}{2} - 26000 \frac{(0.540m)^3}{3} =$
$=8733 J=8.73 kJ$

part b) The resolution of the problem is the same, we just have to use the new length of the barrel (L=0.95 m) inside the final formula, and we find the new value of the work:
$W=16000(0.95m)+10000 \frac{(0.95m)^2}{2} - 26000 \frac{(0.95m)^3}{3} =$
$=12280 J=12.28 kJ$