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Tema [17]
3 years ago
7

A bullet is fired from a rifle pointed 45 degrees above horizontal. The bullet leaves the muzzle traveling 1400 m/s. How many se

conds does it take the bullet to reach the high point of its trajectory?
Physics
1 answer:
Bogdan [553]3 years ago
5 0

The bullet's vertical velocity at time t is

v=1400\dfrac{\rm m}{\rm s}-gt

where g=9.80\frac{\rm m}{\mathrm s^2} is the acceleration due to gravity.

At its highest point, the bullet's vertical velocity is 0, which happens

0=1400\dfrac{\rm m}{\rm s}-gt\implies t=\dfrac{1400\frac{\rm m}{\rm s}}g\approx\boxed{142.857\,\mathrm s}

(or about 140 s, if you're keeping track of significant figures) after being fired.

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Two identical particles, each with a mass of 4.5 mg and a charge of 30 nC, are moving directly toward each other with equal spee
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   r₁ = 20.5 cm

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In this exercise we can use the conservation of energy

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        Em₀ = U_g + U_e + K = -G \frac{m_1m_2}{r} +k \frac{q_1q_2}{r} - 2 ( \frac{1}{2}  m v^2)

the two in the kinetic energy is because they are two particles

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        Em₀ = em_f

the charges and masses of the two particles are equal

         -G \frac{m^2}{r} + k \frac{q^2}{r} + m v^2 = - G \frac{m^2}{r_1} + k \frac{q^2}{r_1}        

         

sustitute the values

-6.67-11 (4.5 10-3) ² / 0.25 - 9, 109 (30 10-9) ² / 0.25 + 4.5 10-3 4² = - 6.67 10- 11 (4.5 10-3) ² / r1 -9 109 (30 10-9) ² / r1

    -5.4 10⁻¹⁵ + 3.24 10⁻⁵ - 7.2 10⁻⁵ = -1.35 10⁻¹⁵ / r₁  + 8.1 10⁻⁶ / r₁

We can see that the terms that correspond to the gravitational potential energy are much smaller than the terms of the electric power, which is why we depress them.

      3.24 10⁻⁵ - 7.2 10⁻⁵ =  8.1 10⁻⁶ / r₁

      -3.96 10⁻⁵ = 8.1 10⁻⁶ / r₁

      r₁ = 8.1 10⁻⁶ /3.96 10⁻⁵

      r₁ = 2.045 10⁻¹ m

      r₁ = 20.5 cm

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Please allow me to know if I helped with a thank you, rating or brainly.

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Download pdf
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